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What is Physics?

Physics is often described as the study ofmatterandenergy. It is concerned with

how matter and energy relate to each other, and how they affect each other over

time and through space.

Physicists ask the fundamental questions how did the universe begin? How and

of what is it made? How does it change? What rules govern its behaviour?

Physicists may be roughly divided into two camps:experimentalphysicists

andtheoreticalphysicists. Experimental physicists design and run careful

investigations on a broad range of phenomena in nature, often under conditions

which are atypical of our everyday lives. They may, for example, investigate what

happens to the electrical properties of materials at temperatures very nearabsolute zero (460 degrees Fahrenheit) or measure the characteristics of energy

emitted by very hot gases. Theoretical physicists propose and develop models and

theories to explain mathematically the results of experimental

observations.Experimentandtheorytherefore have a broad overlap. Accordingly,

an experimental physicist remains keenly aware of the current theoretical work

in his or her field, while the theoretical physicist must know the experimenter's

results and the context in which the results need be interpreted.

It is also useful to distinguishclassicalphysics andmodernphysics.Classical

physicshas its origins approximately four hundred years ago in the studies

ofGalileoandNewtonon mechanics, and similarly, in the work

of Ampere,Faraday,MaxwellandOerstedone hundred fifty years ago in the

fields of electricity and magnetism. This physics handles objects which are6



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neither too large nor too small, which move at relatively slow speeds (at least

compared to the speed of light: 186,000 miles per second!).

The emergence ofmodern physicsat the beginning of the twentieth century wasmarked by three achievements.

The first in1905 wasEinstein's brilliant model of light as a stream of particles

(photons). The second, which followed a few months later, was his revolutionary

theory of relativity which described objects moving at speeds close to the speed

of light. The third breakthrough came in 1910 withRutherford'sdiscovery of the

nucleus of the atom.Rutherford's work was followed byBohr's model of the

atom, which in turn stimulated the work ofde-

Broglie,Heisenberg,Schroedinger,Born,Pauli,Diracand others on the quantum

theory. The avalanche of exciting discoveries in modern physics continues today.

Given these distinctions within the field of physicsexperimentalandtheoretical

classicalandmodern,it is useful to further subdivide physics into various

disciplines, including astrophysics, atomic and molecular physics, biophysics,

solid state physics, optical and laser physics, fluid and plasma physics, nuclearphysics, and particle physics.

Following the discovery of a particle with properties consistent with theHiggs

boson atCERN in 2012 allfundamental articles predicted by the standard

model, and no others, appear to exist; however,physics beyond the Standard

Model, with theories such assupersymmetry, is an active area of research.

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http://en.wikipedia.org/wiki/Higgs_boson


http://en.wikipedia.org/wiki/CERN

http://en.wikipedia.org/wiki/Fundamental_particles

http://en.wikipedia.org/wiki/Physics_beyond_the_Standard_Model


http://en.wikipedia.org/wiki/Supersymmetry

http://en.wikipedia.org/wiki/CERN

http://en.wikipedia.org/wiki/Fundamental_particles



http://en.wikipedia.org/wiki/Supersymmetry





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Scope and Excitement of Physics

We can get some idea of the scope of physics by looking at its various sub-disciplines. Basically, there are two domains of interest : macroscopic and

microscopic. The macroscopic domain includes phenomena at the laboratory,

terrestrial and astronomical scales. The microscopic domain includes atomic,

molecular and nuclear phenomena. Classical Physics deals mainly with

macroscopic phenomena and includes subjects like Mechanics, Electrodynamics,

Optics and Thermodynamics. Mechanics founded on Newton’s laws of motion and

the law of gravitation is concerned with the motion (or equilibrium) of particles,

rigid and deformable bodies, and general systems of particles. The propulsion ofa rocket by a jet of ejecting gases, propagation of water waves or sound waves in

air, the equilibrium of a bent rod under a load, etc., are problems of mechanics.

Electrodynamics deals with electric and magnetic phenomena associated with

charged and magnetic bodies. Its basic laws were given by Coulomb, Oersted,

Ampere and Faraday, and encapsulated by Maxwell in his famous set of

equations. The motion of a current-carrying conductor in a magnetic field, the

response of a circuit to an ac voltage (signal), the working of an antenna, the

propagation of radio waves in the ionosphere, etc., are problems ofelectrodynamics. Optics deals with the phenomena involving light. The working

of telescopes and microscopes, colours exhibited by thin films, etc., are topics in

optics. Thermodynamics, in contrast to mechanics, does not deal with the motion

of bodies as a whole.

Rather, it deals with systems in macroscopic equilibrium and is concerned with

changes in internal energy, temperature, entropy, etc., of the system through

external work and transfer of heat. The efficiency of heat engines andrefrigerators, the direction of a physical or chemical process, etc., are problems of

interest in thermodynamics. The microscopic domain of physics deals with the

constitution and structure of matter at the minute scales of atoms and nuclei

(and even lower scales of length) and their interaction with different probes such

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as electrons, photons and other elementary particles. Classical physics is

inadequate to handle this domain and Quantum Theory is currently accepted as

the proper framework for explaining microscopic phenomena. Overall, the edifice

of physics is beautiful and imposing and you will appreciate it more as youpursue the subject. You can now see that the scope of physics is truly vast. It

covers a tremendous range of magnitude of physical quantities like length, mass,

time, energy, etc. At one end, it studies phenomena at the very small scale of

length (10-14 m or even less) involving electrons, protons, etc.; at the other end,

it deals with astronomical phenomena at the scale of galaxies or even the entire

universe whose extent is of the order of 1026 m. The two length scales differ by a

factor of 1040 or even more. The range of time scales can be obtained by dividing

the length scales by the speed of light : 10–22 s to 1018 s. The range of massesgoes from, say, 10–30 kg (mass of an electron) to 1055 kg (mass of known

observable universe). Terrestrial phenomena lie somewhere in the middle of this

range. Physics is exciting in many ways. To some people the excitement comes

from the elegance and universality of its basic theories, from the fact that a few

basic concepts and laws can explain phenomena covering a large range of

magnitude of physical quantities. To some others, the challenge in carrying out

imaginative new experiments to unlock the secrets of nature, to verify or refute

theories, is thrilling. Applied physics is equally demanding. Application andexploitation of physical laws to make useful devices is the most interesting and

exciting part and requires great ingenuity and persistence of effort. What lies

behind the phenomenal progress of physics in the last few centuries? Great

progress usually accompanies changes in our basic perceptions. First, it was

realised that for scientific progress, only qualitative thinking, though no doubt

important, is not enough. Quantitative measurement is central to the growth of

science, especially physics, because the laws of nature happen to be expressible

in precise mathematical equations. The second most important insight was thatthe basic laws of physics are universal — the same laws apply in widely different

contexts. Lastly, the strategy of approximation turned out to be very successful.

Most observed phenomena in daily life are rather complicated manifestations of

the basic laws. Scientists recognised the importance of extracting the essential

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features of a phenomenon from its less significant aspects. It is not practical to

take into account all the complexities of a phenomenon in one go. A good

strategy is to focus first on the essential features, discover the basic principles

and then introduce corrections to build a more refined theory of thephenomenon. For example, a stone and a feather dropped from the same height

do not reach the ground at the same time. The reason is that the essential aspect

of the phenomenon, namely free fall under gravity, is complicated by the

presence of air resistance. To get the law of free fall under gravity, it is better to

create a situation wherein the air resistance is negligible. We can, for example, let

the stone and the feather fall through a long evacuated tube. In that case, the

two objects will fall almost at the same rate, giving the basic law that acceleration

due to gravity is independent of the mass of the object. With the basic law thusfound, we can go back to the feather, introduce corrections due to air resistance,

modify the existing theory and try to build a more realistic theory of objects

falling to the earth under gravity.

Physics, Technology and Society

Throughout the recent decades, we have been experiencing a dramatic change in

the world of technology.

I-Phones, Blackberries and laptops are becoming an important part of our daily

lives. We are all so dependent on these devices that sometimes we treatthem as

if it is part of our family.

Technology has become a priority for many people. It makes life easier to live on

and less time-consuming. Hence, people do not have to do all the hard labour

anymore.

Although technology has made life more convenient, there is certain amount of

drawbacks in which it has damaged the quality of our life. It separates

individuals from reality. The IPod is one example; by putting into ear buds and

immersing yourself in music while in public, you are disconnecting yourself from

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The Internet detracts from the communication abilities of society, especially the

young. In formative teen years, lack of personal communication due to excessive

Internet usage can have an overall negative effect on mental and physical health.

Social networking inflicts damage on children and pushes them into leadingisolated, antisocial individuals, while living in a virtual world away from real life.

Overuse may ultimately lead to Internet addiction, which negatively affects a

child's scholastic performance, relationship with the family and psychological

health. Children can also access information that is unsuitable and

communicate with people they shouldn't come in contact with.

Protecting children should be a parent's primary concern and they must talk to

their kids about Internet safety and spend more time with them. In turn,

children should attend sporting sections, various educational courses, theatres

etc. Otherwise, they will degrade if spending almost all day near laptops and

other devices.

Technology hinders personal communication. Nowadays, people are starting to

become emotionally attached to technological devices. It is becoming very

difficult to spend a few hours, or even minutes without the usage of computers,

mobile phones etc.

Some experts believe that Internet users will lose the savvy, patience to conduct

social relations in the corporeal world. They stress a serious concern involving

proliferation of inappropriate content, such as violence, bias, hate speech,

profanity and pornography. The Internet gives people an easy way to find

unnecessary information.

These sorts of desperations changed people's social lives. Networking websitessuch as Facebook, Twitter etc. appear as ineffective way to communicate with

people, simply because conversations are very vague.

As a reminder, Twitter has 271 million monthly active users, 78 percent of

Twitter active users are on mobile. As of the first quarter 2014, social network

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Facebook had 1.28 billion monthly active users and 654 million people use

Facebook on mobile on a daily basis. That being said, people cannot imagine

their lives without technology. It seems that human beings will become too

computerized one day.

Society must be able to utilize technology while not allowing it to impede social

interactions, particularly for those who are easily influenced.

Online communication is now in fashion, putting a thick wall between you and

reality. Modern technologies could replace live talking with texting to your

friends. What is the next to change our life?

The connection between physics, technology and society can be seen in manyexamples. The discipline of thermodynamics arose from the need to understand

and improve the working of heat engines. The steam engine, as we know, is

inseparable from the Industrial Revolution in England in the eighteenth century,

which had great impact on the course of human civilisation. Sometimes

technology gives rise to new physics; at other times physics generates new

technology. An example of the latter is the wireless communication technology

that followed the discovery of the basic laws of electricity and magnetism in the

nineteenth century. The applications of physics are not always easy to foresee. Aslate as 1933, the great physicist Ernest Rutherford had dismissed the possibility

of tapping energy from atoms. But only a few years later, in 1938, Hahn and

Meitner discovered the phenomenon of neutron-induced fission of uranium,

which would serve as the basis of nuclear power reactors and nuclear weapons.

Yet another important example of physics giving rise to technology is the silicon

‘chip’ that triggered the computer revolution in the last three decades of the

twentieth century A most significant area to which physics has and will

contribute is the development of alternative energy resources. The fossil fuels ofthe planet are dwindling fast and there is an urgent need to discover new and

affordable sources of energy. Considerable progress has already been made in

this direction (for example, in conversion of solar energy, geothermal energy, etc.,

into electricity), but much more is still to be accomplished. Table1.1 lists some of

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the great physicists, their major contribution and the country of origin. You will

appreciate from this table the multi-cultural, international character of the

scientific endeavour. Table 1.2 lists some important technologies and the

principles of physics they are based on. Obviously, these tables are notexhaustive. We urge you to try to add many names and items to these tables with

the help of your teachers, good books and websites on science. You will find that

this exercise is very educative and also great fun. And, assuredly, it will never

end. The progress of science is unstoppable! Physics is the study of nature and

natural phenomena. Physicists try to discover the rules that are operating in

nature, on the basis of observations, experimentation and analysis. Physics deals

with certain basic rules/laws governing the natural world. What is the nature of

physical laws? We shall now discuss the nature of fundamental forces and thelaws that govern the diverse phenomena of the physical world.

Fundamental Forces in Nature

We all have an intuitive notion of force. In our experience, force is needed to

push, carry or throw objects, deform or break them. We also experience the

impact of forces on us, like when a moving object hits us or we are in a merry-go-round. Going from this intuitive notion to the proper scientific concept of force is

not a trivial matter. Early thinkers like Aristotle had wrong ideas about it. The

correct notion of force was arrived at by Isaac Newton in his famous laws of

motion. He also gave an explicit form for the force for gravitational attraction

between two bodies. We shall learn these matters in subsequent chapters. In the

macroscopic world, besides the gravitational force, we encounter several kinds of

forces: muscular force, contact forces between bodies, friction (which is also a

contact force parallel to the surfaces in contact), the forces exerted bycompressed or elongated springs and taut strings and ropes (tension), the force

of buoyancy and viscous force when solids are in contact with fluids, the force

due to pressure of a fluid, the force due to surface tension of a liquid, and so on.

There are also forces involving charged and magnetic bodies. In the microscopic

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domain again, we have electric and magnetic forces, nuclear forces involving

protons and neutrons, inter-atomic and intermolecular forces, etc. We shall get

familiar with some of these forces in later parts of this course. A great insight of

the twentieth century physics is that these different forces occurring in differentcontexts actually arise from only a small number of fundamental forces in

nature. For example, the elastic spring force arises due to the net

attraction/repulsion between the neighbouring atoms of the spring when the

spring is elongated/compressed. This net attraction/repulsion can be traced to

the (unbalanced) sum of electric forces between the charged constituents of the

atoms. In principle, this means that the laws for ‘derived’ forces (such as spring

force, friction) are not independent of the laws of fundamental forces in nature.

The origin of these derived forces is, however, very complex. At the present stageof our understanding, we know of four fundamental forces in nature, which are

described in brief here :

Gravitational Force

The gravitational force is the force of mutual attraction between any two objects

by virtue of their masses. It is a universal force. Every object experiences thisforce due to every other object in the universe. All objects on the earth, for

example, experience the force of gravity due to the earth. In particular, gravity

governs the motion of the moon and artificial satellites around the earth, motion

of the earth and planets around the sun, and, of course, the motion of bodies

falling to the earth. It plays a key role in the large-scale phenomena of the

universe, such as formation and evolution of stars, galaxies and galactic clusters.

Electromagnetic Force

Electromagnetic force is the force between charged particles. In the simpler case

when charges are at rest, the force is given by Coulomb’s law : attractive for

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magnetic effects and a magnetic field gives rise to a force on a moving charge.

Electric and magnetic effects are, in general, inseparable – hence the name

electromagnetic force. Like the gravitational force, electromagnetic force acts over

large distances and does not need any intervening medium. It is enormouslystrong compared to gravity. The electric force between two protons, for example,

is 1036 times the gravitational force between them, for any fixed distance. Matter,

as we know, consists of elementary charged constituents like electrons and

protons. Since the electromagnetic force is so much stronger than the

gravitational force, it dominates all phenomena at atomic and molecular scales.

(The other two forces, as we shall see, operate only at nuclear scales.) Thus it is

mainly the electromagnetic force that governs the structure of atoms and

molecules, the dynamics of chemical reactions and the mechanical, thermal andother properties of materials. It underlies the macroscopic forces like ‘tension’,

‘friction’, ‘normal force’, ‘spring force’, etc. Gravity is always attractive, while

electromagnetic force can be attractive or repulsive. Another way of putting it is

that mass comes only in one variety (there is no negative mass), but charge

comes in two varieties: positive and negative charge. This is what makes all the

difference. Matter is mostly electrically neutral (net charge is zero). Thus, electric

force is largely zero and gravitational force dominates terrestrial phenomena.

Electric force manifests itself in atmosphere where the atoms are ionised andthat leads to lightning If we reflect a little, the enormous strength of the

electromagnetic force compared to gravity is evident in our daily life. When we

hold a book in our hand, we are balancing the gravitational force on the book

due to the huge mass of the earth by the ‘normal force’ provided by our hand.

The latter is nothing but the net electromagnetic force between the charged

constituents of our hand and the book, at the surface in contact. If

electromagnetic force were not intrinsically so much stronger than gravity, the

hand of the strongest man would crumble under the weight of a feather ! Indeedto be consistent, in that circumstance, we ourselves would crumble under our

own weight !

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The strong nuclear force binds protons and neutrons in a nucleus. It is evident

that without some attractive force, a nucleus will be unstable due to the electric

repulsion between its protons. This attractive force cannot be gravitational since

force of gravity is negligible compared to the electric force. A new basic forcemust, therefore, be invoked. The strong nuclear force is the strongest of all

fundamental forces, about 100 times the electromagnetic force in strength. It is

charge-independent and acts equally between a proton and a proton, a neutron

and a neutron, and a proton and a neutron. Its range is, however, extremely

small, of about nuclear dimensions (10–15m). It is responsible for the stability of

nuclei. The electron, it must be noted, does not experience this force. Recent

developments have, however, indicated that protons and neutrons are built out of

still more elementary constituents called quarks.

Weak Nuclear Force

The weak nuclear force appears only in certain nuclear processes such as the β-

decay of a nucleus. In β-decay, the nucleus emits an electron and an uncharged

particle called neutrino. The weak nuclear force is not as weak as the

gravitational force, but much weaker than the strong nuclear andelectromagnetic forces. The range of weak nuclear force is exceedingly small, of

the order of 10-16 m.

Towards Unification of Forces

We remarked in section 1.1 that unification is a basic quest in physics. Great

advances in physics often amount to unification of different theories and

domains. Newton unified terrestrial and celestial domains under a common law

of gravitation. The experimental discoveries of Oersted and Faraday showed that

electric and magnetic phenomena are in general inseparable. Maxwell unified

electromagnetism and optics with the discovery that light is an electromagnetic

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succeed in this venture. But this did not deter physicists from zealously pursuing

the goal of unification of forces. Recent decades have seen much progress on this

front. The electromagnetic and the weak nuclear force have now been unified and

are seen as aspects of a single ‘electro-weak’ force. What this unification actuallymeans cannot be explained here. Attempts have been (and are being) made to

unify the electro-weak and the strong force and even to unify the gravitational

force with the rest of the fundamental forces. Many of these ideas are still

speculative and inconclusive. Table 1.4 summarises some of the milestones in

the progress towards unification of forces in nature.

1.5 Nature Of Physical Laws Physicists explore the universe. Theirinvestigations, based on scientific processes, range from particles that are

smaller than atoms in size to stars that are very far away. In addition to finding

the facts by observation and experimentation, physicists attempt to discover the

laws that summarise (often as mathematical equations) these facts. In any

physical phenomenon governed by different forces, several quantities may change

with time. A remarkable fact is that some special physical quantities, however,

remain constant in time. They are the conserved quantities of nature.

Understanding these conservation principles is very important to describe theobserved phenomena quantitatively. For motion under an external conservative

force, the total mechanical energy i.e. the sum of kinetic and potential energy of a

body is a constant. The familiar example is the free fall of an object under gravity.

Both the kinetic energy of the object and its potential energy change

continuously with time, but the sum remains fixed. If the object is released from

rest, the initial potential energy is completely converted into the kinetic energy of

the object just before it hits the ground. This law restricted for a conservative

force should not be confused with the general law of conservation of energy of anisolated system (which is the basis of the First Law of Thermodynamics). The

concept of energy is central to physics and the expressions for energy can be

written for every physical system. When all forms of energy e.g., heat, mechanical

energy, electrical energy etc., are counted, it turns out that energy is conserved.

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The general law of conservation of energy is true for all forces and for any kind of

transformation between different forms of energy. In the falling object example, if

you include the effect of air resistance during the fall and see the situation after

the object hits the ground and stays there, the total mechanical energy isobviously not conserved. The general law of energy conservation, however, is still

applicable. The initial potential energy of the stone gets transformed into other

forms of energy: heat and sound. (Ultimately, sound after it is absorbed becomes

heat.) The total energy of the system (stone plus the surroundings) remains

unchanged. The law of conservation of energy is thought to be valid across all

domains of nature, from the microscopic to the macroscopic. It is routinely

applied in the analysis of atomic, nuclear and elementary particle processes. At

the other end, all kinds of violent phenomena occur in the universe all the time. Yet the total energy of the universe (the most ideal isolated system possible !) is

believed to remain unchanged. Until the advent of Einstein’s theory of relativity,

the law of conservation of mass was regarded as another basic conservation law

of nature, since matter was thought to be indestructible. It was (and still is) an

important principle used, for example, in the analysis of chemical reactions. A

chemical reaction is basically a rearrangement of atoms among different

molecules. If the total binding energy of the reacting molecules is less than the

total binding energy of the product molecules, the difference appears as heat andthe reaction is exothermic. The opposite is true for energy absorbing

(endothermic) reactions. However, since the atoms are merely rearranged but not

destroyed, the total mass of the reactants is the same as the total mass of the

products in a chemical reaction. The changes in the binding energy are too small

to be measured as changes in mass. According to Einstein’s theory, mass m is

equivalent to energy E given by the relation E= mc2, where c is speed of light in

vacuum. In a nuclear process mass gets converted to energy (or vice-versa). This

is the energy which is released in a nuclear power generation and nuclearexplosions.

Energy is a scalar quantity. But all conserved quantities are not necessarily

scalars. The total linear momentum and the total angular momentum (both

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derived from Newton’s laws of motion in mechanics. But their validity goes

beyond mechanics. They are the basic conservation laws of nature in all

domains, even in those where Newton’s laws may not be valid. Besides their great

simplicity and generality, the conservation laws of nature are very useful inpractice too. It often happens that we cannot solve the full dynamics of a complex

problem involving different particles and forces. The conservation laws can still

provide useful results. For example, we may not know the complicated forces that

act during a collision of two automobiles; yet momentum conservation law

enables us to bypass the complications and predict or rule out possible outcomes

of the collision. In nuclear and elementary particle phenomena also, the

conservation laws are important tools of analysis. Indeed, using the conservation

laws of energy and momentum for β-decay, Wolfgang Pauli (1900-1958) correctlypredicted in1931 the existence of a new particle (now called neutrino) emitted in

β-decay along with the electron. Conservation laws have a deep connection with

symmetries of nature that you will explore in more advanced courses in physics.

For example, an important observation is that the laws of nature do not change

with time! If you perform an experiment in your laboratory today and repeat the

same experiment (on the same objects under identical conditions) after a year,

the results are bound to be the same. It turns out that this symmetry of nature

with respect to translation (i.e. displacement) in time is equivalent to the law ofconservation of energy. Likewise, space is homogeneous and there is no

(intrinsically) preferred location in the universe. To put it more clearly, the laws of

nature are the same everywhere in the universe. (Caution: the phenomena may

differ from place to place because of differing conditions at different locations.

For example, the acceleration due to gravity at the moon is one-sixth that at the

earth, but the law of gravitation is the same both on the moon and the earth.)

This symmetry of the laws of nature with respect to translation in space gives

rise to conservation of linear momentum. In the same way isotropy of space (nointrinsically preferred direction in space) underlies the law of conservation of

angular momentum*. The conservation laws of charge and other attributes of

elementary particles can also be related to certain abstract symmetries.

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Symmetries of space and time and other abstract symmetries play a central role

in modern theories of fundamental forces in nature.

Exercise

1.1 Some of the most profound statements on the nature of science have comefrom Albert Einstein, one of the greatest scientists of all time. What do you think

did Einstein mean when he said : “The most incomprehensible thing about the

world is that it is comprehensible”?

1.2 “Every great physical theory starts as a heresy and ends as a dogma”. Give

some examples from the history of science of the validity of this incisive remark.

1.3 “Politics is the art of the possible”. Similarly, “Science is the art of the

soluble”. Explain this beautiful aphorism on the nature and practice of science.1.4 Though India now has a large base in science and technology, which is fast

expanding, it is still a long way from realising its potential of becoming a world

leader in science. Name some important factors, which in your view have

hindered the advancement of science in India.

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1.5 No physicist has ever “seen” an electron. Yet, all physicists believe in the

existence of electrons. An intelligent but superstitious man advances this analogy

to argue that ‘ghosts’ exist even though no one has ‘seen’ one. How will you refute

his argument?

1.6 The shells of crabs found around a particular coastal location in Japan seem

mostly to resemble the legendary face of a Samurai. Given below are two

explanations of this observed fact. Which of these strikes you as a scientific

explanation?

(a) A tragic sea accident several centuries ago drowned a young Samurai. As a

tribute to his bravery, nature through its inscrutable ways immortalised his face

by imprinting it on the crab shells in that area.

(b) After the sea tragedy, fishermen in that area, in a gesture of honour to their

dead hero, let free any crab shell caught by them which accidentally had a shape

resembling the face of a Samurai. Consequently, the particular shape of the crab

shell survived longer and therefore in course of time the shape was genetically

propagated. This is an example of evolution by artificial selection.

[Note: This interesting illustration taken from Carl Sagan’s ‘The Cosmos’

highlights the fact that often strange and inexplicable facts which on the firstsight appear ‘supernatural’ actually turn out to have simple scientific

explanations. Try to think out other examples of this kind].

1.7 The industrial revolution in England and Western Europe more than two

centuries ago was triggered by some key scientific and technological advances.

What were these advances ?

1.8 It is often said that the world is witnessing now a second industrial

revolution, which will transform the society as radically as did the first. List somekey contemporary areas of science and technology, which are responsible for this

revolution.

1.9 Write in about 1000 words a fiction piece based on your speculation on the

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1.10 Attempt to formulate your ‘moral’ views on the practice of science. Imagine

yourself stumbling upon a discovery, which has great academic interest but is

certain to have nothing but dangerous consequences for the human society. How,

if at all, will you resolve your dilemma ?

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2.1 Introduction

PHYSICS is the science of objective observation, an observation which is the

same for all individuals. An individual observes through his sense of touch or

sight, but these senses are not always reliable. To illustrate the inaccuracy of our

sense of touch, we consider three pans containing cold, warm and hot water. If

you put your finger in cold water and then in warm water, your sense of touch will tell you that it is hot. But if you put your finger first in hot water and then in

warm water, your sense tells you it is cold. This clearly suggests the necessity of

making a measurement to arrive at the truth. It is necessary to measure the

degree of hotness of water in each pan. In other words, it is not enough to

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describe a phenomenon in a general and qualitative way. A number must be tied

to it. Thus, physics is a science of measurement. Lord Kelvin, a leading physicist

of the 19th century, once said:

“When you measure what you are talking about and express it in numbers, you

know something about it: but when you cannot, your knowledge is of a meagre and

unsatisfactory kind ; it may be the beginning of knowledge, but you have scarcely

in your thoughts advanced to the stage of science.”

2.2 Fundamental and Derived quantities

Physical quantities can be classified into two namely, fundamental quantities and

derived quantities.Fundamental quantities are quantities which cannot beexpressed in terms of any other physical quantity. For example quantities like

length, mass, time, temperature are fundamental quantities.Quantities that can

be expressed in terms of fundamental quantities are called derived quantities.

Area, volume, density etc. are examples of derived quantities.

2.3 Unit and Various Systems

To measure a quantity, we always compare it with some reference standard. To

say that a rope is 10 metres long is to say that it is 10 times as long as an object

whose length is defined as 1 metre. Such a standard is called a unit of the

quantity.

Therefore,unit of a physical quantity is defined as the established standard used

for comparison of the given physical quantity.

The units in which the fundamental quantities are measured are called

fundamental units and the units used to measure derived quantities are called

derived units.

2.3.1 System International de Units (SI system of units)

In earlier days, many systems of units were followed to measure physical

quantities. The British system of foot-pound-second or FPS system, the Gaussian

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system of centimetre - gram - second or CGS system, the metre - kilogram -

second or MKS system were the three systems commonly followed. To bring

uniformity, the General Conference on Weights and Measures in the year 1960,

accepted the SI system of units. This system is essentially a modification overMKS system and is, therefore rationalised MKSA (metre kilogram second ampere)

system.

This rationalisation was essential to obtain the units of all the physical

quantities in physics.

In the SI system of units there are seven fundamental quantities and two

supplementary quantities. They are presented in Table 1.1.

Table 2.1SI System of units

Fundamental quantities:

Physical quantity UnitSymbo

l

Length metre M

Masskilogra

mKg

Time second SElectric current ampere A

Temperature kelvin K

Luminous intensity candela cd

Amount of

substancemole mol

Supplementary quantities:

Physical quantity Unit Symbol

Plane angle radian rad

Solid angle steradian sr

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2.3.2 Uniqueness of SI system

The SI system is logically far superior to all other systems. The SI units have

certain special features which make them more convenient in practice.Permanence and reproducibility are the two important characteristics of any unit

standard. The SI standards do not vary with time as they are based on the

properties of atoms. FurtherSI system of units are coherent system of units, in

which the units of derived quantities are obtained as multiples or submultiples of

certain basic units. Table 1.2 lists some of the derived quantities and their units.

Table 2.2

Physical

quantity

Expression Unit

Area

Volume

Velocity

Acceleration

Angular

velocity

Angular

acceleration

Density

Momentum

Moment ofintertia

Force

Pressure

length × breadth

area x height

displacement/

time

velocity / time

angulardisplacement /

time

angular velocity /

time

mass / volume

mass × velocitymass x

(distance )2

m2

m3

m/s

m/s2

rad/s

rad/s2

kg/m3

kgm/

skgm2

N

N/m2

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mass x

acceleration

force/area

2.3.3 SI standards

Length

Length is defined as the distance between two points. The SI unit of length is

metre.

One standard metre is equal to 1650 763.73 wavelengths of the orange- red light

emitted by the individual atoms of krypton- 86 in a krypton discharge lamp.

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Mass

Mass is the quantity of matter contained in a body. It is independent of

temperature and pressure. It does not vary from place to place. The SI unit of

mass is kilogram.

The kilogram is equal to the mass of the international prototype of the kilogram (a

platinum- iridium alloy cylinder) kept at the International Bureau of Weights and

Measures at Sevres, near Paris, France.

An atomic standard of mass has not yet been adopted because it is not yet

possible to measure masses on an atomic scale with as much precision as on amacroscopic scale.

Time

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Until 1960 the standard of time was based on the mean solar day, the time

interval between successive passages of the sun at its highest point across the

meridian. It is averaged over a year. In 1967, an atomic standard was adopted

for second, the SI unit of time.

One standard second is defined as the time taken for 9192631770 periods of the

radiation corresponding to unperturbed transition between hyperfine levels of the

ground state of caesium –133 atom.

Atomic clocks are based on this. In atomic clocks, an error of one second occurs

only in 5000 years.

Ampere

The ampere is the constant current which, flowing through two straight parallel

infinitely long conductors of negligible cross-section, and placed in vacuum 1 m

apart, would produce between the conductors a force of 2 × 10-7newton per unit

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The Kelvin is the fraction of 1/273.6 of the thermodynamic temperature of the triple

point of water*

* Triple point of water is the temperature at which saturated water vapour, pure

water and melting ice are all in equilibrium. The triple point temperature of water is

273.16 K.

Candela

The candela is the luminous intensity in a given direction due to a source, which

emits monochromatic radiation of frequency 540 × 1012 Hz and of which the

radiant intensity in that direction is 1/683 watt per steradian.

Mole

The mole is the amount of substance which contains as many elementary entities

as there are atoms in 0.012 kg of carbon -12.

2.3.4 Rules and conventions for writing SI units and their symbols

1. The units named after scientists are not written with a capital initial letter.

For example : newton, henry, watt

2. The symbols of the units named after scientist should be written by a capital

letter.

For example: N for newton, H for henry, W for watt

3. Small letters are used as symbols for units not derived from a proper name.

For example: m for metre, kg for kilogram

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4. No full stop or other punctuation marks should be used within or at the end of

symbols.

For example: 50 m and not as 50 m.

5. The symbols of the units do not take plural form.

For example: 10 kg not as 10 kgs

6. When temperature is expressed in kelvin, the degree sign is omitted.

For example: 273 K not as 273o

(If expressed in Celsius scale, degree sign is to be included. For example 100oC

and not 100 C)

7. Use of solidus is recommended only for indicating a division of one letter unit

symbol by another unit symbol. Not more than one solidus is used.

For example: m s-1 or m/s, J/K mol or JK-1mol-1 but not J/K/mol.

8. Some space is always to be left between the number and the symbol of the

unit and also between the symbols for compound units such as force,

momentum, etc.

For example, it is not correct to write 2.3m. The correct representation is 2.3 m;kg m s-2and not as kgms-2

9. Only accepted symbols should be used.

For example : ampere is represented as A and not as ‘amp’ or ‘am’ ; second is

represented as ‘s’ and not as ‘sec’.

10. Numerical value of any physical quantity should be expressed in scientific

notation.

For an example, density of mercury is

1.36 × 104

kg m-3

and not as 13600 kg m-3

2.3.5 Expressing larger and smaller physical quantities

Once the fundamental units are defined, it is easier to express larger and smaller

units of the same physical quantity. In the metric (SI) system these are related to

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the fundamental unit in multiples of 10 or 1/10. Thus 1 km is 1000 m and 1

mm is 1/1000 metre. Table 1.3 lists the standard SI prefixes, their meanings and

abbreviations. In order to measure very

large distances, the following units are used.

(i) Light year

Light year is the distance travelled by light in one year in vacuum.

Distance travelled = velocity of light × 1 year

1 light year = 3×108 ms-1×1 year(inseconds) = 3 × 108 × 365.25 × 24 × 60 × 60

= 9.467 × 10

15

m

1 light year = 9.467 × 1015 m

Table 2.3Prefixes for power of ten

Power of

10

Prefix Abbreviatio

n10-15

10-12

10-9

10-6

10-3

10-2

10-1

101

102

103

106

femto

pico

nano

micro

milli

centi

deci

deca

hecto

kilo

mega

f

p

n

µ

m

c

c

da

h

k

M

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109

1012

1015

giga

tera

peta

G

T

P

(ii) Astronomical unit

Astronomical unit is the mean distance of the centre of the Sun from the centre

of the Earth.

1 Astronomical unit (AU) = 1.496 × 1011m

2.4 Measurement

Physics can also be defined as the branch of science dealing with the study of

properties of materials. To understand the properties of materials, measurement

of physical quantities such as length, mass, time etc., are involved. Theuniqueness of physics lies in the measurement of these physical quantities.

2.4.1 Determination of distance

For measuring large distances such as the distance of moon or a planet from the

Earth, special methods are adopted. Radio-echo method, laser pulse method and

parallax method are used to determine very large distances.

Laser pulse method

The distance of moon from the Earth can be determined using laser pulses. The

laser pulses are beamed towards the moon from a powerful transmitter. These

pulses are reflected back from the surface of the moon. The time interval between

sending and receiving of the signal is determined very accurately.

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Ift is the time interval andc the velocity of the laser pulses, then the distance of

the moon from the Earth isd =ct/2

2.4 Determination of mass

The conventional method of finding the mass of a body in the laboratory is by

physical balance. The mass can be determined to an accuracy of 1 mg. Now-a-

days, digital balances are used to find the mass very accurately. The advantage of

digital balance is that the mass of the object is determined at once.

2.5 Measurement of time

We need a clock to measure any time interval. Atomic clocks provide better

standard for time. Some techniques to measure time interval are given below.

Quartz clocks

The piezo-electric property* of a crystal is the principle of quartz clock. These

clocks have an accuracy of one second in every 10 seconds.

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*When pressure is applied along a particular axis of a crystal, an electric potential

difference is developed in a perpendicular axis.

Atomic clocks

These clocks make use of periodic vibration taking place within the atom. Atomic

clocks have an accuracy of 1 part in 10

2.6 Accuracy and precision of measuring instruments

All measurements are made with the help of instruments. The accuracy to which

a measurement is made depends on several factors.

For example, if length is measured using a metre scale which has graduations at

1 mm interval then all readings are good only up to this value.The error in the

use of any instrument is normally taken to be half of the smallest division on the

scale of the instrument. Such an error is called instrumental error. In the case of ametre scale, this error is about 0.5 mm.

Physical quantities obtained from experimental observation always have some

uncertainty. Measurements can never be made with absolute precision. Precision

of a number is often indicated by following it with± symbol and a second number

indicating the maximum error likely.

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For example, if the length of a steel rod = 56.47±3 mm then the true length is

unlikely to be less than 56.44 mm or greater than 56.50 mm.If the error in the

measured value is expressed in fraction, it is called fractional error and if

expressed in percentage it is called percentage error. For example, a resistorlabelled “470 Ω, 10%” probably has a true resistance differing not more than

10% from 470 Ω. So the true value lies between 423 Ω and 517 Ω.

2.6.1 Significant figures

The digits which tell us the number of units we are reasonably sure of having

counted in making a measurement are called significant figures. Or in other

words,the number of meaningful digits in a number is called the number ofsignificant figures. A choice of change of different units does not change the

number of significant digits or figures in a measurement.

For example, 2.868 cm has four significant figures. But in different units, the

same can be written as 0.02868 m or 28.68 mm or 28680 µm. All these numbers

have the same four significant figures.

From the above example, we have the following rules.

i) All the non-zero digits in a number are significant.

ii) All the zeroes between two non-zeroes digits are significant, irrespective of thedecimal point.

iii) If the number is less than 1, the zeroes on the right of decimal point but to

the left of the first non-zero digit are not significant. (In 0.02868 the underlined

zeroes are not significant).

iv) The zeroes at the end without a decimal point are not significant. (In 23080

µm, the trailing zero is not significant).

v) The trailing zeroes in a number with a decimal point are significant. (The

number 0.07100 has four significant digits).

Examples

i) 30700 has three significant figures.

ii) 132.73 has five significant figures.

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iii) 0.00345 has three and

iv) 40.00 has four significant figures.

2.6.2 Rounding off

Calculators are widely used now-a-days to do the calculations. The result given

by a calculator has too many figures. In no case the result should have more

significant figures than the figures involved in the data used for calculation. The

result of calculation with number containing more than one uncertain digit,

should be rounded off. The technique of rounding off is followed in applied areas

of science.

A number 1.876 rounded off to three significant digits is 1.88while the number1.872 would be 1.87. The rule is that if the insignificant digit (underlined) is

more than 5, the preceding digit is raised by 1, and is left unchanged if the

former is less than 5.

If the number is 2.845, the insignificant digit is 5. In this case, the convention is

that if the preceding digit is even, the insignificant digit is simply dropped and, if

it is odd, the preceding digit is raised by 1. Following this, 2.845 is rounded off to

2.84 where as 2.815 is rounded off to 2.82.

Examples

1. Add 17.35 kg, 25.8 kg and 9.423 kg. Of the three measurements given, 25.8

kg is the least accurately known.

∴ 17.35 + 25.8 + 9.423 = 52.573 kg

Correct to three significant figures, 52.573 kg is written as 52.6 kg

2. Multiply 3.8 and 0.125 with due regard to significant figures.

3.8 × 0.125 = 0.475

The least number of significant figure in the given quantities is 2.

Therefore the result should have only two significant figures.

∴ 3.8 × 0.125 = 0.475 = 0.48

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2.6.3 Errors in Measurement

The uncertainty in the measurement of a physical quantity is called error. It is

the difference between the true value and the measured value of the physical

quantity. Errors may be classified into many categories.(i) Constant errors

It is the same error repeated every time in a series of observations. Constant

error is due to faulty calibration of the scale in the measuring instrument. In

order to minimise constant error, measurements are made by different possible

methods and the mean value so obtained is regarded as the true value.

(ii) Systematic errors

These are errors which occur due to a certain pattern or system. These errorscan be minimised by identifying the source of error.

Instrumental errors, personal errors due to individual traits and errors due to

external sources are some of the systematic errors.

(iii) Gross errors

Gross errors arise due to one or more than one of the following reasons.

(1) Improper setting of the instrument.

(2) Wrong recordings of the observation.

(3) Not taking into account sources of error and precautions.

(4) Usage of wrong values in the calculation.

Gross errors can be minimised only if the observer is very careful in his

observations and sincere in his approach.

(iv) Random errors

It is very common that repeated measurements of a quantity givevalues whichare slightly different from each other.These errors have no set pattern and occur

in a random manner. Hence they are called random errors. They can be

minimised by repeating the measurements many times and taking the arithmetic

mean of all the values as the correct reading. The most common way of

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expressing an error is percentage error. If the accuracy in measuring a quantity

x is∆x, then the percentage error inx is given by -

(x/x)∆ × 100 %

1.6.4 Different types of errors:

(i) Absolute error : Absolute error in the measurement of a physical quantity is

the magnitude of the difference between the true value and the measured value

of the quantity.

Let a physical quantity be measured n times. Let the measured values be a1, a2,

….,an.

The arithmetic mean of these values is

am=a1+a2+…+an

n

Usually, am is taken as the true value of the quantity, if the same is unknown

otherwise.

By definition, absolute errors in the measured values of the quantity are

∆a1 = am – a1,∆a2 = am – a2………

∆an = am – an The absolute errors may be positive in certain cases and negative in certain other

cases.

(ii)Mean absolute error :It is the arithmetic mean of the magnitudes of absolute

errors in all the measurements of the quantity. It is represented by a∆ . Thus

∆a1 +∆a2 + …….. +∆ana∆ = ----------------------------

nHence the final result of measurement may be written as ( a = am ± a∆ ).

This implies that any measurement of the quantity is likely to lie between (a = am+ a∆ ) and (a = am - a∆ ).

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(iii) Relative error or Fractional error : The relative error or fractional error of

measurement is defined as the ratio of mean absolute error to the mean value of

the quantity measured.

Thus Relative error or Fractional error is given by Mean absolute error a∆

= -------------------------- = ----

Mean value am

(iv)Percentage error : When the relative/fractional error is expressed in

percentage, we call it percentage error. Thus

a∆

Percentage error = ------ x 100 %am

1.6.4.1 Propagation of Errors :.

(1)Error in sum of the quantities :

Suppose x = a + b and

a = absolute error in measurement of a∆

b = absolute error in measurement of b∆

x = absolute error in calculation of x, i.e. sum of a and b.∆

The maximum absolute error in x is

x = ± (a +b)∆ ∆ ∆

Percentage error in the value of x is

(a + b)∆ ∆

= --------------- x 100 %

a + b

Example 1

Two resistances R1 = (100 ± 3) ohm and R2 = (200 ± 4) ohm are connected in

series. Find the equivalent resistance.

Solution : Here,

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R1 = (100 ± 3) ohm and

R2 = (200 ± 4) ohm

When resistances are connected in series,

R = R1 + R2= (300 ± 7) ohm.

(2) Error in difference of the quantities : Supposex = a – b and

a=absolute error in measurement of a,∆

b=absolute error in measurement of b,∆

x = absolute error in calculation of x, i.e. difference of ‘a’ and ‘b’.∆

The maximum absolute error in x is

x = ± (a + b)∆ ∆ ∆

Percentage error in the value of x is(a + b)∆ ∆

= --------------- x 100 %

a – b

Example 2

The external and internal diameters of a hollow cylinder are determined with a

Vernier Callipers. The results recorded as 4.23 ± 0.01 cm and 3.89 ± 0.01 cm.

Determine the thickness of the cylinder wall with error limits.

Solution :Here d1= 4.23 ± 0.01 cm andd2= 3.89 ± 0.01 cm

Now, the thickness of the cylinder wall

= d1 – d2= (4.23 ± 0.01) – (3.89 ± 0.01)

= (0.34 ± 0.02) cm.

(3) Error in product of quantities:Supposex = a × b and


b=absolute error in measurement of b∆

x =absolute error in calculation of x i.e. product of ‘a’ and ‘b’.∆

Then, the maximum fractional error in x is

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x∆ a b∆ ∆

---- = ± ----- + -----

x a b

Example 3

A potential difference, V = (100 ± 2) volt, when applied across a resistance R,

gives a current I = (10 ± 0.5) ampere. Calculate the percentage error in R. Given

R = V/I.

Solution:

Here V = (100 ± 2) volt

I = (10 ± 0.5) ampere

R = V/I = 100/10 = 10 ohmNow, relative error in resistance

∆R ∆V ∆I

--- = ---- + ---

R V I

= ± [ 2/100 + 0.5/10]

= ± 0.07

% Error in R = ± 0.07 x 100 = 7%

(4) Error in division of quantitiesSupposex = a/b and

a = absolute error in measurement of a,∆

b = absolute error in measurement of b∆

x = absolute error in calculation of x, i.e. division of ‘a’ and ‘b’.∆

The maximum fractional error in x is x a b∆ ∆ ∆

= ---- = ± ----- + ----- x a b

(5) Error in quantity raised to power:

Suppose x = am/bn and

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b=absolute error in measurement of b∆

x=absolute error in calculation of x i.e. division of ‘a’ and ‘b’.∆

The maximum fractional error in x is x m a n b∆ ∆ ∆

= ---- = ± ------- + -------

x a b

.’. % error in the value of ‘x’ = m (% error in value of a) + n (% error in value of b)

*Note:The quantity which have maximum power must be measured carefully

because it's contribution to error is maximum.

Example 4

An experiment measures quantities a, b and c and X is calculated from the

formula:

X = ab2/c3

The percentage errors in a, b and c are ±1%, ±3% and ±2% respectively. What is

the percentage error in X?

Solution : Here X = ab2/c3

∆a /a = ± 1%,∆b/b = ± 3%,

∆c/c = ± 2%.

% error in X is given by

(∆X/X) =[(∆a/a)+2(∆b/b)+3(∆c/c)]x100%

= ± [1% +2 x 3% + 3 x 2%]

= ± 13%.

1.7 Dimensional Analysis

Dimensions of a physical quantity are the powers to which the symbols of

fundamental quantities must be raised to specify those quantities.

We know that

velocity =displacement / time

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Dimensionally [v]=[L]/[T] =[M0LT-1]

Here [M], [L] and [T] are the dimensions of the fundamental quantities mass,

length and time respectively. Therefore velocity has zero dimensions in mass, one

dimension in length and – 1 dimension in time. Thus the dimensional formulafor velocity is [M0LT-1] or simply [LT-1]. The dimensions of fundamental quantities

are given in Table 1.4 and the dimensions of some derived quantities are given in

Table 1.5

Table 1.4Dimensions of fundamental quantities

Fundamental qunatity Dimensions

Length LMass M Time T Temperature KElectric current ALuminous intensity cd Amount of substance mol

Table 1.5Dimensional formulae of some derived quantities

Physical

Quantity

Mathematical

Formula

Dimension

al Formula Area

Denisty

Acceleration

Momentum

Force

Work/energy

Power

Radius of

gyration

Pressure

Length x Breadth

Mass/Volume

Velocity/Time

Mass x Velocity

Mass x

AccelerationForce x

displacement

Work/time

(Distance)2

[L2]

[L3]

[LT-2]

[MLT-1]

[MLT-2]

[ML2 T-2]

[ML2 T-3]

[L2]

[ML-1 T-2]

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Surface

Tension

Frequency

Torque

Force/area

Force/length

1/time period

Force x

displacement

[MT-2]

[T-1]

[ML

2

T

-2

]

Dimensional quantities

Constants which possess dimensions are called dimensional constants. Planck’s

constant and universal gravitational constant are dimensional constants.

Dimensional variables are those physical quantities which possess dimensions

but do not have a fixed value. Example - velocity, force, etc.

Dimensionless quantities

There are certain quantities which do not possess dimensions. They are called

dimensionless quantities. Examples are strain, angle, specific gravity, etc. They

are dimensionless as they are the ratio of two quantities having the same

dimensional formula.

Principle of homogeneity of dimensions An equation is dimensionally correct if the dimensions of the various terms on

either side of the equation are the same. This is called the principle of homogeneity

of dimensions. This principle is based on the fact that two quantities of the same

dimension only can be added up, the resulting quantity also possessing the same

dimension.

The equation A + B = C is valid only if the dimensions of A, B and C are the

same.

1.7.1 Uses of dimensional analysis

The method of dimensional analysis is used to

(i) convert a physical quantity from one system of units to another.

(ii) check the dimensional correctness of a given equation.

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(iii) establish a relationship between different physical quantities in an equation.

(i) To convert a physical quantity from one system of units to another

The knowledge of physical quantities and their dimensions helps one to change a

particular quantity from one system of units to another as seen from the example

below.

Example 5

Given the value of G in cgs system is 6.67×10-8 dyne-cm2g-2. Calculate its value in

SI units.

Solution : In cgs system, G = 6.67 × 10-8 In SI system, G = ?M1 = 1g M2 = 1 kg

L1 = 1 cm L2 = 1m

T1 = 1s T2= 1 s

The dimensional formula for gravitational constant is [M-1L3 T-2]

In CGS system dimensional formula for G is [M1 xL1 y T1z]

In SI system dimensional formula for G is [M2 xL2 y T2z]

Here x = -1, y = 3 and z = -2

∴ G [M2 xL2 y T2z] = Gcgs [M1 xL1 y T1z]Or G = Gcgs[M1/M2] x [L1/L2] y [T1/T2]z

= 6.67x10-8[1g/kg]-1[1cm/m]-3[1s/s]-2

=6.67x10-8[1/1000]-1[1/100]-3[1/1]-2

= 6.67 x 10-11 N m2 kg-2.

(ii) To check the dimensional correct -ness of a given equation

Let us take the equation of motion

s =ut + ½at2

Applying dimensions on both sides,

[L] = [LT]-1[T] + [LT]-2 [T2]

(½ is a constant having no dimension)

[L] = [L] + [L]

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As the dimensions on both sides are the same, the equation is dimensionally

correct.

(iii) To establish a relationship between the physical quantities in an equation

The concept of dimensions helps us to derive a relation between various physical

quantities. Let us find an expression for the time periodT of a simple pendulum.

The time periodT may depend upon (i) massm of the bob

(ii) length ’l’of the pendulum and (iii)acceleration due to gravityg at the place

where the pendulum is suspended.

(i.e)Tαmxlygz ------------- (1)

OrT =k mxlygz

Wherekis a dimensionless constant of proportionality. Rewriting equation (1)

with dimensions,

[T1] = [Mx][Ly][LT−2]z

[T1] = [MxLy+zT−2z]

Comparing the powers of M, L and T on both sidesx= 0,y+z= 0 and−2z=1

Solving forx,yandz,

x= 0,y= ½ andz= –½

From equation (1),

T=km0l½g−½

T = k [l/g]1/2= k √ l /g

Experimentally the value ofkis determined to be 2π.

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∴ T =2π √ l

g

Do You Know? When dimensions are given, the

physical quantity may or may not be

unique. However, if a physical quantity

is given, its dimensions are unique.

1.7.2 Limitations of Dimensional Analysis

(i) The value of dimensionless constants cannot be determined by this method.

(ii) This method cannot be applied to equations involving exponential andtrigonometric functions.

(iii) It cannot be applied to an equation involving more than three physical

quantities.

(iv) It can check only whether a physical relation is dimensionally correct or not.

It cannot tell whether the relation is absolutely corrector not. For example

applying this techniques = ut + ¼ at2 is dimensionallycorrect whereas the

correct relation is

s = ut + ½ at2

Exercices

Multiple Choice Questions

1.The number of significant figures in 5418000 are –

(i) 4 (ii) 5

(iii)6 (iv) 7

2.The length and breadth of a metal sheet are 3.124m and 3.002mrespectively. The area of this sheet

up to four correct significantfigures is -(i) 9.37 m2 (ii) 9.378 m2

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(iii) 9.3782 m2 (iv) 9.37824 m2

3.The percentage error in the measurement of mass and speed are 2 % and

3% respectively. The error in kinetic energy obtained by measuring mass and

speed will be(i) 12% (ii) 10%(iii) 8% (iv) 5%4.A force of F is applied on to a square plate of side L. If percentage error in

determining L is 2% and that in F is 4%, what is the permissible error in

pressure?(i)2% (ii) 4%(iii)6% (iv) 8%

5.A physical quantity P is related to four variables a, b, c and d as follows:

P = a3 b2/√cd The percentage errors in a, b, c and d are 1%, 3%, 4% and 2% respectively.

What is the error in quantity P ?(i) 12% (ii) 16%(iii) 20% (iv) 24%6.Which of the following pair does not have similar dimensions?(i) Stress and pressure(ii) Angle and strain(iii) Tension and surface tension

(iv) Planck’s constant and angular momentum7.Which of the following is a dimensional constant?(i) Refractive index(ii) Poisson’s ratio(iii) Relative density(iv) Gravitational constant8.A sphere of radius a moves with velocity v in a medium and the force F

acting on it is given by : . F = 6πηav The dimensions of ‘η’ will be

(i) ML-1 T-1 (ii) MT-1(iii) MLT-2 (iv) ML-3

9.Two quantities A and B have different dimensions. Which mathematical

operation given below is physically meaningful?(i) A/B (ii) A + B(iii) A – B (iv) None

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10.In the standard equation:snth = u +1/2 [2n – 1]

What dimensions do you give to Snth?(i) MoLTo

(ii) MLT-1

(iii) MoL-1 T(iv) MoLo T-1

Exercise Questions :

1. Write the dimensional formula for the following a) Impulse b) Kinetic energy.2.Can there be a physical quantity which has no units and no dimensions?

3.Energy density and pressure have the same dimensions. Comment.4.How many ergs are there in 1 kilo-watt hour?5. Write the number of significant figures of the following measuring

quantities :(i) 400 m (ii) 0.008 Kg

6.Give three examples of dimensionless variables?7. The mass of a body is measured by two persons is 10.2 kg and 10.23 kg.

Which one is more accurate and why?8. The value of Stefan’s constant is 5.67x 10-8 Js-1m-2K-4.Find its value in CGS

system.9.Convert a power of 1 MW on a system with fundamental units 10 kg, 1dm

and 1 min.10. Volume V of water which passes any point of a canal during ‘t’ second is

connected with the cross section A of the canal and the velocity u of water

by the relation

v =K Atu

where ‘K’ is dimensionless constant. Verify the correctness of the relation.

Answers to MCQ:

1.(i) 2.(ii) 3.(iii) 4.(iv) 5.(i) 6.(iii) 7.(iv) 8.(i)

9.(i) 10.(ii)

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3.1 Introduction

Rectilinear Motion − What it means?

Motion of a body that moves along a straight line such as the motion of a carmoving on a straight road.

Position − How to Describe it?

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• Locating an object requires finding its position relative to a reference point.

• Reference point is often taken as the origin of a coordinate system.

• The coordinates (x, y, z) of the object describe the position of the object

with respect to the coordinate axes.

• Coordinate system along with time constitutes a frame of reference.

Path Length

• Length of the actual path traversed by a body in a given time

• It is ascalar quantity. Therefore, only magnitude is important, not the

direction of movement. (Implies that path length can never be negative)

Displacement

• A change of position ∆R from coordinateR1 (x1, y1, z1) to coordinateR2(x2, y2, z2)

• This is the shortest distance between the initial and final position of a body.

• It is a vector quantity. Therefore, both magnitude and direction are

important to describe displacement. (Implies that displacement can be

negative depending on the initial and final positions of a body in a coordinate

system).

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How Path Length and Displacement are Different?

Consider this example.

If an object goes from A to B and then B to C in time ‘t’, then

• path length = AB + BC (arithmetic sum of the distances)

• displacement,∆x = AC−→− (shortest distance between points A and C)

Example 3.1

A particle moves along a circle of radius ‘r’. It starts from A and moves clockwise.Calculate the distance travelled by the particle and its displacement in each case. Take centre of the circle as the origin.

i. (i) From A to B (ii) From A to C

ii. (iii) From A to D

iii. (iv) In one complete revolution

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Solution:

(i) Distance travelled by the particle from A to BDisplacement =∣ AB∣=√OA2+OB2

=√r2+r2

=2√r Direction is along negative X and Y axis.

(ii) Distance travelled by particle from A to C is

Displacement =|AC|= 2r Direction is along negative X axis.

(iii) Distance travelled from A to D

Displacement=∣ AD∣=√r2+r2

=2√r Direction is along Positive X and Y axis.

(iv) For one complete revolution i.e., motion from A to A,

Total distance travelled = 2πr

Displacement = 0 [the final position coincides with the initial position]

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Example 3.2

A man is walking from his office to home. He leaves his office and walks 400 m

west. He then turns and walks 700 m north. Determine the magnitude anddirection of his resultant displacement?Solution:

His resultant displacement is given by:

S = √4002+7002

S = √ 160000+490000

S = √ 650000

S = 806.23 meters.

Though total distance travelled is 1100 m but resultant displacement is 806.23 m

in North – West direction.

Example 3.3

A bus is heading city B from city A via city C. The bus started from city A andmoved towards city C in south direction at the distance of 100 km. It then turnsleft and moves for another 200 km before reaching the city B. Find the totaldistance travelled by the bus and the magnitude and resultant of thedisplacement of the bus from city A?

Solution:1. Total distance travelled by the bus is . D = 100 + 200 = 300 km2. Magnitude of the displacement is,|S| = √,1002+2002

= √,50000−−−−−= 223.61 km

3. Direction of resultant displacement is south – west.

Speed

• Speed =

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• It is a scalar quantity, which means that it requires no direction (it impliesthat speed cannot be negative).

• Instantaneous speed is the speed at a particular instant (when the intervalof time is infinitely small).

i.e., instantaneous seed:

Average Speed

• Average speed of a particle is defined as the total distance travelled by theparticle divided by the total time taken during which the motion took place.

• Suppose that a car is covering a distance of 160 km from A to B and coverssuccessive 40 km distances in time 1.2 h, 1.4 h, 1.6 h, and 0.9 h respectively.

The speed of car is different at different at every successive interval. In suchcases, we need to find the average speed. Average speed = Total distance travelled Total time taken

Average Velocity

• Average velocity is the ratio of the change in displacement x∆ to the timeinterval t∆ in which the change in displacement occurred.

• Ifx1 andx2 are the positions of a particle at timest1 andt2 respectively thenthe magnitude of displacement of the particle in time interval∆t = (t2 -t1) is

x→∆

= x2 – x1 Average velocity vavg = Displacement / Time taken

= (x2 – x1) / (t2 - t1)=x /t∆ ∆

Velocity

• Velocity =

• It is a vector quantity. Therefore, the direction of movement is taken intoconsideration (it implies that velocity contains algebraic sign).

• In aposition-time graph, the slope of the curve indicates the velocity andthe angle of the slope with thex-axis indicates the direction.

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• Average velocity is given by :

wherex2 andx1 are the positions of the object att2 andt1respectively• Instantaneous velocity is at a given instant (slope at a particular point onthex – tcurve).

• When the motion is not uniform sometimes instantaneous velocity has moreimportance than average velocity.

Uniform Motion:

In uniform motion, a body undergoes equal displacements in equal intervals oftime.

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Acceleration

• Acceleration is the rate of change of the velocity of an object.• Average acceleration, a=Change in velocity/Time taken

= (v2-v1) /(t2-t1) =v/t∆ ∆

• In a velocity-time graph, the slope of the curve indicates the averageacceleration and the angle of the slope indicates the direction of change of the velocity.

• Instantaneous acceleration is the given at a particular instant (slope at aparticular point on (v – t)curve

In this case, the rate of change of velocity with time remains constant.Graphically, such motion can be represented as :

Graphical Representation of Accelerated Motion

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Terminology

u → Initial velocity ,v → Final velocity

a → Acceleration ,t → Time

x0→ Initial position ,x→ Position at ‘t’

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Equations of MotionCalculus Method

(i) Velocity − Time Relation

Acceleration,

a

or dv =adt

Integrating the above,

⇒ v − u =at

.’.v = u + at - - - - - - - - (1)

(ii) Displacement−Time Relation

Instantaneous velocity,

or dx =vdt

or dx = (u +at)dt [ ‘.’ eqn – 1]

Integrating the above relation,

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or (x − x0) =ut + ½ at2

.’. x =x0 +ut + ½ at2- ------------ (2)

(iii) Velocity−Displacement Relation

The acceleration is given as :

or a dx =v dv

Integrating the above expression,

or

or a ( x− x0 )=v2

2 −

u2

2

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i.e., v2 −u2 = 2a (x −x0) ------------ (3)

Equations of Motion (Graphical method)

Let an object is moving with uniform acceleration.

Let the initial velocity of the object = uLet the object is moving with uniform acceleration ‘a’Let object reaches at point B after time, t and its final velocity becomes ‘v’Draw a line parallel to x-axis DA from point, D from where object starts moving.Draw another line BA from point B parallel to y-axis which meets at E at Y-axis.Let OE = time, t

Now, from the graph,

BE = AB + AE

⇒ v = DC + OD (‘.’ AB = DC & AE = OD)

⇒ v = DC + u (‘.’ OD = u)

⇒ v = DC + u ------------------- (i)

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Above equation is the relation among initial velocity (u), final velocity (u),

acceleration (a) and time (t). It is calledfirst equation of motion.

Now, the distance covered by the object in the given time ‘t’ is given by the area of

the trapezium ABDOE

Let in the given time, t the distance covered by the moving object = s =

The above expression gives the distance covered by the object moving with

uniform acceleration. This expression is known assecond equation of motion.

The distance covered by the object moving with uniform acceleration is given by

the area of trapezium ABDO Therefore,

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The above expression gives the relation between position and velocity and is

called thethird equation of motion.

Motion of an Object under Free Fall:

As object is released from rest, thus,u = 0 The equations of motion become –v = 0 −gt = −9.8t m s−1

y = 0 – ½gt2 = −4.9t2 mv2 = 0 − 2gy = −19.6y m s−1

Note

Distance travelled innth second ofuniformly accelerated motion is given by the relation,

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Example 3.4

A particle starts with an initial velocity 5.0 ms−1along the positivex-direction andit accelerates uniformly at the rate of 2 ms−2.(i) Find the distance travelled by it in the first three seconds.(ii) How much time does it take to reach the velocity 9.0 ms−1?(iii) How much distance will it cover in reaching the velocity 9.0 ms−1?

Solution:Hereu = 5 ms−1,a = 2 ms−2

(i) We have to calculateS, whent = 3 s

(ii) We have to calculatet, when v = 9 ms−1

v=u +at 9 = 5 + (2)t

9 − 5 = 2t

⇒ 2t = 4 t = 2 s(iii) Here,v = 9 m s−2,S = ?

v2

=u2

+ 2aS (9)2 = (5)2 + 2 × 2 ×S 81 = 25 + 4S ⇒ 4S = 81 − 25 ⇒ 4S = 56 ⇒ S = 14 m

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Note: For computation purpose we take the value of acceleration due togravity(g) in the downward motion of a body as positive and in the upwardmotion as negative.

Example 3.5

A ball is thrown vertically upwards with a velocity of 30 ms−1 from the top of amulti-storeyed building. The height of the point from where the ball is thrown is20.0 m from the ground.(i) How high will the ball rise?(ii) How long will it be before the ball hits the ground? Takeg = 10 ms−2

Solution:

(i) For vertical motion,u = 30 ms

−1

a = −10 ms−2 (Upward motion)v = 0 (Highest point)S = ?t =t1 (say)

As v2 =u2 + 2aS,⇒ 0 = (30)2 + 2 (−10)S⇒ S = 45 m As v =u +at,

0 = 30 + (−10)t1⇒ t1 = 3 s(ii) For vertical downward motion,

u = 0 ms− 1 ,a = 10 ms− 2

.’. S = (45 + 20) m = 65 mt=t2 (say)

As ,

⇒

⇒ = 13 s

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⇒

∴ Total time =t1+t2=(3 + 3.6) s = 6.6 s

Relative Velocity

The relative velocity of a body A with respect to another bodyB ( v AB) is the timerate at which Achanges its position with respect toB.

(a) When both bodies move in same direction

If A andB are moving in the same direction, then the resultant relative velocity is: v AB = v A – vB(b) When both bodies move in opposite directions

If A andB are moving in the opposite directions, then the resultant relative

velocity is : v AB = v A + vB

Example 3.6

Two carsX and Yare moving with speeds 60 kmh−1 and 80 kmh−1 respectivelyalong parallel straight paths. Both the cars started from the same position. Whatis the position of carX with respect to Y after 15 minutes?Solution:

Speed of carX,v x = 60 km h−1

Speed of car Y,v y = 80 km h−1

Both the cars are moving in the same direction. Thus, relative velocity of carX with respect to car Y,vxy =vx −vy= (60 − 80) km h−1

= −20 km h−1

∴ Separation of carX with respect to car Y after 15 minutes is

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== −5 km[ Distance = Speed × Time]

Example 3.7 Two trains 110 m and 90 m in lengths are running in opposite directions with velocities 75 km h−1 and 64 km h−1. In what time will they completely cross eachother?Solution:

Here,v A = 75 km h−1

vB = −64 km h−1 [ ‘.’ Trains are moving in opposite directions]Length of train A,l A = 110 mLength of trainB,lB = 90 mRelative velocity of the two trains is :

v AB =v A −vB = 75 − (−64)= (75 + 64) km h−1

= 139 km h−1

= 139 km h−1

= 38.6 m s−1

Total distance to be travelled by each train for completely crossing the other train= (110 + 90) = 200 m∴ Time taken by each train to cross the other train

= 5.2 s

Motion along a Circular Path:

Motion of an object along a circular path is called circular motion. Since, on a

circular path the direction of the object is changing continuously to keep it onthe path, the motion of the object is called accelerated motion.

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If the radius of circle is ‘r’, then Therefore, circumference = 2πrLet time ‘t’ is taken to complete one rotation over a circular path by any object,

Where, v = velocity, r = radius of circular path and t = timeMotion of earth around the sun, motion of moon around the earth, motion of atop, motion of blades of an electric fan, etc. are the examples of circular moti

Centripetal Acceleration

Acceleration is the velocity change with respect to time. When it is constant for a

particular period of time then it is calledconstant acceleration. It can be of

average acceleration, instantaneous acceleration, when an object undergoes

circular motion then the acceleration is of three types that are centripetalacceleration, tangential, and angular acceleration.

The essential part of roller coaster is a curve or a circular path which

undergoescentripetal acceleration. Its direction is in the direction of centre of

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circular path. When we sit in train then the force which feel by us pushing us

toward the outer edge of the circular path, this is calledcentrifugal force. This

is not an actual force while it is inertia of our body or resistance to direction of

train. Here, we discuss about centripetal acceleration in which the direction istowards the centre of circular path and its two main factors that are tangential

and radial and its mathematical formula.

Centripetal Acceleration

"The centripetal acceleration is the rate of change of tangential velocity."

When an object is moving with uniform acceleration in circular direction, it is

said to be experiencing the centripetal acceleration.

To understand this type of acceleration clearly, lets understand three differenttypes of accelerations properly.

1.Linear Acceleration: Speed of the object changing while direction is constant.

2.Centripetal Acceleration: Speed of the object is constant while direction is

changing.

3. Angular Acceleration: Both the speed and direction of the object are changing.

There are a few key points to be noted about the centripetal acceleration :

1. The direction of the centripetal acceleration is always along the radius vector of the circular direction.

2. The magnitude of the centripetal acceleration can be calculated by the

tangential speed and angular velocity.

3. A object traveling in a circular direction experiences both the centripetal

and the angular acceleration.

4. The centripetal acceleration and angular acceleration are always 90o to

each other

Example 3.8 A 5-kg object moves at a constant speed of 10 m/s in a 5.0 m radius circle. Whatis the object's acceleration?Solution:

From the problem it is clear that we have following variables available with us,

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vt = 10 m/s;

r = 5 meters;

m = 5 kg

Now since object is moving in the circle so its acceleration is calculated bycalculating the centripetal acceleration

ac =v2t/r

ac = 102 /5 = 100/5

‘.’ ac = 20m/sec2

So, acceleration of the object is 20 m/s2

Example 3.9

If the tangential velocity of an object is increased by the factor of ‘2’, i.e. doubled,then what is the effect on its centripetal acceleration?

Solution:

From the equation of centripetal acceleration it is clear that the centripetal

acceleration is directly proportional to the square of the tangential velocity so if

the tangential velocity is increased by a factor of ‘2’ the centripetal acceleration is

increased by the factor of ’4’.

Example 3.10

If the radius of the curve on which an object is travelling is decreased by the

factor of ‘2’, i.e. halved, then what is the effect on its centripetal acceleration?

Solution:

From the equation of centripetal acceleration it is clear that the centripetal

acceleration is inversely proportional to the radius of the curvature so if the

radius is decreased by a factor of ‘2’ centripetal acceleration increases by the

factor of ’2’.

Derivation of Centripetal Force

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The force of a moving body is given as : F = ma ..................(1)

From the figure, we can write the expression :

PS = PQ + QS

∆v = v2 - v1 The triangle AOB and PQS are similar. So,

∆v( AB) = v x r

AB = arc AB = v∆t

∆v(v∆t) = vr

∆v / ∆t = v2/r

‘.’ a = v2/r

Substitute this value in (1) we get the centripetal forceF = m v2/r = mrω2

Exercises

Multiple Choice Questions:

1. Which of the following graphs can possibly represent the 1-D motion of a

particle?

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A) B)

C) D)

2. One train A 120m long is moving in a direction with a speed of 20 m/s.

Another train B 130m long is moving with a speed of 30 m/s in the opposite

direction. The train B crosses the train A in a time

A) 5s B) 36s

C) 38s D) None of these

3.Use the following information to answer the next question.

The speed of a motor launch in still water is 8 m/s. The speed of the stream is 4

m/s. A large piece of cork drops from motor launch as it begins travelling

upstream. After travelling 5 km, the motor launch turns back to catch the

floating cork. How long does the motor launch take to reach the cork?

A) 22.45 min B) 27.78 minC) 36.34 min D) 41.67 min

4. What do you infer, if s-t graphs of two cyclists meet at a point?

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(C) They are at rest

(D) They are startng from rest

5. Use the following information to answer the next question.

Consider the following statements regarding the motion of a particle in a plane.

I. The acceleration of a uniform circular motion is constant.

II. At the highest point of a parabolic trajectory, the speed of the particle is

minimum.

III. The horizontal component of velocity of a projected particle is constant

throughout the motion.

Among the given statements,

(A) only statementI is incorrect

(B) only statementIII is incorrect

(C) both statementsI andII are incorrect

(D) both statements II andIII are incorrect

6. The velocity-time graph of a motion starting from rest with uniform

acceleration is a straight line

A) parallel to time axis

B) parallel to velocity axis

C) not passing through origin

D) having none of the above characterstics

Answers :

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1. C 2. A 3. D 4. B

5. A 6. A

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4.1 Introduction

In the last chapter we have learned about the motion of a particle along a

straight line.

• Straight line motion or rectilinear motion is motion in one dimension. Now

in this chapter, we will consider both motion in two dimensions and three

dimensions.

• In two dimensional motion path of the particle is constrained to lie in a

fixed plane. Examples of such motion are projectile shot from a gun, motion ofmoon around the earth, circular motion and many more.

• To solve problems of motion in a plane, we need to generalize kinematic

language of previous chapter to a more general using vector notations in two

and three dimensions.

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4.2 Scalars And Vectors

The mathematical quantities that are used to describe the motion of objects can

be divided into two categories. The quantity is either a vector or a scalar. Thesetwo categories can be distinguished from one another by their distinct

definitions:

• Scalars are quantities that are fully described by a magnitude (or

numerical value) alone. Examples of scalar quantities are: the distance

between two points, mass of an object, the temperature of a body and the time

at which a certain event happened. The rules for combining scalars are the

rules of ordinary algebra. Scalars can be added, subtracted, multiplied and

divided just as the ordinary numbers.• Vectors are quantities that are fully described by both a magnitude and a

direction and they obey the laws of vector algebra like the triangle law of

addition or the parallelogram law of addition. Examples of vector quantities

are: displacement, velocity, acceleration, force and momentum.

To represent a vector, we can use a bold face type or it is often represented by an

arrow placed over a letter. Thus, both v and v represent the velocity vector. The

magnitude of a vector is often called its absolute value, indicated by |v

| =v.

4.2.1 Equality of Vectors

Two vectors, A andB are equal if they have the same magnitude and direction,

regardless of whether they have the same initial points, as shown in fig 4.1(a). In

general, equality is indicated as A = B.

Figure 4.1(a) (b) (c)

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In Fig. 4.1(c), vectors A and B have the same magnitude but they are not equal

because they have different directions.

4.2.2 Multiplication Of Vectors By Real NumbersMultiplying the vector A by the positive scalarc is equivalent to adding togetherc

copies of the vector A.

Thus 3 A = A + A + A.

Multiplying a vector by a scalar will get you a vector with the same direction, but

different magnitude, as the original.

(a) (b)

(c) The result of multiplying A by λ is a vector in the same direction as A, with a

magnitude of λA. If λ is negative, then the direction of A is reversed by scalar

multiplication. The factor λ by which a vector A is multiplied could be a scalar

having its own physical dimension. Then, the dimension of λ A is the product of

the dimensions of λ and A. For example, if we multiply a constant velocity vector

by duration (of time), we get a displacement vector.

4.2.3 Addition And Subtraction of Vectors — Graphical Method

The head-to-tail method is a graphical way to add vectors, ttail of the vector is

the starting point of the vector, and the head (or tip) of a vector is the final,

pointed end of the arrow. The resultant vectorR is defined such that A+B=R The

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magnitude and direction ofR are then determined with a ruler and protractor,

respectively.

Figure 4.3

Head-to-Tail Method: The head-to-tail method of graphically adding vectors is

illustrated for the two displacements of the person walking in a city considered in

Figure 4.3. (a) Draw a vector representing the displacement to the east. (b) Draw

a vector representing the displacement to the north. The tail of this vector shouldoriginate from the head of the first, east-pointing vector. (c) Draw a line from the

tail of the east-pointing vector to the head of the north-pointing vector to form

the sum or resultant vectorD. The length of the arrowD is proportional to the

vector’s magnitude and is measured to be 10.3 units. Its direction, described as

the angle with respect to the east (or horizontal axis)θ is measured with a

protractor to be 29.1º.

If two sides of a triangle represent the two vector taken in the same order, the thirdside of the triangle determines the sum of the two vectors, this method is also

known astriangle method of vector addition.

We can also use theparallelogram methodto find the sum of two vectors.

According to it, if two co-initial vectors represent the adjacent sides of a

parallelogram in same magnitude and direction, then the diagonal of the

parallelogram represents the resultant of the two vectors.

Figure 4.4 (a) (b)

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Both the triangle and the parallelogram rules of addition are procedures that are

independent of the order of the vectors; that is, if we find the resultant ofB+ A,

the same vectorRis obtained.

Thus, vector addition iscommutative: A+B=B+ A The addition of vectors also obeys the associative law as shown Fig. 4.5. The

result of adding vectors AandBfirst and then adding vectorCis the same as

the result of addingBandCfirst and then adding vector A:

( A+B) +C= A+ (B+C)

Figure 4.5 Illustrating the associative law of vector addition.

Vector subtraction is a straightforward extension of vector addition. To definesubtraction (say we want to subtractB from A, written A –B, we must first define

what we mean by subtraction. Thenegative of a vectorB is defined to be –B; that

is, graphicallythe negative of any vector has the same magnitude but the opposite

direction, as shown inFigure 4.2(b). In other words,B has the same length as –B,

but points in the opposite direction. Essentially, we just flip the vector so it

points in the opposite direction.

A – B = A +( –B).If two opposite vectors are added

A+ (– A), the sum will be a zero vector. Since the magnitudes of the two vectors

are the same, but the directions are opposite, the resultant vector has zero

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magnitude with no specified direction and is represented by0called anull

vectoror azero vector:

A – A=0|0|= 0

The null vector also results when we multiply a vector A by the number zero. Themain properties of0are:

A+0= A

λ x0=0

0 x A=0

4.3 Resolution of a Vector

The process of splitting a vector into various parts or components is called

"Resolution of Vector"

These parts of a vector may act in different directions and are called "components

of vector".

We can resolve a vector into a number of components .Generally there are three

components of vector viz.

Component along X-axis called x-componentComponent along Y-axis called Y-component

Component along Z-axis called Z-component

Here we will discuss only two components x-component & Y-component which

are perpendicular to each other. These components are called rectangular

components of vector.

Method of resolving a vector into rectangular components

Consider a vector acting at a point making an angleθ with positive X-axis.

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Vector is represented by a line OA. From point A draw a perpendicular AB on

X-axis. Suppose OB and BA

represents two vectors. Vector OA is parallel to X-axis and vector BA is parallel to

Y-axis.

Magnitude of these vectors are v x and v y respectively. By the method of head to

tail we notice that the sum of these vectors is equal to vectorv .

Thus v x and v y are the rectangular components of vectorv .

vx = Horizontal component ofv

vy = Vertical component ofv

Consider right angled triangle ∆OAB,

and

4.4 Position and Displacement Vectors

To describe the position of on object in space we require aposition vector. A

position, location, or radius vector is a vector which represents theposition of

a point P in space in relation to an arbitrary reference origin O. Let P and P be′

the positions of the object initially and at time t, respectively [Fig. 4.1]. We join O

and P by a straight line. Then, OP is theposition vector of the object at time t.

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position vector, OP denoted by r. The length of the vector r represents the′ ′

magnitude of the vector and its direction is the direction in which P lies as seen

from O. If the object moves from P to P, the vector PP (with tail at P and tip at P)′ ′ ′

is called thedisplacement vector corresponding to motion from point P (at timet = 0) to point P (at time t = t).′

4.5.1 Average velocity

Consider a particle moving along a curved path in x-y plane shown below in the

figure.

Suppose at any time, particle is at the point P and after some time 't' is at pointP’ where points P and P’ represents the position of particle at two different

points.

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Position of particle at point P is described by the Position vectorr from origin O

to P given byr = xi + y j

where x and y are components ofr along x and y axis

As particle moves from P to P’, its displacement would be would be ∆r which isequal to the difference in position vectorsr andr'. Thus,

∆r =r'-r = (x'i+ y' j) – (xi+y j)

= (x'-x)i+ (y'-y) j = xi+ ∆y j ---(1)

where ∆x = (x'-x) and ∆y = (y'-y)

Now, if ∆t is the time interval during which the particle moves from point P to Q

along the curved path then average velocity ( vavg) of particle is the ratio of

displacement and corresponding time interval,

since vavg= ∆r/∆t, the direction of average

velocity is same as that of ∆r

Magnitude of ∆r is always the straight line distance from P to Q regardless of any

shape of actual path taken by the particle.

Hence average velocity of particle from point P to Q in time interval ∆t would be

same for any path taken by the particle.

4.5.2. Instantaneous velocity

We already know that instantaneous velocity is the velocity of the particle at any

instant of time or at any point of its path.

If we bring point Q more and more closer to point P and then calculate average

velocity over such a short displacement and time interval then

where v is known as the instantaneous velocity of the particle.

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Thus, instantaneous velocity is the limiting value of average velocity as the time

interval approaches zero.

As the point Q approaches P, direction of vector ∆r changes and approaches tothe direction of the tangent to the path at point P. So instantaneous vector at any

point istangent to the path at that point.

Figure on the next page shows the direction of instantaneous velocity at point P.

Thus, direction of instantaneous velocity v at any point is always tangent to thepath of particle at that point.

Like average velocity we can also express instantaneous velocity in component

form,

where v x and v y are x and y

components of instantaneous velocity.

Magnitude of instantaneous velocity is

| v|= √(v x)2+(v y)2 -----------(4)

and angle θ which velocity vector makes with x-axis is,

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tanθ = v x/v y -----------(5)

Expression for instantaneous velocity is

Thus, if expression for the co-ordinates x and y are known as function of time

then we use equations derived above to find x and y components of velocity.

4.6 Average and Instantaneous acceleration

Suppose a particle moves from point P to point Q in x-y plane as shown below in

the figure.

Suppose v1 is the velocity of the particle at point P and v2 is the velocity of

particle at point Q

Average acceleration is the change in velocity of particle from v1 to v2 in time

interval ∆t as particle moves from point P to Q. Thus average acceleration is

Average acceleration is the vector quantity having direction same as that of ∆ v.

Again if point Q approaches point P, then limiting value of average acceleration

as time approaches zero defines instantaneous acceleration or simply the

acceleration of particle at that point. Thus, instantaneous acceleration is –

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Figure below shows instantaneous acceleration ‘a’ at point P.

Instantaneous acceleration does not have same direction as that of velocity vector

instead it must lie on the concave side of the curved surface.

Thus velocity and acceleration vectors may have any angle between 0 to 180

degrees between them.

4.7 Motion with constant acceleration

Motion in two dimensions with constant acceleration we know is the motion in

which velocity changes at a constant rate i.e, acceleration remains constant

throughout the motion

We should set up the kinematic equation of motion for particle moving withconstant acceleration in two dimensions.

Equations for position and velocity vector can be found generalizing the equation

for position and velocity derived earlier while studying motion in one dimension.

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Thus velocity is given by equation

. v= v0+at --------- (8)

where v is velocity vector, v0 is initial velocity vector anda is instantaneous

acceleration vector.Similary position is given by the equation :

r –r0= v0t + ½at2 (9)

wherer0 is initial position vector

i,e

r0= x0i+ y0 j

and average velocity is given by the equation

vav =½ ( v+ v0) (10)

Since we have assumed particle to be moving in x-y plane, the x and y

components of equation (8) and (9) are

v x= v0x+ a xt (11a)

x0= v0xt + ½ a xt2 (11b)

and

v y= v0y+ a yt (12a)

y0=v0yt + ½ a yt2 (12b)

From above equations 11 and 12, we can see that for particle moving in (x-y)

plane although plane of motion can be treated as two separate and simultaneous

1-D motion with constant acceleration.

Similar result also hold true for motion in a three dimension plane (x-y-z)

4.8 Projectile Motion

Projectile motion is a special case of motion in two dimensions when accelerationof particle is constant in both magnitude and direction

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An object is referred as projectile when it is given an initial velocity which

subsequently follows a path determined by gravitational forces acting on it. For

example bullet fired from the rifle, a javelin thrown by the athlete etc

Path followed by a projectile is called its trajectory. Acceleration acting on a projectile is constant which is acceleration due to gravity

(g = 9.81 m/s2) directed along vertically downward direction.

We shall treat the projectile motion in a Cartesian co-ordinates system taking y-

axis in vertically upwards direction and x-axis along horizontal directions

Now x , y components of acceleration of projectile are : a x= 0 and a y= –g.

Since a xis zero, the horizontal component of velocity is constant and vertical

motion is simply a case of motion with constant acceleration.

Suppose at time t = 0 object is at origin of co-ordinate system and velocitycomponents v0x and v0y. From above, a x=0 and a y=-g. From equations 11 and 12

in the previous section, the components of position and velocity are

x = v xt (14a)

v x= v0x (14b)

y = v0y – ½ gt2 (15a)

v y= v0y – gt (15b)

Figure below shows motion of an object projected with velocity v0 at an angle θ0

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In terms of initial velocity v0 and angle θ0 components of initial velocity are

v0x= v0cosθ0 (16a)

v0y= v0sinθ0 (16b)

Using these relations in equation 14 and 15 we find :` x = (v0cosθ0) t (17a)

y = (v0sinθ0) t - ( ½ )gt2 (17b)

v x= v0cosθ0 (17c)

v y= v0sinθ0-gt (17d)

Above equations describe the position and velocity of projectile as shown in fig.5

at any time t.

(a) Equation of Path of projectile

From equation 17a,

t = x/(v0cosθ0)

Now putting this value of ‘t’ in equation 17b, we find

y =(tanθ0 )x-[g/2(v0cosθ0)2]x2 (18)

In equation (18), quantities θ0, g and v0 are all constants and equation (18) can

be compared with the equation

y = ax – bx2

where a and b are constants. This is the equation of the parabola. From this we conclude that path of the

projectile is a parabola as shown in figure 5

(b) Time of Flight

At point of maximum height v y= 0.Thus from equation (17d),

v y= v0sinθ0- gt

0= v0sinθ0- gt

or tm=v0sinθ0/g (19)

Time of flight of projectile which is the total time during which the projectile is in

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vertical distance travelled is zero. Thus from equation (17b)

T =2usinθ0

g ------------- (20)

Maximum height reached by the projectile can be calculated by substituting t =tm in equation 17b,

y = Hm= (v0sinθ0)2/g - (g/2)(v0sinθ0/g)2

H ¿u

2sin

2θ0

2g -------- (21)

(c) Horizontal Range of ProjectileSince acceleration g acting on the projectile is acting vertically , so it has no

component in horizontal direction.

So, projectile moves in horizontal direction with a constant velocity v0cosθ0. So

range R is

R = OA = velocity x time of flight

Maximum range is obtained when sin2θ0=1 or θ0=450. Thus when θ0 =

450 maximum range achieved for a given initial velocity is (v0)2/g.

4.9 Uniform Circular Motion

When an object moves in a circular path at a constant speed then motion of the

object is called uniform circular motion.

In our everyday life, we came across many examples of circular motion for

example cars going round the circular track and many more. Also earth and

other planets revolve around the sun in a roughly circular orbits

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Here in this section we will mainly consider the circular motion with constant

speed

If the speed of motion is constant for a particle moving in a circular motion still

the particles accelerates becuase of costantly changing direction of the velocity.Here in circular motion, we use angular velocity in place of velocity we used while

studying linear motion.

(a) Angular velocity

Consider an object moving in a circle with uniform velocity v as shown below in

the figure.

The velocity v at any point of the motion is tangential to the circle at that point.

Let the particle moves from point A to point B along the circumference of the

circle .The distance along the circumference from A to B is s = R θ (23)

Where R is the radius of the circle and θ is the angle moved in radian's

Magnitude of velocity is

v=ds/dt= R (dθ/dt) ---------- (24)

Since radius of circle remains constant

.’. ω=dθ/dt ---------------(25)

is called the angular velocity defined as the rate of change of angle swept by

radius with time. Angular velocity is expressed in radians per second (rads-1)

From equation 24 and 25, we find the following

v = ω R -------- (26)

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Thus for a particle moving ain circular motion, velocity is directly proportional to

radius for a given angular velocity.

For uniform circular motion i.e, for motion with constant angular velocity the

motion would be periodic which means particle passes through each point ofcircle at equal intervals of time

Time period of motion is given by

T=2π/ω --------- (27)

Since 2π radians is the angle θ in one revolution

If angular velocity ω is constant then integrating equation (25) within limits θ0 to

θ, we find

where θ0 is the angular position at time t0 and θ is the angular position at time

t .The above equation is similar to rectilinear motion result (x-x0)=v(t-t0)

(b) Angular acceleration

Angular acceleration is defined as the rate of change of angular velocity moving

in circular motion with time.

Thus

α=dω/dt=d2θ/dt2 ------------(29)

Unit of angular acceleration is rads-2

For motion with constant angular acceleration

or ω = ω0+ α(t-t0) ------------(30) where ω0 is angular velocity at time t0.

Again since, ω = dθ/dt

or dθ = ωdt then from equation 30,

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If in the beginning t0=0 and θ0=0 the

angular position at any time t is given by : θ = ωt+( ½ )αt2

This result is of the form similar to what we find in case of uniformly accelerated

motion while studying rectilinear motion.

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5.1 Introduction

Newton's laws of motion are threephysical laws that together laid the

foundation forclassical mechanics. They describe the relationship between a

body and theforces acting upon it, and itsmotion in response to said forces.

They have been expressed in several different ways over nearly three

centuries and can be summarised as follows:

5.1.1First law: When viewed in aninertial reference frame, an object either

remains at rest or continues to move at a constant velocity, unless acted upon by

an externalforce.

The first law can be stated mathematically as

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Consequently,

• An object at rest will stay at rest unless an external force acts upon it.

• An object in motion will not change its velocity unless an external force acts upon

it.

There are therefore two clauses or parts to Newton’s first law - one that predicts

the behaviour of stationary objects and the other that predicts the behaviour of

moving objects. summarized in the following diagram.

3.1.2Second law: The vector sum of theforces on an object is equal to

themass “m” of that object multiplied by theacceleration vector “a” of the object.3.1.3 Third law: When one body exerts a force on a second body, the second body

simultaneously exerts a force equal in magnitude and opposite in direction onthe first body.

Newton showed that these laws of motion, combined with his law of universal

gravitation explained Kepler’s laws of planetary motion.

Newton's laws are applied to objects which are idealised as single point masses in

the sense that the size and shape of the object's body are neglected to focus on

its motion more easily. This can be done when the object is small compared to

the distances involved in its analysis, or the deformation and rotation of the bodyare of no importance. In this way, even a planet can be idealised as a particle for

analysis of its orbital motion around a star.

Newton's laws hold only with respect to a certain set offrames of

reference calledNewtonian or inertial reference frames. Some authors interpret

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the first law as defining what an inertial reference frame is; from this point of

view, the second law only holds when the observation is made from an inertial

reference frame, and therefore the first law cannot be proved as a special case of

the second. Other authors do treat the first law as a corollary of the second.In the given interpretation, mass, acceleration,momentum, and (most

importantly)force are assumed to be externally defined quantities. This is the

most common, but not the only interpretation of the way one can consider the

laws to be a definition of these quantities.

Whatever draws or presses another is as much drawn or pressed by that other. If

you press a stone with your finger, the finger is also pressed by the stone. If a

horse draws a stone tied to a rope, the horse (if I may so say) will be equallydrawn back towards the stone: for the distended rope, by the same endeavour to

relax or unbend itself, will draw the horse as much towards the stone, as it does

the stone towards the horse, and will obstruct the progress of the one as much

as it advances that of the other. If a body impinges upon another, and by its force

changes the motion of the other, that body also (because of the equality of the

mutual pressure) will undergo an equal change, in its own motion, toward the

contrary part. The changes made by these actions are equal, not in the velocities

but in the motions of the bodies; that is to say, if the bodies are not hindered byany other impediments. For, as the motions are equally changed, the changes of

the velocities made toward contrary parts are reciprocally proportional to the

bodies. This law takes place also in attractions.

5.1.4Examples of inertia of rest:

• If pulled quickly, a tablecloth can be removed from underneath of dishes. The

dishes have the tendency to remain still as long as the friction from the

movement of the tablecloth is not too great.

• If a stopped car is hit by a moving car from behind, the passengers inside mayexperience whiplash as a result of the body moving forward but the head lagging

behind. The head is experiencing inertia.

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• When a car is abruptly accelerated, drivers and passengers may feel as though

their bodies are moving backward. In reality, inertia is making the body want to

stay in place as the car moves forward.

• If an index card is placed on top of a glass with a penny on top of it, the indexcard can be quickly removed while the penny falls straight into the glass, as the

penny is demonstrating inertia.

• If the wind is blowing, a tree’s branches are moving. A piece of ripe fruit that falls

from the tree will fall in the direction the wind is moving because of inertia.

5.1.5Examples of inertia of motion along a straight line:

• One's body moves to the side when a car makes a sharp turn.

• Tightening of seat belts in a car when it stops quickly.• A ball rolling down a hill will continue to roll unless friction or another force

stops it.

• Men in space find it more difficult to stop moving because of a lack of gravity

acting against them.

• If one drove a car directly into a brick wall, the car would stop because of the

force exerted upon it by the wall. However, the driver requires a force to stop his

body from moving, such as a seatbelt, otherwise inertia will cause his body to

continue moving at the original speed until his body is acted upon by some force.• If you jump from a car of bus that is moving, your body is still moving in the

direction of the vehicle. When your feet hit the ground, the grounds act on your

feet and they stop moving. You will fall because the upper part of your body

didn’t stop and you will fall in the direction you were moving.

5.2Momentum:

Inclassical mechanics linear or translational momentum is the product of

themass and velocity of an object. For example, a heavy truck moving quickly

has a large momentum—it takes a large or prolonged force to get the truck up to

this speed, and it takes a large or prolonged force to bring it to a stop afterwards.

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If the truck were lighter, or moving more slowly, then it would have less

momentum.

Like velocity, linear momentum is a vector quantity, possessing a direction as wellas a magnitude

Linear momentum is also aconserved quantity, meaning that if aclosed

system is not affected by external forces, its total linear momentum cannot

change. In classical mechanics,conservation of linear momentum is implied

byNewton's laws; but it also holds inspecial relativity (with a modified formula)

and, with appropriate definitions, a (generalized) linear momentum conservationlaw holds in electrodynamics,quantum mechanics,quantum field theory,

andgeneral relativity.

5.2.1 Impulse

Animpulse J occurs when a forceF acts over an interval of time ∆t, and it is

given by

Since force is the time derivative of momentum, it follows that

This relation between impulse and momentum is closer to Newton's wording of

the second law.

Impulse is a concept frequently used in the analysis of collisions and impacts.

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5.2.2 Relation to force: The second law

If a forceF is applied to a particle for a time interval ∆t, the momentum of the

particle changes by an amount

In differential form, this givesNewton’s second law of motion: the rate of change

of the momentum of a particle is equal to the forceF acting on it:

If the force depends on time, the change in momentum (orimpulse) betweentimest1 andt2 is

The second law only applies to a particle that does not exchange matter with its

surroundings, and so it is equivalent to write

so the force is equal to mass timesacceleration.

5.2.2.1Limitation of the second law:

The net force should be defined as the rate of change ofmomentum; this

becomes

F = ma

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only if the mass is constant. Since themass changes as the speed approaches

thespeed of light, F=ma is seen to be strictly a non-relativistic relationship which

applies to the acceleration of constant mass objects. Despite these limitations, it

is extremely useful for the prediction of motion under these constraints.

5.2.3Law of conservation of linear momentum:

One of the most powerful laws in physics is the law ofmomentum conservation.

The law of momentum conservation can be stated as follows.

For a collision occurring between object 1 and object 2 in anisolated system, thetotal momentum of the two objects before the collision is equal to the total

momentum of the two objects after the collision. That is, the momentum lost by

object 1 is equal to the momentum gained by object 2.

The above statement tells us that the total momentum of a collection of objects

(asystem) isconserved - that is, the total amount of momentum is a constant or

unchanging value. This law of momentum conservation will be the focus of the

remainder of Lesson 2. To understand the basis of momentum conservation, let's

begin with a short logical proof.

5.2.4 The Logic behind Momentum Conservation

Consider a collision between two objects - object 1 and object 2. For such a

collision, the forces acting between the two objects are equal in magnitude and

opposite in direction (Newton's third law). This statement can be expressed in

equation form as follows.

F1 =-F2 The forces are equal in magnitude and opposite in direction

The forces act between the two objects for a given amount of time. In some cases,

the time is long; in other cases the time is short. Regardless of how long the time

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is, it can be said that the time that the force acts upon object 1 is equal to the

time that the force acts upon object 2. This is merely logical. Forces result from

interactions (or contact) between two objects. If object 1 contacts object 2 for

0.050 seconds, then object 2 must be contacting object 1 for the same amount oftime (0.050 seconds). As an equation, this can be stated as

t1 = t2

Since the forces between the two objects are equal in magnitude and opposite in

direction, and since the times for which these forces act are equal in magnitude,

it follows that theimpulses experienced by the two objects are also equal in

magnitude and opposite in direction. As an equation, this can be stated as

F1 x t = -F2 x t

The impulse before impact is equal to the impulse after impact but in the

opposite direction.

Butthe impulse experienced by an object is equal to the change in momentum of

that object (the impulse-momentum change theorem). Thus, since each object

experiences equal and opposite impulses, it follows logically that they must also

experience equal and opposite momentum changes. As an equation, this can bestated as

M1 x v ∆ 1 = M2 x v ∆ 2

The change in momentum is equal in magnitude and opposite in direction

Example 5.1

Consider a collision in football between a fullback and a linebacker during agoal-line stand. The fullback plunges across the goal line and collides in midair with

the linebacker. The linebacker and fullback hold each other and travel together

after the collision. The fullback possesses a momentum of 100 kgm/s, East

before the collision and the linebacker possesses a momentum of 120 kg*m/s,

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West before the collision. The total momentum of the system before the collision

is 20 kg*m/s, West. Therefore, the total momentum of the system after the

collision must also be 20 kg*m/s, West. The fullback and the linebacker move

together as a single unit after the collision with a combined momentum of 20kg*m/s. Momentum is conserved in the collision. A vector diagram can be used

to represent this principle of momentum conservation; such a diagram uses an

arrow to represent the magnitude and direction of the momentum vector for the

individual objects before the collision and the combined momentum after the

collision.

Example 5.2

Now suppose that a medicine ball is thrown to a clown who is at rest upon the

ice; the clown catches the medicine ball and glides together with the ball across

the ice. The momentum of the medicine ball is 80 kg*m/s before the collision.

The momentum of the clown is 0 m/s before the collision. The total momentum

of the system before the collision is 80 kg*m/s. Therefore, the total momentum of

the system after the collision must also be 80 kg*m/s. The clown and the

medicine ball move together as a single unit after the collision with a combinedmomentum of 80 kgm/s. Momentum is conserved in the collision.

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Momentum is conserved for any interaction between two objects occurring in an

isolated system. This conservation of momentum can be observed by a total

system momentum analysis or by a momentum change analysis. Useful means of

representing such analyses include a momentum table and a vector diagram.

Later in Lesson, we will use the momentum conservation principle to solveproblems in which the after-collision velocity of objects is predicted.

Example 5.3

When fighting fires, a fire fighter must use great caution to hold a hose that

emits large amounts of water at high speeds. Why would such a task be difficult?

Solution: The hose is pushing lots of water (large mass) forward at a high speed. This means the water has a large forward momentum. In turn, the hose must

have an equally large backwards momentum, making it difficult for the fire

fighters to manage.

Example 5.4

A large truck and a Volkswagen have a head-on collision.

a. Which vehicle experiences the greatest force of impact?

Ans: Both experience the same force of impact

b. Which vehicle experiences the greatest impulse?

Ans: Both experience the same impulse.

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Ans: Both experience the same change in momentum.

d. Which vehicle experiences the greatest acceleration?

Ans: Due to smaller mass the Volkswagen experience larger force.

Example 5.5

The velocity of a freely falling body increases continuously. Is it a violation of the

law of conservation of energy? Explain

Solution:

No. The velocity of the freely falling body increases towards the earth, conversely

the velocity of the earth increases towards the body.

Example 5.6

Explain recoil of a gun on firing.

Ans: p ∆ gun = - p ∆ bullet

Action by gun on bullet. Reaction by bullet on the gun.

5.3 Equilibrium of concurrent forces

Equilibrium means "no acceleration".

Since a force is a "push" or "pull" exerted on a body, equilibrium means that

the total of all forces acting on a body must be zero.

According to Newton's second law,

F = m x a

Ifa = 0 thenF must be zero.

Concurrent means that the forces intersect through a single point.

If forces are concurrent, we can add them together as vectors to get the

resultant.If the body is not accelerating, it must be in equilibrium, so that means the

resultant is zero.

For concurrent forces, the body is a point.

So for concurrent forces in equilibrium, the forces should all add up to give zero.

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5.3.1 The Free Body Diagram (FBD)

The Free Body Diagram is a strict diagram that isolates the body for study.

SeeFree Body Diagrams for more information. The idea of the FBD is to focus on

one particular part or group of parts (called the body) and replace every external

member with the force they would apply.

1. Isolate the body. (An outline is best because we are supposed to forget about

the inside of the body).

2. Locate border crossings. Identify the contact points where forces are crossing

the boundary. Gravity acts through centre.

3. Line of Action. Some types of connections have a known direction. E.g.

Cables have force running through the centreline.

4. "To the Body" Since Newton's 3rd law has every action with an opposite

reaction, we must eliminate half the forces. Identify those forces that are applied

"to the body", and eliminate those done "by the body".

If the FBD were drawn to scale, the body might be length (mm, m etc) and the

forces might be another scale (N, kN etc).

5.3.2 The Force Polygon (FP)

The force polygon must be drawn strictly to scale, and everything is a Force. The

only information coming from the FBD is;

• Force magnitudes

• Force Angles

Warning! Do not attempt to bring any FBD Lengths into the FP. There are no

metres in the Force Polygon.

Example Diagrams

These cranes are not accelerating, so they are in equilibrium. Therefore all the

forceson any body should add up to zero. The body is actually the connection

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point which is probably a lifting eye of a hook. The FBD shows as much as we

know from the Space diagram - in this case angles are known but only one

magnitude. The force polygon should form a closed loop (since resultant = 0), so

this defines the lengths (and hence the magnitudes) of F1 and F2.

CAD programs are very helpful when working with force polygons.

The Equilibrium Equations

Equilibrium simply says the resultant is zero. Mathematically, this can be stated

that the Fx and Fy components are zero.

So, for concurrent forces in 2 dimensions (planar), equilibrium means that...

Very often we know the angle of the forces but not the magnitudes. When solving

mathematically, this means we will need to use simultaneous equations.

5.4Friction

Friction is theforce resisting the relative motion of solid surfaces, fluid layers,

and material elements sliding against each other.

5.4.1 Static friction

Static friction is friction between two or more solid objects that are not moving

relative to each other. For example, static friction can prevent an object from

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sliding down a sloped surface. The coefficient of static friction, typically denoted

asμs, is usually higher than the coefficient of kinetic friction.

5.4.2Kinetic friction

Kinetic (or dynamic) friction occurs when two objects are moving relative to each

other and rub together (like a sled on the ground). The coefficient of kinetic

friction is typically denoted asμk, and is usually less than the coefficient of static

friction for the same materials.

5.4.3 Limiting friction

The maximum value of static friction, when motion is impending, is sometimes

referred to as limiting friction. It is also known as traction.

5.4.4Laws of friction

Law 1 : When two bodies are in contact the direction of the forces of Friction onone of them at it's point of contact, is opposite to the direction in which the point

of contact tends to move or moves relative to the other.

Law 2 : If the bodies are in equilibrium, the force of Friction is just sufficient to

prevent motion and may therefore be determined by applying the conditions of

equilibrium of all the forces acting on the body.

The amount of Friction that can be exerted between two surfaces is limited and if

the forces acting on the body are made sufficiently great, motion will occur.Hence, we definelimiting friction as the friction which is exerted when

equilibrium is on the point of being broken by one body sliding on another. The

magnitude of limiting friction is given by the following three laws.

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Law 3: The ratio of the limiting friction to the Normal reaction between two

surfaces depends on the substances of which the surfaces are composed, and

not on the magnitude of the Normal reaction.

This ratio is usually denoted by . Thus if the Normal reaction is R, the limiting friction is .

For given materials polished to the same standard is found to be constant and

independent of R.

is calledThe Coefficient of friction

Law 4 : The amount of limiting friction is independent of the area of contact

between the two surfaces and the shape of the surfaces, provided that the

Normal reaction is unaltered.

5.4.5 Angle of friction

For the maximum angle of static friction between granular materials,

For certain applications it is more useful to define static friction in terms of the

maximum angle before which one of the items will begin sliding. This is calledtheangle of friction or friction angle. It is defined as:

whereθ is the anglefrom horizontal andµs is the static coefficient of friction

between the objects. This formula can also be used to calculateµs from empirical

measurements of the friction angle.

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5.4.6 Lubricated friction:

Lubricated friction is a case of fluid friction where a fluid separates two solid

surfaces. Lubrication is a technique employed to reduce wear of one or both

surfaces in close proximity moving relative to each another by interposing a

substance called a lubricant between the surfaces.In most cases the applied load is carried by pressure generated within the fluid

due to the frictional viscous resistance to motion of the lubricating fluid between

the surfaces. Adequate lubrication allows smooth continuous operation of

equipment, with only mild wear, and without excessive stresses or seizures at

bearings. When lubrication breaks down, metal or other components can rub

destructively over each other, causing heat and possibly damage or failure.

5.4.7Rolling friction

Rolling resistance, sometimes calledrolling friction orrolling drag, is the

force resisting themotion when a body (such as a ball,tire, or wheel) rolls on a

surface. It is mainly caused bynon-elastic effects; that is, not all the energy

needed for deformation (or movement) of the wheel, roadbed, etc. is recovered

when the pressure is removed. Two forms of this are hysteresis losses, and

permanent(plastic) deformation of the object or the surface (e.g. soil). Another

cause of rolling resistance lies in theslippage between the wheel and the surface, which dissipates energy. Note that only the last of these effects involvesfriction,

therefore the name "rolling friction" is to an extent a misnomer.

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http://en.wikipedia.org/wiki/Tire

http://en.wikipedia.org/wiki/Wheel

http://en.wikipedia.org/wiki/Plasticity_(physics)

http://en.wikipedia.org/wiki/Plastic_deformation

http://en.wikipedia.org/wiki/Frictional_contact_mechanics

http://en.wikipedia.org/wiki/Friction


http://en.wikipedia.org/wiki/Ball

http://en.wikipedia.org/wiki/Tire

http://en.wikipedia.org/wiki/Wheel

http://en.wikipedia.org/wiki/Plasticity_(physics)

http://en.wikipedia.org/wiki/Plastic_deformation

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In analogy withsliding friction, rolling resistance is often expressed as a

coefficient times the normal force. This coefficient of rolling resistance is

generally much smaller than the coefficient of sliding friction.

5.5Uniform Circular Motion

5.5.1Centripetal Acceleration

Before discussing the dynamics of uniform circular motion, we must explore its

kinematics. Because the direction of a particle moving in a circle changes at a

constant rate, it must experience uniform acceleration. But in what direction isthe particle accelerated? To find this direction, we need only look at the change

in velocity over a short period of time:

A particle in Uniform Circular Motion

The diagram above shows the velocity vector of a particle in uniform circular

motion at two instants of time. By vector addition we can see that the change in

velocity,∆v , points toward the center of the circle. Since acceleration is the

change in velocity over a given period of time, the consequent acceleration points

in the same direction. Thus we define centripetal acceleration as an acceleration

towards the center of a circular path. All objects in uniform circular motion must

experience some form of uniform centripetal acceleration. We find the magnitude of this acceleration by comparing ratios of velocity and

position around the circle. Since the particle is travelling in a circular path, the

ratio of the change in velocity to velocity will be the same as the ratio of the

change in position to position. Thus:

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= =

Rearranging the equation,

= Thus,

ac = We now have a definition for both the magnitude and direction of centripetal

acceleration: it always points towards the center of the circle, and has a

magnitude ofv 2/r .

5.5.2Centripetal Force

Centripetal force is the force that causes centripetal acceleration. By using

Newton's Second Law in conjunction with the equation for centripetal

acceleration, we can easily generate an expression for centripetal force.

F c =ma =Remember also that force and acceleration will always point in the same

direction. Centripetal force therefore points toward the center of the circle.

There are many physical examples of centripetal force. In the case of a car

moving around a curve, the centripetal force is provided by thestatic frictional

force of the tires of the car on the road. Even though the car is moving, the force

is actually perpendicular to its motion, and is a static frictional force. In the case

of an airplane turning in the air, the centripetal force is given by the lift provided by its banked wings. Finally, in the case of a planet rotating around the sun, the

centripetal force is given by the gravitational attraction between the two bodies.

With the knowledge of physical forces such as tension, gravity and friction,

centripetal force becomes merely an extension of Newton's Laws. It is special,141



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however, because it is uniquely defined by the velocity and radius of the uniform

circular motion. All of Newton's Laws still apply, free body diagrams are still a

valid method for solving problems, and forces can still be resolved into

components. Thus the most important thing to remember regarding uniformcircular motion is that it is merely a subset of the larger topic of dynamics.

Example 5.7

A 2 kg ball on a string is rotated about a circle of radius 10 m. The maximum

tension allowed in the string is 50 N. What is the maximum speed of the ball?

Solution: The centripetal force in this case is provided entirely by the tension in

the string. If the maximum value of the tension is 50 N, and the radius is set at

10 m we only need to plug these two values into the equation for centripetal

force:

T =F c = implies thatv =

thus

v = = 15.8 m/s

Example 5.8

During the course of a turn, an automobile doubles its speed. How much

additional frictional force must the tires provide if the car safely makes around

the curve?

Solution:SinceF c varies withv 2 , an increase in velocity by a factor of two must

be accompanied by an increase in centripetal force by a factor of four.

Example 5.9

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A satellite is said to be in geosynchronous orbit if it rotates around the earth

once every day. For the earth, all satellites in geosynchronous orbit must rotate

at a distance of 4.23×107 meters from the earth's center. What is the magnitude

of the acceleration felt by a geosynchronous satellite?Solution:

The acceleration felt by any object in uniform circular motion is given bya = .

We are given the radius but must find the velocity of the satellite. We know that

in one day, or 86400 seconds, the satellite travels around the earth once. Thus:

v = =

== 3076 m/s

Thus

a = = = 0.224 m/s2

Example 5.10

The maximum lift provided by a 500 kg airplane is 10000 N. If the plane travels

at 100 m/s, what is its shortest possible turning radius?Solution: Again, we use the equationF c = . Rearranging, we find

thatr = . Plugging in the maximum value for the lift of the plane, we find

that

r min = = 500m

Example 5.11

A popular daredevil trick is to complete a vertical loop on a motorcycle. This trickis dangerous, however, because if the motorcycle does not travel with enough

speed, the rider falls off the track before reaching the top of the loop. What is the

minimum speed necessary for a rider to successfully go around a vertical loop of

10 meters?

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Solution:

During the entire trip, the rider experiences two different forces: the normal force

from the track, and the gravitational force. At the top of the loop, both these

forces point down, or towards the center of the loop. Thus the combination ofthese forces provides the centripetal force at that point. At the minimum speed of

the motorcycle, however, he experiences no normal force. One can see this by

envisioning that if the rider had gone any slower, he would have fallen off the

track. Thus, at the minimum speed, all the centripetal force is provided by

gravity. Plugging into our equation for centripetal force, we see that

mg =

Rearranging the equation,v min = = = 9.9 m/s

Thus the rider must be traveling at least 9.9 m/s to make it around the loop.

5.5.3Cyclist Negotiating a Curved Level Surface

While negotiating a curved level surface, a cyclist has to lean inwards which give

the necessary centripetal force which prevents him from falling down. Diagram

shows a cyclist leaning at an angle θ with the vertical. N is the normal reaction

which is shown by:N = mg

Where m is the mass of the cyclist plus the bicycle. The tension of friction

between the road and the tyres is

F = μN = mg

The cyclist will skid if the centripetal force mv2/R exceeds the frictional tension F,

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Where R is the radius of the curved surface. Thus skidding occurs if

5.5.4Banking of Curves:

The large amount of friction between the tyres and the road would affect the

tyres. To minimize the wearing out of tyres the road bend is banked, i.e. the

outer part of the road is raised a little so that the road slopes towards the center

of the curved track. Take as a car of mass m is moving around a banked track ina circular path of radius R as shown in Fig. Let N1 and N2 be the reaction at each

tyre due to the road. Then the total reaction is N = N1 + N2 acting in the middle of

the car. Ifθ is the angle of the banking, the vertical component Ncosθ supports

the weight mg of the car while the horizontal component Nsinθ provides the

necessary centripetal force.

Thus,

and

Also F = μN Where F is the force of friction action acting radically inwards on the car. These

equations provide

and

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The first equation calculate the proper banking angle for given v, R and μ, and

the second equation the maximum speed at which the car can successfully

negotiate the curve for given R, μ and θ.

For given θ and R, there is an optimum (best) speed for negotiating a bankedcurve at which there will be the least tear and wear, i.e. when friction is not

needed at all ( μ= 0). If μ= 0, this speed is : v = (Rg tanθ )1/2

The car will not skid if the angle of banking of the track satisfies the equation:

5.5.5Circular motion in a vertical plane

It is interesting to consider an object that is being swung round on a string in a

vertical circle.

Let the tension in the string be T1 at the bottom of the circle, T2 at the sides and

T3 at the top.

At the bottom of the circle:

T1 - mg = mv2/r so T1 = mv2/r + mg

At the sides of the circle:

T2 = mv2/r

At the top of the circle:

T3+ mg = mv2/r so T3 = mv2/r - mg

So, as the object goes round the circle the tension in the string varies being

greatest at the bottom of the circle and least at the top. Therefore if the string is

to break it will be at the bottom of the path where it has to not only support the

object but also pull it up out of it straight-line path.

An example of this would be an aircraft looping the loop or some of the funfair

rides already mentioned. The g force on the pilot would be greatest as they try

and pull out of the bottom of the loop, and this is the place they would

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expect to black out if they are travelling round the loop at constant speed.

5.6Connected motion:

1. The two mass arem1 andm2 and the pulley system is arranged as in Figure

If m1<m2 then m1 gets accelerated upward and m2 gets accelerated downwards.

Since the string is taut T1 = T2 = T (say) and acceleration be a

Therefore,

T- m1g = m1a

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And m2g –T = m2a

Hence a = (m2-m1)g/(m1+m2)

And tension T = 2m1m2/(m1+m2)

2.Consider two masses, m1 and m2, connected by a light inextensible string.

Suppose that the first mass slides over a smooth, frictionless, horizontal table,

whilst the second is suspended over the edge of the table by means of a light

frictionless pulley. See Figure. Since the pulley is light, we can neglect its

rotational inertia in our analysis. Moreover, no force is required to turn africtionless pulley, so we can assume that the tension of the string is the same

on either side of the pulley.

Let us apply Newton's second law of motion to each mass in turn. The first mass

is subject to a downward force m1g, due to gravity. However, this force is

completely cancelled out by the upward reaction force due to the table. The mass

m1 is also subject to a horizontal force , due to the tension in the string, which

causes it to moverightwards with acceleration

The second mass is subject to a downward force m2g, due to gravity, plus an

upward force due to the tension in the string. These forces cause the mass to

movedownwards with acceleration.

Now, the rightward acceleration of the first mass must match the downward

acceleration of the second, since the string which connects them is inextensible.

Thus, equating the previous two expressions, we obtain

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Note that the acceleration of the two coupled masses isless than the full

acceleration due to gravity ‘g’, since the first mass contributes to the inertia of

the system, but does not contribute to the downward gravitational force which

sets the system in motion.

5.7Motion in an elevator:

1. No acceleration of elevator:

If the acceleration of the elevator is zero, then there are two possible scenarios;

the elevator can be at rest (stationary, zero velocity) or moving with a constant

speed (no acceleration if velocity does not change). In this case, the action and

reaction force pair between the person and the scale is just the weight. The

person pushes down on the scale with a force of

-W=-mg (negative direction)

and the scale pushes back up against the man with a Normal Force of

FN = +W = +mg.

Because the reading on the scale is the magnitude of the normal force, the scale

will read the true weight when the elevator is NOT accelerating.

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2. Elevator speeding upwards:

In this case, the elevator and the person are starting from rest at a lower floor.

The elevator accelerates upward. The inertia of the person would prefer to staystationary, so the elevator floor and scale must push up on the person to

accelerate him upward long with the elevator. (The person doesn't sink into the

floor when the elevator accelerates up. The elevator and the scale and the person

all move together.)The scale therefore has to push upward with extra force on the

person to accelerate the person's mass upward. This results in a greater contact

force between the scale and the person. Therefore the Normal Force is larger, so

the reading on the scale is a number that is GREATER than the true weight.

Let's consider Newton's 2nd

Law (ΣF=ma) acting on the person. The overallacceleration of the person is upward (with the elevator). So ma is positive

(upward). The only external forces acting on the person are the force of gravity

acting down (-W=-mg) and the supporting Normal Force FN that the scale applies

upward on the person.

So ΣF = ma = -mg + FN .

We want to know FN because that is the number that we read off the scale. So

FN = mg + ma,

which is GREATER than the true weight.

3. Elevator slowing down upwards:

In this case, the elevator and the person are initially moving upward at a

constant speed and slowing down to rest at a higher floor. The acceleration of theelevator is downward (opposite to the upward motion, which causes a reduction

of the velocity). The inertia of the person would prefer to keep moving upward at

a constant speed, so the elevator floor and scale effectively drop out a little bit

from underneath the personas the elevator slows down. The person doesn't float

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upward, because again the elevator and the person move together, but the

contact force between the person and the scale is reduced. The scale therefore

has to push upward with less force on the person to support the person's weight.

Therefore the Normal Force is smaller, so the reading on the scale is a numberthat is LESS than the true weight. Let's consider Newton's 2nd Law (ΣF=ma)

acting on the person. The overall acceleration of the person is downward (with

the elevator). So ma is negative (downward). The only external forces acting on

the person are the force of gravity acting down (-W=-mg) and the supporting

Normal Force FN that the scale supplies upward on the person.

So ΣF = -ma = -mg + FN

We want to know FN because that is the number that we read off the scale. Now

FN = mg – ma

which is LESS than the true weight.

3. Elevator slowing down downwards

In this case, the elevator and the person are initially moving downward at aconstant speed and then slow to rest at a lower floor. The elevator accelerates

upward (opposite direction to negative/downward velocity to reduce velocity

magnitude). The inertia of the person would prefer to keep moving downward at

the constant speed, so the elevator floor and scale must push up on the person

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to accelerate him upward, slowing him down. (The person doesn't sink into the

floor here either. Elevator and scale and person move together.)The scale

therefore has to push upward with extra force on the person to accelerate the

person's mass upward. This results in a greater contact force between the scaleand the person. Therefore the Normal Force is larger, so the reading on the scale

is a number that is GREATER than the true weight. Let's consider Newton's 2nd

Law (ΣF=ma) acting on the person. The overall acceleration of the person is

upward (with the elevator). So ma is positive (upward). The only external forces

acting on the person are the force of gravity acting down (-W=-mg) and the

supporting Normal Force FN that the scale applies upward on the person.

So ΣF = ma = -mg + FN

(Note that this is the same equation as we got in case 2.) We want to know FN because that is the number that we read off the scale.

Now FN = mg + ma

which is GREATER than the true weight.

4. Elevator speeding up downwards

In this case, the elevator and the person are initially at rest at a higher floor. The

elevator then speeds up in the downward direction towards a lower floor. The

acceleration of elevator is negative/downward (increasing the velocity magnitude

in the downward direction). The inertia of the person would prefer to stay at rest,

so the elevator floor and scale effectively drop out a little bit from underneath the

person as the elevator accelerates down. The person doesn't float upward herealso, because again the elevator and the person move together, but the contact

force between the person and the scale is reduced. The scale therefore has to

push upward with less force on the person to support the person's weight.

Therefore the Normal Force is smaller, so the reading on the scale is a number

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that is LESS than the true weight. Let's consider Newton's 2nd Law (ΣF=ma)

acting on the person. The overall acceleration of the person is downward (with

the elevator). So ma is negative (downward). The only external forces acting on

the person are the force of gravity acting down (-W=-mg) and the supportingNormal Force FN that the scale applies upward on the person.

So ΣF = -ma = -mg + FN

(Note that this is the same equation that we got for Case 3.) We want to know FN because that is the number that we read off the scale.

Now FN = mg – ma

which is LESS than the true weight.

5. The case of Free Fall ( a = - g )

If the elevator cable were to break, the whole elevator-scale-person system would

all begin to accelerate downward due to the force of gravity. All objects in freefall

accelerate downward with the same magnitude (acceleration due to gravity, g). The scale and the person are free falling together, so there is NO contact force

(Normal Force) between the scale and the person. (When they are both falling

together, there is no way that the scale can support any of the person's

weight.)Note that this is a special case of downward acceleration which we

discussed in Case 3 and Case 5.

Just as in Cases 3 and 5, the apparent weight (which is zero when a = -g) is less

than the true weight.

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Exercise


1. A small block B is placed on block A of mass 1 kg and length 20 cm. If initially

the block is placed at the right end of block A. A constant horizontal force of 10 N

is applied on the block A. All the surfaces are assumed frictionless. Find the timein which B separates from A.(a) 0.2 s(b) 0.32 s(c) 0.39 s

(d) 0.45 s

2. A block of massm is suspended with 2 ropes making angles α & β w.r.t to the

horizontal. Find the ratio of tension in the ropes.(a)sin α/sin β(b)tan α/tan β(c)cos α /cos β(d)cosβ/cos α

3. What must the acceleration of a lift going up be so that a block of mass "M"

exerts force 4Mg on the floor?(a) 4g(b) 2g(c) 3g(d) g

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4. A lamp hangs vertically from a cord in an elevator which is descending with a

downward acceleration of a=2.0m/s2. The tension in the cord is T=10.0N. What is

the mass m of this lamp? (Take g = 10ms-2)

(a)1.25kg(b)4.00 kg(c)0.64kg(d)2kg

5. A toy train consists of three identical compartment A, B and C. It is being

pulled by a constant force F along C. The ratio of the tensions in the string

connecting AB and BC is(a) 2 : 1

(b) 1 : 3(c) 1 : 1(d) 1 : 2

Numericals

1.A mass of 5 kg is suspended by a rope of length 2 m from the ceiling. A forceof 45 N in the horizontal direction is applied at the midpoint R of the rope, asshown. What is the angle the rope makes with the vertical in equilibrium?(Take g = 10 ms-2). Neglect the mass of the rope.

2.A mass of 3 kg rests on a horizontal plane. The plane is gradually inclineduntil at an angle θ = 20° with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface?

3. A small block B is placed is placed on another block A of mass 7 kg andlength 15 cm. Initially the block B is near the right end of block A. A constanthorizontal force of 10 N is applied to the block A. All the surfaces are assumedfrictionless. Find the time elapsed before the block B separates from A.

4.A ball of mass 5 kg and a block of mass 12 kg are attached by a lightweightcord that passes over a frictionless pulley of negligible mass as shown in the

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figure. The block lies on a frictionless incline of angle 30o. Find the magnitudeof the acceleration of the two objects and the tension in the cord. Take g = 10ms-2.

5.Two blocks of masses 5 kg and 7 kg are placed in contact with each other on africtionless horizontal surface. A constant horizontal force 20 N is applied tothe block of mass 7 kg.(a) Determine the magnitude of the acceleration of the two-block system.

(b) Determine the magnitude of the contact force between the two blocks.

6.A 75.0 kg man stands on a platform scale in an elevator. Starting from rest,the elevator ascends, attaining its maximum speed of 1.20 m/s in 1.0 s. Ittravels with this constant speed for the next 10.0 s. The elevator thenundergoes a uniform acceleration in the negativeydirection for 1.70 s andcomes to rest. What does the scale register -(a) before the elevator starts to move?(b) during the first 1.00 s?(c) while the elevator is traveling at constant speed?(d) during the time it is slowing down? Take g = 10 ms-2.

7.Two masses 5 kg and 7 kg situated on a frictionless, horizontal surface areconnected by a light string. A force of 50 N is exerted on one of the masses tothe right. Determine the acceleration of the system and the tensionTin thestring.

8.A block of mass m1 = 5kg on a frictionless horizontal table is connected to a block of mass m2 = 3kg by means of a very light pulley P1 and a light fixedpulley P2as shown in figure. Ifa1 anda2 are accelerations ofm1 andm2,

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respectively. (a) Find the relation betweena1 anda2 (b) the tensions in thestrings (c) the accelerationsa1 and a2. Take g=10ms-2.

Answers to Practice Exercise

1. 480

2. µs = tan200= 0.36

3. t= 0.46s

4. T = 52.94N , a = 0.59s

5. a = 1.67ms-2, F1 = 8.33N, F2=11.67N

6. (a) 750N (b) 660N (c) 750N (d) 802.5N

7. a= 4.17ms-2, T=20.8N

8. (a) a1=2a2 (b) 2T1 = T2, T1 = 13N,

T2 = 26N (c) a1=2.6ms-2, a2 = 1.3ms-2

9. (a)M = 3msn!

(b) T1 = 3m"sn!, T2 = 6m"sn!

Answers to MCQ1. (a)

2. (d)

3. (c)

4. (a)

5.(d)

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6.1 Introduction

In the previous chapter we discussed

Newton's laws and analyzed the

motion of objects. In this unit the

effect that work has upon the energy

of an object (or system of objects) will

be investigated; the everyday

definition of “work” and the one that

we use in physics are quite different

from each other. When most people

think about “work” they think of the

job that they have. A farmer

ploughing the field, a teacher

teaching a class, a student preparing

for the examination all are said to be

working. A person who can put in

long hours of work is said to have a

large stamina or energy. So energy is

the capacity to do work The word ‘power’ is used in everyday

life with different shades of meaning.

In karate or boxing we talk of

‘powerful’ punches. These are

delivered at a great speed. This shade

of meaning is close to the meaning of

the word ‘power usedin physics. We shall find that there is

at best a loose correlation between

the physical definitions and the

physiological pictures these terms

generate in our minds. The aim of

this chapter is to develop an

understanding of these three physical

quantities. Before we proceed to thistask, we need to develop a

mathematical prerequisite, namely

the scalar product of two vectors.

6.2 The Scalar Product

The scalar product of two vectors can be constructed by taking

thecomponent of one vector in the

direction of the other and multiplying

it times the magnitude of the other

vector. This can be expressed in the

form:

If the vectors are expressed in terms

of unit vectors i, j, and k along the x,

y, and z directions, the scalar product

can also be expressed in the form:

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The scalar product is also called the"inner product" or the "dot product"

in some mathematics texts. A.A = A x. A x +A y.A y + Az.Az A2 = A x2 + A y2 + Az2

A.B = 0 if A & B are perpendicular

Properties of the Dot Product

(1)(Commutative Property) For any

two vectors A and B, A.B = B. A.

(2) (Scalar Multiplication

Property) For any two vectors A and B and any

real number c,(cA).B = A. (cB) =c (A.B)

(3) (Distributive Property) For any 3

vectors A, B and C, A.(B+C) = A.B + A.C.

6.3 Unit Vectors

It is convenient to use vectors of unit

length to specify the directions of vector quantities in the various

coordinate systems. InCartesian

coordinates it is typical to use i, j and

k to represent unit vectors in the x, y

and z directions respectively. A vector

which specifies a position in space

with respect to the origin of the

coordinate system could then be written.For unit vectors i, j, k we have

i.i = j.j =k.k = 0i.j = j.k =k.i = 0

Example 6.1 A body constrained to move along

the z-axis of a coordinate system is

subject to a constant forceF given

by F= (-i +2j +3k) N Where i, j, k are unit vectors along

the X, Y and Z axes of the system

respectively? What is the work done

by this force in moving the body a

distance of 4 m along the z-axis?Solution:

Force exerted F= (-i +2j +3k) Nd= 4 k m

Work done = F. d= (-i +2j +3k). (4 K)

= 0 + 0 + 3x4= 12J

Hence 12 J work is done by the force

on the body.

Try this:

Find the angle between force F = (3i+

4j +5k) and displacement d = (5i + 4j

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+3k) unit. Also find the projection of

F on d?

6.4 Work

When a force acts upon an object

to cause a displacement of the

object, it is said that work was

done upon the object. There are

three key ingredients to work - force

displacement and cause. There are

several good examples of work thatcan be observed in everyday life - a

horse pulling a plough through the

field, a man pushing a grocery cart

down the aisle of a grocery store.

6.4.1 Work Equation

Mathematically, work can be

expressed by the following equation. W = F d cos Θ

whereF is the force,d is the

displacement, and the angle (theta)

is defined as the angle between the

force and the displacement vector.

6.4.2 Positive Negative & Zero

work

Case A: A force acts rightward upon

an object as it is displaced rightward.

In such an instance, the force vector

and the displacement vector are inthe same direction. The angle

between F and d is 0 degrees and

cos0 = +1

Examples of positive work:

(i) When a body falls freely undergravity, force of gravity and

displacement are in the same

direction and the work done by force

of gravity is positive. (ii) When a spring is stretched the

work done by the stretching force is

positive.

Case B: A force acts leftward upon an

object that is displaced rightward. In

such an instance, the force vector

and the displacement vector are in

the opposite direction. Thus, the

angle between F and d is 180 degrees

and cos 180 is –1.

Examples of negative work

(i) A car skidding to a stop on a

roadway surface or a baseball runner

sliding to a stop on the infield dirt. In

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direction opposite the objects motion

in order to slow it down. The force

doesn't cause the displacement but

ratherhindersit. These situations involve what is

commonly callednegative work.

Case C:(i) A force acts upward on an

object as it is displaced rightward. In

such an instance, the force vector

and the displacement vector are at

right angles to each other. Thus, theangle between F and d is 90 degrees.

Then the work one is zero as cos900 =

0(ii)When the force applied or the

displacement or both are zero

Examples of Zero work

Waiter who carried a tray full of

meals above his head by one arm

straight across the room at constant

speed. The force supplied by the

waiter on the tray is an upward force

and the displacement of the tray is a

horizontal displacement. As such,

the angle between the force and the

displacement is 90 degrees. If the

work done by the waiter on the tray

were to be calculated, then the

results would be zero.(ii) When we push hard against a

wall, the force we exert on the wall

does no work, because s = 0.

However, in this process, our muscles

are contracting and relaxing

alternately and internal energy isused. That is why we get tired.

Example 6.2

The sign of work done by a force on a body is important to understand.

State carefully if the following

quantities are positive or negative?a) Work done by a man in lifting a

bucket out of a well by means of a

rope tied to the bucket.(b) Work done by gravitational force

in the above case,(c) Work done by friction on a body

sliding down an inclined plane,(d) Work done by an applied force on

a body moving on a rough horizontal

plane with uniform velocity,(e) Work done by the resistive force of

air on a vibrating pendulum in

bringing it to rest.

Try these:

A body of mass 2 kg initially at rest

moves under the action of an applied

horizontal force of 7 N on a table with

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coefficient of kinetic friction = 0.1

Compute the(a) Work done by applied force

in 10 s(b) Work done by friction in 10 s(c) Work done by the net force on the

body in 10 s,(d) Change in kinetic energy of the

body in 10 s and interpret your

results.

6.4.3 Units of Work

Jouleis the SI unit of work.1 Joule = 1 newton x 1 meter = 1Nm

Dimension formula for work:

[M1 L2 T-2]C.G.Sunit of work iserg.

1 erg = 1 dyne-cm Joule: Work done is 1 joule if a force

of 1N displaces a body through a

distance of 1 m along the direction of

force.1 J = (1N)(1 m) = 1 kg m2s-2

Erg: Work done is 1 erg if a force of 1

dyne displaces a body through a

distance of 1 cm along the direction

of force.1 erg = (1 dyne) (1 cm) = 1 g cm2s-2

Try This:

A man pushes a roller with a force of

50 N through a distance of 20 m.

What is the work done by him if the

handle of the roller is inclined at anangle of 600 with the ground? (Ans.

500J)

6.4.4 Conservative and Non-

Conservative Forces

The work a conservative force does on

an object in moving it from one point

to another is path independent - itdepends only on the end points of

motion Eg: Force of gravity and the

spring force. The total work done by a conservative

force on a particle is zero when it

moves round any closed path and

returns to the initial positionFor non-conservative force the work

done in moving from one point to

another depends on path taken.Eg: Friction and air resistance

Fig 6.1 (a)

6.5 Work done by Constant Force

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(Graphical Method)

Determine the work done by a

constant force by interpreting a force

–displacement graph. As work is the product of force and

distance, it can be represented as the

area under a graph of force as a

function of distance. In the below

graph the shaded region represents

the work done on an object that

undergoes a constant force.Fig 6.2

Thus, the work done (area of theshaded region) can be calculated by

multiplying distance (base) by force

(height).

6.6 Work Done by a Variable Force

Fig 6.3

Consider a body being moved, along

an arbitrary path, by a force which

continuously changes in magnitude

as well as direction. We first consider a force, acting on a

body that varies in magnitude only.

Assume that the magnitude of force

is a function of position ‘x’ that is

F(x), acting in the positive X

direction. We should like to find out

the work done, by F in moving a body

from x1 to x2. Here W is given by W =

The physical significance of definite

integral is area under the curve.Hence, W is nothing but Area under

the curve F(x) between the lines x =

x1 and x = x2 as shown in figure given

below.Now, consider that the force F

changes its magnitude as well as

direction from point to point. The

work done by the variable force F on

the body for two closer points on the

curve. If the displacement ∆x is

small, we can take the force F(x) as

approximately constant and the work

done is∆W =F(x).∆x

Adding successive rectangular areas

in Fig 6.3 we get the total work done

as

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W=∑ x i

xf

F ( x ) . Δ x

where the summation is from the

initial position xi to the final position xfIf the displacements are allowed to

approach zero, then the number of

terms in the sum increases without

limit, but the sum approaches a

definite value equal to the area under

the curve in Fig. Then the work done

is

W = lim∆x→0 ∑ x i

xf

F ( x ) . Δ x

= ∫ xi

x f

f ( x ) dx

Where ‘lim’ stands for the limit of the

sum when ∆x tends to zero. Thus, for

a varying force the work done can beexpressed as a definite integral of

force over displacement.

6.7 Energy

A body which has the capacity to do

work is said to possess energy.For example, water in a reservoir is

said to possess energy as it could be

used to drive a turbine lower down

the valley. There are many forms of

energy e.g. electrical, chemical heat,

nuclear, mechanical etc.

The SI units are the same as those

for Work, Joules J.In this module only purely

mechanical energy will be considered. This may be of two

kinds,potential andkinetic.

6.7.1 Potential Energy

There are different forms of potential

energy Two examples are :1. Gravitational Potential Energy

It may be described as energy due to

position relative to a standard

position (normally chosen to be the

earth's surface.) The potential energy of a body may be

defined as the amount of work it

would do if it were to move from itscurrent position to the standard

position.

Eg: A pile driver raised ready to fall

on to its target possesses

gravitational potential energy.

Fig 6.4

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A body is at rest on the earth's

surface. It is then raised a vertical

distance h above the surface. The

work required to do this is the forcerequired times the distance h.Since the force required is its weight,

and weight, W = mg, then the work

required is mgh. The body now possesses this amount

of energy - stored as potential energy

- it has the capacity to do this

amount of work, and would do so ifallowed to fall to earth.Potential energy is thus given by:

Potential Energy = mgh

where h is the height above the

earth's surface.Gravitational potential energy of an

object, as a function of the height h,

is denoted by V(h) and it is thenegative of work done by the

gravitational force in raising the

object to that height. V(h) = mgh

If h is taken as a variable, it is easily

seen that the gravitational force F

equals the negative of the derivative

of V(h) with respect to h. Thus,F = -dV(h)/dh = -mg The negative sign indicates that the

Gravitational force is downward.Physically, the notion of potential

energy is applicable only to the class

of forces where work done against the

force gets ‘stored up’ as energy. When external constraints are

removed, it manifests itself as kineticenergy. Mathematically,Potential energy V(x) is defined if the

force F(x) can be written asF(x) = - dV/dx

The dimensions of potential energy

are [ML2 T –2] and the unit is joule (J),

the same as kinetic energy or work.

To reiterate, the change in potential

energy, for a conservative force, ∆V is

equal to the negative of the work

done by the force∆V = − F(x)∆x

2. Elastic Potential Energy

Fig 6.5

Elastic potential energy is the

potential energy observed due to

stretching or compression by an

external force to a given elastic

object.Formula for elastic potential

energy

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Fig 6.6

According to Hooke's law,Force given to stretch the spring is

directly proportional to its

displacement. It is given by

Where k is the spring constant, and x

is the amount by which the spring isstretched (x > 0) or compressed (x <

0). When a moving object runs into a

relaxed spring it will slow down, come

to rest momentarily, before

accelerating in a direction opposite to

its original direction. While the object

is slowing down it will compress the

spring. As the spring is compressedthe kinetic energy of the block is

gradually transferred to the spring

where it is stored as potential energy. The potential energy of the spring in

its relaxed position is defined to be

zero. The potential energy of the

spring in any other state can be

obtained from Hooke's law

TheElastic Potential Energyfor the

stretched spring is given by

where P.E is the elastic potential

energy.

Elastic Potential energy formula isused for the problems where

displacement, elasticity or elastic

force are mentioned. It is expressed

in Joule.

Example 6.3

The vertical spring is attached to the

load of mass 5 kg and is compressed by 8 m. Calculate the Force constant

of the spring.Solution: Mass m = 5kg,Distance x = 8 cmForce is given by F = ma = 5 kg × 9.8 m/s2

= 49 NForce in the stretched spring is given

by : F=kxForce Constant k= F/x = 49N/8m

= 6.125 N/m.

Try this

When a 4 kg mass is hung vertically

on a certain light spring that obeys

Hooke's law, the spring stretches

2.5cm. If the 4 kg mass is removed,(a) how far will the spring stretch if a

1.5 kg mass is hung on it, and

(b) how much work must an external

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agent do to stretch the same spring 4

cm from its un-stretched position?

HOTS Question A 1 kg block situated on a rough

incline is connected to a spring of

spring constant 100 N m –1 as shown

in Fig. The block is released from rest

with the spring in the un-stretched

position. The block moves 10 cm

down the incline before coming to

rest. Find the coefficient of friction between the block and the incline.

Assume that the spring has a

negligible mass and the pulley is

frictionless.

Fig 6.8

6.7.2 Kinetic Energy

Kinetic energy may be described as

energy due to motion.

Fig 6.7

The energy possessed by an object by

virtue of its motion is called kinetic

energy

For example when a hammer is usedto knock in a nail, work is done on

the nail by the hammer and hence

the hammer must have possessed

energy.Only linear motion will be considered

here.

Formula for kinetic energy

Let a body of mass m moving with

speed v be brought to rest with

uniform deceleration by a constant

force F over a distance s.Using Equation : v2 = u2 +2as We get 0 = u2+ 2as

s = v2/2a And work done is given by

Work done = Force x Displacement = F x s =F x (v2/2a) The force is F = ma so Work done = max v2/2a

= ½ mv2

Thus the kinetic energy is given by

6.7.3 Kinetic Energy and Work

Done (Work-Energy Theorem)

When a body with mass m has its

speed increased from u to v in a

distance s by a constant

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force F which produces an

acceleration a, then from Equation of

motion we know

v2

= u2

+2as v2-u2 = 2asMultiplying by ½ ½ v2 -½ u2 = asmultiplying this by ‘m’ gives an

expression of the increase in kinetic

energy (the difference in kinetic

energy at the end and the start)

Thus since F = ma

but also we know

So the relationship between kinetic

energy can be summed up as Work done by forces acting on a

body = change of kinetic energy in

the body.

W = K f- Ki = ∆K

This is known as the work-energy

theorem.

6.7.4 Work – Energy Theorem for a

Variable Force (Calculus Method)

So far we have looked at the work

done by a constant force. In the

physical world, however, this often is

not the case. Consider a force acting

on an object over a certain distance

that varies according to the

displacement of the object. Let us call

this force F(x), as it is a function of x. Though this force is variable, we can

break the interval over which it acts

into very small intervals, in which

the force can be approximated by a

constant force. Let us break the force

up into N intervals, each with

length dx. Also let the force in each of

those intervals be denoted by F1, F2…FN. Thus the total work

done by the force is given by: W = F1 dx + F2 dx + F3 dx + ……

+ FNdx

Thus W = Fn dx

This sum is merely an approximationof the total work. Its degree of

accuracy depends on how small the

intervals dxare. The smaller they

are, the more divisions ofF arise,

and the more accurate our

calculation becomes. Thus to find an

exact value, we find the limit of our

sum as dx approaches zero. Clearlythis sum becomes an integral, as this

is one of the most common limits

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Fn dx = F(x)dx

Thus W = ∫ xi

x f

F ( x ) dx

We have generated an integral

equation that specifies the work done

over a specific distance by a position

dependent force. It must be noted

that this equation only holds in the

one dimensional case. In other

words, this equation can only be

used when the force is always parallel

or anti-parallel to the displacement of

the particle. The equation we derived

for work done by a variable force, we

can use it to yield the work-energy

theorem.Fnet= ma = m(dv/dt)

= m (dv/dx) x (dx/dt)= mv (dv/dx)

Now we plug in our expression for

force into our work equation:

Wnet = Fnetdx

= mv (dv/dx) dx = mvdv

Integrating from vo to vf

Wnet = mv dv

= ½ mvf2 – ½ mvo2

This result is precisely the Work

-Energy theorem. Since we have

proven it with calculus, this theoremholds for constant and non-constant

forces alike.Example 6.4

A car of mass 1000 kg travelling at

30m/s has its speed reduced to

10m/s by a constant breaking force

over a distance of 75m.Find:

(a)The cars initial kinetic energy(b)The final kinetic energy(c)The breaking force

Solution:

a) Initial Kinetic Energy = ½ mv2

= 500x 302 =450000J = 450 kJ b) Final Kinetic Energy

= ½ mv2= 50 kJc) Change in kinetic energy

= 450 - 50 = 400 kJ Work done = change in kinetic energySo F x s = 400 kJBreaking force x 75 = 400 kJBreaking force = 400000/75 = 5333N

Try this

A body of mass 0.5 kg travels in a

straight line with velocity v=ax3/2

where a=5m-1/2s-1. What is the work

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done by the net force during its

displacement from x = 0 to x = 2 m?

6.8 Law of Conservation of

Mechanical Energy

The total amount of mechanical

energy, in a closed system in the

absence of dissipative forces (e.g.

friction, air resistance), remains

constant.For problems involving only

conservative forces, we can now quite

easily derive conservation of

mechanical energy. As we already know ∆U = - W

∆K = W

Together this says that∆K+ ∆U =0

This says that K+U doesn’t change,so it must be constant:

K + U = a constant

We callK+U the total mechanical

energy .

6.8.1 Examples of Conservation of

Energy

Energy conservation of a simplependulum

Fig 6.9

At the two extreme points A &B the

potential energy is maximum and the

kinetic energy is minimum. At the

mean postion the energy is

completely kinetic and the pendulam

possessess zero potential energy.In between te extreme ponts and the

mean position the energy is partly

kinetic and partly potential.The

graph for energy variation at different

points is shown below

Fig:6.10

Energy conservation of an

oscillating spring

Note that the amount of work done by the spring on the block after it

returns to its original position is zero.

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Fig 6.11

The kinetic energy of the mass on the

spring increases as it approaches the

equilibrium position; and it decreases

as it moves away from theequilibrium position.Every time the spring is compressed

or stretched relative to its relaxed

position, there is an increase in the

elastic potential energy. The amount

of elastic potential energy depends on

the amount of stretch or compression

of the spring. The equation thatrelates the amount of elastic potential

energy to the amount of compression

or stretch (x) is PEspring = ½kx2

wherek is the spring constant(in N/m) andx is the distance that

the spring is stretched or compressed

relative to the relaxed, un-stretchedposition. A vibrating mass on a spring has the

greatest potential energy in the

maximum stretched or compressed

position. And it also has the smallest

potential energy at the mean position

when the velocity of vibrations is

maximum.

Fig6.12

Energy conservation in the case of

a freely falling body.

Let a body of mass 'm' placed at a

height 'h' above the ground, start

falling down from rest.In this case we have to show that the

total energy (potential energy +

kinetic energy) of the body at A, Band C remains constant, i.e, potential

energy is completely transformed into

kinetic energy.

Fig 6.13Body of mass 'm' placed at a height

'h'

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At A,

Kinetic energy =1/2mv2 = 0[the

velocity is zero as the object isinitially at rest]

Total energy at A = PE + KE = mgh +

0 Total energy at A = mgh … (1)

At B,

Potential energy = mgy= mg (h -x) [Ht from ground is (h -

x)]Potential energy = mgh - mgx

Kinetic energy =1/2mv2

The body covers the distance x with a

velocity v. We make use of the third

equation of motion to obtain velocityof the body.

v2 - u2 = 2aS

Here, u = 0, a = g and s = x

v2 - 0 = 2gx v2 = 2gx

Kinetic energy = ½ x m x 2gx= mgx Total energy at B = PE + KE

= mgh –mgx + mgx

Total energy at B = mgh … (2)

At C,

Potential energy = m x g x 0 [h =

0]

Potential energy = 0


The distance covered by the body is h

v2 - u2 = 2aS

Here, u = 0, a = g and S = h


= 1/2 x m x 2gh = mgh Total energy at C = PE + KE

= 0 + mgh

Total energy at C = mgh … (3)

It is clear from equations 1, 2 and 3

that the total energy of the body

remains constant at every point.

Thus, we conclude that law of

conservation of energy holds good inthe case of a freely falling body.

Example 6.5

A bob of mass m is suspended by a

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horizontal velocity vo at the lowest

point A such that it completes a

semi-circular trajectory in the

vertical plane with the string becoming slack only on reaching the

topmost point, C. This is shown in

Fig. Obtain an expression for (i)vo ;

(ii) the speeds at points B and C; (iii)

the ratio of the kinetic

energies(KB/KC) at B and C. Comment

on the nature of the trajectory of the

bob after it reaches the point C

Fig 6.14

Solution:

(i) There are two external forces on

the bob: gravity and the tension (T )

in the string. The latter does no work

since the displacement of the bob is

always normal to the string. The

potential energy of the bob is thus

associated with the gravitational

force only. The total mechanicalenergy E of the system is conserved.

We take the potential energy of the

system to be zero at the lowest point

A.

Thus,at A :(i)E = 1/2mv02………………… (1) T A – mg = mv02/L … (2) (Newton’s

second law) where T A is the tension inthe string at A. At the highest point C, the string

slackens, as the tension in the string

(TC) becomes zero.E = 1/2mvc2+2mgL………. (3) mg = mvc2/L………………….(4)[Newton’s Second Law] where vC is

the speed at C.

From eqn (3) & (4); E = 5/2mgLEquating this to the energy at A5/2mgL = 1/2mv02

or v0 =√5 gL

(ii) It is clear from eqn (4) that

vc =√ gL

At B, the energy isE=1/2 mvB2 + mgL

Equating this to the energy at A andemploying the result from (i),

namely,v02 = 5gL1/2 mvB2 + mgL= 1/2mv02= 5/2 mgL

vB =√3 gL

(iii)The ratio of the kinetic energies at

B and C is: KB/Kc=[1/2mvB

2]/[1/2mvc2]=3:1

At point C, the string becomes slack

and the velocity of the bob is

horizontal and to the left. If the

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projection akin to a rock kicked

horizontally from the edge of a cliff.

Otherwise the bob will continue on

its circular path and complete therevolution.Example 6.6

A cyclist and his bicycle has a mass

of 80 kg. After 100m he reaches the

top of a hill, with slope 1 in 20

measured along the slope, at a speed

of 2 m/s. He then free wheels the

100m to the bottom of the hill wherehis speed has increased to 9m/s.How much energy has he lost on the

hill?Solution:

Fig 6.15

If the hill is 100m long then height is:h = 100x1/20 = 5mSo potential energy lost ismgh=80x9.81x5 =394JIncrease in kinetic energy is= ½ mv2 – 1/2mu2=1/2m (v2 –u2 )=40(81-4) =3080JBy the principle of conservation of

energyInitial energy = Final energy +loss of

energyLoss of energy =3924 – 3080 = 844 J

Try this

The potential energy function for a

particle executing linear simpleharmonic motion is given by

V(x) =kx2/2, where k is the force

constant of the oscillator. For k = 0.5

N m –1, the graph of V(x) versus x is

shown in Fig below. Show that a

particle of total energy 1 J moving

under this potential must ‘turn back’

when it reaches x = ± 2 m.

Fig 6.16

6.9 Power

Power is the rate at which work is

done. It is the work/time ratio.Mathematically, it is computed usingthe following equation.Power = Work / time = W / t The standard metric unit of power is

the Watt. As is implied by the

equation for power, a unit of power is

equivalent to a unit of work divided

by a unit of time. Thus,a Watt is

equivalent to a Joule/second.Kilowatt and megawatt are higher

units of power.1 kilowatt (KW) = 103 watts1 Megawatt (MW) = 106 watts

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Another unit of power is horse power

(h.p) and 1 h.p. = 746 watts

6.9.1Power Based on Force andSpeed

We can calculate power based on

force and speed. Because work

equals force times distance, you can

write the equation for power the

following way, assuming that the

force acts along the direction of

travel:

where s is the distance travelled.

However, the object’s speed, v, is

just s divided by t, so the equation

breaks down to

Example 6.7

Calculate the power of an enginerequired to lift 105kg of coal per hour

from a mine 100m deep. Given

g=10m/s2.Solution:

Power = workdone/time The work done in lifting coal is

nothing but gravitational potential

energy.Power = mgh /tMass of coal = 105 kgg=10m/s2

h=100m Time = 1 hour=60x60 =3600s

P =105 x 10 x 100/3600 = 100x104/36 =27777.78 W

Try This :

A pump on the ground floor of a

building can pump up water to fill a

tank of volume 30 m3 in 15 min. If

the tank is 40 m above the ground,

and the efficiency of the pump is

30%, how much electric power is

consumed by the pump?

6.10 Different Forms Of Energy

Thermal energy

Thermal energy or heat energy

reflects the temperature difference

between two systems.Example: A cup of hot coffee has

thermal energy. You generate heat

and have thermal energy with respectto your environment.Nuclear energy

Nuclear energy is energy resulting

from changes in the atomic nuclei

orfrom nuclear reactions.

Example:Nuclear fission,nuclear

fusion, and nuclear decay are

examples of nuclear energy. An

atomic detonation or power from a

nuclear plant is specific example of

this type of energy.

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Chemical energy

Chemical energy resultsfrom

chemical reactions between atoms or

molecules. There are different typesof chemical energy, such as

electrochemical energy and chemi-

luminescence.

Example: A good example of chemical

energy is an electrochemical cell or

battery.

Electromagnetic energyElectromagnetic energy is energy

from light or electromagnetic waves.Example: Any form of light

haselectromagnetic energy,including

parts of the spectrum we can't see.

Radio, gamma rays, x-rays,

microwaves, andultraviolet light are

some examples of electromagnetic

energy.

Fig

6.17

6.11 Collisions

Elastic Collision

An elastic collision is that in which

the momentum of the system as well

as kinetic energy of the system beforeand after collision is conserved.Inelastic Collision

An inelastic collision is that in which

the momentum of the system before

and after collision is conserved but

the kinetic energy before and after

collision is not conserved.

Elastic Collision in One Dimension

Consider two non-rotating spheres of

mass m1 and m2 moving initially

along the line joining their centers

with velocities u1and u2in the

same direction. Let u1is greater than

u2. They collide with one another

and after having an elastic collision

start moving with velocities v1and

v2 in the same directions on the

same line.

Fig 6.18

Momentum of the system before

collision = m1u1 + m2u2Momentum of the system after

collision = m1 v1 + m2 v2 According to the law of conservation

of momentum:

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m1u1 + m2u2= m1 v1 + m2 v2m1 v1 – m1u1= m2u2 – m2 v2

m1(v1 – u1) = m2 (u2 – v2) ------- (1)

SimilarlyK.E of the

system before collision= ½ m1u12 + ½ m2u22

K.E of the system after collision= ½ m1 v12 + ½ m2 v22

Since the collision is elastic, so the

K.E of the system before and after

collision is conserved. Thus,½ m1 v12 + ½ m2 v22 = ½ m1u12 + ½ m2u22

½ (m1 v12 + m2 v22) = ½ (m1u12 + ½ m2u22) And

m1 v12- m1u12= m2u22-m2 v22

m1(v12-u12) = m2(u22-v22) ------- (2)Dividing equation (2) by equation (1)

v1+u1 = u2+v2From the above equation v1= u2+ v2-u1_________ (a) v2= v1+ u1-u2_________ (b)Putting the value of v2 in equation (1)m1 (v1-u1) =m2 (u2-v2)m1 (v1-u1) =m2u2-(v1+u1-u2)

m1(v1-u1) =m2u2-v1-u1+u2m1(v1-u1) = m2 2u2-v1-u1m1 v1-m1u1=2m2u2-m2 v1-m2u1m1 v1+m2 v1=m1u1-m2u1+2m2u2 v1(m1+m2)=(m1-m2)u1-2m2u2

In order to obtain v2 putting the value

of v1 from equation (a) in equation (i)m1(v1-u1) = m2(u2-v2)

m1(u2+v2-u1-u1)=m2(u2-v2)m1(u2+v2-2u1)=m2(u2-v2)m1u2+m1 v2-2m1u1=m2u2-m2 v2m1 v2+m2 v2=2m1u1+m2u2-m1u2 v2(m1+m2)= 2m1u1+(m2-m1)u2

Example 6.8

Two balls, each with mass 2 kg, and velocities of 2 m/s and 3 m/s collide

head on. Their final velocities are 2

m/s and 1 m/s, respectively. Is this

collision elastic or inelastic?Solution: To check for elasticity, we

need to calculate the kinetic energy

both before and after the collision.

Before the collision, the kineticenergy is 1/2(2)(2)2 + 1/2(2)(3)2 = 13 .

After, the kinetic energy is1/2(2)

(2)2 + 1/2(2) (1)2 = 5. Since the kinetic

energies are not equal, the collision is

inelastic.

Try this

Which of the following potentialenergy curves in Figure cannot

possibly describe the elastic collision

of two billiard balls? Here ‘r’ is the

distance between centres of the balls.

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Fig 6.19

Collision in Two Dimensions A collision in two dimensions obeys

the same rules as a collision in one

dimension:

Total momentum in each direction is

always the same before and after the

collision Total kinetic energy is the same

before and after an elastic collisionConsider a particle A of mass m1moving with a velocity v1i in the X

direction. Let it collide with a particle

B of mass m2 at rest .After collisions

let A move with a velocity v1f in a

direction making an angle θ1 along

with the X axis and B with a velocity

v2f in a direction making an angle θ2 with the X axis.

Fig

6.20

Applying the law of conservation of

momentum in the X and Y direction

X directionm1 v1i= m1 v1fcos θ1 +m2 v2fcosθ2

Y direction0 = m2 v2fsin θ2 –m1 v1fsin θ1

Assuming the collision to be perfectly

elastic. Conservation of kineticenergy yields,1/2m1 v1i2 =1/2m1 v1f2+1/2m2 v2f2

i.e., m1 v1i2=m1 v1f2 +m2 v2f2

Coefficient of Restitution

When there is a head on Collision

between two bodies, the ratio of their

relative velocity after collision and

their relative velocity before collision

is called the Coefficient of restitution. Thus e =∣ v1- v2/u1-u2∣

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For a perfectly elastic collision,the value of e is 1

If 0 < e < 1, then the collision is

inelastic.For a perfectly inelastic collision,e = 0.

If e > 1, then the collision is a super-

elastic collision. The comparison between Elastic and

inelastic collision is given below:

Example 6.9

A luxury car with a mass of 1800Kg

stopped at a traffic light is struck

from the rear by a car with a mass of

900kg. The two cars become

entangled as a result of the collision.

If the car was moving at 20m/s

before the collision, what will be the

velocity of the entangled mass after

the collision?

Solution:

Use the law of conservation of

momentum for a perfectly inelastic

collision:m₁ v₁ (i) + m₂ v₂ (i) = (m₁ + m₂) v (f)

(1800kg) (0) + (900kg) (20m/s) =

(1800kg+900kg)v(f)

v(f) = 18,000kg•m/s / 2700kg

= 6.7m/s

Try This: A trolley of mass 200kg moves with a

uniform speed of 36km/h on a

frictionless track. A child of mass

20kg runs on the trolley from one

end to the other (10m away) with a

speed of 4m/s relative to the trolley

in a direction opposite to its motion

and jumps out of the trolley. What isthe final speed of the trolley? How

much has the trolley moved from the

time the child begins to run?

(Ans: 25.9m)

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7.1 Introduction

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Rotational motion or we can say

circular motion can be analyzed in

the same way of linear motion. In thisunit we will examine the motion of

the objects having circular motion.

For example, we will find the velocity,

acceleration and other concepts

related to the circular motion in this

section. Uniform circular motion is

one of the example of this subject. In

uniform circular motion speed of theobject is always constant and

direction is changing. Thus, velocity

of the object is changing and as a

result object has acceleration. Some

concepts will be covered in this units;

rotational speed (angular speed),

tangential speed (linear speed),

frequency, period, rotational inertia ofthe objects, torque, angular

momentum and its conservation.

Inphysics, a rigid body is an

idealization of a solid body in

whichdeformation is neglected. In

other words, thedistance between

any two givenpoints of a rigid bodyremains constant in time regardless

of externalforces exerted on it. Even

though such an object cannot

physically exist due torelativity,

objects can normally be assumed to

be perfectly rigid if they are not

moving near thespeed of light.In classical mechanics a rigid body is

usually considered as a continuousmass distribution, while inquantum

mechanics a rigid body is usually

thought of as a collection of point

masses. For instance, in quantum

mechanicsmolecules (consisting of

the point masses: electrons and

nuclei) are often seen as rigid bodies

In this chapter, we will study aboutthe special kind of motion of a system

of particles that is rotation. We see

examples of rotational motion in our

everyday life for example rotation of

earth about its own axis create the

cycle of day and night. Motion of

wheel, gears, motors, planet, blades

of the helicopter etc are all theexample of rotational motion. To

understand the rotational motion as

a whole we are first required to

understand the concept of angular

position, velocity, acceleration,

centripetal acceleration. Till now in

our study of dynamics we have

always analyzed motion of an object by considering it as a particle even

when the size of the object is not

negligible. Here in this chapter we

will consider rigid bodies having

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definite shape and size and are

capable of having both rotational and

translational motion.

7.2 Rotational Motion of a Rigid

Body

Rotational motion is more

complicated than linear motion, and

only the motion of rigid bodies will be

considered here. Arigid body is an

object with a mass that holds a rigidshape, such as a phonograph

turntable, in contrast to the sun,

which is a ball of gas. Many of the

equations for the mechanics of

rotating objects are similar to the

motion equations for linear motion.7.3 Angular velocity and angular

acceleration Theangular displacement of a

rotating wheel is the angle between

the radius at the beginning and the

end ofa given time interval. The SI

units are radians. The average

angular velocity (ω, Greek letter

omega), measured in radians per

second, is

Theangular acceleration (α, Greek

letter alpha) has the same form as

the linear quantity

and is measured in

radians/second/second or rad/s2.

The kinematics equations for

rotational motion at constant angular

acceleration are

Consider a wheel rolling without

slipping in a straight line. The

forward displacement of the wheel is

equal to the linear displacement of a

point fixed on the rim. As can be

shown in Figure ,d =S =rθ

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Figure 1

A wheel rolling without slipping.In this case, the average forward

speed of the wheel isv =d/t = (rθ)/t =rω wherer is the distance from the

center of rotation to the point of the

calculated velocity. The direction of

the velocity is tangent to the path of

the point of rotation. The average forward acceleration of

the wheel isa T =r(ω f − ωo)/t =rα This component of the acceleration is

tangential to the point of rotation and

represents the changing speed of the

object. The direction is the same as

the velocity vector. The radial component of the linear

acceleration isa r =v 2/r = ω2 r.

Consider a rigid body of arbitraryshape rotating about a fixed axis

through point O and perpendicular to

the plane of the paper as shown

below in the figure-1 FIG.

while the body is rotating each and

every point in the body moves in a

circle with their center lying on the

axis of rotation and every pointmoves through the same angle

during a particular interval of time.

Consider the position of a particle

say ith particle at point P at a

distance ri from point O and at an

angle θi which OP makes with some

reference line fixed in space say OX

as shown below in the figure 2.

If particle moves an small distance

dsi along the arc of the circle in small

amount of time dt thendsi=vidt ----(1)

where vi is the speed of the particle.dθ is the angle subtended by an arc

of length dsi on the circumference of

a circle of radius ri,so dθ( in radians)

would be equal to the length of the

arc divided by the radius

i.e.

dθ=dsi/ri =vidt/ri ----(2)

distance dsi would vary from particleto particle but angle dθ swept out in

a given time remain same for all the

particles i.e. if particle at point P

moves through complete circle such

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that

dθ=2π radian

Then all the other particles of the

rigid body moves through the angulardisplacement dθ=2π. So rate of

change of angle w.r.t time i.e. dθ/dt is

same for all particles of the rigid

body and dθ/dt is known as angular

velocity ω of the rigid body so

ω=dθ/dt ----(3).Putting equation (3) in equation (2)

we find vi=ri(dθ/dt) =riω ---(4)

This shows that velocity of ith

particle of the rigid body is related to

its radius and the angular velocity of

the rigid body. we know that 2π

radians =3600 or π radians/1800=1

And this relation can be used for

expressing angular velocity in degreeto that of angular velocity in terms of

radian. Angular acceleration is the

rate of change of angular velocity

w.r.t time. Thus for rigid body

rotating about a fixed axis

Differentiating equation (4) w.r.t to t

we find

where ai=ait=riα is the tangential

component of linear acceleration of a

point at a distance rifrom the axis.Each particle in the rigid body also

has the radial linear acceleration

component v2/r, which can also be

expressed in terms of an of angular

velocity i.e.

and this acceleration aic pointinginwards towards the radial line is

also known as centripetal

acceleration.

7.4 Kinetic energy of Rotation

From equation (4) we know that

magnitude of velocity of the ith

particle in a rigid body rotating abouta fixed axis is vi=ri(dθ/dt) =riω where

ri is the distance of particle from the

axis of rotation and ω is the velocity

of the particle. Kinetic energy of the

ith particle of mass mi is given by

The total kinetic energy of the rigid

body as a whole would be equal to

the sum of KEs of all particles in the

body thus

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Since angular velocity ω is same for

all the particles in the body so

tell us how the mass of the rotating

body is distributed about the axis of

rotation is known as rotational

inertia or moment of inertia of the

rotating body. Moment of inertia ofthe rigid body can be obtained by

imaging the body to be subdivided

into large number of particles ,the

mass of the each particle is then

multiplexed by its squared distance

from the axis and then summing over

these products for all the particles in

the body.Si unit of moment of inertiais Kgm2. So rotational kinetic energy

of a body can now be written as:

Above expression of rotational kinetic

energy KE of a rotating rigid body is

analogous to the translational kinetic

energy where moment of inertia isanalogous to mass m (or inertia) and

angular velocity is analogous to

velocity v. Moment of inertia not only

depends on the mass but also on

how this mass is distributed about

the axis of rotation and it must be

specified first before calculating

moment of inertia of any body.

7.5 Calculation of moment of

inertia

We already know that the moment of

inertia of a system about axis of

rotation is given as:

where mi is the mass of the ith

particle and ri is its perpendicular

distance from the axis of rotation. For

a system consisting of collection of

discrete particles, above equation can

be used directly for calculating the

moment of inertia. For continuous

bodies ,moment of inertia about a

given line can be obtained using

integration technique. For this

imagine dividing entire volume of the

rigid body into small volume

elements dV so that all the points in

a particular volume element are

approximately at same distance from

the axis of rotation and le r be this

distance if dm is the mass of this

volume element dV, then moment of

inertia may be given by

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Since density ρ of the element is

defined as mass per unit volume so

ρ=dm/dV hence equation (13) may be

written as,

For homogeneous body density ρ is

uniform hence ρ can be taken out of

the integral sign i.e.,

above integration can be carried out

easily for bodies having regular

shapes as can be seen from examples

given below.Moment of inertia of uniform rod

about a perpendicular bisector

Consider a homogeneous and

uniform rod of mass M and length L

as shown below in the figure:

we have to calculate the moment of

inertia of the rod about the bisector

AB. Consider middles point O to be

the origin of the rod .Also consider an

element of the rod between thedistance x and x+dx from the origin.

Since the rod is uniform so its

density ρ=M/L

Hence mass of the element

dm=(M/L)dx. Perpendicular distance

of this element from line AB is x,so

that moment of inertia of this

element about AB is:

For x=-L/2, the element is at the left

end of the rod ,As x changes from

-L/2 to +L/2 ,the element covers the

whole rod. Thus the moment of

inertia of the entire rod about AB is

Moment of inertia of the uniform

circular plate about its axis

Consider a uniform circular plate of

mass M and radius R as shown below

in the figure:

Let O be the center of the plate and

OX is the axis perpendicular to the

plane of the paper. To find the

moment of inertia of the plate about

the axis OX draw two concentriccircles of radii x and x+dx having

these centers at O, so that they form

a ring. Area of this ring is equal to its

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circumference multiplied by its width

i.e. Area of the ring =2πxdx.

Mass of the ring would be

Moment of inertia of this ring about

axis OX would be:

Since whole disc can be supposed to

be made up of such like concentric

rings of radii ranging from O to R ,wecan find moment of inertia I of the

disc by integrating moment of inertia

of the ring for the limits x=0 and x=R

Moment of inertia of a uniform

sphere of radius R about the axis

through its center,

Consider a sphere of mass M and

radius R .Let us divide this sphere

into thin discs as shown in the

figure

If r is the distance of the disc then

Volume of the disc would be

and its mass would be

dm= ρdV. Moment of inertia of this

disc would be

Moment of inertia of the whole

sphere would be

Factor 2 appears because of

symmetry considerations as the right

hemisphere has same MI as that of

left one. Integration can be carried

out easily by expanding (R2-x2)2. On

integrating above equation we find

Now mass of the sphere is

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Hence,

7.6 Theorems of Moment of Inertia

There are two general theorems

which proved themselves to be of

great importance on moment of

inertia. These enable us to determine

moment of inertia of a body about an

axis if moment of inertia of body

about some other axis is known as

Perpendicular Axis Theorem.

This theorem is applicable only to the

plane laminar bodies. This theorem

states that, the moment of inertia of

a plane laminar about an axis

perpendicular to its plane is equal to

the sum of the moment of inertia of

the lamina about two axis mutually

perpendicular to each other in its

plane and intersecting each other at

the point where perpendicular axis

passes through it. Consider plane

laminar body of arbitrary shape lying

in the x-y plane as shown below in

the figure.

The moment of inertia about the z-

axis equals to the sum of the

moments of inertia about the x-axis

and y axis. To prove it consider the

moment of inertia about x-axis

where sum is taken over all the

element of the mass mi. The moment

of inertia about the y axis is

Moment of inertia about z axis is

where ri is perpendicular distance of

particle at point P from the OZ axis.

For each elementri2=xi2 + yi2

Parallel Axis Theorem:

This theorem relates the moment of

inertia about an axis through the

center of mass of a body about a

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second parallel axis. Let Icm be the

moment of inertia about an axis

through center of mass of the body

and I be that about a parallel axis ata distance r from C as shown below

in the figure

Then according to parallel axis

theorem I=Icm+Mr2 where M is the total mass of the

body. Consider a point P of the body

of mass mi at a distance xi from O

From point P drop a perpendicular

PQ on to the OC and join PC.So that OP2=CP2+ OC2+ 2OC.CQ(From geometry) andmiOP2=miCP2+ miOC2+ 2miOC.CQ

Since the body always balances

about an axis passing through center

of mass, so algebraic sum of the

moment of the weight of individual

particles about center of mass must

be zero. Here

which is the algebraic sum of such

moments about C and therefore eq as

g is constant Thus we have I = Icm + Mr2 ---(17)

7.7 Rotation with constant angular

acceleration

We have already studied motion with

constant acceleration while studying

translational motion. Now we will

study the rotational motion withconstant angular acceleration. When

a rigid body rotates with constant

acceleration we have

Again we have angular velocity

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Equation (8), (9) and (10) are the

equation of motion with constant

angular acceleration.

7.8 Torque

It is easier to open a door by pushing

on the edge farthest from the hinges

than by pushing in the middle. It isintuitive that the magnitude of the

force applied and the distance from

the point of application to the hinge

affect the tendency of the door to

rotate. This physical

quantity,torque, is t = r × F sin θ,

whereF is the force applied,r is the

distance from the point of applicationto the center of the rotation, and θ is

the angle fromr toF.Consider two forces F1 and F2 having

equal magnitude and opposite

direction acting on a stick placed on

a horizontal table as shown below in

the figure

Here note that line of action of forces

F1 and F2 is not same. So they tend torotate the stick in clockwise

direction. This tendency of the force

to rotate an object about some axis is

called torque. Torque is the rotational

counterpart of force. torque tends to

rotate an body in the same way as

force tends to change the state of

motion of the body. Figure belowshows a rigid body pivoted at point O

so that point O is fixed in space and

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the body is free to rotate

Let P be the point of application of

force. This force acting at point P

makes an angle θ with the radius

vectorr from point O to P. This forceF can be resolved into two

components

F⊥=Fsinθ

F||=Fcosθ

as they are perpendicular and

parallel tor Parallel component of force does not

produce rotational motion of bodyaround point O as it passes through

O. Effect of perpendicular

components producing rotation of

rigid body through point O depends

on magnitude of the perpendicular

force and on its distance r from O.

Mathematically, torque about point O

is defined as product ofperpendicular component of force

and r i.e.τ=F⊥r = Fsinθ r = F(rinθ) = Fd ---

(18)

where d is the perpendicular distance

from the pivot point ) to the line of

action of force F. Quantity d=rinθ is

called moment arm or liner arm offorce F. If d=0 the there would be no

rotation. Torque can either be

anticlockwise or clockwise depending

on the sense of rotation it tends to

produceUnit of torque is Nm

Consider the figure given below

where a rigid body pivoted at point O

is acted upon by the two forces

F1 and F2. d1 is the moment arm of

force F1 and d2 is the moment arm of

force F2

Force F2 has the tendency to rotaterigid body in clockwise direction and

F1 has the tendency to rotate it in

anti clockwise direction. Here we

adopt a convention that anticlockwise

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moments are positive and clockwise

moments are negative. Hence

moment τ1 of force F1 about the axis

through O isτ1=F1d1 And that of force F2 would beτ2=-F2d2.Hence net torque about O isτtotal= τ1+ τ2=F1d1-F2d2Rotation of the body can be

prevented if : τtotal=0 or or τ1=-τ2, We earlier studied that when a body

is in equilibrium under the action ofseveral coplanar force, the vector

sum of these forces must be zero i.e.

ΣF x=0 and ΣF y=0 We know state our second condition

for static equilibrium of rigid bodies

that is

"For static equilibrium of rigid body

net torque in clockwise directionmust be equal to net torque in

anticlockwise direction w.r.t some

specified axis”.i.e. Στ=0. Thus for static equilibrium of an rigid

body, the resultant external force

must be zero.

i.e, ΣF=0. The resultant external torque about

any point or axis of rotation must be

zero. i.e., Στ=0.

7.9 Work and Power

We know that when we apply force on

any object in direction of the

displacement of the object, work issaid to be done. Similarly force

applied to the rotational body does

work on it and this work done can be

expressed in terms of moment of

force (torque) and angular

displacement θ. Consider the figure

given below where a force F acts on

the wheel of radius R pivoted at pointO. So that it can rotate through point

O.

This force F rotates the wheel

through an angle dθ and dθ is small

enough so that we can regard force to

be constant during corresponding

time interval dt

Work done by this force isdW=Fds

But ds=RdθSo, dW=(FR)dθNow FR is the torque ‘τ’ due to force

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dW= τ dθ ----(19)if the torque is constant while angle

changes from θ1 to θ2 then

W= τ (θ2-θ1)= τ ∆θ ---(20) Thus work done by the constant

torque equals the product of the

torque and angular displacement. we

know that rate of doing work is the

power input of torque soP=dW/dt= τ (dθ/dt)

= τ ωIn vector notation

P=τ.ω7.10 Torque and angular

acceleration

While discussing and defining torque

or moment of force, we found that

necessary condition for a body not to

rotate is that resultant torque aboutany point should be zero. However

this condition is necessary but not

sufficient for a rigid body to be static

for example in absence of resultant

torque a body once set in rotation will

continue to rotate with constant

angular velocity. Analogous to

translation motion when torque actson a rigid body rotating about a point

with constant angular velocity then

angular velocity of the body does not

remain constant but changes with

angular acceleration α which is

proportional to the externally applied

torque. Consider a force Fi acting on

the ith particle of mass mi of the rigid body pivoted about an axis through

point O as shown below in the figure:

This force Fi as discussed earlier hastwo components one parallel to the

radius vectorri and one

perpendicular to theri. Component of

force parallel to radius vector does

not have any effect on the rotation of

the body. Component of force

Fi perpendicular does affect the

rotation of the body and produces

torque about point O through which

the body is pivoted which is given by

τi=Fi⊥ri ---(21).if Fi⊥ is the resultant force acting on

the ith particle ,then from Newton’s

second law of motion:Fi⊥=miai⊥ = miriα ----(22)

where ai⊥ is the tangentialacceleration of the body. From

equation (21) and (22)

τi=miri2α

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And taking sum over all the particles

in the body we have ∑τi=∑(miri2α)

=α∑(miri2

) ---(23)as angular acceleration is same for

all the particles of the body. We know

that

∑(miri2) = I where I is the moment of inertia of

the rigid body .Hence in terms of

moment of inertia equation 23

becomes ∑τ=Iα ---(24) We have denoted resultant torque

acting on the body ∑τsub>i as ∑τ. Both

the torque and angular acceleration

are vector quantities so in vector form

∑τ=Iα ---(25) Alternatively equation (24) which is

rotational analogue of Newton second

law of motion (∑F=ma) can be writtenas

∑τ= Iα = I(dω/dt)=d(Iω)/dt ---

(26) which is similar to the equation F=d(m v/dt=dp/dt

wherep is the linear momentum. The

quantity Iω is defined as the angular

momentum of the system of particles

Angular momentum =IωL=Iω.

From equation 26 we see that

resultant torque acting on a system

of particles equal to the rate of

change of the angular momentum,∑τ=dL/dt

7.11 Angular momentum andtorque as vector product

In any inertial frame of reference the

moment of linear momentum of a

particle is known as angular

momentum or, angular momentum of

a particle is defined as the moment of

its linear momentum. In rotational

motion angular momentum has thesame significance as linear

momentum has in the linear motion

of a particle. Value of angular

momentum of angular momentum is

equal to the product of linear

momentum andp(=m v) and the

position vectorr of the particle from

origin of axis of rotation.

Angular momentum vector is usually

represented byL. If the linear

momentum of any particle

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momentum of the particle is given by

L =r×p = m(r× v) ------- (1) Angular momentum is a vector

quantity and its direction isperpendicular to the direction

ofrandp and could be found out by

right hand screw rule. From equation

1 scalar value or magnitude of

angular momentum is given as

|L|=rpsinθ --------(2)

where θ is the angle betweenr andp.

For a particle moving in a circularpath,

v=ω×r; (3)

whereω is the angular velocity.

Therefore

L=m[r×(ω×r)]= mω(r.r)-r(r.ω)= mr2ω=Iω -------- (4)

But (r.ω)=0 because in circular

motionr andω are perpendicular toeach other. Here I is the moment of

inertia of the particle about the given

axis also the direction ofL andω is

same and this is a axial vector.

Writing equation (1) in the

component form we get

Writing angular momentum in

component form we get

writing equation 5 again we get

Comparing unit vectors on both thesides we get

Unit of angular momentum in CGS is

gm.cm2/sec and in MKS system it is

Kgm.m2/sec or Joule/sec.7.12 Angular momentum and

torque of the system of particles

Consider a system of particles made

up of number of particles, moving

independently to each other. Let L1,

L2, L3, .............etc. be the angular

momentum of different particles of

the system w.r.t. to a given point.

The angular momentum of theparticle system w.r.t. a given point is

equal to the vector sum of angular

momentum of all the particles of the

system.

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IfL is the angular momentum of the

system of particles or the body as a

whole then,

Torque acting on system of

particles

From equation above angular

momentum of the system of particles

given as

When particles of the system are in

motion then their motion is due toexternal and interaction due to

internal forces so force acting on any

particle of the system is given by

Here Fiext is the external force acting

on the ithparticle and ∑Fij is the sumof the force acting on the particle due

to internal interaction of different

particles. Putting the value ofFi in

the equation we get,

RHS of the equation 2 shows that

summation of the moment ofinteracting force (internal) .Here

internal interaction forces balance

each other so torque due to internal

forces adds to zero hence

The total torque acting on a system of

particles is equal to the vector sum of

the torque acting on the different

particles due to external force on the

particle and its value is also equal to

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the rate of change of angular

momentum

7.13 Law of conservation of

angular momentum

Torque acting on any particle is given

by

If external torque acting on any

particle is zero then,

Hence in absence of external torque

the angular momentum of the

particle remains constant or

conserved. Total torque acting on any

system is given by

If total external force acting on any

particle system is zero or,

If total external torque acting onanybody is zero, then total angular

momentum of the body remains

constant or conserved.

7.14 Kinetic Energy of rolling

bodies (rotation and translation

combined)

Let us now calculate the kineticenergy of a rolling body. For this

consider a body with circular

symmetry for example cylinder,

wheel, disc, sphere etc. When such a

body rolls on a plane surface, the

motion of such a body is a

combination of translational motion

and rotational motion as shown below in the figure.

At any instant the axis normal to the

diagram through the point of contact

P is the axis of rotation. If the speed

of the centre of mass relative to an

observer fixed on the surface is Vcmthen the instantaneous angular

speed about an axis through P would

be

ω=vcm/R

where R is the radius of the body. To

explain this consider the figure given

below

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At any instant different particles of

the body have different linear speeds.

The point P is at rest vcm=0

instantaneously, the centre of mass

has speed vcm=Rω and the highest

point on the circumference P has

speed vcm=2Rωrelative to point P. Now again

consider the first figure the top of the

cylinder has linear speed vcm+ Rω=vcm+vcm=2vcm which is greater than the linear

speed of any other point on the

cylinder. We thus note that the centerof mass moves with linear speed

vcm while the contact point between

the surface and rolling object r has a

linear speed of zero. Therefore at that

instant all particles of the rigid body

are moving with the same angular

speed ω about the axis through P and

the motion of the body is equivalentto pure rotational motion. Thus total

kinetic energy is

K=½(IPω2)

where IP is the moment of inertia of

the rigid body about point P.

From parallel axes theorem IP=Icm+MR2

where Icm is the moment of inertia of

the body of mass M about parallel

axis through point O. Therefore,

K=½(Icmω2)+½(MR2ω2) =½Icmω2+½MVcm2

here the first term represents the

rotational kinetic energy of the

cylinder about its center of mass, and

the second term represents the

kinetic energy the cylinder would

have if it were just translating

through space without rotating. Thus, we can say that the total

kinetic energy of a rolling object is

the sum of the rotational kineticenergy about the center of mass and

the translational kinetic energy of the

center of mass. If k is the radius of

gyration of the body about a parallel

axis through O then I=Mk2 and total

kinetic energy would then be,

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The moments of inertia for different

regular shapes are shown in Figure

below.

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8.1 Introduction

The discovery of universal law of

gravitation by Isaac Newton is one of

the greatest events in the history of

Physics. Until the 17th century the

tendency of a body to fall to the earth

ie, its weight was considered as an

inherent property of all bodies which

needed no further explanation. ButNewton and his contemporary Robert

Hooke regarded the weight of a body

as the result of the attraction

between the body and the earth.

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Till Newton’s time the laws governing

the motion of planets were

considered different from those

governing motion of bodies falling tothe earth. The motion of planets and

the sun was a subject of great

interest at that time. This subject

was discussed by the students of

material philosophy at Cambridge in

1664. Newton was one among them.

In 1665, the institution was closed

temporarily on account of plague andthe students were sent home. Newton

continued his thoughts on

gravitational force.One day, when he was sitting under

an apple tree in a contemplative

mood, the fall of an apple caught his

attention. He compared the

acceleration of a body like the applefalling to the earth with the

acceleration of the moon falling

towards the earth while revolving

round the earth in its orbit. This lead

him to the universal law of

gravitation. The steps which led him

to this important law can be

summarized as follows.

The acceleration of an apple falling

near the earth’s surface a0 = g = 9.8

m/s2. The radial acceleration of the

moon when it revolves round theearth is am = v2/rm where v is the

velocity of the moon and rm is the

radius of the moon’s orbit.

v =2 π rm

period of t h e moon

rm = 3.84 x 108mPeriod =27.3 days = 27.3 x24 x60 x60

s v =

2 πX 3.84 X 108

2.73 X 24 X 60 X 60

= 1.02 X 103 m/s= 1.02 km/s

am = v2/rm =(1.02 X 10

3 )2

3.84 X 108

= 2.72 X 10-3m/s2

am/aa =2.72 X 10

−3

9.8 = 2.77 x 10-4

At the time of Newton, it was known

that the moon’s orbit had a radius of

about 60 times that of the radius of

the earth. ie, rm = 60rere = radius of the earth ie, distance of

the apple from the centre of the

earth.(

re

rm

¿¿2 =

( 1

60

¿¿2 = 2.77 x 10-4

Thus the ratio of the accelerations is

in the inverse ratio of the square of

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the radii of the orbits. In other words,

the ratio of the forces producing the

acceleration of the moon and the

apple is in the inverse ratio of thesquare of the distances from the

centre of the earth. This led Newton

to conclude that the force which

makes the apple fall to the earth and

the force which makes the moon

revolve round the earth are one and

the same. The force varies inversely

as the square of the distance. Newtonmade these calculations and arrived

at the conclusion in 1666. But he did

not publish it till 1687 for two

reasons.

(1) Non – availability of accurate values

of the distances involved.(2) In the above calculation, the distance

of the apple was taken from thecentre of the earth.

This assumes that the entire mass of

the earth is concentrated at the

centre of the earth. Such an

assumption could not be justified till

Newton developed calculus and

proved this result. In 1687, he

published the inverse square law ofgravitation in his book “Principia

Mathematica”

8.2 Newton’s universal law of

gravitation

From further analysis, Newton found

that the force of attraction by a body A exerted on another body B depends

on the mass of the body a. By

Newton’s III law, the body B exerts an

equal force on A. It must be

proportional to the mass of the body

B as well. Therefore the force between

two bodies must be proportional to

the product of their masses. The universal law of gravitation was

stated as follows:“Every material particle attracts every

other material particle with force

which is directly proportion to the

product of the masses and inversely

proportional to the square of the

distance between them”

Let m1 and m2 be the masses of two

particles separated by a distance ‘d’.

Let F be the attraction between them.

Then F ∞ m1m2 --------- (1)

F ∞1d

2 ------------ (2)

Combining (1) and (2) F ∞m1m2

d2

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where G is a constant called

universal constant of gravitation,

which is the same for all particlesirrespective of their masses, nature,

position and time. The law is known as universal law of

gravitation because it is true for all

particles, at all times, and at all

places. If two large sized bodies are

involved ‘d’ is the distance between

their centres of mass.Did you know?

8.3 Expression for acceleration due

to gravity

Consider a body of mass ‘m’ on the

surface of the earth of mass M and

radius R. The force of gravity acting

on the body

F =GMm

R2

Acceleration due to gravity

g = Fore of gravit!

Mass

g = GMm / R2

m = GM R

2

Variation of g with height ‘h’ above the

ground:

If we go to the top of a mountain, the

distance from the centre of the earth

will increase. Therefore the value of g

will decrease. The variation of ‘g’ with

height can be found out as follows.

Let ‘g’ be the value of acceleration

due to gravity on the surface of earth. Then

g =GM

R2 --------------(1)

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Let g1 be the acceleration due to

gravity at a height ‘h’ from the

surface of the earth. Then

g1=GM

( R+h)2 --------------

(2)

g

g1 =( R+h)2

R2

= (1 +h

R )2 = 1+2 (h

R ¿

------- (3)Uing (1+x)n= 1 + nx where x <<<< 1

And neglecting higher termh2

R2

g

g1 = (1 -2h R )

Therefore g1= g (1 -2h R )

Example 1

At what height from the surface of

the earth will the value of be 1% less

than that on the surface of the

earth?

h = R2 (

g−g1

g1 ¿

Here R = 6400 km g = 100 (say) g1 = 100 – 1 = 99

Therefore h =6400

2 (1

99¿=32"m

nearly

Example 2

At what height from the surface of

the earth will the value of ‘g’ be the

half the value of g on the surface ofthe earth?

Solution: We know that g =GM

R2

---- (1)

Let the value of g be,g2 at a height

h above the ground.

Therefore g2 = GM ( R+h)2 --------------

(2)

Dividing 2 =( R+h)2

R2

√2 = R+h

R = R+h = √2R

h = √2R – R = (√2 -1)R = 0.414 R

Example 3

If g is the acceleration due to gravity

on the surface of the earth, what will

be its value at a height equal to the

radius of the earth?Solution:

g =GM

R2 and g1=

GM

( R+h)2

But h = R

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g1 =GM

(2 R )2

g1

g =

1

4

.’. g1 =g4

Variation of g with depth below the

surface:

We know that

g =GM

R2

But M =4

3 πR3D

where D is the mean densitySo g =

GM

R2 x

4

3 πR2D

g =4

3 πGDR --------- (1)

where R is the distance from the

centre of the earth. Let g1 be the

acceleration due to gravity at the

bottom of a mine of depth ‘h’. Theng1 =

4

3 πGD(R – h)

-------- (2)

g1

g = R−h

R = 1 -h

R

---------- (3)

h = R (g−g1g ¿

From equation (2) we find that

acceleration due to gravity decrease

as the depth of the mine increases.

At the centre of the earth h of R. g1 =

0.

Thus acceleration due to gravity atthe centre of the earth is 0.

From (3)

g1= g (1 -hg ¿

Example 4

At what depth below the surface of

the earth will the value of g be 1%less than the value of g on the

surface?Solution:

We know that h = R(g−g1

g ¿

R = 6400 kmg = 100 (say)

g1

= 99h = 6400 x

1

100 = 64 km

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g = R #02 GM

R2 =

R #02

Multiplying both sides by mGMm

R2 = mR #0

2

Thus a body appears to be

weightless, when the gravitational

force is completely utilized to supply

centripetal force. This is what

happens in a satellite. In a satellite

the gravitational force on the body

inside the satellite will be completely

utilized for making it revolve round

the earth.No portion of the gravitational force

will be available to keep the bodies in

satellite pressed on the floor of the

satellite. Therefore the body will have

a tendency to float. This is what is

known as weightlessness in space

flight.

8.4 Gravitational Potential Energy

Around the earth there is a

gravitational force field. A mass

placed in this field possesses

gravitational potential energy due toits position in the force field. The

work done in lifting a body in this

field is stored as gravitational P.E of

the body. Near the surface of the

earth, the gravitational force acting

on the body of mass m is a constant

= mg. Therefore the work done in

lifting a body through a small heighth = mgh. It resides on the body as

the increase in P.E. The increase in

P.E. = mgh.

On the other hand if we consider

distances comparable with the radius

of the earth, the variation of

gravitational force is to be

considered. Suppose we want to find

the work done in lifting a body of

mass ‘m’ from a radial distance r1, to

radial distance r2. The procedure is as follows;Consider a body of mass ‘m’ at P at a

distance ‘r’ from the centre of the

earth. The gravitational force acting

on the body is F =GMm

r2

Consider a neighbouring point P1 at a

distance dr. Over this small distance,

the gravitational force can be

considered to be a constant.

Therefore the small amount of work

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dw = Fdr =GMm

r2

r

Therefore total work done in lifting

the bodyFrom r1 to r2

W = ∫r1

r2

GMm

r2

dr

= GMm∫r1

r2

r−2 dr = GMm[ r−1

−1 ]r1

r2

= -GMm[ 1r ]r1

r2

= -

GMm[( 1r 2 )−( 1r1

)] W =

GMm[ 1r1

−1

r 2 ]

This work done is converted into P.E. Therefore the increase in P.E. of the

body when lifted from r1 to r2 is equalto:

V(x)=GMm[ 1r1

−1

r 2 ]

Example 5

Starting from the expression for the

increase in PE when a body is taken

from r1 to r2 obtain expression forincrease in PE near the surface of the

earth.Solution:

Expression for increase in PE is

GMm[ 1r1

−1

r 2 ]

Suppose a body of mass ‘m’ is liftedfrom the surface of the earth through

a height ‘h’ Then r1 = R and r2 = R + h

So increase in PE =GMm[ 1r1

−1

r 2 ]

= GMm[ 1 R− 1

R+h ]

=GMm[ R+h− R

R ( R+h ) ]= GMm[ h

R( R+h) ]h is small compared to R. Therefore

(R+h) is nearly equal to R.

Increase in P.E. =GMm

h

R2

=GM

R2

XmX h

= g X m X h = mgh

Example 6

Assuming the expression for the

increase in PE, obtain expression forthe PE of a body at a radial distance r

from the centre of the earth.Solution:

Increase in PE when a mass m is

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r1 to r2 is equal toGMm[ 1r1

−1

r 2 ]

PE at r2 – PE at r1 =GMm

[ 1

r1

−1

r 2

]Let us find the increase in P.E. when

the body is lifted from r to ∞ Then r1 =r; r2 = ∞P.E. at infinity – PE at r =

GMm[ 1r − 1

$ ] =

GMm[ 1r −0]=

GMmr

The zero level of P.E is taken at

infinity because at infinity the

gravitational force acting on the body

will be zero. Therefore 0 – P.E at r =

GMmr

P.E at r =−GMm

r

Thus the gravitational P.E. of a mass

‘m’ at a distance r from a mass M is

equal to−GMm

r

Because P.E. at infinity is zero, the

P.E. at any finite distance is negative.

The negative sign shows that work is

to be done in increasing the distance.

The negative sign in the P.E also

shows that the force involved is one

of attraction.

If two masses are separated by a

distance and if the masses are free to

move, they will approach each other

under the mutual force of attraction.

In doing so the bodies are doing

work. Therefore their P.E. will

decrease.

If the PE is found to be positive then

the force involved will be one of

repulsion.

Example 7

Represent graphically the variation of

gravitational PE with distance apart.Solution :

P.E. = = −GMmr

P.E. ∞1

r

P.E x r = constant Therefore the graph between the P.E.

and the radial distance is a

rectangular hyperbola.

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Did You Know ?

Expression for the intensity of

gravitational field on the surface of the

earth:

Consider a body of mass ‘m’ placed

on the surface of the earth of mass M

and radius R. Gravitational force on

the mass m is

F =GMm

r2

Intensity of gravitational field E =

F m

= ( GMm

R2

m ) =GM

R2

---------------- (1) Acceleration due to gravity on the

surface of earth

g =GM

R2

------------------ (2) Therefore from (1) and (2) E = gIntensity of gravitational field is equal

to the acceleration due to gravity.

E = g =GM

R2

8.5 Escape Velocity

If we throw a stone, it first rises upand then comes down. It is because

of the gravitational force of the earth

that it comes down. If we increase the

speed of projection, it rises to greater

height before falling.If we project it with sufficient speed it

will escape from the earth’s

gravitational force of attraction andreach infinity.“This minimum velocity with which a

body is to be projected, so that it may

escape from the earth’s gravitational

force of attraction altogether and218

Gravitational Field:

“The !"#e "ro$%& " '" (here

)r"*i+"+io%" "++r"#+io% i e-!erie%#e& i

#"e& )r"*i+"+io%" e&.Strength of Gravitational Field:

“The +re%)+h or i%+e%i+/ o

)r"*i+"+io%" e& "+ " !oi%+ i &e%e&

" +he or#e "#+i%) i% $%i+ '" "+ +he



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never return to the earth is called

Escape velocity. It is denoted by ‘ve’.

Expression for Escape VelocityIf we neglect air resistance an

expression for escape velocity can be

obtained as follows.

Consider a body of mass m on the

surface of the earth of mass M and

radius R.

PE on the ground =−GMm

R

Let Vc be the escape velocity. Then KE

on the ground =1

2 m v

2

Total energy on the ground=

−GMm R +

1

2 m v

2

---------- (1)If we project the body with escape

velocity vc it will just reach infinity

with zero kinetic energy.KE at infinity = 0PE at infinity = 0

Total energy at infinity = 0 -----------(2)By law of conservation of energy

−GMm

R +1

2 m v

2

= 0

1

2 m v

2

=GMm

R v

2

= 2

GM

R

vc = √2GM

R

--------------- (3)

vc = √2GM

R2 R = √2gR

--------------- (3) Where g is acceleration due to gravity

on the surface of earthg = 9.8 m/s2

R = 6.38 x 106m Vc = √2 x 9.8 x 6.38 x 106

= 11.2 x 103 m/s= 11.2 km/s

Escape velocity from the Moon can be

calculated from the relation

vc = √2GM

R

Where M is the moon’s mass and R is

Moon’s radiusM = 7.4 x 1022 kgR = 1740 km = 1740 x 1000 mG = 6.65 x 10-11

vc = √

2 X 6.67 X 10−11

X 7.2 X 1022

1740 X 1000

= 2.38 km/s vc for Moon = 2.38 km/s

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The escape velocity of a body on

earth from the solar system can be

calculated from the relation

vc = √2GM

R

Where M is the mass of the sun Ms =

2 x 1033 kg, R is radius of the earth’s

orbit R = 1.5 x 1011 m

vc = √2 X 6.67 X 10

−11 X 2 X 10

30

1.5 x1011 = 42

km/s

This is the escape velocity of a body

on earth with respect to the sun. But

the earth has an orbital velocity 30

km/s. Everybody on earth possesses

this speed. Therefore the escape

velocity of a body on earth from the

solar system with respect to the earth

itself is equal to 42 – 30 =12 km/s.Escape velocity from the sun = 618

km/s

vc = √2GM

R

The escape velocity is independent of

the mass of the body.

The escape velocity is

independent of the angle ofprojection

vc = √2GM

R

M =4

3 πR3D

vc = √2G R x

4

3 πR3D

= R √2Gx4

3 πD = Constant x

REscape velocity from a planet is

directly proportional to the radius of

the planet if the mean density is a

constant.

Do you know?

Example 8

220

Gravitational Escape Velocityfor a Black Hole

Black Hole s a *$ mass* s&n #$ sta$

t/at /as c#as*d #n ts*+, s&c/ t/at ts

"$atat#na +*d s s# st$#n" t/an n#t **n

"/t can *sca* ts &. t s ca*d a bac

/#* b*ca&s* t/at s /# t a*a$s t#

t**sc#*s.

T/* *&at#n +#$ t/* "$atat#na *sca*

*#ct #+ a ac #* &s*s t/* s**d #+

"/t as t/* /"/*st #ssb* *#ct +#$

mat*$a t$n" t# *sca* t/* sta$. T/*

d*+nn" s*a$at#n +$#m t/* c*nt*$ #+ a

ac #* s ca*d t/* event

horizon #$ Schwarzschild radius and can b*



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Explain why the moon does not have

an atmosphere while the earth

continues to have it?

Solution:Escape velocity from the moon is

nearly 2.38 km/s. The R.M.S velocity

of gas molecules is greater than this.

Therefore the gas molecules escaped

from the moon one by one and the

moon does not have an atmosphere

at present.

The escape velocity from the earth is very large equal to 11.2km/s. The

R.M.S velocity of gas molecules like

oxygen and nitrogen is much less

than this. Therefore, they cannot

escape from the earth. The earth still

continues to have an atmosphere.

Example 9

Explain why earth’s atmosphere has

no hydrogen in it? Why the sun has

plenty of it?Solution:

The RMS velocity VRMS= √3 RT M

For hydrogen and helium M is small.

Therefore their RMS velocity is larger

than 11.2 km/s. Therefore theearth’s atmosphere does not have

hydrogen. Though helium is escaping

it is being replenished by the

radioactive decay process.

Though there is no hydrogen in the

earth’s atmosphere, the sun has

plenty of it, because, the escape

velocity from the sun is very largeequal to 618 km/s.

8.6 Satellites

A satellite is a body which revolves

round a planet. A planet is a body

which revolves round the sun. The

moon is revolving round the earth.

Hence it is a natural satellite of theearth. Any man made body which

revolves round the earth just like

moon is called an artificial satellite.Newton was the first to think of an

artificial satellite. His arguments may

be summarized as follows:

Imagine a tower extending above the

earth’s atmosphere. Imagine a gun

being fixed horizontally. If a bullet is

fired, the bullet follows a curved path

and strikes the point A1 on thesurface of the earth. If the speed is

increased progressively, it strikes at.

A2, A3, A4 etc. But for a certain speed

of projection, the bullet will not be

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velocity and the curvature of the

earth, it will continue to fall to the

earth, but not able to reach it. It will

reach the point of projection andcontinue its motion round the earth.

Thus for a satellite to be launched, it

must be raised to a sufficient height

and then given a sufficient horizontal

velocity. It will then revolve round the

earth in a fixed orbit.

8.7 Orbital Velocity

Consider a satellite in its orbit of

radius r. Let m be the mass of the

satellite. Let V0 be the orbit velocity. Then centripetal force required for

circular motion =m% 0

2

r

This is supplied by the gravitational

forceGMm

r2

m% 02

r =GMm

r2 % 02

=GM

r

v0 = √GM

r --------- (1)

v0 ∞1

√ r v0 ∞ rn

where n =

−1

2

r = R+h, where h is the

height of the satellite from the

surface of the earth.

ie v0 = √GM

R+h -------- (2)

R is nearly equal to 6400 km. For a

close orbit h = 200 km. Therefore h

can be neglected compared to R.

v0 = √GM R (3)

v0 = √ R

GM

R2

v0 = √Rg (4)Equations 3 and 4 are for close orbits

v0 = √Rg =

√6400x1000x10= 8000 m/s = 8 km/sExpression for Period of a

Satellite:

Consider a satellite in an orbit of

radius ‘r’.

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Distance covered for one revolution =

2πr

Period =

2 πr

% 0 =

2 πr

√

GM

r =

2 πrX r1/2

√ GM

T=2 πX r

3/2

√ GM

T = r3 /2

T2∞ r3

This is Kepler’s third law. Another expression for period:

T=2 π r

3/ 2

√ GM

r= R + hGM = r2g

Substituting the value,

T =

R+h¿

¿¿1/22π ¿¿

=

R+h¿¿¿2

2π √ ¿¿

T = 2π

R+h

¿¿¿3¿

√ ¿

For a satellite in a close orbit h is

nearly = 200 km

T = 2π

6400+200¿ X 10(¿¿ 3 ]2¿¿

√ ¿

=1.46 hour

Thus a satellite at about 200 km

above the surface of the earth will

make one revolution in every 1.46

hours. Such a satellite could be seen

from a place on earth only

intermittently. It can be seen for

some time every 1.46 hours. If h

increases ‘T’ also will increase. For a

certain height, period T will become

equal to 24 hours. The earth’s period

of rotation about its own axis is also

24 hours. If the direction of the

rotation of satellite is same as that of

the earth, the satellite will appear to

be stationary above the surface of the

earth. Such a satellite is called

Geostationary satellite or parking

satellite or synchronous satellite.

8.7.1 Geostationary or

Synchronous Satellite:

A geostationary satellite is a satellite

which appears to be stationary above

a place on the surface of the earth.its period is same as the period of

rotation of the earth, about its own

axis i.e, 24

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hours. Its direction of rotation is

same as that of the earth ie from

west to east. All satellites are

projected from west east. This is totake advantage of the inertia of

motion caused by the rotation of the

earth. Its plane of rotation is the

equatorial plane.(The plane of rotation of every

satellite passes through the centre of

the earth.)

Its height from the ground is about36000 km.Synchronous satellites are used for

relaying radio, television, telephone

signals from one place to another.

Therefore stationary satellites are

also known as communication

satellite Three such satellites arranged at the

corners of an equilateral triangle can

be used to take TV and radio signals

to all in habituated areas of earth. They are used for weather forecastingSatellites are used for studying

atmosphere and cosmic rays. They are used in warfare and also for

spy work.

Some communication Satellites:1)Insat IA2)Insat IC3)Insat I-D4)Telstar (the first communication

satellite)

Example 10

Calculate the height of synchronous

satellite.

Solution:

We know that T2 =4 π

2( R+h)2

R2

g

( R+h)3 = ( T 2 R2 g

4 π 2 )

R+h¿ ) = ( T

2 R

2g

4 π 2 )

1/3

=

( (24 X 60 X 60 )2 X (6400 X 1000 )2 X 10

4 π 2 )

1 /3

= 42650 km R + h = 42650 km h = 42650 – R

= 42650 – 6400= 36250 km

Example 11 What is the relation between orbital

velocity and escape velocity?Solution: The escape velocity is

ve = √2GM

R

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For a close orbit

ve = √GM

R

Dividing both1

√2 = 0.707 .’. v0 = 0.707 ve

8.8 Kepler’s Laws of Planetary

Motion

I Law – Law of Elliptical Orbits:

Every planet revolves round the sun

the elliptical orbit, with the sun, at

one of the foci.II Law – Law of equal areas:

The areal velocity of a planet is a

constant ie., the radius vector joining

the sun and the planet sweeps out

equal areas in equal intervals of time.III Law – Harmonic Law:

The square of the period of revolution

of a planet is directly proportional tothe cube of the mean distance

between the sun and the planet.

Explanation of First Law:

The shape of the ellipse is as shown

in the figure. AB and CD are the

major and the minor axis

respectively. F1 and F2 are the foci O

is the centre. The sun is at F1. When

the planet is at B it is farthest from

the sun. B is known as APOGEE.

When the planet is at A it is nearestto the sun. A is known as PERIGEE.F1O/AO = e. It is known as

eccentricity of the ellipse.

Second Law:

Area SAP is the area traced by the

radius vector in one month when the

planet is near perigee and area SBQ

is the area traced by the radius

vector in one month when the planetis near apogee. These two areas are equal. It means

that the linear velocity of a planet is

not a constant. It varies with position

on the orbit. It is maximum at

perigee and minimum at apogee. If vpis the velocity at perigee and va is the

velocity at apogee then.

%p%e =

1+e1−e

where e is the eccentricity of the

ellipse.

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Third Law:

Let ‘T’ be the period of revolution and

‘r’ the mean distance between the

sun and the planet. Then T2

∞ R3

.Let T1 and T2 be the period of revolution

of two planets and r1 and r2 b e their

mean distance from the sun. Then T12∞ r13

T22∞ r33

SoT 1

2

T 22 =

r13

r2

3

Deducing Newton’s law of gravitation

from Kepler’s laws.

Consider a planet of mass m

revolving round the sun of mass M in

a circular orbit of radius ‘r’ with

orbital velocity v0.

Centripetal force F =m% o

2

r .

But v0=2 πr

T .

F = ( mr ) x ( 4 π

2r2

T 2 ) =

4 π 2

m

r2 x

r3

T 2

By Kepler’s III law

r3

T 2 = constant, K

F =4 π

2m&

r2 F = 4 π

2 & x (

m

r2¿

The source of this force lies in the

sun and therefore the factor 4 π 2 &

depends only on the sun. Thesimplest assumption is that the

factor 4 π 2 & is directly proportional

to the mass of the sun.4 π

2 & ∞ M

4 π 2 & = GM where G is a constant

ie, F =

GMm

r2

Example 12

What is the ratio¿T'∨ ¿

&'¿ for a

satellite?Solution:

¿T'∨¿ =1

2

GMmr

&' =1

2

GMmr

¿T'∨ ¿ &'

¿ = 1

Example 13

What is the ratio of¿ ('∨ ¿

&'

¿ in

any orbit?Solution:

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¿ ('∨ ¿ &'

¿ =( GMm

r )1

2 (GMmr ) = 2

Example 14

A body of mass m is taken to a height

h = nR form the surface of the earth.

What is the increase in PE?Solution:

Initial PE = -GMm

R

Final PE = -GMmnR+ R = -

GMm

(n+1) R

Increase in PE = Final PE – Initial PE

=-(GMm)(n+1) R -

(−GMm R )

= GMm R -

(GMm)(n+1) R

=GMm

R [1 -

1

n+1 ]

=GMm

R [

n+1−1

n+1 ]

=1

n+1 ( GMm R )

=n

n+1 (mgR)

Example 15

If the increase in PE is1

2 mgR what

is the height to which a body is

taken?Solution:

12 mgR = nn+1 mgR

nn+1 =

1

2

n = 1 Therefore h = nR = 1Rh = R

Example 15

What is the ratio of the acceleration

due to gravity?Solution:

g p

ge =

m

n2

Where m = M p M e = 2

n = R p

Re = 2

thereforeg p

ge =

2

22 =

1

2

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Example 16

Planets Neptune and Saturn are at

1013

m and 1012

m respectively. Whatis the ratio of their periods?Solution:

T2 ∞ r3

T n2 ∞ rn

3

T s2

$ rs3

T n2

T s2 =

rn3

rse3 = ( 10

13

1012 )

3

= 103

T 0T e = 10√10

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9.1 Introduction

In Chapter 7, we studied the rotationof the bodies and then realised that

the motion of a body depends on howmass is distributed within the body. We restricted ourselves to simplersituations of rigid bodies. A rigid body generally means a hard solidobject having a definite shape andsize. But in reality, bodies can bestretched, compressed and bent.Even the appreciably rigid steel bar

can be deformed when a sufficientlylarge external force is applied on it. This means that solid bodies are notperfectly rigid. A solid has definite shape and size. Inorder to change (or deform) the shapeor size of a body, a force is required.If you stretch a helical spring bygently pulling its ends, the length of

the spring increases slightly. When you leave the ends of the spring, itregains its original size and shape. The property of a body, by virtue of which it tends to regain its originalsize and shape when the applied

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force is removed, is known aselasticity and the deformationcaused is known as elastic

deformation. However, if you applyforce to a lump of putty or mud, theyhave no gross tendency to regaintheir previous shape, and they getpermanently deformed. Suchsubstances are called plastic andthis property is called plasticity.

Putty and mud are close to idealplastics.

The elastic behaviour of materialsplays an important role inengineering design. For example, while designing a building, knowledgeof elastic properties of materials likesteel, concrete etc. is essential. Thesame is true in the design of bridges,automobiles, ropeways etc.One could also ask — Can we design an aeroplane which is very light but sufficiently strong?Can we design an artificial limb which is lighter but stronger? Whydoes a railway track have a particularshape likeI? Why is glass brittle while brass is not? Answers to suchquestions begin with the study ofhow relatively simple kinds of loadsor forces act to deform differentsolids bodies. In this chapter, weshall study the elastic behaviour andmechanical properties of solids which would answer many such questions.

9.2 Some Important definitions

Deforming force

The applied force on a body whichchanges the shape or size of the bodyis called deforming force

Restoring force

When a deforming force is removedfrom the body, the internal force which brings back the body tooriginal shape is called restoringforce.

Elasticity

Property of body by virtue of which ittends to regain its original shape orsize on removal of deforming force iscalled elasticity.Elastic body

Those bodies which regain its original

shape/size on removal of deformingforce are called elastic body.eg. Steel, rubber (steel is more elasticthan rubber)Plastic body

Those bodies which do not regain itsoriginal shape/size on removal ofdeforming force is called plastic bodies. eg. Clay, putty, plastic etc.

Perfectly elastic body

Those bodies which regain its originalshape/size immediately and

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completely after removal of deformingforce from it.eg. quartz, phosphor-bronze string.

Perfectly plastic body When a small deforming force isapplied on a body, on its removal, itdoes not regain its original shape orsize.eg. putty, paraffin wax etc.Elastic limit

The limit up to which the body showsthe property of elasticity is called

elastic limit.

9.3 Stress

Stress- restoring force acting on a

body per unit area is called stress.

Stress =forearea

The SI unit of stress is N/m2 or

Pascal whereas its CGS unit isdyne/cm2

9.3.1Types of stress

There are three types of stress

depending upon which parameter of

the body changes.1. Longitudinal stress

When a deforming force is appliednormal to the body, then therestoring force per unit area is callednormal stress.It can be divided into two types -

a)Tensile stress

When a deforming force is applied on

a body such that there is increase in

length of body, then restoring force

per unit area is called tensile stress. b)Compressive stress

When a deforming force is appliedsuch that there is decrease in length

or compression of body, then

restoring force per unit area is called

compressive stress.

Tensile and compressive stresses aretogether calledlongitudinal stress.

Fig 9.1 Showing Longitudinal Stress

2. Tangential or Shearing stress

When a deforming force is appliedtangential to the surface of the bodyso that it produces a change in the

shape of the body without anychange in volume then the restoringforce per unit area is calledtangential stress. It is also calledshearing stress where

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Shearing stress = Tangential Force/A

Fig 9.2 Showing Shearing Stress

3. Hydraulic or Normal stress

When a solid body undergoes changein volume without change ingeometric shape on applying forceperpendicular on every point onsurface of body then restoring forceper unit area is hydraulic stress.eg. Bubble in a water filled container.

Fig 9.3 Showing Shearing Stress

9.4 Strain

When the deforming force is applied

on a body then the ratio of change in

dimension to original dimension is

strain.Strain=

hange∈dimensionoriginaldimension

It has no units.

There are 3 types of strain

1)Longitudinal strain

When a deforming force is applied on

a body such that length changesthen ratio of change in length tooriginal length is longitudinal strain.

Longitudinal strain =) L L

2)Bulk/Volumetric strain- When a deforming force is applied ona body such that volume changes

then ratio of change in volume tooriginal volume is bulk strain.Bulk strain=V/V∆

3)Shearing strain

When a deforming force produces achange in shape of body withoutchange in volume then ratio ofchange in dimension to originaldimension is sheering strain. It ismeasured by angle of sheer.

tanθ = x/L∆

which for small angles becomesθ = x/L∆

Therefore shearing strain is also

defined as ratio of displacement of

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surface under a tangential force to

the perpendicular distance of

displaced surface to the fixed surface.

9.5 Hooke’s Law

Within the elastic limit, the stressdeveloped is directly proportional tostrain produced in a body.

Stress ∝ strain,

Stress = (E) strain, where E is modulus of elasticity

E =stress

strain

9.5.1Types of modulus of

elasticity

1. Young’s modulus of elasticity:

within the elastic limit, the ratio of

normal stress to longitudinal strain

Y=normal stress/longitudinal strain =

F / *) L/ L =

FL * ) L

SI unit is N/m2 or pascalIf material is rod/wire, A=area of

cross-section= π r2

2. Bulk modulus:

Within the elastic limit, the ratio ofnormal stress to volumetric strain.

B=nor mal stress

volumetri strain

B= F / *

−) % /% = F%

− * ) %

-ve sign shows that with increase in

pressure, a decrease in volumeoccurs.Here F/A=P(pressure)

B=-PV/V∆

Compressibility is defined as the

reciprocal of bulk modulus

K=1/B = - V/PV∆

3. Shear modulus or Modulus ofrigidity

Within the elastic limit, ratio of

tangential/ sheering stress to

sheering strain.

G=

t angential∨s h ear stresssheeringstrain

G= F / *) x / L or

F / *tanθ

= FL

* ) x or F

*tanθ

If σs is sheering stress thenG = σs/θ

9.6 The Stress-Strain Curve

A wire is suspended vertically from arigid support by the side of a scaleand load is applied to its lower endand is increased gradually, till the wire breaks. Stress and strain is

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noted for each load and a graph isdrawn as shown

OA-Hooke’s law, OB = elastic regionCE = plastic region, D = max. stressE = fracture point

• Part OA is a straight line showing

that stress α strain. At point A, if

load is removed the body regains

its original dimension i.e. it obeys

Hooke’s Law and point A is called

proportional limit.• Part AB-In this region, stress and

strain are not directly proportional

but still the body returns to its

original shape when load is

removed. Then the point B is called

elastic limit/yield point and

corresponding stress is yield

stress. (σ y)• Part BC- In this region, even for

small stress, the strain increases

rapidly. At the point C if load is

removed the wire will not return to

its original dimension. It takes a

new path CC’ and OC’ is called

permanent set. The deformation is

called plastic deformation.• Part CD- If stress is further

increased slowly, strain increases

more rapidly and finally at point D

it experiences max. stress called

ultimate tensile strength.• Point E - Beyond point D, body

breaks by the application of stress

and the point is called fracturepoint. If the point E is close to D,

the material is said to be brittle

and if it is away, it is called ductile.

9.7 Application of Elastic

Behaviour of Materials

1. Knowing Poisson’s ratio

Thus

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The negative sign is put in because

the change in diameter is in oppositesense to the change in length of the wire.

2. Calculating Elastic Energy

Density

The force ‘F’ needed to stretch a wire

of length L by an amount‘l’ is given

by :F = YAl / L where A is the area of cross section of

wire and Y is the Young’s modulus.Small work done in stretching the

wire by ‘dl’ is :

dW=Fdl

dW= ∫0

l

Fdl

.’. W = ∫0

l +*ldl L =

+* L ∫

0

lldl

Or W = YAl2

2 L2

Which is stored in the wire as its

Elastic Potential Energy

.’. P.E. U =+*l L

l2 =

1

2 x F x

l

Dividing by volume of wire Al, we get

the energy density as :u = U / Al

3.Calculating thickness of the

rope used in Cranes

Cranes used for lifting and moving

heavy loads from one place to

another have a thick metal rope to

which the load is attached. The ropeis pulled up using pulleys and

motors.Suppose we want to make a crane,

which has a lifting capacity of 10

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1000 kg). How thick should the steel

rope be? We obviously want that the

load does not deform the rope

permanently. Therefore, theextension should not exceed the

elastic limit. We find that mild steel

has a yield strength (Sy) of about 300

× 106 Nm –2. Thus, the area of cross-section ( A) of

the rope should at least be A≥W /Sy=Mg/Sy= (104kg × 10 ms-2)/(300 × 106 Nm-2)= 3.3 × 10-4 m2

corresponding to a radius of about 1

cm for a rope of circular cross-

section. Generally a large margin of

safety (of about a factor of ten in the

load) is provided. Thus a thicker rope

of radius about 3 cm is

recommended. A single wire of this

radius would practically be a rigid

rod. So the ropes are always made of

a number of thin wires braided

together, like in pigtails, for ease in

manufacture, flexibility and strength. A bridge has to be designed such that

it can withstand the load of the

flowing traffic, the force of winds and

its own weight. Similarly, in thedesign of buildings use of beams and

columns is very common. In both the

cases, the overcoming of the problem

of bending of beam under a load is of

prime importance. The beam should

not bend too much or break.Let us consider the case of a beam

loaded at the centre and supportednear its ends as shown in figure

below.

A bar of length l, breadth b, anddepthd when loaded at the centre by a load

Wsags by an amount given byδ =W l3/(4bd3Y)

This relation can be derived using

what you have already learnt and a

little calculus. From above equation

we see that to reduce the bending for

a given load, one should use a

material with a large Young’s

modulusY. Thus, for a given material, increasing

the depthdrather than the breadth

bis more effective in reducing the

bending, since δ is proportional tod-3

and only tob-1(of course the lengthl

of the span should be as small as

possible).

4. Understanding the Shape of

Beams and Girders

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As we see from above, on increasing

the depth, unless the load is exactly

at the right place (difficult to arrange

in a bridge with moving traffic), thedeep bar may bend. This is called

buckling. To avoid this, a common

compromise is the cross-sectional

shape shown in Figure below.

This section provides a large load

bearing surface and enough depth to

prevent bending. This shape reduces

the weight of the beam without

sacrificing the strength and hence

reduces the cost.

5. Using Pillars and Columns

Use of pillars or columns is also very

common in buildings and bridges. A

pillar with rounded ends as shown in

Fig (a) supports less load than that

with a distributed shape at the endsFig (b).

The precise design of a bridge or a

building has to take into account the

conditions under which it will

function, the cost and long period,

reliability of usable materials etc.

6. Calculating Max Height

attained by any earthen structure

on earth

The answer to the question why the

maximum height of a mountain on

earth is ~10 km can also be provided

by considering the elastic propertiesof rocks. A mountain base is not under

uniform compression and this

provides some shearing stress to the

rocks under which they can flow. The

stress due to all the material on the

top should be less than the critical

shearing stress at which the rocksflow. At the bottom of a mountain of height

h,the force per unit area due to the

weight of the mountain ishρg where

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ρ is the density of the material of the

mountain andgis the acceleration

due to gravity. The material at the

bottom experiences this force in the vertical direction, and the sides of the

mountain are free. Therefore this is not a case of

pressure or bulk compression. There

is a shear component, approximately

hρgitself. Now the elastic limit for a

typical rock is 30 × 107 Nm-2.

Equating this tohρg with ρ=3×10

3

kgm-3 gives,hρg= 30 × 107 Nm-2 Orh=30×107 Nm-2/(3 × 103 kgm-3×10ms-2) = 10 km which is slightly more than the

height of Mt. Everest. So no structure

made of rock can ever rise beyond 10km.

Exercise

1 MARK QUESTIONS1. A wire is stretched by a force such

that its length becomes double. How

will the Young’s modulus of the wire be affected?2. How does the Young’s modulus

change with rise in temperature?

3. Which of the three modulus of

elasticity – Y, K and η is possible in

all the three states of matter (solid,

liquid and gas)?4. The Young’s modulus of steel is

much more than that for rubber. For

the same longitudinal strain, which

one will have greater stress?5. Which of the two forces –

deforming or restoring is responsible

for elastic behavior of substance?6. Define malleability of a substance.7. What are elastomers?

TWO MARKS QUESTIONS1. Distinguish ductile and brittle

substances.2. Explain why bridges are declared

unsafe after long use? THREE MARKS QUESTIONS1. Derive an expression for the elastic

potential energy of a stretched wire.2. Show that the elastic potential

energy stored per unit volume of a

stretched wire is equal to 1/2 x stress

x strain.3. Define stress and mention its

types. 4. Define strain and

mention its types.

5. Which are the different moduli ofelasticity? Define each and write

mathematical expression for each.

FIVE MARKS QUESTIONS

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1. Draw the Stress vs Strain graph of

a metallic wire subjected to a

gradually increasing stretching

force and explain the differentregions of the graph.

2. Describe the experiment to

determine the Young’s modulus of

a metallic wire.

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10.1 Introduction

In this chapter we will study about

the common properties of liquid and

gases in contest to their fluid

behaviour which distinguish them

from solid.In our daily life we deal with fluid in

one or more ways like breathing,

blood circulation in our veins andarteries, the envelop around the

earth surface, major constituent part

of earth surface, processes for life

occurrence in both plants and

animals are based on fluidic

behaviour. A fluid cannot resist the deforming

force, it flows under the action offorces. Its shape change continuously

as long as force act on it. This

deformation is caused by shearing

force/stress which acts tangentially

on its surface. But solid resists any

deformation force while at rest. This force may cause some

displacement but the deformationdoes not take place easily. Hence we

can say

“ A fluid is a substance which

deforms in continuous manner or

flows, when act upon by shearing

forces/stress”

But converse to this if fluid at rest which means no shearing forces act

tangentially, it must be normal to the

planes to which they are acting. This

is condition when relative velocity of

particles of fluid is zero.Let us consider an example of flow o

liquid in a pipe. At the wall of pipe

the velocity of water will be zero, velocity will increases as we move

downwards centre of pipe. But as the

boundary widen or we can say if a

fluid is a long way front boundary,

then all particles move with same

velocity and there will be no shear

forces acts as all particles have zero

relative velocity as shown infigure10.1a below.

v3>v2>v1

V1 V2

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V3 V (max)

Figure 10.1a

In practice, flow of fluids depends

upon solid boundaries like airplanes,

cars, pipes, river channels etc.

10.2 PRESSURE: In our daily life, we deals with

pressure on many ways like fix nail in

wall, drag objects, drive vehicles, cut

fruits and vegetables, while walking,

lifting weights, etc.From all examples we conclude that

pressure depend on area of contact

on which force applied.For uniform force, we notice/observe

that less the area more will be theimpact or more the area less will be

the impact. This concept is pressure

i.e. Pressure α1

*rea for

constant/uniform force.How pressure acts on a body in a

static fluid? To know it let us

consider a small pressure –sensing

device be suspended inside a fluid –

filled vessel as shown in figure (c).

the sensor consist of a piston of area

of cross section ∆A resting against a

spring with scale in a closely fitted

cylinder.

Figure 10.2a A readout arrangement reveals that

∆F force act normal to the piston.

Hence we can define pressure on

piston from fluid as P= ∆F/ ∆A for

uniform force act over a flat area A,

we can write above equation as P =

F/A.Here we involve only magnitude of

force F not the direction as it is

uniformly distributed over every point

of the area, hence it gives same value

of pressure whatever the orientation

of sensor.Pressure is a scalar quantity. Its

dimension is ML-1 T-2. S.I unit is Nm-

2. Also named as Pascal (Pa), in the

honour of French scientist Blaise-

Pascal (1623-1662) who is known for

his contributed studies on fluidpressure. A common unit is atmosphere i.e. the

average pressure exerted by the

atmosphere at the sea level. 1 atm =

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1.01 x 105 Pa = 760 torr = 14.7

l.b/in2. Some unit of pressure is psi

(pound per square inch).

Question 10.1: Explain why the

blood pressure in humans is greater

at the feet than at the brain. The

blood pressure in humans is greater

at the feet than at the brain.

Atmospheric pressure at a height of

about 6 km decreases to nearly half

of its value at the atmosphericthough the height of the atmosphere

is more than 100 km. Hydrostatic

pressure is a scalar quantity even

though pressure is force divided by

area. Answer: As liquid pressure is given

by the relation,

P = h ρ gIt can be inferred that pressure is

directly proportional to height.

Hence, the blood pressure in human

vessels depends on the height of the

blood column in the body. The height

of the blood column is more at the

feet than it is at the brain. Hence, the

blood pressure at the feet is more

than it is at the brain.Density of air decreases with increase

in height from the surface. At a

height of about 6 km, density

decreases to nearly half of value at

the sea level. Atmospheric pressure is

proportional to density. Hence, at a

height of 6 km from the surface, it

decreases to nearly half of its value atthe sea level. When force is applied

on a liquid, the pressure in the liquid

is transmitted in all directions.

Hence, hydrostatic pressure does not

have a fixed direction and it is a

scalar physical quantity.

Question 10.2: A 50 kg girl wearinghigh heel shoes balances on a single

heel. The heel is circular with a

diameter 1.0 cm. What is the

pressure exerted by the heel on the

horizontal floor? Answer: Mass of the girl, m = 50 kg.

Diameter of the heel, d = 1 cm = 0.01

mRadius of the heel, r Area of the heel= π (0.005)2= 7.85 × 10 –5 m2

Force exerted by the heel on the floor:F = mg= 50 × 9.8= 490 N

Pressure exerted by the heel on thefloor:= 6.24 × 106N m –2

Therefore, the pressure exerted by

the heel on the horizontal floor is

6.24 × 106 N m –2

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Some common pressure values

Body / object Pressure(Pa)

Center of sun 2x1016

Center of earth 4x1011

Atmosphere at

sea level

1x105

Normal blood

pressure

1.6x104

Best lab vacuum 10-12

Table 10.2a

10.3 DENSITY:Property of fluid which describes

nature of fluid to find the density of

fluid at any point, we isolate a small

volume (∆V) around that point andthe mass (∆m) of fluid contained in it.

The density of fluid then,ρ= ∆m/ ∆V

Or, ,= lim

% -0

dm/d%

And Ρ = m/V, for uniform density Where m is mass and v is volume of

fluid sample. It is positive scalar

quantity. Its dimension is ML-3 T0 andS.I unit is kg-m3. Variation of density

for incompressible liquid is nearly

constant at all pressure but variable

for gases with pressure. Why?

NOTE:

1.The density of water at 4oC is 1 x

103 kg m3.

2.Relative density of any substanceis density of substance with

respect to density of water at 4oC.3.Relative density is dimensionless

positive scalar quantity.

Density of Common Fluids At 4o C

Table 10.3a

COMMONFLUID

DENSITY(kgm-3)

Blood (whole at

37oC)

1060

Ethyl Alcohol 806Mercury 13600Oil (Hydraulic) 800 Water 998

Question 10.3 The body of a man

weight is about 720 N contains 4.6 x

10-3 m3 blood. Find blood weight and

express it as a percentage of the body

weight. Answer:

a.)The mass of body of man is

m = ρ V = 1060 X 4.6 x 10-3kg = 4.87 Kg For weight of blood W = mg

= 4.87 X 9.8 = 47.73 Newton

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b.)The percentage of body weight in

comparison to blood weight

Percentage =47.73

720 X 100

= 6.63 %

10.4 Pascal’s Principal

As pressure due to fluid increases

with depth due to weight of the fluid

above the point of observationConsider following cases

1.Open tube manometer: in this case

we are seeing that a U-tube which isopen at one end and fixed with the

chamber at other end, chamber is

closed be lid at its open end. As

pressure at point A and point B are

equal, but forces are different. At

point B pressure exerted due to

chamber, which is given as

PB = P2 --------------10.1

Similarly at point AP A = P1 + ρ g h

----------------10.2But point A and B both are at same

level so are in equilibrium conditionHence P A = PB

From equation 1 and 2 we can writeabove equation as :P2 = P1 + ρ g hP2 = P1 + ρgh.’. P2 - P1 = ρ g h

This difference in pressure exerted by

chamber at atmospheric pressure is

balanced by the pressure due to

fluid. This behavior is described byPASCAL’S Principal.

PASCAL’S PRINCIPAL: Any change in the pressure applied to

a completely enclosed fluid is

transmitted undiminished to all parts

of the portion of fluid and its enclosing

walls.NOTE: In the absence of gravity

pressure exerted by fluid of any

submerged object will be equal on all

of its part. In other word we can say

the pressure in a fluid in equilibrium

is same where.Experimental Proof of Pascal’s Law

Figure 10.4aConsider a vessel as given in above

figure with Incompressible uniform

liquid with pistons at different

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positions. Suppose that force FD

applied on piston D. It has been

observed that equal force on area

exerted on positions A,B and C suchthat FD/AD = FB/AB = FC/AC = F A/A A This equation demonstrates that the

pressure is transmitted

undiminished in all directions as per

PASCALS Law.Question 10.4: Torricelli’s barometer

used mercury. Pascal duplicated itusing French wine of density 984

kgm –3. Determine the height of the

wine column for normal atmospheric

pressure. Answer:

Density of mercury, ρ1= 13.6 × 103

kg/m3

Height of the mercury column h1=

0.76 mDensity of French wine, ρ2 = 984

kg/m3

Height of the French wine column = Acceleration due to gravity, g = 9.8

m/s2

The pressure in both the columns is

equal, i.e.,

Now, Pressure in the mercury column= Pressure in the French wine

column

= 10.5 m.Hence, the height of the French wine

column for normal atmospheric

pressure is 10.5 m.

Question 10.5: A U-tube contains

water and methylated spirit

separated by mercury. The mercurycolumns in the two arms are in level

with 10.0 cm of water in one arm and

12.5 cm of spirit in the other. What is

the specific gravity of spirit? Answer

The given system of water, mercury,

and methylated spirit is shown as

follows:

Height of the spirit column, h1= 12.5cm = 0.125 mHeight of the water column, h2= 10

cm = 0.1 mP0 = Atmospheric pressureρ1 = Density of spirit

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ρ2 = Density of waterPressure at point B is;

Pressure at point D is;

Pressure at points B and D is the

same.

So,

Therefore, the specific gravity of spirit

is 0.8.

10.5 Effect of Gravity

Figure 10.5aConsider a vessel filled with fluid. Let

us consider a volume element in form

of cylindrical column of mass m and

height h inside it as given in above

figure 10.5a.In the absence of gravity the pressure

exerted by liquid on each point ofcylindrical column will be equal. But

in the presence of gravity pressure

will be different.

We see that cylindrical column is at

rest i.e. it is in equilibrium state. How

it is in equilibrium state. How it is

possible? On lateral surface ofcylinder the pressure exerted by fluid

in opposite direction balance each

other i.e. cancel effect of pressure of

one another but for cross sectional

area’s at point O1 and O2, the

downward force (force exerted by

liquid column above point O1 plus

force due to gravity) is balanced bythe upward force exerted by liquid

column below point O2.Let P1 pressure exerted on area A

vertically downward on top face of the

column and P2 is the pressure

exerted on bottom face of area

vertically upward. So, under

equilibrium condition we have,∑

F!=0

P1 A + mg + P2 A = 0P2 A = P1 A + mgP2 A = P1 A + ρ V g

P2 = P1 + ,% g

*

P2 = p1 + ρ g h

Form this equation we clearly note

1.The pressure is same at all points

at same depth.

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2.Pressure is independent of the

shape of vessel in which object is

placed.

3.As the depth increase thedifference in pressure on object

also increase and it’s vice versa.

NOTE: This difference in pressure if P1

= Pa such that P2– Pa is called gauge

pressure at that point.

As we discussed that pressure is

unaffected by the shape of vessels or

containers, Consider three cases of

different shaped vessels filled with

same liquid to same height.Pressure experience on object of

mass m placed at the bottom of each

Vessel are same by measuring

instrument. This is example of

Hydrostatic Paradox.

10.6 ATMOSPHERIC PRESSURE

(Pa)

Just as ocean water exerts pressure

on bottom of sea, same like this

envelope of gaseous surrounds earth

atmosphere also exerts pressure due

to atmosphere is called atmosphere

pressure. At sea level the value of one

atmospheric pressure is 1.013 X 105

Pa. This amount of pressure

corresponds to 14.70 lb in-2and is

referred to as one atmosphere (atm).

10.7 Measurement of Atmospheric

Pressure [Torricelli’s Experiment]

The first experiment to measureatmospheric pressure was given by E.

Torricelli an Italian Scientist in year

1643. He filled a strong glass tube of

nearly one meter by closing its open

end with thumb in a dish containing

mercury. He observed that mercury

in the tube fall down first and then

stopped. He inclined or raise the tubehigher or lower, the vertical height of

mercury in tube above mercury level

in dish remain constant i.e. 76cm.

Torricelli explained his observation

by saying that the mercury column is

supported by the pressure of the

atmospheric air acting on the free

surface of mercury in the dish. Inother word we can say that

atmospheric pressure in dish is

balanced by 76cm of Hg in tube. The

pressure exerted by 76cm of mercury

column of tube at base is same to

that exerted by entire atmosphere

does.

10.7.1 Applications of Pascal’s Law

A) Hydraulic Lift:

Hydraulic machines like hydraulic

brakes, hydraulic lift, hydraulic press

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etc work on the principal of Pascal’s

law. Let us look at a glance how these

machines work. In case of hydraulic

lift which is used to support or liftheavy objects.

Figure 10.6a

Above figure shows a liquid container which has two opening ends narrow

and broader fitted with pistons say A

and B when force F A is applied on

piston A of cross sectional area Aaraise another piston B. how much

force applied on piston B to lift it up.

As per Pascal’s law when height of

both columns are same then,Pa = P bSo, FB/A b = F A/AaOr, FB = (F A/Aa) x A bLet us take values for these terms

towards right side of above equation.F A= 20 N Aa = 2 x 10-2 m2

Ab = 5 x 10-1 m2

Putting above terms in aboveequation we getFB= 500N

This is much higher than F A.Hence

by applying very small force at one

end can exert huge force to lift a

heavy object without doing any extra

work.

10.7.2. Hydaulic Brakes

Figure 10.6b

In a car, trucks, cranes or any other

automobile a smooth worked brakes

system is very important in order to

avoid skidding, slipping. By proper

retarding force (frictional force) one

need an efficient braking system which help vehicle to balance itself

under such conditions. Let us see

how it works form above sketch of

such breaking system, when peddle

is pressed the lever operates. The

piston P is pressed, this pressure

transmitted to P1 and P2 as per

Pascal’s law. These pistons expandthe rim on which brakes shoe’s S1

and S2 are attached which touched

the inner rim of moving wheel and

retard the motion but the area of

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pistons P1 andP2 are more than area

of P. So a small force applied to the

brake pedal produces a large thrust

on the wheel as P ∞ 1/A.

Question 10.8: A hydraulic

automobile lift is designed to lift cars

with a maximum mass of 3000 kg.

The area of cross-section of the

piston carrying the load is 425 cm

would the smaller piston have to

bear? Answer:The maximum mass of a car

that can be lifted, Area of cross-

section of the load -carrying piston, A

= 425 cm2 = 425 × 10–4m2= 3000 ×

9.8 = 29400 N

The maximum pressure exerted on the

load,

= 6.917 × 105 Pa

Pressure is transmitted equally in all

directions in a liquid. Therefore, the

maximum pressure that the smaller

piston would have to bear is 6.917 ×

10

5

Pa

10.8Streamline Flow

The fluid in motion is known as fluid

dynamics. Flow of fluids is of various

types, we come to that later on. The

flow of liquid we see is very smooth

somewhere, violent and very fast

specially flow of water in river. The

nature of fluid, physical conditions ofmedium through which it passes also

play vital role for the motion of fluid.

The flow of liquid is said to be

streamline flow if flow of liquid is

steady in nature what is this steady

flow? In such flow the particle

passing through one particular point

has same velocity, means it is a pointthrough which when each particle of

medium passes will attain same

velocity, which doesn’t mean that at

other particles must have same

velocity, it may have different

velocities.

Figure 10.7a In above figure 10.7a at position 1

and 2 all particles pass through

same points will have only 10m/s

and 5m/s velocity. Such path taken by the fluid/particle under a steady

flow is termed as streamline. A streamline is a line drawn in the

fluid such that a tangent to the

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streamline at any point is parallel to

the fluid velocity at that point. Two streamline never cross each

other if it is so then velocity atcrossing point would change from

moment to moment it is condition

that does not exist in steady flow.

10.8 (a)Equation Of Continuity:

Have you ever observed that when

we use our thumb/finger to control

over the velocity of fluid by reducing

the area of cross section, but the

volume of liquid flow out before and

after remain same. This behavior of

fluid is described by equation of

continuity.If fluid enters at some x kgs-1 then it

must have leave at same rate

whatever be the area of cross section

of inlet or outlet point is. This floe

through tube is termed as mass flow

rate.

Figure 10.7aConsider a pipe/tube in which

incompressible fluid (uniformly

distribution molecules) flow as given

in above figure 10.7a. During same

interval of time the distance covered

by fluid section P and Q is V2∆t and

V1∆t respectively. Also the volumeflow in this time period is

At point P:

V2 = A2 v2∆t At point Q:

V1= A1 v1∆tHence the mass of fluid will be;

At point P;

∆m2 = V2∆X2

Or, ∆m2 = ρ2 A2 v2∆t Also at point Q;

∆m1 = V1∆X1Or, ∆m1 = ρ1 A1 v1∆t And the rate of floe of mass will be At point P;

∆m2/∆t = ρ2 A2 v2 At point Q;

∆m1/∆t = ρ1 A1 v1Since no fluid can cross the sidewallsof the tube, the mass flow rates at

the positions P and Q must be equal.So,∆m1/∆t = ∆m2/∆tρ1 A1 v1= = ρ2 A2 v2For incompressible fluid the density

of fluid are same i.e. ρ1 = ρ2.

So, above equation becomes A1 v1= A1 v1 This is equation of continuity which

proves that volume of fluid per

second at both tends for an

incompressible fluid will be same

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irrespective of the area of cross -

section of the tube. It also prove that

mass is conservative (neither created

nor be destroyed) as it flow along atube.

Question 10.6 The cylindrical tube

of a spray pump has a cross 40 fine

holes each of diameter 1.0 mm. If the

liquid flow inside the tube is 1.5 m

min. what is the speed of ejection of

the liquid through the holes? Answer :

Area of cross-section of the spray

pump,Number of holes,n= 40Diameter of each hole,d= 1 mm = 1

× 10 –3 mRadius of each hole,r=d/2 =0.5×10 –

3m

Area of cross-section of each hole, Total area of 40 holes, A2 =n= 40 × π (0.5 × 10–3)2m2

= 31.41 × 10 –6m2

Speed of flow of liquid inside the

tube,Speed of ejection of liquid through

the holes = v2 According to the law of continuity, we

have:

= 0.633 m/s Therefore, the speed of ejection of the

liquid through the holes is 0.633 m/s

10.8(b)Bernoulli’s Equation :

In 1738 Daniel Bernoulli (1700-1782)

formulated the famous equation for

fluid flow that bears his name. The

Bernoulli Equation is a statement

derived from conservation of energy

and work-energy ideas that come

from Newton's Laws of Motion. An important and highly useful

special case is where friction is

ignored and the fluid is

incompressible. This is not as unduly

restrictive as it might first seem. The

absence of friction means that the

fluid flow is steady. That is, the fluid

does not stick to the pipe sides and

has no turbulence. Most common

liquids such as water are nearly

incompressible, which meets the

second condition.Consider the case of water flowing

though a smooth pipe. Such a

situation is depicted in the figure

below. We will use this as our

working model and obtain Bernoulli'sequation employing the work-energy

theorem and energy conservation. The phenomenon described by

Bernoulli's principle has many

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practical applications; it is employed

in the carburettor and the atomizer,

in which air is the moving fluid, and

in the aspirator, in which water is themoving fluid. In the first two devices

air moving through a tube passes

through a constriction, which causes

an increase in speed and a

corresponding reduction in pressure.

As a result, liquid is forced up into

the air stream (through a narrowtube that leads from the body of the

liquid to the constriction) by the

greater atmospheric pressure on the

surface of the liquid. In the aspirator

air is drawn into a stream of water as

the water flows through a

constriction. Bernoulli's principle can

be explained in terms of the law ofconservation of energy.

As a fluid moves from a wider pipe

into a narrower pipe or a

constriction, a corresponding volume

must move a greater distance forward

in the narrower pipe and thus have a

greater speed.

At the same time, the work done by

corresponding volumes in the wider

and narrower pipes will be expressed

by the product of the pressure and

the volume. Since the speed is

greater in the narrower pipe, the

kinetic energy of that volume is

greater.

Then, by the law of conservation of

energy, this increase in kinetic energy

must be balanced by a decrease in

the pressure-volume product, or,

since the volumes are equal, by a

decrease in pressure.10.8(c) Proof of Bernoulli’s

Equation

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10.8(d) Venturimeter:

Venturimeter is a device used for

measuring the rate of flow of a fluid

through a pipe. The basic principle

on which a venturimeter works isthat by reducing the cross sectional

area of the flow of passage, a

pressure difference is created and the

measurement of the pressure

difference enables the determination

of the discharge through a pipe. The

inlet section of the venturimeter is of

the same diameter as that of thepipe, which is followed by a

convergent cone. The convergent cone

is a short pipe, which tapers from the

original size of the pipe to that of the

throat of the venturimeter. The throat

of the venturimeter is a short

parallel-sided tube having uniform

cross sectional area smaller thanthat of the pipe. The divergent cone of

the venturimeter is a gradually

diverging pipe with its cross sectional

area increasing from

that of the throat to the original size

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pressure taps are provided to

measure the pressure difference. By

applying the Bernoulli equation to

the inlet section and at the throat,(i.e., section 1 and 2) an expression

for the discharge is obtained. Water is

allowed to flow through the meter at

different rates ranging from zero to

the maximum and the corresponding

pressure differences shown in the

manometer are noted. The actual

discharge A is determined using themeasuring tank and the stop watch.

Applying Bernoulli along the

streamline from point 1 to point 2 inthe narrowthroatof the Venturimeter

we have

By the using the continuity equation

we can eliminate the velocityu2,

Substituting this into and

rearranging the Bernoulli equation

we get

To get the theoretical discharge this

is multiplied by the area. To get the

actual discharge taking in to account

the losses due to friction, we include

a coefficient of discharge

This can also be expressed in terms

of the manometer readings

Thus the discharge can be expressed

in terms of the manometer reading.

Notice how this expression does not

include any terms for the elevation or

orientation (z1 or z2) of the

Venturimeter. This means that the

meter can be at any convenient angle

to function.

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10.8(e) Torricelli’s Theorem:

The word efflux means fluid out flow.

Torricelli discovered that speed of

efflux from an open tank is given by a

formula identical to that of a freely

falling body.Consider a tank containing a liquid

of density r with a small hole in its

side at a height Y1 from the bottom.

The air above the liquid is at height

Y2 is at a pressure P. From the

equation of continuity weHave v1 A1 =v2 A2Now A2 > > A1 , v2 » 0

i.e the fluid to be approximately atrest at the top.Now applying Bernoulli’s equation at

points 1 and 2 & noting that at the

whole P1 = Pa, the atmospheric

pressure we have

Take Y2- Y1 = h then v1 = √2g h+

2( (− (a) ,

If the tank is open to atmosphere

than P = Pa

Then v1 = √ 2gh

This is known as Torricelli’s law

10.9 Viscosity

Viscosity is an internal property of a

fluid that offers resistance to flow.For liquids, it corresponds to the

informal concept of "thickness".For example,honey has a much

higher viscosity than water. Viscosity is a property arising

fromfriction between neighbouring

particles in a fluid that are moving at

different velocities. When the fluid is forced through a

tube, the particles which comprise

the fluid generally move faster near

the tube's axis and more slowly near

its walls: therefore somestress, is

needed to overcome the friction

between particle layers and keep the

fluid moving. For the same velocity

pattern, the stress required is

proportional to the fluid's viscosity.

Figure 10.9a

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Thedynamic (shear) viscosity of a

fluid expresses its resistance to

shearing flows, where adjacent layers

move parallel to each other withdifferent speeds.It can be defined through the

idealized situation known as

aCouette flow, where a layer of fluid

is trapped between two horizontal

plates, one fixed and one moving

horizontally at constant speed .

If the speed of the top plate is smallenough, the fluid particles will

moveparallel to it, and their speed

will varylinearly from zero at the

bottom to at the top. Each layer of

fluid will move faster than the one

just below it, and friction between

them will give rise to aforce resisting

their relative motion. In particular,the fluid will apply on the top plate a

force in the direction opposite to its

motion, and an equal but opposite

one to the bottom plate. An external force is therefore required

in order to keep the top plate moving

at constant speed.

The magnitude of this force is found

to be proportional to the speed and

the area of each plate, and inversely

proportional to their separation :

TheSI unit of dynamic viscosity is

thePascal-second (Pa·s), (equivalent

to (N·s)/m2, or kg/(m·s)). If

afluid with a viscosity of one Pa·s is

placed between two plates, and one

plate is pushed sideways with ashear

stress of onePascal, it moves a

distance equal to the thickness of the

layer between the plates in

onesecond. Water at 20 °C has a

viscosity of 0.001002 Pa·s, while a

typical motor oil could have a

viscosity of about 0.250 Pa·s. Thecgs physical unit for dynamic

viscosity is the poise (P), named

after Jean Leonard Marie Poiseuille.It is more commonly expressed

ascentipoise (cP). Water at 20 °C has

a viscosity of 1.0020 cP.1 P = 0.1 Pa·s,1 cP = 1 m Pa·s = 0.001 Pa·s = 0.001

N·s/m2.

10.10 STOKE’S LAW:

In 1851,George Gabriel

Stokes derived an expression, now

known asStokes' law, for thefrictional force – also calleddrag

force – exerted onspherical objects

with very smallReynolds

numbers (e.g., very small particles) in

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a continuous viscous fluid. Stokes'

law is derived by solving theStokes

flow limit for small Reynolds numbers

of theNavier–Stokes equations:

WhereFd is the frictional force –

known asStokes' drag – acting on

the interface between the fluid and

the particle (inN),μ is thedynamic

viscosity (kg /m*s),R is the radius of

the spherical object (in m), andv is

the particle's velocity (in m/s).Stokes' law makes the following

assumptions for the behaviour of a

particle in a fluid:• Laminar Flow• Spherical particles• Homogeneous (uniform in

composition) material• Smooth surfaces

• Particles do not interfere witheach other.

Figure 10.10a

Stokes's law is the basis of thefalling-sphere viscometer, in which

the fluid is stationary in a vertical

glass tube. A sphere of known size

and density is allowed to descend

through the liquid as shown in above

figure 10.10a.If correctly selected, it reaches

terminal velocity, which can bemeasured by the time it takes to pass

two marks on the tube. At terminal (or settling) velocity, the

excess forceFg due to the difference

of the weight of the sphere and

the buoyancy on the sphere, (both

caused bygravity :

With ρ p and ρ f themass density of

the sphere and the fluid, respectively,

andg thegravitational acceleration.

Demanding force balance:Fd =Fg and

solving for the velocityV gives the

terminal velocityVs.Note that since buoyant force

increases asR and Stokes dragincreases asR, the terminal velocity

increases asR and thus varies

greatly with particle size as shown

below. If the particle is falling in the

viscous fluid under its own weight

due togravity, then aterminal

velocity, or settling velocity, is

reached when this frictional forcecombined with the buoyant

force exactly balances

thegravitational force. The resulting terminal velocity (or

settling velocity) is given by:262

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wherevs is the particle's settling

velocity (m/s) (vertically downwards

if ρ p > ρ f upwards if ρ p < ρ f ) g isthegravitational acceleration (m/s2)

ρ p is themass density of the particles

(kg/m3), ρ f is the mass density of the

fluid (kg/m3) andμ is thedynamic

viscosity (kg /m*s).

10.11 Reynold’s Number:

Influid mechanics, theReynolds

number (Re) is adimensionless

quantity that is used to help predict

similar flow patterns in different fluid

flow situations. The concept was

introduced byGeorge Gabriel

Stokes in 1851, but the Reynolds

number is named afterOsborneReynolds (1842–1912), who

popularized its use in 1883. The Reynolds number is defined as

theratio of inertial forces

to viscous forces and consequently

quantifies the relative importance of

these two types of forces for given

flow conditions.Reynolds numbers frequently arise

when performing scaling of fluid

dynamics problems, and as such can

be used to determinedynamic

similitude between two different

cases of fluid flow.

The Reynolds number is defined below for each case.

Where:• is the mean velocity of the

object relative to the fluid (SI

units: m/s)• is a characteristic linear

dimension (m).• is thedynamic viscosity of

thefluid (Pa·s or N·s/m² or kg/

(m·s))• is thekinematic viscosity (

) (m²/s)• is thedensity of the fluid

(kg/m³).Reynolds number, in fluid

mechanics, a criterion of whetherfluid (liquid or gas) flow is absolutely

steady (streamlined, orlaminar) or on

the average steady with small

unsteady fluctuations (turbulent). Whenever the Reynolds number is

less than about 2,000, flow in a pipe

is generally laminar, whereas, at

values greater than 2,000, flow isusually turbulent. Actually, the transition between

laminar and turbulent flow occurs

not at a specific value of the Reynolds

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number but in a range usually

beginning between 1,000 to 2,000

and extending upward to between

3,000 and 5,000.

Question 10.6: Glycerine flows

steadily through a horizontal tube of

length 1.5 m and radius 1.0 cm. If

the amount of glycerine collected per

second at one end is 4.0 × 10-3 kg m –3

pressure difference between the two

ends of the tube? (Density ofglycerine = 1.3 × 103 kg m –3 and

viscosity of glycerine = 0.83 Pa s).

[You may also like to check if the

assumption of laminar flow in the

tube is correct]. Answer:

Length of the horizontal tube,Radius of the tube, r = 1 cm = 0.01 mDiameter of the tube, d = 2r = 0.02 mGlycerine is flowing at a rate of

4.0×10 kg s –1

M = 4.0 × 10 –3 kg s –1

Density of glycerine, ρ = 1.3 × 10 kg

m –3

Viscosity of glycerine, η = 0.83 Pa s Volume of glycerine flowing per sec:

= 10 –3 kg s –1.

According to Poiseville’s formula, we

have the relation for the rate of flow:

Where, p is the pressure difference

between the two ends of the tube

= 9.8 × 102 PaReynolds’ number is given by the

relation:

Reynolds’ number is about 0.3.

Hence, the flow is laminar.

10.12 Dynamic Lift:

Afluid flowing past the surface of a

body exerts aforce on it.Lift is the

component of this force that

isperpendicular to theoncoming flow

direction.It contrasts with thedrag force,

which is the component of thesurface forceparallel to the flow

direction. If the fluid is air, the force

is called anaerodynamic force.

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In water, it is called ahydrodynamic

force. Lift is most commonly

associated with the wing of afixed-

wing aircraft, although lift is alsogenerated

bypropellers kites,helicopter, rotors,

rudders,sails andkeels onsailboats,hydro-foils, wings onauto racing cars

wind turbines, and other streamlined

objects. When an aircraft isclimbing

descendingor banking in a turn thelift is tilted with respect to the

vertical. Lift may also be entirely

downwards in someaerobatic

maneuvers, or on the wing on a

racing car. In this last case, the

termdown force is often used. Aerodynamic lift is distinguished

from other kinds of lift in fluids.

Aerodynamic lift requires relative

motion of the fluid which

distinguishes it fromaerostatic lift

or buoyancy lift as used by balloons,

blimps, and dirigibles. Aerodynamic lift usually refers to

situations in which the body is

completely immersed in the fluid, and

is thus distinguished fromplanning

lift as used by motorboats,

surfboards, and water-skis, in which

only a lower portion of the body is

immersed in the lifting fluid flow.

Figure10.11a

Figure 10.11b

Movement of object from its original

path is due to Magnus effect as

depict in figure 10.11a. TheMagnus

effect is the commonly observed

effect in which a spinning ball (orcylinder) curves away from its

principal flight path. It is important

in many ball sports. Under the

Magnus effect, top spin produces a

downward swerve of a moving ball,

greater than would be produced by

gravity alone, and back spin has the

opposite effect.

10.11(a) Blood And Heart Attack:

Bernoulli’s principle says that the

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incompressible, non viscous fluid

remains constant. Bernoulli’s

principle helps in explaining blood

flow in artery. The artery may get constricted due to

the accumulation of plaque on its

inner walls. In order to drive the

blood through this constriction a

greater demand is placed on the

activity of the heart. The speed of the flow of the blood in

this region is raised which lowers thepressure inside and the artery may

collapse due to the external pressure.

The heart exerts further pressure to

open this artery and forces the blood

through. As the blood rushes through the

opening, the internal pressure once

again drops due to same reasons

leading to a repeat collapse. This may

result in heart attack. Question 10.11a; What is the largest

average velocity of blood flow in an

artery of radius 2 × 10-1 mm flow

must remain laminar? (b) What is the

corresponding flow rate? (Take

viscosity of blood to be 2.084 × 10 –3

Pa s). Answer

(a)Radius of the artery,r= 2 × 10 mDiameter of the artery,d= 4 × 10 – 3m

Viscosity of blood,

Density of blood, ρ= 1.06 × 103

kg/m3

Reynolds’ number for laminar flow,

NR = 2000 The largest average velocity of blood

is given by the relation:

Therefore, the largest average velocity

of blood is 0.983 m/s. Flow rate is

given by the relation:R= πr2 vavg

Therefore, the corresponding flow

rate is=

10.12 Surface Energy and Surface

Tension:

Thesurface energy is defined as the

sum of all intermolecular forces that

are on the surface of a material, the

degree of attraction or repulsion force

of a material surface exerts on

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intermolecular bonds that occur

when a surface is created.In thephysics ofsolids, surfaces

must be intrinsically lessenergetically favorable than the bulk

of a material (the molecules on the

surface have more energy compared

with the molecules in the bulk of the

material), otherwise there would be a

driving force for surfaces to be

created, removing the bulk of the

material. The surface energy may therefore be

defined as the excess energy at the

surface of a material compared to the

bulk. In the case of liquids this same

definition is applied to define the

surface tension as a result of this

surface tension liquid with low

surface tends to contract and formdroplets.Surface tension can be defined as the

resistance of a fluid to deform or

break, such resistance is defined

directly by the intermolecular forces

that are on the surface.

Figure 10.12a

When the substrate has a high

surface energy, i.e. it tends to attract,

and the adhesive has a low surface

tension, has little resistance to

deformation or rupture, a good

wetting of the adhesive on the

substrate is produced as shown in

figure 10.12a above. For example,

silicone adhesives have a low surface

tension; this is the main reason why

these adhesive bond on a wide range

of substrates.On the other hand, those substrates

which contain silicone surfaces havea low surface energy and because of

this reason, they are very difficult to

get a good wet ability and a good

adhesion with any material unless

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you do a surface treatment to remove

the silicone and increase surface

energy of the substrate surface.

10.12(a) Angle Of Contact

Thecontact angle is the angle,

conventionally measured through the

liquid, where

aliquid/ vapour interface meets

asolid surface. It quantifies the wet

ability of a solid surface by a liquid

via the Young equation. A given

system of solid, liquid, and vapour at

a given temperature and pressure

has a unique equilibrium contact

angle. However, in practicecontact

angle hysteresis is observed, ranging

from the so-called advancing(maximal) contact angle to the

receding (minimal) contact angle. The equilibrium contact is within

those values, and can becalculated

from them. The equilibrium contact

angle reflects the relative strength of

the liquid, solid, and

vapourmolecular interaction.

Figure10.12a

Figure 10.12b

The theoretical description of contact

arises from the consideration of

athermodynamic equilibrium betwee

n the threephases: theliquid phase

(L), thesolid phase (S), and the

gas/ vapour phase (G) (which could

be a mixture of ambient atmosphereand an equilibrium concentration of

the liquid vapour). The “gaseous” phase could also be

another (immiscible) liquid phase. If

the solid–vapourinterfacial energy is

denoted by , the solid–liquid

interfacial energy by , and the

liquid–vapour interfacial energy (i.e.

thesurface tension) by , then the

equilibrium contact angle is

determined from these quantities

by Young's Equation:

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10.13 Drops and Bubbles

The fact that air has to be blown into

a drop of soap solution to make a

bubble should suggest that the

pressure within the bubble is greater

than that outside. This is in fact the

case: this excess pressure creates a

force that is just balanced by the

inward pull of the soap film of the

bubble due to its surface tension.

Figure10.13a

Consider a soap bubble of radius r as

shown in Figure 10.13a. Let the

external pressure be Po and the

internal pressure P1. The excess pressure P within the

bubble is therefore given by:Excess pressure (P) = P0 -P1

Consider the left-hand half of the bubble. The force acting from right to

left due to the internal excess

pressure can be shown to be PA,

where A is the area of a section

through the centre of the bubble. If

the bubble is in equilibrium this

force is balanced by a force due to

surface tension acting from left to

right. This force is 2 x 2πrT (the factor of 2

is necessary because the soap film

has two sides) where T is the

coefficient of surface tension of the

soap film. Therefore2 x 2πr T = PA = Pπr2 giving:Excess pressure in a soap bubble(P) = 4T/r A bubble of air within a liquid hasonly one liquid-air surface and the

excess pressure within such a bubble

is simply:

Excess pressure in an air bubble(P) = 2T/rBoth these formulae show that the

excess pressure within a small

bubble is greater than that withinlarger bubble.

Bubble diameter

(2r) (µm)

((Pa

)

(at

m)

1000 286 0.002823.0 95800 0.9480.3 958000 9.477

Table 10.13a

Question 10.12a:Figure 10.24 (a) shows a thin liquid

film supporting a small weight = 4.5

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× 10-2N. What is the weight

supported by a film of the same

liquid at the same temperature in

Fig. (b) and (c)? Explain your answerphysically.

Answer: Take case (a): The length of the liquid film

supported by the weight, The weight supported by the film, A liquid film has two free surfaces.

∴ Surface tension

In all the three figures, the liquid is

the same. Temperature is also the

same for each case.Hence, the surface tension in figure

(b) and figure (c) is the same as in

figure (a), i.e., 5.625 × 10 –2 N m –1.

Since the length of the film in all the

cases is 40 cm, the weight supported

in each case is 4.5 × 10 –2 N.

10.14 Capillary Rise

Capillary action (sometimes

capillarity capillary motion,

or wicking ) is the ability of

aliquid to flow in narrow spaces

without the assistance of, and in

opposition to external forceslikegravity. The effect can be seen in

the drawing up of liquids between the

hairs of a paint-brush, in a thin tube,

in porous materials such as paper, in

some non-porous materials such as

liquefiedcarbon fiber, or in a cell. It

occurs because ofintermolecular

forces between the liquid andsurrounding solid surfaces. If the

diameter of the tube is sufficiently

small, then the combination

ofsurface tension (which is caused

bycohesion within the liquid)

andadhesive forces between the

liquid and container act to lift the

liquid. In short, the capillary actionis due to the pressure of cohesion

and adhesion which cause the liquid

to work against gravity. A common

apparatus used to demonstrate the

first phenomenon is thecapillary

tube. Adhesion occurs between the

fluid and the solid inner wall pulling

the liquid column up until there is asufficient mass of liquid

forgravitational forces to overcome

these intermolecular forces. The

contact length (around the edge)

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between the top of the liquid column

and the tube is proportional to the

diameter of the tube, while the weight

of the liquid column is proportionalto the square of the tube's diameter.

So, a narrow tube will draw a liquid

column higher than a wider tube will.

Figure 10.14aLet the radius of the glass capillary

tube be r, the coefficient of surface

tension of the liquid he T, the density

of the liquid be ρ, the angle of

contact between the liquid and the

walls of the tube be θ and the height

to which the liquid rises in the tube

be h.Consider the circumference of the

liquid surface where it meets the

glass. Along this line the vertical

component of the surface tension

force will be 2πr cosθ T

This will draw the liquid up the tubeuntil this force by the downward force

due to the column of liquid of height

h, that is just balanced at

equilibrium:

Therefore,

2πr cosθ T = πr2ρgh which gives capillary rise

h = 2Tcosθ/ rρg

This for an angle of contact of

0o becomes:Capillary rise (h) = 2T / rρg

We will assume that if the radius of

the tube is small the shape of the

liquid surface is very nearly

hemispherical.

The pressure at A must beatmospheric, but since A is within a

hemispherical surface the pressure

at B must be less than A by an

amount 2T/r. The pressure at C is also

atmospheric but it is greater than the

pressure at B by the hydrostatic

pressure hρg. Therefore atequilibrium we have : h = 2T/rρgBoth these methods show that the

rise is greater in tubes with a narrow

bore and for zero angles of contact.In fact when the coefficient of surface

tension is measured by capillary rise

in the laboratory the values obtained

are nearly always too small becauseof the difficulty of getting perfectly

clean apparatus.

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10.15 Detergent And Surface

Tension

Molecules of most detergents and

soaps are long chain hydrocarbonmolecules with an ionic group at one

end, usually carrying a negative

charge, thus making it an anion. This

charge is balanced by the opposite

charge of a soluble cation, for

example Na+. The long hydrocarbon

chains do not interact well with water

molecules, and many of them areeffectively ‘squeezed out’ to the

interfaces between the water and the

air or the glass sides of the beaker.

The effect of these molecules on the

water surface is to considerably

weaken the forces between water

molecules there, thus lowering the

surface tension.

Figure 10.15a

When the drop of detergent is added

to the powdered surface, the initial

effect is to draw the powder back to

the edges very rapidly as thedetergent molecules form their own

surface layer with a lower surface

tension than the water. As the

detergent gradually mixes with the

water, the powder begins to sink, and

a needle will now pass through the

surface with ease under its own

weight. However, if lycopodiumpowder is used, which is less dense

than water, it remains at the edges.

Other powders may clump into

nodules if they are not wetted by the

detergent solution. In hard water

there is a significant concentration of

calcium, Ca2+, and/or magnesium,

Mg2+ cations. These cations form aninsoluble compound with soap

anions, so instead of forming a

surface layer, they are precipitated

out, leaving the surface tension little

changed.2/\/\/\/\/\/\/COO−(aq) + Ca2+

(aq) →

(/\/\/\/\/\/\/COO)2Ca(s)

However, the calcium and

magnesium salts of many detergent

molecules are soluble, so detergents

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still lower the surface tension of hard

water.

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11.1Introduction:

Typically temperature may be defined

as• The degree of hotness or

coldness of a body or

environment.• A measure of the warmth or

coldness of an object or substance

with reference to some standard

value.• A measure of the average

kinetic energy of the particles in a

sample of matter, expressed in

terms of units or degrees

designated on a standard scale.• Any of various standardized

numerical measures of this ability,such as the Kelvin, Fahrenheit,

and Celsius scale• Temperature is a measure of

the ability of a substance, or more

generally of any physical system,

to transfer heat energy to another

physical system

On the other hand heat is the energy

present in a substance. If two bodies

at different temperatures are brought

in contact with each other then heat

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energy flows from the body at higher

temperature to the body at lower

temperature.

While temperature is measured in

kelvin (S.I units), heat energy is

measured in joule (S.I units). While

temperature of a substance is the

average kinetic energy of its

molecules, heat energy present in it

is the sum total of the kinetic energy

and potential energy present in it.

11.2Measurement of temperature

Many methods have been developed

for measuring temperature. Most of

these rely on measuring some

physical property of a working

material that varies with

temperature.One of the most common devices for

measuring temperature is theglass

thermometer.E.g.(a) Expansion of liquid with

temperature: Mercury or alcohol

thermometer(b) Increase in resistance with

increase in temperature: Thermistor(c) Measurement of heat radiation:

Pyrometers

The scales of temperature are• Celsius scale

• Fahrenheit scale• Kelvin scale

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C = (9/5)(F-32) = K-273

11.3Ideal gas equation

The ideal gas law is theequation of

state of a hypotheticalideal gas. It is

a good approximation to the behaviour of manygasesunder many

conditions, although it has several

limitations. It was first stated

byÉmile Clapeyron in 1834 as a

combination ofBoyle's

law andCharles's law. The ideal gas

law is often introduced in its common

form:

whereP is thepressure of the

gas,V is the volume of the gas,n is

theamount of substance of gas

(measured inmoles),R is the ideal,

or universal,gas constant, andT is

thetemperature of the gas.It can also be derived microscopically

fromkinetic theory, as was achieved

(apparently independently) by August

Krönig in 1856 andRudolf

Clausius in 1857.

The above graph shows that for all

gases the pressure i.e. intermolecular

forces becomes zero at -273.150C or 0

K. This temperature is called

absolute zero.

11.4 Thermal expansion:

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response to a change in

temperature throughheat transfer. When a solid has prominent linear

dimensions like a wire then theincrease in volume is manifested in

the form of linear expansion and

when a solid has prominent

superficial dimensions like a sheet

then the increase in volume is

manifested in the form of superficial

expansion. Temperature is a

monotonic function of the averagemolecular kinetic energy of a

substance. When a substance is heated, the

kinetic energy of its molecules

increases. Thus, the molecules begin

moving more and usually maintain a

greater average separation. Materials

which contract with increasingtemperature are unusual; this effect

is limited in size, and only occurs

within limited temperature ranges

(see examples below). The degree of

expansion divided by the change in

temperature is called the

material's coefficient of thermal

expansion and generally varies withtemperature.

11.4.1Factors affecting thermal

expansion

Unlike gases or liquids, solid

materials tend to keep their shape

when undergoing thermal expansion.

Thermal expansion generallydecreases with

increasing bond energy, which also

has an effect on themelting point of

solids, so, high melting point

materials are more likely to have

lower thermal expansion. In general,

liquids expand slightly more than

solids. The thermal expansion of glasses is

higher compared to that of

crystals. At the glass transition

temperature, rearrangements that

occur in an amorphous material lead

to characteristic discontinuities of

coefficient of thermal expansion or

specific heat. These discontinuities allow detection

of the glass transition temperature

where a super-cooled liquid

transforms to a glass. Absorption or desorption of water (or

other solvents) can change the size of

many common materials; many

organic materials change size muchmore due to this effect than they do

to thermal expansion. Common

plastics exposed to water can, in the

long term, expand by many percent.

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11.4.2 Coefficient of thermal

expansion

The coefficient of thermal expansion

describes how the size of an objectchanges with a change in

temperature. Specifically, it measures

the fractional change in size per

degree change in temperature at a

constant pressure.Several types of coefficients have

been developed: volumetric, area, and

linear which are used depending onthe particular application and which

dimensions are considered

important. For solids, one might only

be concerned with the change along a

length, or over some area.

The volumetric thermal expansion

coefficient is the most basic thermalexpansion coefficient. In general,

substances expand or contract when

their temperature changes, with

expansion or contraction occurring in

all directions. Substances that

expand at the same rate in every

direction are calledisotropic.

For isotropic materials, the area and

linear coefficients may be calculated

from the volumetric coefficient by

taking the square root or cubic root

of volume thermal expansion

coefficient. Mathematical definitions

of these coefficients are defined below

for solids, liquids, and gases.

11.4.2.1 General volumetricthermal expansion coefficient:

In the general case of a gas, liquid, or

solid, the volumetric coefficient of

thermal expansion is given by γ = (V/V T)∆ ∆ P

The subscript p indicates that the

pressure is held constant during the

expansion, and the subscript "V"

stresses that it is the volumetric (not

linear) expansion that enters this

general definition. In the case of a

gas, the fact that the pressure is held

constant is important, because the

volume of a gas will vary appreciably

with pressure as well as temperature.

For a gas of low density this can be

seen from theideal gas law.

11.4.2.2 General superficial

thermal expansion coefficient:

The area thermal expansion

coefficient relates the change in amaterial's area dimensions to a

change in temperature. It is the

fractional change in area per degree

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of temperature change. Ignoring

pressure, we may write:β = (A/A T)∆ ∆ P

11.4.2.3 General linear thermal

expansion coefficient

To a first approximation, the change

in length measurements of an object

("linear dimension" as opposed to,

e.g., volumetric dimension) due to

thermal expansion is related totemperature change by a "linear

expansion coefficient". It is the

fractional change in length per degree

of temperature change. Assuming

negligible effect of pressure, we may

write: = (l/lT)∆ ∆ P

It is seen that =β/2 = γ/3

11.4.2.4 Effects on strain:

For solid materials with a significant

length, like rods or cables, an

estimate of the amount of thermal

expansion can be described by the

materialstrain, given by and

defined as:

thermal =∆l/l where∆l is the change in length

before the change of length andl is

the original length. For most solids,

thermal expansion is proportional to

the change in temperature:

Thus, the change in either

thestrain or temperature can be

estimated by: thermal = T∆

Where,

is the difference of the temperature

between the two recorded strains,

measured indegrees

Celsius orKelvin, and is thelinear coefficient of thermal

expansion in "per degree Celsius" or

"per Kelvin", denoted by °C−1 or K−1,

respectively. In the field of continuum

mechanics, the thermal expansion

and its effects are treated aseigen

strain and eigen stress.

11.4.5 Consequences of thermal

expansion in solids:

1. Gaps are left when laying rail

tracks: This is to cope for thermal

expansion in summers.

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Thermal expansion of long

continuous sections of rail tracks is

the driving force forrail buckling.

This phenomenon resulted in 190train derailments during 1998 –2002

in the US alone.2. Overhead telephone cables are

kept sagging between two poles:

This is to cope for contraction of the

cables in winter.3. When you pour a hot liquid into

a thick glass tumbler, it cracks: When the hot water is poured into

the tumbler, some parts of the glass

warm up quickly (like the sides which

are thin.) These parts expand.

However, they are still attached to the

thicker parts of the glass (like the

bottom), which have not expanded.

This difference in expansion sets upstrong forces in the glass which can

break it.

11.5 Contraction effects/negative

thermal expansion / Anomalous

expansion

A number of materials contract onheating within certain temperature

ranges; this is usually callednegative

thermal expansion, rather than

"thermal contraction". For example,

the coefficient of thermal expansion

of water drops to zero as it is cooled

to 3.983 °C and then becomes

negative below this temperature; this

means that water has amaximumdensity at this

temperature, and this leads to bodies

of water maintaining this

temperature at their lower depths

during extended periods of sub-zero

weather. Also, fairly pure silicon has

a negative coefficient of thermal

expansion for temperatures betweenabout 18 and 120. Anomalous expansion of water takes

place because when water is heated

to 277K hydrogen bonds are formed.

Though ice is supposed to expand

when it is converted into water, this

gradual formation of hydrogen bonds

causes it to contract, i.e. thecontraction caused due to the

formation of hydrogen bonds is

greater than the actual expansion of

ice. At 277K water has the maximum

density because all the hydrogen

bonds are formed by 277K beyond

which water obeys the kinetic theoryof molecules, an increase in volume

when heated and the reverse when

cooled. The same thing happens in

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the reverse when water is cooled

beyond 277K.

Graph showing variation of volume

with temperature as water is heated

from 00C

Graph showing variation of density with

temperature as water is heated from 00C

11.5.1Consequence of anomalous

expansion of water:1.Survival of aquatic life under

extreme cold conditions: The peculiar expansion of water has

an important bearing on the

preservation of aquatic life during

very cold weather. As the

temperature of a pond or lake falls,

the water contracts, becomes denserand sinks. A circulation is thus set up until all

the water reaches its maximum

density at 4° C. In due course, ice

forms on the top of the water, and

after this the lower layers of water at

4° C can lose heat only by

conduction.Only very shallow water is thus liable

to freeze solid. In deeper water there

will always be water beneath the ice

in which fish and other creatures can

live.

Survival of aquatic life under sub zero

conditions

2. Unlogged water pipes burst

under severe cold conditions:

This is due to the fact that ice

expands on freezing. Hence the

volume of the ice exceeds the volume

of the pipe and the pipe may burst.

To overcome this pipes are logged i.e.

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covered with a heat insulator like

asbestos, glass wool of glass fibre.3. Under severe frost conditions

people in cold countries may sufferof frost bites:

The water in the veins freezes and

expands. Thus the veins burst and

the person loses sensation of limbs.

This is called frost bites.

11.6 Thermal expansion of liquids:

Liquids do not have a definite shape.

They take the shape of the container.

Thus, we can specify a liquid by its

volume. Hence, we can speak of

volume expansion only for liquids.

Expansion of liquids is much greater

than that of solids. A liquid is heated in a container. Heat

flows through the container to the

liquid. This means that the container

expands first, due to which the level

of the liquid falls. When the liquid

gets heated, it expands more and

beyond its original level. We cannot

observe the intermediate state. We

can only observe the initial and the

final levels. This observed expansion

of the liquid is known as the

apparent expansion of the liquid.If we consider the expansion of the

container also and measure the total

expansion in volume of the liquid,

then the expansion is termed as the

absolute expansion of the liquid.

Expansion of container with liquid

In the figure L1 represents the

original level at the level of the

liquid first falls to L2, because the

container gets heat first. When the liquid gets heated, it

expands much more than the

container and its level rises to L3. We can only observe the increase in

level from L1 to L3. Intermediate level

L2 goes unnoticed. The expansion we

measure is the apparent expansion of

the liquid. The corresponding

coefficient is coefficient of apparent

expansion. The coefficient of apparent expansion

is defined as the ratio of apparent

increase in volume of the liquid to its

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original volume for every degree rise

in temperature.Coefficient of apparent expansion

If we evaluate the increase in volume

of the liquid taking into account the

expansion of the vessel also, then we

say it is absolute expansion of the

liquid. We can show that,

The coefficient of absolute expansionof a liquid = coefficient of apparent

expansion + coefficient of cubical

expansion of the material of the

container.

11.7 Specific Heat Capacity:

i. The quantity of heat given to the body is directly proportional to the

mass ‘m’ of the body and the rise in

temperature ‘T’ of the body∆

Hence Q m T∆

or Q = mcT∆

where c is the constant of

proportionality called the specific

heat capacity of the substance. Therefore c = Q/(m T)∆

If m=1 and T = 1∆

we have c = Q Thespecific heat (also called specific

heat capacity) is the amount

ofheat required to change the

temperature of unit mass (or unit

quantity, such as mole) of a

substance by one degree. Therefore, unlike theextensive

variable heat capacity, which

depends on the quantity of material,

specific heat is anintensive

variable and has units of energy per

mass per degree (or energy per

number of moles per degree).

The S.I unit of specific heat capacity

is Jkg-1K-1. The other unit is calg-10C-1.1 Jkg-1K-1= 4186 calg-10C-1

Substance ‘c’ in Jkg-1K-1

Water 4186

Wood 1700Sand 830Ice 2093Methyl Alcohol 2530Ethyl Alcohol 2440Copper 385

11.8Calorimetry:

Calorimetry is the science associated

with determining the changes inenergy of a system by measuring the

heat exchanged with the

surroundings.

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Any process that results in heat

being generated and exchanged with

the environment is a candidate for a

calorimetric study.Hence it is not surprising to discover

that calorimetry has a very broad

range of applicability, with examples

ranging from drug design in the

pharmaceutical industry, to quality

control of process streams in the

chemical industry, and the study of

metabolic rates in biological (peopleincluded) systems. Indeed if the full

range of applications were to be

mentioned, the allocated disk space

on this site would soon be used up.

Acalorimeter is a device used to

measure heat of reaction. The principle of calorimetry states

that neglecting heat losses due to

radiation, the heat energy lost by ahot body is equal to the heat energy

gained by a cold body.

11.9Latent Heat:

Latent heat is the energy released or

absorbed by a body or

athermodynamic system during a

constant-temperature process. Atypical example is a change ofstate of

matter, meaning aphase

transition such as the melting of ice

or the boiling of water.Obviously latent heat would be of two

kinds:• Latent heat of fusion• Latent heat of vaporizationSpecific latent heat of fusion is the

heat energy required by unit mass of

a solid to completely change its state

from solid to liquid at its melting

point. The specific latent heat of

vaporization is heat energy required

by unit mass of a liquid to completely

change its state from liquid to gas at

its boiling point. The S.I unit of either is Jkg-1.

Table showing latent heat of some

substances

Substance Latent heat

of fusion

(J/kg)

Latent heat vaporisation

(J/kg)

Ethyl alcohol 108000 855000

Water 334000 2268000284

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Methylalcohol98800 1100000

Silver 88000 2390000

Copper 207000 4730000

11.10Heat transfer:

Heat can be transferred through a

medium in three different ways:

11.10.1 Conduction

The transfer of energy between

objects that are in physicalcontact. Thermal conductivity is the

property of a material to conduct

heat.Heat can be conducted between two

bodies which are in contact with each

other; heat "flows" from one to the

other.

• Materials which conduct heat well are calledconductors of heat.

Electrical conductors (such as

metals) are good conductors of

heat.• Materials which do not conduct

heat well are calledinsulators.

Electrical insulators (for example,

wood or glass) are usually goodinsulators of heat. Materials with

low density, such as air or foamed

plastic, are normally also good

insulators unless they happen to

be electrical conductors. To

prevent heat from moving from

one place to another, we usually

place an insulator between.Once a good insulator becomes hot,

however, it stays that way for a long

time, because it is difficult for the

material to lose heat by conduction.

Think of a hot ceramic pan and a hot

metal pan: which cools faster?

11.10.2 Convection: The transfer of energy between an

object and its environment, due to

fluid motion. This is a different kind of heat

transfer than conduction. In

conduction, heat itself is moving; in

convection, hot portions of a fluid

move through the body of the fluid. The hot fluidmixes with the cold

fluid, and heat is transferred more

quickly than by conduction. What we commonly call a "rolling

boil" results from convection. Hot

fluids rise through surrounding,

cooler fluid because they are less

dense; cooler fluids sink through warmer fluids because they are more

dense. This causes circular motion of the

fluid away from a source of heat.

Convection in water drives ocean285

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currents; convection in air drives

weather patterns; and convection of

molten rock inside the earth is

thought to drive plate tectonics.

Convection leads to the

counterintuitive fact that good

insulators (like air) can transfer heat

efficiently -- as long as the air is

allowed to move freely. Trapped air, as between panes of a

double window, cannot transfer heat

well because it cannot mix with air of

a different temperature.

11.10.3 Radiation

Radiation is the simplest means of

heat transfer. Heat radiation is

carried not by moving atoms (as inconduction or convection) but by

electromagnetic waves. Radiation is

the only way that heat can move

through a vacuum, and is the reason

that even a closed thermos bottle

(which has a vacuum between the

inner and outer parts) will eventually

come to the same temperature as itssurroundings.Heat transfer is most efficient by

convection, then by conduction;

radiation is the least efficient and

slowest means of heat transfer. Low

efficiency of heat transfer means that

vacuums make excellent insulation.

11.11 Black body radiation:

Black-body radiation is the type

ofelectromagnetic radiation within or

surrounding a body

inthermodynamic equilibrium with

its environment, or emitted by

a black body (an opaque and non-

reflective body) held at constant,

uniform temperature.

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The radiation has a specific spectrum

and intensity that depends only on

the temperature of the body.

The thermal radiation spontaneouslyemitted by many ordinary objects can

be approximated as blackbody

radiation.

A perfectly insulated enclosure that

is in thermal equilibrium internally

contains black-body radiation and

will emit it through a hole made in its

wall, provided the hole is small

enough to have negligible effect upon

the equilibrium.

A black-body at room temperature

appears black, as most of the energy

it radiates isinfra-red and cannot be

perceived by the human eye. Because

the human eye cannot perceivecolour at very low light intensities, a

black body, viewed in the dark at the

lowest just faintly visible

temperature, subjectively appears

grey, even though its objective

physical spectrum peaks in the

infrared range.

When it becomes a little hotter, it

appears dull red. As its temperature

increases further it eventually

becomes blindingly brilliant blue-

white. Although planets and stars are

neither in thermal equilibrium withtheir surroundings nor perfect black

bodies, black-body radiation is used

as a first approximation for the

energy they emit.

Black holes are near-perfect black

bodies, in the sense that they absorb

all the radiation that falls on them. It

has been proposed that they emit

black-body radiation (calledHawking

radiation), with a temperature that

depends on the mass of the black

hole.

However, the zero principle of

thermodynamics requires thermal

equilibrium of any two subsystems toimply the equality of temperatures,

not of their masses.

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As the temperature decreases, the

peak of the black-body radiation

curve moves to lower intensities and

longer wavelengths. The black-bodyradiation graph is also compared

with the classical model of Rayleigh

and Jeans.

11.11.1 Wien's displacement law:

Wien's displacement law shows how

the spectrum of black-body radiationat any temperature is related to the

spectrum at any other temperature. If

we know the shape of the spectrum

at one temperature, we can calculate

the shape at any other temperature.

Spectral intensity can be expressed

as a function of wavelength or of

frequency. A consequence of Wien's

displacement law is that the

wavelength at which the intensity per

unit wavelength of the radiation

produced by a black body is at a

maximum, , is a function only of

the temperature

where the constant ‘b’ known as

Wien's displacement constant, is

equal to 2.8977721×10−3 K-m.

11.11.2Stefan–Boltzmann law

TheStefan–Boltzmann law states that

the power emitted per unit area of

the surface of a black body is directly

proportional to the fourth power of its

absolute temperature:

where ‘ j’is the total power

radiated per unit areaT istheabsolute temperature and

σ = 5.67×10−8 W m−2 K−4 is theStefan–

Boltzmann constant.

11.12 Greenhouse Effect Green

The greenhouse effect is a process by

which thermal radiation from a

planetary surface is absorbed byatmosphericgreenhouse gases, and

is re-radiated in all directions.Since part of this re-radiation is back

towards the surface and the lower

atmosphere, it results in an elevation

of the average surface temperature

above what it would be in the

absence of the gases.Solar radiation at the frequencies

of visible light largely passes through

the atmosphere to warm the

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ofinfrared thermal radiation.

Infrared radiation is absorbed by

greenhouse gases, which in turn re-

radiate much of the energy to thesurface and lower atmosphere. The mechanism is named after the

effect of solar radiation passing

through glass and warming

agreenhouse, but the way it retains

heat is fundamentally different as a

greenhouse works by reducing

airflow, isolating the warm air insidethe structure so that heat is not lost

byconvection.If an ideal thermally

conductive blackbody were at the

same distance from the Sun as the

Earth is it would have a temperature

of about 5.3 °C. However, since the

Earth reflects about 30% of theincoming sunlight, this idealized

planet'seffective temperature (the

temperature of a blackbody that

would emit the same amount of

radiation) would be about −18 °C. The surface temperature of this

hypothetical planet is 33 °C below

Earth's actual surface temperature ofapproximately 14 °C. The mechanism that produces this

difference between the actual surface

temperature and the effective

temperature is due to the atmosphere

and is known as the greenhouse

effect.

Earth’s natural greenhouse effectmakes life as we know it possible.

However, human activities, primarily

the burning of fossil fuels and

clearing of forests, have intensified

the natural greenhouse effect,

causingglobal warming.

11.12.1 Greenhouse gases

By their percentage contribution to

the greenhouse effect on Earth the

four major gases are:• water vapour, 36–70%

• carbon dioxide, 9–26%•

methane, 4–9%• ozone, 3–7%

The major non-gas contributor to the

Earth's greenhouse effect,clouds,

also absorb and emit infrared

radiation and thus have an effect on

radiative properties of the

atmosphere

Practice exercise

1.Two scientists are doing an

experiment designed to identify

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the boiling point of an unknown

liquid. One scientist gets a result

of 120°C; the other gets a result of

250°F. Which temperature ishigher and by how much?

2.Dry ice, carbon dioxide, sublimes

(i.e., goes from a solid to a gas

without melting) at -78.5°C.

Express this temperature in K.3.A steel rod has a length of exactly

20 cm at 30C. How much longer

is it at50

0

C? [Use αSteel = 11 ×10−6/0C]4.By how much does the volume of

an Al cube 5.00 cm on an edge

increase when the cube is heated

from 10.00C to 60.00C? [Use α Al =

23 × 10−6 0C-1.]5. A certain substance has a mass

per mole of 50 g/mole. When 314

J of heat is added to a 30.0 g

sample of this material, its

temperature rises from 25.0C

to45.0C. (a) What is the specific

heat of this substance? (b) How

many moles of the substance are

present? (c) What is the molar

specific heat of the substance?6.It takes 487.5 J to heat 25 grams

of copper from 25 °C to 75 °C.

What is the specific heat capacity

of copper in Joules/g·°C?

7.How much heat does a refrigerator

need to remove from 1.5 kg of

water at 20.0 °C to make ice at

0°C?


1. Tatiana is researching vacation

destinations, and she sees that

the average summer temperature

in Barcelona, Spain is around

26°C. What is the averagetemperature in degrees

Fahrenheit? A) 79°F

B) -3°FC) 45°F

D) 58°F

2. The apparatus shown below is

used to measure the specific heatcapacity of a liquid. Which of the

following can be done to improve

the accuracy of the experiment?(1) Take the final temperature of

the liquid immediately after the

power supply is switched off. (2) Cover the cup with a lid.

(3) Totally immerse the heatingelement of the heater into the

liquid. A) (1) onlyB) (1) and (3) only

C) (2) and (3) only

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D) (1), (2) and (3)3.In general, the land cools down

faster than the sea at night

because A). the rise in temperature of the

sea is higher than that of the

land in the daytime.B). water has a high specific heat

capacity.C). wind blows from the sea

towards the land.D). wind blows from the land

towards the sea.4. Which of the following statements

is/are correct?(1) No radiation is emitted from

hot gases because there is no

free electron in gases.(2) The density distribution of the

heated fluid determines

whether there is convection.(3) When conduction occurs,

atoms can only vibrate in their

original positions. A). (1) onlyB). (3) onlyC). (1) and (2) onlyD). (2) and (3) only

Answers to Practice Exercise

1.2500F, 20F2.194.5 K

3.4.4 x 10-3cm4.0.43cm3

5.a) 532Jkg-1K-1

b) 0.6molc) 26.6cal mol-1K-1

6. 0.39Jg-1 K-17. 1.25 x 105 J

Answers to MCQ

1. A 2. D 3. B

4.D

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12.1INTRODUCTION

We have studied thermal properties

of matter in the previous chapter. In

the present chapter we will studysome laws that govern thermal

energy.In winter when we rub our palms

together, we feel warmer because

work done in rubbing produces heat.

Conversely a heat engine uses the

heat of the steam in running the

train.Thermodynamics is the branch of

physics which deals with the

study of transformation of heat

into other forms of energy and

vice versa

It does not deal with the microscopic

constituents of the matter. It deals

with various macroscopic variableslike pressure, temperature etc. The

results of thermodynamics are useful

for other branches and fields of

physics and engineering like

mechanical, chemical, physical,

biomedical etc.

Some Important Terms(i) Thermodynamic system:-

An assembly of extremely large

numbers of particles having certain

value of pressure, volume and

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temperature is called thermodynamic

system.

Eg: gas enclosed in a container.

(ii)Surroundings: - Anything which effects the

thermodynamic system is called

surroundings.

(iii) Thermodynamic variable: -

Variable which determine the

thermodynamic behavior of a system

are called thermodynamic variable.

Eg: pressure, volume, temperature,

internal energy, entropy etc

(iv) Equation of State

The equation of state represents the

connection between the state

variables of a system. For example,

the equation of state of an

ideal/perfect gas is represented as

Where µ is number of moles of the gas

and R is gas constant for one mole of

gas.

12.2

THERMAL EQUILIBRIUM

We know that thermo dynamical

system is the name given to anassembly of an extremely large

number of particles (atoms or

molecules). The thermo dynamical

system may exist in the form of a

solid, a liquid or a gas or a

combination of two or more states. A thermo dynamical system is said to

be in thermal equilibrium whenmacroscopic variables (like pressure,

volume, temperature, mass,

composition etc.) that characterise

the system do not change with time.

For example, a gas is enclosed in a

rigid container and insulated

completely from its surroundings is

in a state of thermo dynamicalequilibrium.On a multiple systems case, when

two systems are in contact with each

other, they can be in contact by way

of a wall that can be perfectly

adiabatic all the way through a wall

that is perfectly diathermic.

Adiabatic Walls

A wall is called adiabatic if the wall

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(heat) between the systems. Materials

such as concrete, asbestos, and

styrofoam represent a good

approximation of adiabatic walls.

Diathermic Walls

It is a conducting wall that allows

flow of heat. In this case, thermal

equilibrium can be attained.

12.3

THERMODYNAMIC EQUILIBRIUM

A system is said to be in

thermodynamic equilibrium if the

macroscopic variables describing the

thermodynamic state of the system

do not change with the time.Consider a gas inside a closed rigid

container completely insulated fromthe surroundings. If the pressure,

volume, temperature, mass and

composition of the gas do not change

with time, then it is in a state of

thermodynamic equilibrium.

Conditions for Thermodynamic

Equilibrium

The system is said to be in

thermodynamic equilibrium if the

conditions for following three

equilibrium is satisfied:

(i)Mechanical equilibrium: When there are no unbalanced

forces within the system and

between the system and thesurrounding, the system is said to

be under mechanical equilibrium.(ii) Chemical equilibrium: The system is said to be in

chemical equilibrium when there

are no chemical reactions going on

within the system or there is no

transfer of matter from one part ofthe system to other due to

diffusion.(iii) Thermal equilibrium: When the temperature of the

system is uniform and not

changing throughout the system

and also in the surroundings, the

system is said to be thermal

equilibrium. Two systems are said

to be thermal equilibrium with

each other if their temperatures

are same.

12.4

ZEROTH LAW OF

THERMODYNAMICS

It defines the exchange of heat

as well as defines the

temperature. The law deals with

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the concepts used for designing

the thermometer. Temperature is basically associated

with the ability to distinguish hotfrom cold. When two bodies having

different temperatures come in

contact with each other, they attain a

common temperature and hence the

state of thermal equilibrium. According to the Zeroth law of

thermodynamics:

When a given object or body A is in

thermal equilibrium with a body or

object B, and separately with a body

or object C also, then B and C would

necessarily be in thermal equilibrium

with each other.

A simple experiment that illustrate

the scope of the Zero Law of

Thermodynamics

Figure 8.1

To understand the law, we have

shown two systems A and B

separated by an adiabatic wall which

does not allow any transfer of heat. The two systems A and B are in

contact with a third system c through

a dia-thermic wall which allows

transfer of heat .The macroscopic

variable of the system A and B will

change until both A and B come to

thermal equilibrium with C.

After this is achieved, suppose the

adiabatic wall between A and B is

replaced by a conducting dia-thermic

wall; and C is insulated from A and B

by an insulating adiabatic wall.

Experiments show that the states of

A and B change no further, i.e, they

are found to be in thermalequilibrium with each other.The physical quantity that determines

whether or not given system A is in

thermal equilibrium with another

system B is called temperature.

Obviously, if two systems are in

thermal equilibrium, their

temperature must be same, and if

they are not in thermal equilibrium,

their temperature s must be

different.

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Thus if a thermometer records one

particular value of temperature for

system A and system B , it will record

the same reading when the twosystems are brought in contact with

each other . The Zerothlaw of thermodynamics is

the basis for temperature

measurement of a body or object. A body at a lower temperature is

called a cold body and a body at a

higher temperature is called a hot body.

12.5 Heat Internal Energy And

Work

Heat

It is a form of energy. Heat is an

energy transfer due to thetemperature difference between a

system and its surroundings. The flow of heat stops when the

temperatures equalise; the two

bodies are then in thermal

equilibrium.

Internal Energy Work is the energy transfer brought

about by other means, such as

moving the piston of a cylinder

containing gas.

Internal energy of a system is the

sum of the kinetic and potential

energies of the molecular

constituents of the system.It includes the energy associated with

the random motion of molecules of

the system.Internal energy (as state variable)

depends on the given state of the

system, and not on the path taken to

reach the state.Internal energy and work are

equivalent.

Work

Work is the mode of energy transfer

that produces organized motion,

brought by means that does not

involve temperature difference, but it

is due to the work done by thesystem or work done on the system.

Sign conventions used:

(i)Heat absorbed by a system is

positive. Heat given out by a

system isnegative.

(ii) Work done by a system is

positive. Work done on a system is

negative.

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(iii) The increase in internal energy

of a system is positive. The

decrease in internal energy of a

system isnegative.

Zeroth Law of Thermodynamics

12.6

THERMODYNAMIC PROCESSES

A thermodynamic process is said to

take place when some changes occur

in the state of a thermodynamicsystem i.e , the thermodynamic

parameters of the system change

with time .

Following are some of the important

thermodynamic processes:

1.Isothermal process is that whichoccurs at a constant temperature.

2. Adiabatic process is that process

in which no heat leaves or enters

the thermodynamic system during

the change.

3.Isobaric process is that whichoccurs at constant pressure.

4.Isochoricprocess is that which

occurs at a constant volume.

Table 8.1 given below summaries

the four special thermodynamic

processes.

S.

no

Types of

processes

Feature

1 IsothermalTemperature

constant2 Adiabatic No heat flow

between the system

and

surroundings ,ie,Q=0∆

3 Isobaric Pressure constant

4 Isochoric Volume constant

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Quasi-Static Process

An idealised process in which at

every stage the system is in anequilibrium state.

Such a process is infinitely slow and

hence, named quasi-static (i.e.,

nearly static).

Changes in the system are so slow

that it remains in thermal and

mechanical equilibrium with its

surroundings.For example, to take a gas from the

state (P, T) via a quasi-state process,

we change the external pressure

/temperature by a very small amount

and allow the system to equalize its

pressure /temperature with the

surroundings .Continue the process

infinitesimally slowly till the finalstate (P’,T’) is attained .Obviously a quasi-state process is a

hypothetical construct. The process

must be infinitely slow, should not

involve large temperature differences

or accelerated motion of the piston of

the container.

12.7P-V INDICATOR DIAGRAM

P-V Indicator Diagram is just a graph

between pressure and volume of a

system undergoing an operation.

When a system undergoes anexpansion from state A (P1 V1) to a

state B (P2 V2) its indicator diagram is

shown as follows.

In case of compression system at

state A(P1 V1) goes to a state B(P2 V2)

its indicator diagram is as follows.

Intermediate states of system are

represented by points on the curve.

Importance of P-V diagram is that

the area under the P-V diagram is

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numerically equal to the work done

by a system or on the system.

12.8ISOTHERMAL OPERATION

(CHANGE)

A change in pressure and volume of

gas without any change in its

temperature is called an isothermal

change. In such a change, there is a

free exchange of heat between the gas

and its surroundings. The two essential conditions for a

perfect isothermal changes are:

1.The walls of the container must be

perfectly conducting, to allow free

exchange of heat between the gas

and its surroundings.2.The process of compression or

expansion should be slow so as toprovide time for the exchange of

heat.

For Example

1)Melting process is an isothermal

change, because temperature of a

substance remains constant

during melting.2)Similarly, boiling process is also

an isothermal operation.

Equation of an isothermal change

As in all isothermal operations

temperature is kept, constant,

pressure (P) and volume (V) are

related to each other by Boyle’s law,i.e,

-------------- (1)

Isothermal Curve

If we plot a graph between V and P

keeping temperature constant, we get

a curve called an isothermal curve or

an isotherm for the given mass of gasat a given temperature. In fig 8.3

isotherms for three differentT 1,T 2∧T 3, are shown. The curves

move away from the origin at higher

temperatures.

Figure 8.3

12.9

ADIABATIC OPERATION (CHANGE)

A change in pressure and volume of a

gas in which temperature also

changes is called an adiabatic

change. In such a change, no heat is

allowed to enter into or escape from

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the gas (i.e. , there is no exchange of

heat between the gas and its

surroundings).

The two essential conditions for a

perfect adiabatic change are:

1.The walls of the container must be

perfectly non-conducting in order

to prevent any exchange of heat

between the gas and its

surroundings and2.The process of compression or

expansion should be sudden so

that there is no time for the

exchange of heat.

A few examples of nearly perfect

adiabatic changes are:

1.Sudden compression or expansion

of gas in a container with perfectly

non conducting walls.2.Sudden outburst of the tube of a

bicycle tire3.Propagation of sound eaves in air

and other gas4.Expansion of hot gases in an

internal combustion engine5.Expansion of steam in the cylinder

of steam engine

Equations of adiabatic change

1.The adiabatic relation between P

and V for a perfect gas is

....(2) Where

. =speifi heatof t h e gas at onstant pressurespeifi heatof t h e gasat onstant volume

2.From standard gas equation ,

PV=RT or P= RT

%

Putting in (2), we get RT

% %

=K

T % ( −1) = & R = another constant.

….(3)

This is the adiabatic relation between

T and V for a perfect gas.

3.Again, from standard gas equation

P =

RT

%

Putting in (2) , we get

P R

T

p = K

((1−. )T .

= &

R = another constant

….(4)

This is the adiabatic relation betweenP and T for a perfect gas12.10

SLOPES OF ISOTHERMAL AND

ADIABATIC CURVES

303%

T % ( −1)

=

((1−. )T .

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We can show that the slope of an

adiabatic curve is . times the slope

of an isothermal curve. As . >1

therefore adiabatic curve at any point

is steeper than the isothermal curve

at any point as shown in figs 8.4(a)

and (b). Note that the isothermal

curves lie above the adiabatic curve

on case of expansion as shown in fig

8.4(a)

Figure :8.4 (a) Figure :8.4 (b)

However, in case of compression, theisothermal curve lies below the

adiabatic curve as shown in fig

8.4(b).

12.11

WORK DONE IN AN ISOTHERMAL

EXPANSION

Consider one gram mole of an idealgas enclosed in a cylinder with

perfectly conducting walls and fitted

with a perfectly frictionless and

conducting piston. Let (1,% 1,∧T / Be

the initial pressure, volume and

temperature of the gas. Let the gas

expand to a volume % 2, when

pressure reduces to (2 and

temperature remains constant at T.

at any instant during expansion, let

the pressure of the gas be P.If A is area of cross section of the

piston, then force exerted by the gas

on the piston is

F = P x A …. (5)If we assume that the pressure of the

gas during an infinitesimally small

outward displacement dx of the

piston remains constant, then small

amount of work done during

expansiondW = F .dx = P × A × dx

dW = P(dV)

where dV = A(dx) = small increase in volume of gas . Total work done by the gas in

expansion from initial volume % 1,

to final volume % 2, is

W = ∫% 1

% 2

( d%

From standard equation,

PV = RT or P= RT

%

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W= ∫% 1

% 2 RT %

d% = RT ∫% 1

% 21

% d%

W = RT [ loge % ] v2 v1

W= RT [loge % 2−loge % 1¿

= RT loge

% 2% 1 ….. (6)

.’.…. (7)

Now as (1% 1 = (2% 2

% 2

% 1 =

(1

(2

W =2.3026 RT log10

(1

(2 …. (8)

Amount of heat spent in isothermal

expansion (in calories) is

..... (8a)

This much heat must have been

gained by the gas from surroundings.

That is why temperature of the gas

remains constant at T.

12.12

WORK DONE IN ADIABATIC

EXPANSIONConsider one gram mole of an ideal

gas enclosed in a cylinder with

perfectly non conducting piston. Let (1 / % 1 /∧T 1 / be the initial pressure,

volume and temperature of the gas. If

A is area of cross section of the

piston, then force exerted by the gas

on the pistonF = P × A … (9) Where P is pressure of the gas at any

instant during expansionIf we assume that the pressure of the

gas during an infinitesimally small

outward displacement dx of the

piston remains constant, then small

amount of work during expansion, dW = F x dx = (P x A)dx dW= P dV …… (10)

Where dV = A(dx) = small increase in

volume of the gas . Total work done by the gas in

adiabatic expansion from volume V1to V2 is

W = ∫% 1

% 2

( d%

….. (11)

The equation of adiabatic changes is

P %

= K (a constant) ..… (12)

From(12), P = &

% = K %

−

Put in (11) , we get

W= ∫% 1

% 2

& % −

d% = K

[

% (1− )

1−

] W= & 1−

[% 2(1− )−% 1

(1− )] ….. (13)

Equation of adiabatic changes can be

put as

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(2% 2

= (1% 1

= K

W =1

1−. [ (2% 2

. % 2(1−. )− (1% 1

. % 1(1−. )]

W=1

1− [ (2% 2− (1% 1 ] …… (14)

If T 2 is final temperature of the gas

in adiabatic expansion, then from

standard gas equation, (1% 1 = R T 1 , (2% 2= R T 1

Putting these values in (14) we get

….. (15) Amount of heat is spend in adiabatic

expansion (in calories) would be

H =0 1 =

1

T 2−T ¿¿

R ¿¿

…..

(16) That is why temperature of gas would

fall (T 2 < T 1 ¿ in adiabatic

expansion. from equation (16) W

shall be positive.On the contrary, in adiabatic

compression, temperature will riseT 2 > T 1 . Therefore, W will be

negative.

12.13

WORK DONE IN AN

ISOTHERMAL /ADIABATIC

PROCES IN TERMS OF INDICATOR

DIAGRAM

Consider one gram of mole of an ideal

gas enclosed in a cylinder fitted witha perfectly frictionless piston. Let

(1 / % 1∧T be the initial pressure,

volume and temperature of the gas. If dV is an infinitesimally small

increase in volume of the gas ,

through which pressure exerted by

the as is assumed to remain constant

at P then the small amount of work

done by the gas. dW = P dV …. (17) Total work done by the gas in

expansion from volume % 1¿% 2 is

W = ∫% 1

% 2

(d% …. (18)

The entire operation of the expansionis represented by the indicator

diagram i.e. , P – V diagram in fig 8.5

Figure :8.5 The relative variation of volume V

and pressure is represented by the

curve AB, where A (% 1 / (1¿

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represents initial state of the system

and B (% 2 / (2¿ represents final state

of the system. At any time, the state of the system is

represented by point p on AB. Here

pq=P=pressure and Oq = v, the

volume of the gas. The points p and s

are so close to each other that the

shaded strip pqrs may be regarded asa rectangle of P. dV . Therefore, area

of the strip gives us amount of work

done by the gas in expansion from

volume Oq to volume Or.Proceeding in this way, we can

calculate total work done by the gas

in expansion from volume % 1¿% 2 as

W= ∫% 1

% 2

(d% = ∫ 2

3

area of t h e strip paper

W = area of ABCDA .......(19)

Thus work done by the gas in

isothermal /adiabatic expansion is

equal to area between the PV curveand volume axis.

12.14COMPARISON BETWEEN

ISOTHERMAL AND ADIABATIC

CHANGES

Table 8.2 below gives the main points

of distinction between isothermal andadiabatic changes.

12.15 WORK DONE IN ISOCHORIC

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ISOTHERMALCHANGES

ADIABATICCHANGES

1. Temperature (T)

remains constant

i.e. T=0∆

Heat content (Q)

remain constant

i.e. , Q=0∆

2. System is

thermally

conducting to the

surroundings.

System is

thermally

insulated from the

surroundings3. The changes occurslowly The changes occursuddenly

4. Internal energy (U)

remains constant

i.e. , U=0∆

Internal energy

Changes U ≠

constanti.e. U=0∆

5. Specific heat

becomes infinite

Specific heat

becomes 06. Equation of

isothermal changes

is PV = constant

Equation of

adiabatic changes

is P % =

constant7. Slope of isothermal

curve :d(d% = -

(P/V)

Slope of adiabatic

curve :d(d% = -

. (P/V)

8. Coefficient of

isothermalelasticity 'i= (

Coefficient of

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PROCESS

In an isochoric process, volume V is

constant. No work of done by the gas.

Heat absorbed by the gas goesentirely to change its internal energy

and temperature.

Figure 8.6

Since volume is kept constant.

dW = PdV = 0

W = 0

12.16

WORK DONE IN ISOBARIC

PROCESS

In an isobaric process, pressure P of

fixed. Work done by the gas is

W = P(% 2 - % 1 ) = µR(T 2−T 1¿ ….

(20) As temperature changes, so does the

internal energy of the gas. The heat absorbed increases partially

the internal energy is spent partially

in doing work.

Figure :8.7

12.17CYCLIC AND NON CYCLIC

PROCESSES

A cyclic process consists of a series of

changes which return the system back to its initial state. For example,

suppose a system consists of a gas at

a pressure P, volume V and

temperature T. Let the system be

subjected to a number of changes

including isothermal and adiabatic

expansions and compression s. In

final state, if the system has thesame pressure P, the same volume V

and the same temperature T, the

succession of changes involved are

said to form a cyclic process.On the contrary, in a non – cyclic

process, the series of changes

involved do not return the system

back to initial state.

Work Done During A Non Cyclic

Process

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Conventionally, the work done is

taken as positive, when the gas is

expanding, and it is taken as

negative, when the gas iscompressed. As the two points representing the

initial and final state of a system can

be joined by any number of different

curves, and the area enclosed by

each curve is different. Hence we

conclude that the work done in a

non-cyclic process depends upon thepath chosen or the series of changes

involved.

Work Done In A Cyclic Process

Let A (% 1 / (1 ¿ represent the initial

state of gas , fig 8.8 . Let it subjected

to a number of changes in volume

and pressure such that it acquiresthe final state B (% 2 / (2¿ , via the

path AEB. Therefore, work done by the gas in

going from A to B along AEB is0 1=area *'432*

Figure: 8.8

Let the system be now subjected to

another succession Of changes in

volume and pressure, which return

the system back to the initial state A(% 1 / (1¿ along the path BFA. The

total number of changes involves

obviously constitute a cyclic process. Therefore, work done on the gas in

going from B to A via the path BFA is0 2 = - area BFADCB

Net work done by the gas in the cyclic

process is :0 *- 4 - *=0 1+0 2

= area AEBCDA – area BFADCB= area AEBFAHence we conclude that work done

per cycle is numerically equal to the

area enclosed by the loop

representing the cycle.

12.18

FIRST LAW OF THERMODYNAMICS

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This law is simply the general law of

conservation of energy applied to any

system

We know that the internal energy Uof a system can change through two

modes of energy transfer: heat and

workLet dQ = a small amount os heat

supplied to the system by the

surroundingsdW = small amount of work done by

the system on the surroundingsdU= small change in internal energy

of the system According to general principle of

conservation of energy…. (21)

......(21)

i.e. Energy supplied (dQ) to the

system is spent partially inincreasing the internal energy of the

system (dU) and the rest in doing

work on the surroundings .Eqn (21) is known as the first law of

thermodynamics. As internal energy U of the system is

a state variable, the change in

internal energy of the system (dU)depends only on initial state and final

state of the system. It would not

depend upon the path taken by gas

to go from one state to the other.

However, the combination (dQ – dW),

which is equal to dU shall be

independent of the path taken by the

system to go from initial state to finalstate. The following points are worth noting

regarding the first law of

thermodynamics:

(i) This law , which is basically the law

of conservation of energy, applies to

every process in nature

(ii) The law applies equally to all thethree phases of matter i.e. solid,

liquid and gas .(iii) While applying the first law of

thermodynamics, care must be taken

that all the three quantities dQ, dW ,

dU are expressed either in joules or

in calorie .

(iv) Just as Zeroth law ofthermodynamics introduces the

concept of temperature, the first law

of thermodynamics introduces the

concept of internal energy.(v) From first law of thermodynamics, we

learn that it is impossible to get work

from any machine, without giving an

equivalent amount of energy to themachine.

(vi) The first law establishes the

essential equivalence between work

and heat , as according to the law ,

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internal energy (and hence

temperature) of a system can be

increased either by supplying heat to

it or by doing work in the system or both

12.19 APPLICATIONS OF THE

FIRST LAW OF THERMODYNAMICS

(a)Isothermal process: In a process,

temperature remains constant. If the

system is an ideal gas, whose

internal energy U depends only ontemperature the internal energy shall

remain constant i.e. dU = 0 . The first law of thermodynamics

implies :dQ = dU +dW = 0 + dW = dW i.e. ……………….(22)

(b)Adiabatic process : In an adiabaticprocess, no heat energy enters or

leaves the system as or t is well

insulated i.e, dQ = 0 The first law of thermodynamics

impliesdQ = dU + dW =0

or

When a gas expands adiabatically,

dW is positive. Therefore, dU must be

negative i.e. internal energy of the

system would decrease and the gas

will be cooled. The reverse is true.

(c)Isochoric process: When the volumeof a system (such as gas) is kept

constant ,dW =P(dV)=0. The first law of thermodynamics

impliesdQ = dU + dW

(d)Cyclic process: In such a process,the system is returned to its initial

state after any number of changes. In

that case, no intrinsic property of the

system would change i.e. dU = 0. The first law of thermodynamics

impliesdQ=dU + dW

1)Net work done during a cyclic process

must exactly equal the amount of

heat energy transferred(e)Melting process: When a solid melts

to a liquid , its internal energy

increases , which can be calculated

from first law of thermodynamics

Let m = mass of solid . L = latent heat of the solid Amount of heat absorbed during the

melting process, dQ = mL……………….(23)

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When a substance melts, the change

in its volume dV is very small and

therefore, it can be neglected

dW = P (dV) = P x 0 = 0 According to first law of

thermodynamicsdQ = dU + dWmL = dU + 0 or

………….(24)

Hence the internal energy during the

melting process by (mL) .

Note that the increase in internalenergy during melting process is at

constant temperature. Therefore, it

must be due to increase in internal

potential energy (5 (¿ . The internal

kinetic energy (5 & ) remains

constant.

(f)Boiling process: Using first law of

thermodynamics, we can calculate

increase in its internal energy of a

liquid in the process of boiling. In

boiling, a liquid changes into vapours

at constant temperature and

pressure. Suppose ,

m = mass of liquid at its boiling point% 1 = volume of liquid

( =constant pressure at which

boiling process occurs

% 2 = volume of the same liquid in

the vapour form under same

pressure L = latent heat of vaporizations of

the liquidHence, work done in expansions,

dQ = P (dV) = P (% 2−% 1 ¿ …. (25)

Heat absorbed by the liquid in the

boiling processdQ = mL …. (26) According to the first law of

thermodynamics,dQ = dU + dWdU = dQ – dW

…. (27)

Knowing m, L, P, % 1 and % 2 we

can calculate dU, i.e. increase in

internal energy of the liquid in the

boiling process . This is again due toincrease in internal potential energy.

(g)Relation between two Specific

heats of a gas (Mayer’s formula )

First law of thermodynamics can be

used for establishing between two

specific heats of gas, as detailed

below:Consider one gram of molecule of an

ideal gas enclosed in a cylinder fitted

with a piston, which is perfectly

frictionless. Let P, V, T be the initial

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pressure, volume and temperature of

the gas.Let the gas be heated at a constant

volume through a small range oftemperature dT Amount of heat energy supplied to

the gas

dQ= 3 % .1.dT …. (28)

Where 3 % is the molar specific heat

of the gas at constant volume. As volume remains constant, dV = 0

dW = P (dV) = 0 According to first law of

thermodynamicsdQ = dU + dW3 % .dT = dU + 0

dU = 3 % .dT …. (29)

Let the gas be now heated at

constant pressure through the same

range of temperature dT, when its volume increase by a small amount

dV. Amount of heat energy supplied to

the gas

d 67

= 3 ( . dT …. (30)

where 3 ( is molar specific amount

of the gas at constant pressuredW = P (dV) …. (31)

If d 5 7

is the currosponding

increase in internal energy of the gas,

then according to first law of

thermodynamics.dQ’ = dU’ + dW’

3 ( . dT = dU’ + PdV …. (32)

As rise in temperature of the gas in

the two cases is the same (=dT),

therefore, increase in its internal

energy (which depends only on

temperature in case of ideal gas)

must be same in two cases, i.e.,dU’ =dU

Using (29),dU’ = dU = 3 % dT …. (33)

Putting in (32) we get3 ( dT = 3 % dT + PdV

( 3 (−3 % ) dT = PdV ….. (34)

According to standard gas equationPV = RTDifferentiating both sides, we get

PdV = RdT(since P and R are constant)Putting in (34) , we get

( 3 (−3 % ) dT = RdT

…. (35)

which is the required relation.

12.20LIMITATIONS OF THE FIRST LAW

OF THERMODYNAMICS

The first law of thermodynamics

establishes equivalence between the

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and says that the two can be

converted into each other. Further,

4.18 joule of mechanical work is

required to produce one calorie ofheat and vice-versa. However, this

law has the following limitations:1.The first law does not indicate the

direction in which the change can

occur :For example:

(i) When two bodies at different

temperatures are put in thermalcontact either each other, heat flows

from the body at higher temperature

to the body at lower temperature. We

now know that the heat cannot flow

from the body au lower temperature

to the body at higher temperature,

although first law of thermodynamics

is not violated.(ii) When a moving car is stopped by

applying brakes, work done against

friction is converted into heat. When

the car cools down , it does not start

moving with the conversion of all its

heat energy into mechanical work(iii) When a bullet strikes a target, kinetic

energy of the bullet is converted intoheat energy. But heat energy

developed in the target cannot be

converted back to mechanical energy

of the bullet enabling it to fly back.

2.The first law gives no idea about

the extent of change :

Our observation and experience tell

that there appears to be norestriction on conversion of

mechanical work into heat. But there

are sever restrictions in the reverse

process i.e. conversion of heat energy

into mechanical energy. We know that heat is not converted

into mechanical energy by itself. An

external energy called the heatengine is required for the purpose.No heat engine convert all the heat

energy received from the source into

mechanical energy .The first law of

thermodynamics is silent about all

this.

3.The first law of thermodynamicsgives no information about the

source of heat i.e. whether it is a

hot or cold body.

Solved ExamplesExample 1

A sample of gas ( =1) is taken

through an adiabatic process in

which the volume is compressed from

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1600 cc to 400 cc. If the initial

pressure is 150 KPa, what are the

final pressure and how much work is

done on the gas in the process?Solution:

Here = 1.5% 1=1600 .

% 2 = 400 c.c.

(1=¿ 150 k Pa = 150 x 103 Pa

(2=8

In an adiabatic process, (2% 2

= (1% 1

Therefore,

% 1% 2¿¿

(2= (1¿

= 150

1600

400

¿¿¿

=1200 k Pa Work done on the gas in the

process

W = (1 % 1− (2 % 2

−1

=

(150 x 103 ) (1600 x 10

−6 )−(1200 x103)(400 x10−6)1.5−1

=240−480

0.5

= - 480 J

Example 2

A gram molecule of a gas at 127o

C

expands isothermally until its volumeis doubled. Find the amount of work

done and heat absorbed. Take R =

8.31 J mole−1

& −1

.

Solution:

Here , T = 127o

C = 127 + 273 = 400

K

% 2 = 2 % 1

W = 2.3026 RT log10

% 2% 1

= 2.3026 x 8.31 x 400 log102

= 2.3026 x 8.31 x 400 x 0.3010 1=2.3026 x 103 joule Amount of heat absorbed =

2.303 X 103

4.2 cal. = 548.5 cals.

Example 3

At 27°C, two moles of an ideal mono-

atomic gas occupy a volume V. The

gas is adiabatically expanded to a

volume 2V.

(i) Calculate the ratio of finalpressure to the initial pressure(ii) Calculate the final temperature(iii) Change in internal energy

(iv) Calculate the molar specific heat

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Solution:

Given : n=2 , T = 27 °C = 300 K

V1=V and V2=2V

Now PV y=constantP1 V1 y=P2 V2 y

P2/P2=(V1/V2) y=.55/3

Also

T1 V1 y-1=T2 V2 y-1

or T2=300/25/3=189K

Change in internal energy=nC v∆T

For monoatomic gas C v=3R/2

Substituting all the valuesChange in internal energy=-2764J

As in adiabatic process ∆Q=0,molar

specific heat capacity=0

Example 4

One mole of an ideal gas does 3000 J

of work on its surroundings as itexpands isothermally to a final

pressure of 1 atm and volume of 25

L.

Determine –

(a) the initial volume and

(b) the temperature of the gas.

Solution:

(a) For an isothermal process thetemperature is constant.

Therefore PV = nRT = constant.

PV = 101000 Pa*25*10-3m3= nRT

for an isothermal process.

W/(nRT) = 3000 J/(101000Pa*25*10-3m3) = 1.19

V2/V1= exp(1.19) = 3.28

V1= (25*10-3m3)/3.28 = 7.62*10-3m3=

7.62 liter

(b) For an ideal gas PV = nRT.

101000Pa*25*10-3m3= (8.31 J/K)T.

T = 303.85 K.

Example 5

Calculate the fall in temperature

when a gas initially at 720C is

expanded suddenly to eight times its

original volume. Given γ= 5/3.Solution:

Let V1 = x cc

V2 = 8x cc T1 =273+72 =345 K γ= 5/3. T2 = ?Using the relation

T 1% 1( −1)=T 2% 2

( −1)

Therefore T2 = 345x(1/8)2/3

Taking log of both sides, we get

Log T2 = log 345 – 2/3 log 8=2.5378 – 2/3(0.9031)= 2.5378 -0.6020 = 1.9358Or T2 =86.26 K Therefore the fall in temperature

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= 345 - 86.26 = 258.74 KExample 6

An ideal monoatomic gas goes

through the cyclic process A→B→C→ A as shown to the right.

The temperature of the gas at A is

600K. Calculate the work done on

the gas, the heat absorbed by the

gas, and the change in internal

energy for each process and for the

total cycle.

Solution:

From the ideal gas equation, PV =nRT, we can calculate that TB = 600K

and TC = 200K. Since this is a

monoatomic gas, then we have U =

3/2 nRT. We keep in mind that 1

liter = 10-3 m3. A→B:

Since the gas is expanding, then the work doneon the gas is negative and W AB = area under PV curve = (2x104

Pa)(4x10-3m3) = 80 J∆U AB = 3/2 nR∆ T AB = 0 (T A = TB)From the 1st law, Q AB =∆U AB + W AB = 0

+ 80 J = 80 J (heat isabsorbed)B→C:Since the gas is being compressed,

the work doneon the gas is negativeandsince the area under PV curve =

(1x104Pa)(4x10-3m3) = 40 J WBC= -40J∆UBC = 3/2 nR∆ TBC = 3/2 nR(TC – TB)

= 3/2 PC VC – 3/2 PB VB=3/2 (1x104 Pa)(2 – 6 )x10-3 m3 = -60

JQBC =∆UBC + WBC = -60 J+ (– 40) J = -100 JC→ A: This is an iso-volumeric process, so

WCA = 0.∆UCA = 3/2 nR∆ TCA = 3/2 nR(T A – TC)= 3/2 P A V A – 3/2 PC VC

=3/2 (3x104

Pa – 1x104

Pa)(2x10-3

m3

)= 60 JQCA =∆UCA + WCA = 60 JSummary results:∆U = QNET - WNET

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Note that the total work done in the

cycle is the enclosed area. The

negative value means that the gas

does work on its environment duringthe cycle. Since the gas returns to its

original state, the net change in

internal energy is zero. The net heat absorbed is equal to the

net work doneby the gas. Heat is

absorbed during the processes A→B

and C→ A and rejected during the

process B→C.

Example 7

A thermodynamic system undergoes

a process in which its internal energy

decreases by 500 J. If at the same

time 220 J of work is done on the

system, find the thermal energy

transferred to or from it.Solution:

Increase in internal energy of a

system

= heat put into the system - work

done by the system on its

surroundings,

or

∆U = ∆Q - ∆W.

-500J = ∆Q + 220 J

∆Q = -720 J

Example 8

In a certain process, 600 J of work is

done on the system which gives off

250 J of heat. What is the change in

internal energy for the process?Solution:-

The absorption of heat by the system

tends to raise the energy of the

system. The performance of work by

the system, on the other hand, tends

to lower the energy of the system

because performance of workrequires expenditure of energy. Therefore the change in internal

energy ∆U, of a system is equal to the

head added to the system minus the

work done by the system:∆U = UB – UA = q + w

Here q = – 250 J and W = 600 J Therefore, change in internal energy

∆U = – 250 J + 600 J= 350 JHence, the change in internal energy

for the system is equal to 350 J

Example 9

A sample of ideal gas ( = 1.4) is

heated at constant pressure .If an

amount 140 J of heat is supplied to

the gas , find (a) change in internal

energy of the gas (b) work done by

the gas .Solution:-

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Let the sample of gas contains ‘n’

moles At constant pressure, let the state of

gas change from (% 1 /T 1¿ to (% 2 /T 2¿ Heat supplied dQ = n 3 ( (

T 2−T 1 ¿

= 140 JChange in internal energy of gas

dU =n 3 % (T 2−T 1 ¿

=

n3 %

3 (3

( (T 2−T 1¿

=3 %

3 ( dQ =140

1.4

= 100 J Work done by the gas

dW = dQ – dU= 140 –100 = 40 J

Example 10

At 0o C, and normal atmospheric

pressure, the volume of 1 gram of

water increase from 1 c.c to 1.091 c.c

on freezing. What will be the change

in its internal energy? Normal

atmospheric pressure of 1.013 x 10-5

N/m2 and latent heat of melting of ice

= 80 cal/gramSolution:-

Here , m = 1gram . dV =1.091 -1.000 = 0.091 c.c

= 0.091 x 10 -6 m 3 P = 1.013 x 105 N/ m 2

L = 80 cal/g ; dU = ?Heat released by 1 g of ice on freezing dQ = mL = 1 x 80 = 80 cal. As this heat is given out by water, it

is taken as negative i.e.dQ = -80 cals.

External work done by water, dW = P(dV)= 1.013 x 10 5 x 0.091 x 10-6 J

= 0.0092 joule

=0.0092

4.2 = 0.0022 cal .

As dU = dQ – dW = -80 – 0.0022= -80.0022 cal.Negative sign indicates that internal

energy of water decreases on freezing

Example 11

Calculate the final volume of one

mole of an ideal gas initially at 0oC

and 1 atm pressure. If it absorbs

2000 cal of heat during reversible

isothermal expansion.

Solution:-

The gas is in the standard

temperature and pressure condition

i.e. at S.T.PHence V1 = 22.4 dm3 and V2 have to

be calculated. As given expansion is isothermal and

reversible therefore, U = 0∆

We know that U = q + w∆

But U = 0∆

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Hence q = – w = 2000 cal

= 2000 X 4.184 J = 8368 J As work done in reversible isothermal

expansion is given by: W= -nRT ln (V2/ V1)

ThereforenRTln (V2/ V1) = – w = 8368 J

(I mol) X (8.314 J K- mol-) X (273 K) ln

(V2/ 22.4 dm3) = 8368 J

V2 = 242.50 dm3

Hence the final volume of one mole of

an ideal gas initially at 0oC and 1 atm

pressure is equal to 242.50 dm3

12.21 HEAT ENGINE

Any device which convents heat

continuously into mechanical work is

called a heat engine.

For any heat engine there are three

essential requirements.

1.SOURCE: A hot body at fixed

temperature T1 from which heat

engine can draw heat.2.SINK: A cold body, at a fixed lower

temperature T2, to which any

amount of heat can be3.WORKING SUBSTANCE: The

material, which on being supplied

with heat will do mechanical work.

In heat engine, working substances,

could be gas in cylinder with a

moving piston.

In heat engine working substancetakes heat from the source, convents

a part of it into mechanical work,

gives out rest to the sink and returns

to the initial state. This series of

operations constitutes a cycle. This cycle is represented in fig below

Figure 8.9

Work from heat engine can becontinuously obtained by performing

same cycle again and again.Consider,

Q1 - heat absorbed by working

substance from source

Q2 - heat rejected to the since

W - net amount of work done by

working substance

Q1-Q2 - net amount of heat

absorbed by working substance.

∆U = 0 since in the cycle Working

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According to first law of

thermodynamicsdQ = dU + dW

Therefore, dQ = dw Q1-Q2 = W Thermal efficiency of heat engineIt is denoted by . Thus

= 9et :or" done !le(0 )

total amount if heat a;sor;ed !le ….

(36)

=61−62

61

= 1 -62

61

As some heat is always rejected to the

sink , . 62<0 , therefore efficiency is

always less than 1 i.e. thermal

efficiency of heat engine is always

less than 100 % .

For 62 = 0 efficiency = 1 =100% i.e.

the engine will have 100% efficiency

in converting heat into work . The first law of thermodynamics does

not rule out a heat engine with 100%

efficiency. However practical

experience shows that efficiency is

less than 100% even, if we eliminate

all sorts of losses associated withactual heat engines.

12.22PRINCIPLE OF A HEAT PUMP

or REFRI-GERATOR

A refrigerator as we all know is a

device used for cooling things. It is

also called a heat pump.

An ideal refrigerator can be regardedas Carnot’s ideal heat engine working

in the reverse direction. Thus in a refrigerator, the working

substance would absorb a certain

quantity heat 62 from the sink at

lower temperature. T 2 and reject a

larger amount of heat6

1 to the

source at higher temperature T 1

with the help of an external agency

supplying the necessary energy to

the system.In a refrigerator, the working

substance (usually in gaseous form)

goes through the following steps :

Figure :8.10

(a)Sudden expression of the gas from

high pressure to low pressure,

resulting in cooling of the gas

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converting it into vapour–liquid

mixture.(b)This mixture absorbs heat from the

region to be cooled and getsconverted into vapours

(c)The vapours get heated in account of

external work done in the system by

the supply of electric power to the

refrigerator.(d)The heated vapours then release the

heat to the surroundings, bringing

them to the initial state after

completing the cycle.

In the cycle of changes the working

substance returns to its initial state,

therefore, there is no change in its

internal energy. i.e. dU = 0

If 62 = amount of heat extracted per

cycle from the cold reservoir at lowertemperature T 2 .

W = work done /cycle on the system –

the refrigerant.61=¿ amount of heat released

/cycle to the source (surrounding air

at higher temperature T 1 )

ThenNet amount of heat absorbed,

dQ = 62−61

Work done on the system dW = -W

According to the first law of

thermodynamicsdQ = dU + dW62−61

= 0 –WOr

…. (37)

Coefficient of performance of a

refrigerator

Coefficient of performance of a

refrigerator (β ) is defined as the ratio

of quantity of heat removed per cycle

from the contents of refrigerator (62¿

to the energy spent per cycle (W) to

remove this heat .

i.e. ,

β =62

0

Or

β =

62/61

1−62

61

In a Carnot Cycle,

62

61 =

T 2T 1

Therefore, β =

T 2T 1

1−T 2T 1 …. (39)

Or

…. (40)

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If is efficiency of Carnot cycle, then

1 -T 2T 1=ƞ

Therefore 1 – = T 2T 1

Discussion

1.Coefficient of performance of a

refrigerator stands for efficiency of

refrigerator .thus higher the value of

β more efficient is the refrigerator.

2.Eqn

(40) shows that smaller is the value of (T 1−T 2¿ greater is the

value of β i.e. smaller is the difference

in temperature of atmosphere and

food stuff to be cooled, more efficient

will be the refrigerator.

As the refrigerator works T 2 goes on

decreasing due to formation of ice.T 1 is almost constant. Therefore β

decreases when the refrigerator is

defrosted,T 2 increase and hence β

increases.Thus defrosting is

necessary for better working of

refrigerator.

3.A refrigerator cannot work withoutsome external work done on the

system, i.e. W is never zero. Hence β

= 62 /W cannot be infinite, i.e.

coefficient of performance of a

refrigerator can never be infinite.

12.23SECOND LAW THERMODYNAMICS

Kelvin−Planck Statement

It is not possible to design a heat

engine which works in cyclic process

and whose only result is to take heat

from a body at a single temperature

and convert it completely into

mechanical work.Clausius Statement

It is impossible for a self-acting

machine, unaided by any external

agency, to transfer heat from a body

at lower temperature to another at

higher temperature.

12.24REVERSIBLE AND IR-

REVERSIBLE PROCESSES

• Reversible process is the one which

can be retraced in opposite order

by changing external conditions

slightly.• Those processes which cannot be

retraced in opposite order byreversing the controlling factors are

known as irreversible process.

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• It is a consequence of second law

that all the natural processes are

irreversible process.

• Conditions for reversibility of aprocess are(i) Process is performed quasi-

statically(ii) it is not accompanied by any

dissipative effects.• It is impossible to satisfy these two

conditions perfectly, thus

reversible process is purely an

ideal abstraction.

12.25CARNOT'S HEAT ENGINESadi Carnot devised an ideal cycle of

operation for a heat engine, in the

year 1824 .This is cycle came to be

known as Carnot cycle. The machine

used for realizing this ideal cycle of

operation is called an ideal heat

engine or Carnot heat engine.

Construction

The essential parts of an ideal heat

engine or Carnot engine are as shown

in fig 8.12

Figure 8.11

(I) Source of heat: The source is

maintained at a fixed higher

temperature T1, from which the working substance draws heat .

The source is supposed to

possess infinite thermal capacity

and as such any amount of heat

can be draw from it without

changing its temperature.(II) Sink of heat: The sink is

maintained at a fixed lowertemperature T 2 to which any

amount of heat can be rejected

by the working substance. It has

only infinite thermal capacity

and as such its temperature

remains constant at T 2 , even

when any amount of heat isgiven to it .

(III) Working substance: A perfect

gas acts as the working

substance. It is contained in a

cylinder with non – conducting

sided but having a perfect

conducting base. This cylinder is

fitted with perfectly non –conducting and frictionless

piston.

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Apart from these essential parts,

there is a perfectly insulating stand

or pad on which the cylinder can be

placed. It would isolate the workingsubstance completely from the

surroundings. Hence the gas can

undergo adiabatic changes.

Theory

The Carnot cycle consist of the

following four stages:

1.Isothermal expansion2.Adiabatic expansion3.Isothermal expansion4.Adiabatic expansion

The cycle is carried out with the help

of the Carnot engine as detailed

below:Consider one gram mole of an ideal

gas enclosed in the cylinder. Let% 1

, (1 , T 1 be the initial volume,

temperature of the gas . The initial

state of the gas is represented by the

point A on P-V diagram, fig 8.13. We

shall assume that all the four

processes are quasi-state and non-

dissipative the two conditions for

their reversibility. According to second law of

thermodynamics,no heat engine can

have 100% efficiency

Carnot’s heat engine is an idealized

heat engine that has maximum

possible efficiency consistent with

the second law.Cycle through which working

substance passed in Carnot’s engine

is known as Carnot’s Cycle.Carnot's engine works between two

temperatures

T1 - temperature of hot reservoir

T2 - temperature of cold reservoirIn a Complete Carnot's Cycle system

is taken from temperature T1 to

T2 and then back from temperature

T2 to T1. We have taken ideal gas as the

working substance of Carnot's

engine.Figure below is an indicator diagram

for Carnot's Cycle of an ideal gas

Figure8.

12

1. Isothermal expansion

In step b→c isothermal expansion of

gas taken place and thermodynamic

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variables of gas changes from (P1,

V1,T1) to (P2,V2,T1)If Q1 is the amount of heat absorbed

by working substance from thesource and W1 the work done by the

gas then

61=0 1=∫% 1

% 2

( d%

= R T 1 loge

% 2% 1 ……………….(41)

2. Adiabatic expansion

Step c→d is an adiabatic expansion

of gas from (P2, V2, T1) to (P3,V3,T2). Work done by gas in adiabatic

expansion is given by

0 2 = ∫% 2

% 3

(d% =

1−¿¿

R (% 2−% 2)¿

……………….(42)3. Isothermal compression

Step d→a is iso-thermal compression

of gas from (P3,V3,T2) to (P4,V4,T2).Heat Q2 would be released by the gas

to the at temperature T2 Work done on the gas by the

environment is

62=0 3=∫% 3

% 4

− (d% =- R T 2 loge

% 4% 3

= R T 2 loge

% 3% 4 ……………….

(43)

4. Adiabatic compression

Step a→b is adiabatic compression of

gas from (P4, V4, T2) to (P1, V1, T1)

Work done on the gas is0 4 = ∫

% 4

% 1

− (d%

=

1−¿− RT (T 2−T 1)

¿ ……………….

(44)

Work done by engine per cycle

Net work (W) done by the gas in acomplete cycle

W = 0 1+0 2−(0 3+0 4)

From (42) and (44)0 2=0 4 in magnitude

0 = 0 1−0 3

12.26EFFICIENCY OF CARNOT

ENGINE

It is defined as the ratio of net

mechanical work done per cycle y the

gas (W) from the source (61¿ i.e.

== :

61

Using (36)

==61−62

61 = 1 -

62

61 …

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Referring to fig 8.13, as A (% 1 / (1 )

and B (% 2 /(2 ) lie on the same

isothermal, Therefore, (1% 1= (2% 2 … (46)

As, B(% 2 / (2 ) and C(% 3 /(3 ) lie on

the same adiabatic , (2% 2 = (3% 3 … (47)

Again C(% 3 / (3¿∧ 2 (% 4 / (4) lie on the

same isothermal ,

Therefore (3% 3= (4 % 4 … (48)

Finally, D(% 4 / (4 ¿ and A(% 1 / (1 ¿ lie

on the same adiabatic (4 % 4

= (1 % 1

… (49)

Multiplying (52) ,(53),(54) and (55)( (1% 1 ) ( (2 % 2

) ( (3% 3 ) ( (4% 4)

(2% 2

¿¿ )( (3% 3

¿( (4% 4)

( (1% 1

¿

% 1% 2

% 3% 4

= % 2% 3

% 4 % 1

Or (% 2% 4 )−1

= (% 1% 3)−1

Or% 2% 4=% 1 % 3

i.e.,% 2% 1

=% 3% 4

so that log% 2% 1 =

log% 3% 4 … (50)

Dividing (47) and (45) we get

62

61

= R T 1 loge

% 3% 4

R T 1loge

% 2% 1

=T 2T 1

i.e. … (51)

==1−T 2T 1 ..... (52)

Discussion

1.Eqn (52) reveals that the efficiency

of Carnot heat engine depends inthe temperature of source T 1

and temperature of sink T 2 . 4 ut

the efficiency does not depend

upon the nature of the working

substance.2. As RHS of eqn (52) is less than 1 ,

therefore efficiency of Carnot heatengine is less than 100%

3.Forƞ = 1(i.e. 100 %) eitherT 1=$∨T 2=0 &

As source at infinite temperature

or sink at 0K are not attainable,

Carnot heat engine cannot have

100 % efficiency

4.If T 2=T 1 , then from (52) , = 0Ƞ

i.e. the Carnot engine shall not

work. Hence it is not possible to

convert heat energy into

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mechanical work unless source

and sink of heat ate at different

temperatures.

5.From eqn (51) we find that62

6 1

=T 2T 1

When T 2 ≠0

i.e. some heat must be rejected to

the sink. Hence efficiency of even

an ideal heat engine can never be

100 %

6.Note that Carnot engine is a

reversible engine. Therefore, each

step of Carnot cycle can be

reversed.

12.27CARNOT’S THEOREM

• Working between the two given

temperatures of the hot (T1) and cold(T2) reservoirs, efficiency of no engine

can be more than that of a Carnot

engine.• Efficiency of the Carnot engine is

independent of the nature of the

working substance• For a Carnot cycle

…….(53)It is independent of the nature of the

system.

Example 12

Carnot engine takes in a thousand

kilo calories of heat from a reservoir

at a 627o

C and exhausts it to a

sink at 627o

3 . How much work

does it perform? What is the

efficiency of the engine ?Solution:-

Here 61 = 1000Kcal =

106 cal

T 1=627o C = 627 + 273 = 900 K

T 2=627o

C = 300 K ; W = ?

*s 62

6 1

=T 2T 1

=300

900 x 106 cals

=

1

3

x

10

3

cal

Work done per cycle W = 61 -

62 =2

3 x 10

6

cal .

=2 x 4.2

3 x10

−6

J = 2.8 x 10-6

J

Also ƞ=1−T 2T 1 = 1 -

300900

= 66.667 %

Example 13

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A Carnot engine has its reservoir at

900°C and sink at 3°C.What is the

efficiency of the Carnot engine?

Solution:- Temperature of the reservoir,T1 = 900°C = 900 + 273 =1173 K. Temperature of the sink,T2 = 3°C = 3 + 273 = 276 KEfficiency of Carnot engine,

Therefore, the efficiency of the Carnot

engine is 76.5 %.

Example 14

A diatomic ideal gas is used as a

working substance in a Carnot

engine. The gas undergoes an

adiabatic expansion as a part of the

cycle. This increases its volume from

V to 16 V. What is the efficiency ofCarnot engine?Solution:-

The equation for an adiabatic process

is given as TV γ−1 = constantSo, T1L1 γ−1= T2L2 γ−1

Using the given information V1 = V

V2= 16 V We have T1 V γ−1 = T2(16 V) γ−1

Here, as gas is diatomic γ = 7/5So

or

T1 = 3.03 T2 The efficiency of the Carnot engine

will be

or

So, the efficiency will beη = 0.66 or 66%Example 15

In a typical cannot engine the

temperature of two heat reservoirs

are T1 = 1000° C and T2 = 250° C. If

the temperature T2 is decreased by

50° and T1 is increased by 100° C

then the new efficiency of the Carnot

engine will be?Solution:-

The efficiency of a Carnot engine is

given as

Here, T1 = 1000° C + 100° C = 1100°and T2 = 250° C – 50° C = 200° C

Example 16

Calculate the least amount of work

that must be done to freeze one gram

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of water at 0o

3 by means of

refrigerator. Temp. of surrounding is

27o

3 .How much heat is passed on

to the surroundings in this process ?Solution:-

Here, W = ? , m = 1 gT 2 = 0

o3 = 273 K

T 1 = 27o

3 = 300 K , 61 = ?

62 = mL = 1 x 80 = 80 cal = 80 x

4.2 JCoeff. of performance =

62

0 =

T 2T 1−T 2

W =80 x 4.2 (300−273 )

273

= 33. 22 joule =

33.22

4.2 = 7.91

calHeat passed on to surroundings,

61=62+0 = 80 + 7.91 = 87.91 cal.

Exercise

1 Mark questions

1.Which Thermo dynamical variable

is defined by the first law of

thermodynamics?

2.What is the amount of work donein the Cyclic process?

3.Out of the parameters-

temperature, pressure, work and

volume, which parameter does not

characterize the thermodynamics

state of matter?4.What is the nature of P-V diagram

for isobaric and isochoric process?5.On what factors does the

efficiency of Carnot engine

depends?

2 Marks questions

6.If hot air rises, why is it cooler at

the top of mountain than near the

sea level?

7.What happen to the internal

energy of a gas during (i)

isothermal expansion (ii) adiabatic

Expansion?8.Air pressure in a car increases

during driving. Explain Why?

9.The efficiency of a heat enginecannot be 100%. Explain why?

10.In an effort to cool a kitchen

during summer, the refrigerator

door is left open and the kitchen

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door and windows are closed. Will

it make the room cooler?

3 Marks questions.

11.State the first law of

thermodynamics and discussed

the application of this law to the

boiling process.12.What is thermodynamic

system? Prove that work done by

thermodynamic system is equal

to the area under P-V diagram.13.Prove that Cp - Cv = R , for an

ideal gas .

5 Marks questions

14. What is isothermal process?

State two essential conditions for

such a process to takes place.

Show analytically that the work by

one mole of an ideal gas during

volume expansion from V1 to V2 at

temperature T is given by : W = RT logeV2/V115. Define an adiabatic process.

State two essential conditions for

such a process to takes place.

Derive an expression for adiabatic

process to takes place.

MULTIPLE CHOICE QUESTIONS

1.The thermal state of a body is

defined bya)Heat

b)Temperaturec)Coldd)Specific heat

2.The heat energy required to raise

the temperature of 1 kg of

substance through 1°C is calleda)Heat b)Specific heatc)Latent heat

d)Temperature3.The specific heat of a solida)Is a constant independent of

temperature b)Varies linearly with

temperaturec)First increases with

temperature and then attains a

constant value

d)First decreases withtemperature and then attains a

constant value4.The specific heat of a gas hasa)Only one unique value b)Only two values CP and C Vc)Any value lying between 0 and

infinity depending on the

process.d)Is always zero5.The branch of physics which

deals with relation between heat

and mechanical energy is calleda)Heat

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b)Thermodynamicsc)Thermoelectricityd)Calorimetry

6.In an isothermal processa)Pressure remains constant b)Thermal energy remains

constantc)Volume remains constantd)Temperature remains constant

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13.1 Introduction

Thegas laws developed by Boyle,

Charles, and Gay-Lussac are based

upon empirical observations and

describe the behavior of a gas inmacroscopic terms, that is, in terms

of properties that a person can

directly observe and experience. An

alternative approach to

understanding the behavior of a gas

is to begin with the atomic theory,

which states that all substances are

composed of a large number of verysmall particles (molecules or atoms).

In principle, the observable

properties of gas (pressure, volume,

temperature) are the consequence of

the actions of the molecules making

up the gas.

13.2 BEHAVIOUR OF GASESOne of the most amazing things

about gases is that, despite wide

differences inchemical properties, all

the gases more or less obey thegas

laws. The gas laws deal with how

gases behave with respect to

pressure, volume, temperature, and

amount.Gases are the only state of matter

that can be compressed very tightly

or expanded to fill a very large space.

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13.2.1 Boyle's Law: The Pressure-

Volume LawBoyle's law or the pressure-volume

law states that the volume of a givenamount of gas held at constant

temperature varies inversely with the

applied pressure when the

temperature and mass are constant.

Another way to describing it is saying

that their products are constant.

P1V1 =P2V2 =P3V3 = constant …(eq.

1) The graph (fig. 13.1) below shows

that Boyle's law is strictly not obeyed

by gases at all values of P and T but

it obeys this law only at low pressure

and high temperature.

Fig. 13.1 Graph between P and V at

temperature T1 and T2 such that T1<T2

13.2.2 Charles' Law: The

Temperature-Volume Law

Charles' law or the temperature-

volume law the states that the

volume of a given amount of gas held

at constant pressure is directlyproportional to the Kelvin

temperature.V T

V1/T1 =V2/T2 =V3/T3 =constant (eq.

2) The graph (fig. 13.2) below shows

that experimental graph deviates

fromstraight line. Theoretical and

experimental graphs are in

agreement at high temperature.

Fig. 13.2 Graph between T and V

13.2.3 Gay-Lussac's Law: The

Pressure Temperature LawGay-Lussac's law or the pressure

temperature law states that the

pressure of a given amount of gas

held at constant volume is directly

proportional to the Kelvintemperature.

P TP1/T1 =P2/T2 =P3/T3 =constant (eq.

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13.2.4 Avogadro's Law: The

Volume Amount Law

It gives the relationship between

volume and amount when pressure

and temperature are held constant.

Remember amount is measured in

moles. Also, since volume is one of

the variables, that means the

container holding the gas is flexible

in some way and can expand orcontract.If the amount of gas in a container is

increased, the volume increases. If

the amount of gas in a container is

decreased, the volume decreases.V n This means that the volume-amount

fraction will always be the same value

if the pressure and temperature

remain constant.

V1/n1 =V2/n2 =V3/n3=constant (eq.

4)

13.2.5 The Ideal Gas Law

The previous laws all assume that

the gas being measured is anideal

gas, a gas that obeys them all

exactly. But over a wide range of

temperature, pressure, and volume,

real gases deviate slightly from ideal.

Since, according to Avogadro, the

same volumes of gas contain the

same number of moles, chemistscould now determine the formulas of

gaseous elements and their formula

masses. The ideal gas law is:

PV =nRT …(eq. 5) Wheren is the number of moles of

the number of moles andR is a

constant called theuniversal gas

constant and is equal to

approximately8.314 JK−1mol−1

The graph between is almost

a straight line parallel to pressure

axis as shown in the graph (fig.13.3)

below.

However, the behaviour of real gasesapproaches the behaviour of ideal gas

at low pressure and high

temperatures.

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Fig. 13.3 graph (T1>T2>T3)

13.2.6 Dalton's Law of Partial

Pressures

Dalton's Law of Partial

Pressures states that the total

pressure of a mixture of non-reacting

gases is the sum of their individual

partial pressures.Ptotal =Pa +Pb +Pc + ... …(eq. 6)

Or

Or

The pressure in a flask containing a

mixture of 1 mole of 0.20 mole

O2 and 0.80 mole N2 would be thesame as the same flask holding 1

mole of O2.Partial pressures are useful when

gases are collected by bubbling

through water (displacement). The

gas collected is saturated in water

vapour which contributes to the total

number of moles of gas in thecontainer.

13.2.7 Non-Ideal Gases

The ideal gas equation ( PV= nRT )

provides a valuable model of the

relations between volume, pressure,

temperature and number of particlesin a gas. As an ideal model it serves

as a reference for the behaviour of

real gases. The ideal gas equation

makes some simplifying assumptions

which are obviously not quite true.

Real molecules do have volume and

do attract each other. All gases

depart from ideal behaviour underconditions of low temperature (when

liquefaction begins) and high

pressure (molecules are more crowed

so the volume of the molecule

becomes important). Refinements to

the ideal gas equation can be made to

correct for these deviations.

In 1873 J. D. Van der Waalsproposed his equation, known as the

Van der Waals equation. As there are

attractive forces between molecules,

the pressure is lower than the ideal

value. To account for this the

pressure term is augmented by an

attractive force term a/V2. Likewise

real molecules have a volume. The volume of the molecules is

represented by the term b. The term

b is a function of a spherical

diameter d known as the van der

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Waals diameter. The van der Waals

equation forn moles of gas is:

…(eq. 7)

Example 13.1

Figure shows plot ofPV/T versusP

for 1.00×10 –3 kg of oxygen gas at two

different temperatures.

(a) What does the dotted plot signify?(b) Which is true:T1 >T2 orT1 <T2?

(c) What is the value ofPV/T wherethe curves meet on they-axis?(d) If we obtained similar plots for

1.00×10 –3 kg of hydrogen, would we

get the same value ofPV/Tat the

point where the curves meet on the

y-axis? If not, what mass of hydrogen

yields the same value ofPV/T(for low

pressure high temperature region ofthe plot) ? (Molecular mass of H2 =

2.02 u, of O2 = 32.0 u,R= 8.31 J

mo1 –1 K –1.)Solution:

(a)The dotted line in the given graph

indicates that the ideal behavior of

the gas, i.e., the ratio PV/T is

equal to nR (n is the number ofmoles and R is the universal gas

constant) which is a constant

quantity.(b)T1 > T2 is true for the given graph.

The dotted line in the given graph

represents an ideal gas. The curve

of the gas at temperature T1 is

closer to the dotted line than thecurve of the gas at temperature T2.

A real gas approaches the behavior

of an ideal gas when its

temperature increases.(c)The value of PV/T, where two

curves meet is nR.For an ideal gas PV = nRTPV/T = nRMolecular mass of oxygen =

32.0 gMass of oxygen = 1 x 10-3 kg

= 1 gR = 8.314 J mole-1 K-1

Thus, the value of PV/T =

0.26 J K-1

Example 13.2 Two thermally insulated vessels 1

and 2 are filled with air at

temperatures (T1and T2), volumes (V1and V2) and pressures (P1 and P2)

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two vessels is opened, what will be

the temperature inside the vessel at

equilibrium?

Solution: Total number of moles µ = µ1 + µ2

Or

Or

using Boyle’s law,

Example 13.3

A vessel contains two non reactive

gases : neon (monatomic) and oxygen

(diatomic). The ratio of their partial

pressures is 3:2. Estimate the ratio of

(i) number of molecules and (ii) massdensity of neon and oxygen in the

vessel. Atomic mass of Ne = 20.2 u,

molecular mass of O2 = 32.0 u.Solution:

Partial pressure of a gas in a mixture

is the pressure it would have for the

same volume and temperature if it

alone occupied the vessel. (The totalpressure of a mixture of non-reactive

gases is the sum of partial pressures

due to its constituent gases.) Each

gas (assumed ideal) obeys the gas

law. SinceVandTare common to the

two gases, we haveP1V= μ1 RTand

P2V= μ2 RT, i.e. (P1/P2) = (μ1 / μ2).

Here 1 and 2 refer to neon andoxygen respectively. Since (P1/P2) =

(3/2) (given), (μ1/ μ2) = 3/2.

13.3 KINETIC THEORY OF AN

IDEAL GAS

The kinetic theory of gases can

explain many everyday observations.

Have you ever wondered why water

boils faster at higher altitudes? Or

why inflatable pool toys seem flat

after sitting in a cold garage? How

about why you can smell a candle all

throughout the house? All of these

phenomena and many more can beexplained by the kinetic theory of

gases. The kinetic theory of gases (also

known as kinetic-molecular theory) is

a law that explains the behavior of a

hypothetical ideal gas. According to

this theory, gases are made up of tiny

particles in random, straight line

motion. They move rapidly and

continuously and make collisions

with each other and the walls. This

was the first theory to describe gas

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the walls of the container, rather

than from static forces that push the

molecules apart. Kinetic theory also

explains how the different sizes of theparticles in a gas can give them

different, individual speeds.

Postulates of Kinetic Theory

Kinetic theory of gases makes many

assumptions in order to explain the

reasons gases act the way they do.

According to kinetic theory of gases:

1.Gases consist of particles in

constant, random motion. They

continue in a straight line until they

collide with something—usually each

other or the walls of their container.

Fig. 13.4

2. Particles

are point masses with no volume.

The particles are so small comparedto the space between them, that we

do not consider their size in ideal

gases.

3. No molecular forces are at work.

This means that there is no

attraction or repulsion between the

particles.4. Gas pressure is due to the

molecules colliding with the walls of

the container. All of these collisions

are perfectly elastic, meaning that

there is no change in energy of either

the particles or the wall upon

collision. No energy is lost or gained

from collisions.5. The time it takes to collide is

negligible compared with the time

between collisions.6. The kinetic energy of a gas is a

measure of its Kelvin temperature.

Individual gas molecules have

different speeds, but the temperature

and kinetic energy of the gas refer to

the average of these speeds.7. The average kinetic energy of a gas

particle is directly proportional to the

temperature. An increase in

temperature increases the speed in

which the gas molecules move.8. All gases at a given temperature

have the same average kinetic energy.

9. Lighter gas molecules move fasterthan heavier molecules.

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An ideal gas does not exist - the

nearest we have to it is helium remote

from its liquefying point.

13.4Pressure of an Ideal Gas

Consider a fixed mass ‘M’ of gas

enclosed in a cubical container of

side ‘L’. Let each particle of the gas

have mass ‘m’ and let there be ‘N’

particles. Considering one of the

millions of particles in the gas is

moving towards the wall shaded ingrey, as shown in the figure 13.5.

Let, the x-component of velocity is u1towards the wall and the x-

component of momentum is mu1towards the wall. When the particle hits the wall it will

bounce back with the same speed

but in the opposite direction (because

kinetic energy is conserved in the

elastic collision). The x-component of

momentum after collision will then

be –mu1. The change in momentum∆p = mu1 - (-mu1 ) = 2m u1

The particle has to travel a distance

2L (from the grey wall to the opposite

face and back again) before it next

collides with the grey wall.Now, time = distance / velocity =

2L / u1so the rate of change of momentum

due to a collision with a wall will be: ∆p / ∆t = 2m u1 / (2L/ u1) = m u12 /

LBy Newton's second law, the rate of

change of momentum is equal to

force, so mu12/L is the force exerted

on the particle by the wall during the

collision. By Newton's third law, the

particle exerts an equal but

oppositely directed force on the wall,

and so we can say that: The force on the wall during the

collision

=m u12 / LNow it isn’t only one particle that hits

the wall – there are N particles so

their x-components of velocity are

going to beu1, u2,…. uN. The total force, F, will therefore be

F =mu12/L + mu22/L + mu32/L….. +

muN2/L

= m/ L ( u12+ u22+ u32... + uN2)

Now, pressure on that wall will be

force/contact area =F/A, but A = L2

Therefore pressure on the wall due to

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p = m/ L3( u12+ u2

2+ u32... + uN

2)

We can work out the mean square

velocity is the x direction from this:

( u12

+ u22

+ u32

... + uN2

) = Nū2

So now we know that the total

pressure due to the u-components of

the velocities of the particles will be:

p = m/ L3 Nū2

So far we have only looked at the x

component of the velocities. It is now

time to look at the other two

components (v- along y axis and w-along z axis ) and to add them into

the equation.

c2 = u2 + v2 + w 2

So, it follows that this is true for the

mean square velocities also.

<c2> = <u2 > + <v2 > + <w 2>

Since there are large numbers ofmolecules and they are moving

randomly the mean square

components are equal to each other

(random movement means there is no

preferred direction). We can therefore

say that

<c2> = 3 <u2> or 1/3 <c2> = <u2>

We can now replace our mean square velocity component with one related

to the mean square velocity itself.P = m N<c2> / 3 L3 and as L3= V

We get pV = 1/3 Nmc2rms…(eq. 8)

Now, density = mass / volume, M =

Nm is the total mass and L3 is the

volume,

Therefore P = 1/3 ρ<c2> …(eq. 9)

13.5 Kinetic Interpretation of

Temperature

To investigate the relationship

between molecular kinetic energy and

temperature, let us multiply both

side by V pV = 1/3 ρ V<c2>

as ρ = M/V

so pV = 1/3 M<c2> =1/3 Nm<c2>

=2/3 N( ½ m<c2>)

Now using the ideal gas equation

or …(eq. 10)

Both R and N A are universal

constants, and therefore so also is

R/NA; it is called Boltzmann'sconstant, ‘k’ and it is the gas

constant per particle. The left-hand

term is the average translational

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kinetic energy of a single molecule,

and therefore Average translational kinetic energy

of a particle of gas = 3/2 KT …(eq.11)

Example 13.4

Calculate the root mean square

speed, urms, in m/s of helium at 30oC.Solution:

M= (4.00g/mol)×1kg/1000gM= 4.00×10−3kg/mol

Now, using the equationfor urms substitute in the proper

values for each variable and perform

the calculation.urms = √(3RT/M)urms=1.37×103m/s

Example 13.5

An ideal gas A is there. Initial

temperature is 27°C. The

temperature of the gas is increased to

927°C. Find the ratio of final Vrms to

the initial Vrms Solution:

Vrms=√3RT/MSo it is proportional to Temperature

Now

T1=27°C=300K T2=927°C=1200K

So

Intial vrms = k√300

Final Vrms= k√1200

Ratio of final Vrms to intial Vrms = 2:1

13.6 Degrees of freedom

The degrees of freedom of a

dynamical system are defined as the

total number of independent

quantities or coordinates required to

describe completely the position and

configuration [arrangement of the

constituent atoms in space] of thesystem.

[A] System of one particle:Consider

a system of one particle [point mass]

moving along a straight line, say

along X-axis.its position at any

instant is completely known by its

displacement along X-axis. Hence, it

has one translational degree of

freedom.

If the particle moves in a plane, say

X-Y plane, then two coordinates

[x,y]are required to know its positionat any instant , which can be

determined by knowing the

displacement of the particle along X-

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axis and Y-axis. Hence, it has two

translational degrees of freedom.If the particle is moving in a space,

its position at any instant can bedetermined by knowing the

displacement of the particle along X-

axis, Y-axis and Z-axis .Hence such a

particle has three translational

degrees of freedom.

[B] System of two particles:Let us

now consider a system consisting oftwo particles [two point masses].Let

[x1,y1,z1] and [x2,y2,z2]be the

coordinates of the two particles

respectively .

Each particle has three degrees of

freedom [coordinates].Thus to

describe this system, six coordinates

are required. If two particles remain

at a fixed distance from each other,

there is a definite relation between

them. Hence only five independentcoordinates are required to describe

the configuration of the system.

Therefore, degrees of freedom of this

system is 6-1=5.

In general we can say that the

number of degrees of freedom of a

system is equal to the total number

of coordinates required to specify theposition of its constituent particles

minus the number of independent

relations among the particles.If N=number of particles in the

systemK=number of independent relations

among the particles Then the number of degrees of

freedom of the system isF = 3N – K …(eq. 12)

13.6.1 Calculation of Degrees of

freedom of a gas molecule

[A]Monatomic gas molecule: The

molecule of a monatomic gas

molecule like Helium, neon, Argon,

etc consists of a single atom. It iscapable of translator motion only and

hence has three degrees of freedom.

Here N = 1, K = 0 from eq.(12) f = 3x1 – 0 = 3

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[B]Diatomic gas molecule: The

molecule of a diatomic gas molecule

like Hydrogen, Oxygen, Nitrogen, etcconsists of two atoms separated by a

fixed distance. Considering the

translator motion of its centre of

mass, it has three translational

dergree of freedom. In addition to this

the the molecule can also rotate

about three axes x, y and z. (Figure

……)Rotation along the line joining the

atoms has very small moment of

inertia, thus the molecule has two

rotational degrees of freedom. Hence

the diatomic molecule in all has five

degrees of freedom.

Here N = 2, K = 1 from eq …(12) f

= 3x2 – 1 = 5

At high temperature ( 5000 K), a

diatomic molecule has two additional

degrees of freedom due to vibrational

motion. Each vibrational motion is

associated with kinetic and potential

energies, so one degree of freedom of

vibrational motion is taken as two.

[C] Triatomic gas molecule: They

are of two types linear and non

linear.

In a linear molecule like Carbon

dioxide, Hydrogen Cyanide, etc, the

three atoms are arranged along a

straight line such that they have twoindependent relations.Here N = 3, K = 2

from eq …(12) f = 3x3 – 2 = 7In a non linear molecule like water,

Sulphur dioxide, etc, the three atoms

are located at the three vertices of a

triangle such that that there are

three fixed distances amongst thethree atoms.Here N = 3, K = 2 (as the distances

between the atoms are fixed).

from eq …(12) f = 3x3 – 3 = 6

13.6.2 Degrees of freedom of a

solid

Atom in a solid cannot move from oneposition to another and cannot rotate

about the given axis of rotation.

Therefore, the atom in a solid does

not possess translational energy and

rotational energy. At high temperature, each atom can

only vibrate about its equilibrium

position so it has only Vibrationalenergy. Vibrational energy of an atom is the

sum of kinetic and potential energy.

Since it can vibrate along the three

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axis, therefore the atom in a solid has

six degrees of freedom at high

temperature.

13.6.3 Degrees of freedom of arigid body

A rigid body of finite size can undergo

translation as well as rotational

motion about any axis. Rotational

motion can also be resolved into

three mutually perpendicular

components. Hence, a rigid body will

have six degrees of freedom, three fortranslatory motion and three for

rotatory motion.

13.7 Law of equi-partition of

energy

Consider a molecule of a gas of mass

m, moving with velocity v. the kinetic

energy of translational of themolecule isEt = ½ mv2 = ½ mv x2 + ½ mv y2 + ½

mvz2 (... v2 = v x2 + v y2 + vz2)

…(eq. 13)In thermal equilibrium at

temperature T, the average value of

translational energy of the molecule

is< Et > = < ½ mv x2> + < ½ mv y2 > + <½

mvz2> …(eq. 13) As there is no preferred direction of

motion, so the average kinetic energy

of the gas molecule along all the

three directions is equal. Thus from equation (13)

<½ mv x2

>= <½ mv y2

> = <½ mvz2

>=½kT A gas molecule moving in space has

three degrees of freedom. So, the

energy associated with each gas

molecule per degree of freedom is ½

kT, which is the law of equipartition

of energy.

In case of diatomic gas molecule, in

addition to three translational

degrees of freedom, each molecule

has two rotational degrees of

freedom. Thus total energy of a

diatomic gas molecule is the sum of

transnational energy Et and

rotational energy Er, i.e.,Et + Er = (½ mv x2 + ½ mv y2 + ½ mvz2)

+ ( ½ I1 w12 + ½ I2 w22) …(eq.14)

here and are angular speeds

about the Y and Z axes and the

corresponding moment of inertia

about these axes are I1 and I2.If the two atoms of a diatomic

molecule vibrate along the inter-atomic axis like one-dimensional

oscillator and contribute Vibrational

energy E v. It involves kinetic energy of

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vibration as well as potential energy

of the pair of atoms.E v = ½ mv2 + ½ kr2

where m is the mass of the molecule,k is the force constant of the

oscillator, r is the separation between

the atoms and v = dr/dtHence the total energy of a molecule

isE = Et + Er + Ev =(½ mv x2 + ½ mv y2 + ½ mvz2) + ( ½

I1 w12 + ½ I2 w22) + ( ½ mv2 + ½ kr2 )

…(eq. 15)

An important feature in this equation

is that while each translational and

rotational degree of freedom has

contributed only one squared term in

equation(15), one vibrational mode

contributes two squared terms :

kinetic and potential energies. Eachquadratic term occurring in

equation---( 15 ), is a mode of

absorption of energy by molecule. We

have also proved that in thermal

equilibrium at absolute temperature

T, for each translational mode (i.e.,

degree of freedom) the average energy

is ½ kB T. The most importantprinciple of classical statistical

mechanics (first proved by Maxwell)

states that:

In thermal equilibrium at absolute

temperature T, the total energy is

equally distributed in all possible

energy modes, with each modehaving an average energy equal to ½

kB T. This is called the law of

equipartation of energy.Hence, according to this law, each

translational and rotational degree of

freedom of a molecule contributes ½

kB T to the energy, while each

vibrational frequency contributes 2 x ½ kB T = kB T, since a vibrational

mode has both kinetic and potential

energy modes.

Example 13.6

Calculate the total number of degrees

of freedom possessed by the

molecules in one cm3 of H2 at NTP.Solution:

22400 cm3 of every gas contains 6.02

X 1023 molecules.... Numbers of molecules in 1cm3 of

H2 gas = 6.02 X 1023 / 22400 =

0.26875 X 1020

Number of degrees of freedom of a H2

gas molecule = 5 ... Total number of degrees of

freedom of 0.26875 X 1020 molecules= 0.26875 X 1020 X 5

= 1.34375 X 1020

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13.8 Specific Heat Capacity of

gases

The specific heat of gases can be

predicted theoretically on the basis oflaw of equipartation of energy.

Consider one mole of an ideal gas at

temperature T whose each molecule

has f degrees of freedom. According

to the law of equipartation of energy,

energy associated with each molecule

per degree of freedom = ½ kB T.

Total internal energy of one mole ofthe gas,

U = f (½ kB T)N A = RT (as kB = ) Thus, molar specific heat capacity at

constant volume,

C v =dU/dt = d( RT)/dt = R …(eq.

16)Now, the molar specific heat capacity

at constant pressure,

Cp = C v+ R = R + r …(eq. 17)(since for an ideal gas Cp – Cv = R) The ratio of two molar specific heat

capacities is

…(eq. 18)a)Monatomic gas :

As in this case, f = 3From equation (16),(17) and (18),

= 2.98 cal/mol0C = 12.5 J/mol K

= 4.96 cal/mol0C = 20.8 J/mol K

andIt has been observed that the values

of obtained experimentally for

mono-atomic gases like Argon,

Helium agree very well with the

theoretical value obtained above. b)Diatomic gas:

(i)When the diatomic gas molecule is a

rigid rotator like a dumbbell, f = 5

Thus cal/mol0C= 4.96 cal/mol0C = 20.8 J/mol K

cal/mol0C= 6.95 cal/mol0C = 29.1 J/mol K

and(ii) When the diatomic gas molecule is

not a rigid rotator but has an

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additional vibrational mode. In this

case, f = 7

Thuscal/mol0C= 6.95 cal/mol0C = 29.1 J/mol

K

cal/mol0C= 8.93 cal/mol0C = 37.4 J/mol

K

andc)Polyatomic gas:

In general a polyatomic gas

molecule has three translations,

three rotational degrees of freedom

and number of vibrational modes.

So f = 6+2

Thus

and The theoretical predictions for

specific heats of gases ignoring the

vibrational modes are in good

agreement with experimental values

of specific heats of several gases.

Although, there are discrepancies between the predicted and actual

values of specific heats of several

other gases like Cl2, C2H6 and many

other polyatomic gases. Usually the

experimental values are greater than

the predicted values of specific heats.

However, by including vibrational

modes of motion in the calculations,the values are in close agreement

with each other. Thus, the law of

equipartition of energy is verified at

ordinary temperatures.

Specific Heat Capacity of Solids

In 1819 Dulong and Petit found

experimentally that for many solidsat room temperature, the molar

specific heat is 3R. This is consistent with equi-partition

theory: energy added to solids takes

the form of atomic vibrations and

both kinetic and potential energy is

associated with the three degrees of

freedom of each atom.

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Average vibrational energy per atom= 6 x ½kBT = 3kBT The total vibrational energy or the

internal energy of one mole of atoms

of solid, U = NA x 3 kB T= 3 RTNow at constant pressure ∆Q = ∆U +

P ∆V, since for a solid ∆V is negligible.

Hence,C = ∆Q/ ∆T = ∆U/ ∆t = 3R≈ 25 J/K mol

It signifies that the heat required toraise the temperature of a sample of

a solid depends on the total number

of atoms present in the sample and

not on their atomic masses. So the

molar specific heat of solids is same

near the room temperature. Table showing specific heats of some

elements

At low

temperatures, however, the molar

heat falls below 3R, and eventually

approaches zero at 0 K. At very low

temperatures, the molar heat

capacity varies roughly as the cube of

the temperature. This behaviour was first explained by

Einstein in 1906 by using quantum

theory and further improved byDebye in 1915.The temperature at

which the molar specific heat of a

solid at constant volume becomes

equal to 3R is called as Debye

temperature.

Specific Heat Capacity of Water

We treat water like a solid made up ofthree atoms: two hydrogen and one

oxygen. For each atom average energy is 3kBT.So it hasU= 3 × 3kBT×NA= 9RT

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andC= ∆Q/ ∆T= ∆U/ ∆T= 9R= 74.7 JK-1mol-1

In cgs system, water is defined to

have unit specific heat,i.e., C = 1 cal g-1 K-1 = 4.179 J g-1 K-

1

For 1 mole of water, C = 18 x 4.179

J mole-1 K-1 = 75.22 J mole-1 K-1

Clearly, the predicted value and

observed value of molar specific heat

of water are in good agreement with

each other.

Lastly, we should note an importantaspect specific heat capacities

predicted on the basis of law of

equipartion of energy are

independent of temperature.

However at low temperaturesspecific

heats of all substances approach zero

as temperature approaches zero. Thisis related to the fact that degrees of

freedom get frozen and ineffective at

low temperatures.

According to classical physics degrees

of freedom must remain unchanged

at all times. The behaviour of specific

heats at low temperatures shows theinadequacy of classical physics.

Quantum mechanics requires a

minimum, nonzero amount of energy

before a degree of freedom comes into

play. This is also the reason why

vibrational degrees of freedom come

into play only in some cases.

Example 13.8

A cylinder of fixed capacity 44.8 litres

contains helium gas at standard

temperature and pressure. What is

the amount of heat needed to raise

the temperature of the gas in the

cylinder by 15.0 °C? (R= 8.31 Jmo1 –

1

K –1

).Solution:

Using the gas law PV= nRT, it can

easily be shown that 1 mol of any

(ideal) gas at standard temperature

(273 K) and pressure (1 atm = 1.01 ×

105 Pa) occupies a volume of 22.4

litres. This universal volume is called

molar volume. Thus the cylinder in this example

contains 2 mol of helium.

Further, since helium is monatomic,

its predicted (and observed) molar

specific heat at constant volume,Cv

= (3/2)R, and molar specific heat at

constant pressure,Cp= (3/2)R + R=

(5/2)R.

Since the volume of the cylinder is

fixed, the heat required is determined

byCv. Therefore,

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Heat required = no. of moles × molar

specific heat × rise in temperature

= 2 × 1.5R× 15.0 = 45R

= 45 × 8.31 = 374 J.13.9 Mean free path

On the basis of kinetic theory of

gases, it is assumed that the

molecules of a gas are continuously

colliding against each other.

Molecules move in straight line with

constant speeds between two

successive collisions. Thus path of a

single molecule is a series of zig-zag

paths of different lengths. Thesepaths of different lengths are called

free paths of the molecule. Mean Free

Path is the average distance traversed

by molecule between two successive

collisions.

For finding the mean free path let us

consider a gas containing n

molecules per unit volume.

We assume that only one molecule

which is under consideration is in

motion while all others are at rest. If

d is the diameter of each moleculethen the moving molecule will collide

with all these molecules whose

centers lie within a distance d from

its centre as shown in figure.If v is the velocity of the moving

molecule then in one second it will

collide with all molecules with in a

distance d between the centers. Inone second it will cover a volume

πd2 v.If n is the total number of

molecules per unit volume, then

number of collisions a molecule

suffers in one second= nπd2 vIf s is the total path travelled in

N collisions, then mean free pathλ = s / NIf v is the distance traversed by

molecule in one second then mean

free path is given byλ = total distance traversed in one

second / number of collision suffered

by the molecules = v/πσ2 vn

= 1/πd2n This expression was derived with the

assumption that all the molecules are

at rest except the one which is

colliding with the others.

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More exact statement can be derived

considering that all molecules are

moving with all possible velocities in

all possible directions. More exactrelation found using distribution law

of molecular speeds isλ=1/(√2)πd2n= 1/(√2) πd2N/V …(eq. 19)

its derivation is beyond the scope of

this book.

From the above relation, we can say as gas density increases,the molecules become closer to each

other. Therefore, they are more likely

to run into each other, so the mean

free path decreases.Increasing the number of molecules

or decreasing the volume causes

density to increase. This decreases

the mean free path. On increasingthe radii of the molecules the space

between them decreases, causing

them to run into each other more

often. Therefore, the mean free path

decreases. Pressure, temperature,

and other factors that affect density

can indirectly affect mean free path.

Example 13.9

Determine the mean free path of

argon molecules under normal

conditions.Solution:

Given: The molecule diameter is0.4 nm. The mean free path of molecules

is given by

Here k = 1.38 X 10-23 T = 273.15 K

P = 1.01325 X 10

5

Pad = 4 X 10-10 m

Exercise


1.The kinetic - molecular theory

explains the behavior of

(a)Gases only.(b)Solids and liquids.(c)Liquids and gases.(d)Solids, liquids and gases.

2.If the temperature of 50.0 L of a

gas at 40.0ºC falls by 10.0Cº, what

is the new volume of the gas if the

pressure is constant?

(a)45 l(b)48.4 l(c) 52 l(d)55 l

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3.Which one is not the assumption

in kinetic theory of gases?(a) The molecules of the gas are in

continuous random motion(b) The molecules interact during

the collison

(c) The molecules are tiny hard

sphere undergoing inelastic

collision

(d) The collison are of short

duration

4.Which of the following is trueaccording to Vander Waal gas

equation?

(a) The attractive forces between

the molecules is not negligible

(b) Volume of the molecules is

negligible as compared to the

volume occupied by the gas

(c) Volume of the molecules is notnegligible as compared to the

volume occupied by the gas

(d) The attractive forces between

the molecules is negligible

5.Mean square velocity of five

molecules of velocities 2 m/s, 3

m/s, 4 m/s, 5 m/s and 6 m/s is(a) 10 m2/s2

(b) 18 m2/s2

(c) 20 m2/s2

(d) 15 m2/s2

6.If for a gas this gas is

made up of molecules which are

(a) Diatomic(b) Mixture of diatomic and

polyatomic(c) Mono atomic(d) Polyatomic

7.The average translational kinetic

energy of the molecules of a gas

depends on

(a) the number of moles and thetemperature.

(b) the pressure and the

temperature(c) the pressure only. (d) the temperature only.

8.An ideal gas is enclosed in a

sealed container. Upon heating,

which property of the gas does

not change?a)Volume b)Pressurec)The rate of collisions of the

molecules with each otherd)The rate of collisions of the

molecules with the walls of

the container

9.Under which conditions do real

gases most resemble ideal gases?

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(b)high pressure and low

temperature(c) low pressure and high

temperature(d)high pressure and high

temperature

Questions

10.Air pressure in a car tyre increases

during driving? Why?11.What is the rms speed of a nitrogen

molecule at room temperaturepredicted by the kinetic theory of

gases? (mass of a molecule of N2 is

4.68 x 10 –26 kg )12.If a vessel contains 1 mole of O2 gas

(molar mass 32) at temperature T.

The pressure of the gas is P. What is

the pressure if an identical vessel

contains 1 mole of He at a

temperature 2 T?13.The mean free path for O2

molecules at a temperature of 300 K

and at 1.00 atm pressure is 7.10 ×

10 –8 m. Use this data to estimate the

size of an O2 molecule.14.An ideal gas in a cylinder fitted

with a piston (Figure 17-20) is held

at fixed pressure. If the temperatureof the gas increases from 50° to

100°C, by what factor does the

volume change?

15.Which speed is greater, the

speed of sound in a gas or the

rms speed of the molecules of the

gas? Justify your answer, usingthe appropriate formulas, and

explain why your answer is

intuitively plausible.

16.What are the mole fractions and

partial pressures of each gas in a

2.50 mL container into which

100.00 g of nitrogen and 100.00 g of

carbon dioxide are added at 25°C? What is the total pressure?18.If air is filled in a vessel at the

temperature of 600C. To what

temperature should it be heated in

order that 1/3rd of air may escape

out of vessel?19.The density of water is 1000 kg m –

3

. The density of water vapour at 100°C and 1 atm pressure is 0.6 kg m–

3. The volume of a molecule

multiplied by the total number gives,

what is called, molecular volume.

Estimate the ratio (or fraction) of the

molecular volume to the total

volume occupied by the water

vapour under the above conditions

of temperature and pressure.20.A vessel contains two nonreactive

gases: neon (monatomic) and oxygen

(diatomic). The ratio of their partial

pressures is 3:2. Estimate the ratio

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of (i) number of molecules and (ii)

mass density of neon and oxygen in

the vessel. Atomic mass of Ne = 20.2

u, molecular mass of O2 = 32.0 u.21.A flask contains argon and chlorine

in the ratio of 2:1 by mass. The

temperature of the mixture is 27 °C.

Obtain the ratio of (i) average kinetic

energy per molecule, and (ii) root

mean square speed vrms of the

molecules of the two gases. Atomic

mass of argon = 39.9 u; Molecular

mass of chlorine = 70.9 u.22.Uranium has two isotopes of

masses 235 and 238 units. If both

are present in Uranium hexafluoride

gas which would have the larger

average speed? If atomic mass of

fluorine is 19 units, estimate the

percentage difference in speeds at

any temperature.23.A cylinder of fixed capacity 44.8

litres contains helium gas at

standard temperature and pressure.

What is the amount of heat needed

to raise the temperature of the gas

in the cylinder by 15.0°C? (R= 8.31

J mo1 –1 K –1).

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14.1 Introduction The world is full of things that move.

The motion of physical systems can

be classified into two broad

categories-translatory and vibratory

or oscillatory. If the position of a body

varies linearly with time, its motion is

translatory, e.g. a train moving on a

straight track or a ball rolling on the

ground. A motion that repeats itself

in equal intervals of time is called

periodic motion. e.g. the motion of

the hands of a clock. If a particle in

periodic motion moves back and forth

over the same path, its motion is

called vibratory or oscillatory. Some

examples of oscillatory motion are

the oscillation of the hands of a

walking person, the balance wheel of

a watch, the bob of a pendulum

clock, a mass attached to a spring,

the prongs of a tuning fork, the

piston of an automobile engine, etc.

Oscillations may be very complex,

such as those of a piano string or

those of the earth during an

earthquake. It may be remarked thatmechanical systems are not the only

ones that can oscillate. The atoms in

a solid vibrate. The electrons in a

radiating or receiving antenna are in

oscillation. A tuned circuit in a radio

can oscillate electromagnetically.

Radiowaves, microwaves and visible

light are simply oscillating electric

and magnetic fields. Thus a study ofoscillations is essential for the

understanding of various physical

systems- mechanical, acoustical,

electrical and atomic. Any motion that repeats itself after

regular intervals of time is known as

a periodic motion. The examples of

periodic motion are the motion ofplanets around the Sun, motion of

hands of a clock, motion of the

balance wheel of a watch, motion of

Halley’s comet around the Sun

observable on the Earth once in 76

years.If a body moves back and forth

repeatedly about a mean position, itis said to possessoscillatory motion.

Vibrations of guitar strings, motion

ofa pendulum bob, vibrations of a

tuning fork, oscillations of mass

suspended from a spring, vibrations

of diaphragm in telephones and

speaker system

and freely suspended springs are fewexamples of oscillatory motion. In all

the above cases of vibrations of

bodies, the path of vibration is

always directed towards the mean or

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equilibrium position. The oscillations

can be expressed in terms of simple

harmonic functions like sine or

cosine function.

Harmonic oscillation

Harmonic oscillation is that

oscillation which can be expressed in

terms of single harmonic function

(sine function or cosine function).

Fig. 1.1 Representation of a SHM wave.

A harmonic oscillation of constant

amplitude and single frequency is

calledsimple harmonic motion (SHM).

Periodic and non-periodic function

Any function which repeats itself

after regular intervals of time is

called periodic function.f(t) = sin ωtf(t + T) = sin ω(t +T) = sin 2π/T (t + T) = sin (2πt/T + 2π) = sin 2π/T = sin

ωt.Non-periodic functions do not repeat

after regular intervals of time.

Simple Harmonic Motion

A particle is said to execute simple

harmonic motion if its acceleration is

directly proportional to the

displacement from a fixed point and isalways directed towards that point.

Consider a particle P executing SHM

along a straight line between A and B

about the mean position O (Fig. 1.2).

Fig.1.2 simple harmonic motion of aparticle.

The acceleration of the particle is

always directed towards a fixed point

on the line and its magnitude isproportional to the displacement of

the particle from this point. a ∞yBy definitiona = -ω2y where ω is a constant known as

angular frequency of the simple

harmonic motion.

The negative sign indicates that theacceleration is opposite to the

direction of displacement. Ifmis the

mass of the particle, restoring force

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that tends to bring back the particle

to the mean position is given byF = -m ω2y or F = -k y

The constantk =m ω

2

, is called forceconstant or spring constant. Its unit

isNm-1. The restoring force is directed

towards the mean position.

Thus, simple harmonic motion is

defined as oscillatory motion about a

fixed point in which the restoring force

is always proportional to the

displacement and directed always

towards that fixed point.

Simple Harmonic Motion

From our concept of a simple

harmonic oscillator we can derive

rules for the motion of such a system.

We start with our basic force

formula,F = -kx UsingNewton's Second Law, we can

substitute for force in terms of

acceleration:ma = -kxHere we have a direct relation

between position and acceleration.

For you calculus types, the above

equation is a differential equation,and can be solved quite easily.Note: The following derivation is not

important for a non- calculus based

course, but allows us to fully

describe the motion of a simple

harmonic oscillator.Deriving the Equation for Simple

Harmonic MotionRearranging our equation in terms of

derivatives, we see that:

m = -kx

or

+ x = 0

Let us interpret this equation. The

second derivative of a function

ofx plus the function itself (times a

constant) is equal to zero. Thus the

second derivative of our function

must have the same form as the

function itself. What readily comes tomind is the sine and cosine function.

Let us come up with a trial solution

to our differential equation, and see if

it works. As a tentative solution, we write:x =a cos(ωt) wherea andω are constants.

Differentiating this equation, we see

that

= -aω sin(ωt)and

= -aω 2cos(ωt)

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Plugging this into our original

differential equation, we see that:

-aω 2cos(ωt) + a cos(ωt) = 0

It is clear that, ifω 2 = , then the

equation is satisfied. Thus the equation governing simple

harmonic oscillation is:

x =a cos t

The Equation for Simple Harmonic

Motion

From the equation for simple

harmonic motion we can tell a lot

about the motion of a harmonic

system. First of all,x is maximum

when the cosine function is equal to

1, or whenx =a . Thus a in this

equation is the amplitude of

oscillation, which we have already

denoted byx m . Secondly, we can find

the period of oscillation of the

system. Att = 0 ,x =x m .

Also, att = 2Π ,x =x m .Since both these instances have the

same position, the time between the

two gives us our period of oscillation.

Thus:

T = 2Π and

ν = =Finally,

ω= 2Πν =

Note that the values of period and

frequency depend only on the mass

of the block and the spring constant.

No matter what initial displacement

is given to the block, it will oscillate

at the same frequency. This concept

is important. A block with a small

displacement will move with slower

velocity, but with the same frequency

as a block with a large displacement.So now we know thata =x m. In

addition we can take the time

derivative of our equation to generate

a full set of equations for simple

harmonic motion:

x=x mcos(ωt) v=-ωx msin(ωt) a=-ω2 x mcos(ωt)

The projection of uniform circular

motion on a diameter is SHM

Consider a particle moving along the

Circumference of a circle of radiusaand centre O, with uniform speedv,

in anticlockwise direction as shown

in Fig. 1.1.

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Let XX’ and YY’ be the two

perpendicular diameters.

Fig. 1.3 Projection of circular motion.

Suppose the particle is at P after a

time ‘t’. If ω is the angular velocity,

then the angular displacement θ in

time t is given by θ = ωt.FromP drawPN perpendicular toYY.

As the particle moves from X toY,

foot of the perpendicularN moves

fromO toY. As it moves further from

Y to X ‘,then from X’ toY’ and back

again to X, the pointN moves fromY

toO, from O toY' and back again to

O. When the particle completes one

revolution along the circumference,

the pointN completes one vibration about the

mean positionO. The motion of the

pointN along the diameterYY’ is

simple harmonic.Hence, the projection of a uniform

circular motion on a diameter of a

circle is simple harmonic motion.

Displacement in SHM

The distance travelled by the

vibrating particle at any instant of

time t from its mean position isknown as displacement. When the particle is atP, the

displacement of the particle alongY

axis isy. Then, inOPN,∆ sin θ = ON =y = OPsin θy = OPsinωt (θ = ωt)Since OP =a, the radius of the circle,

@ the displacement of the vibratingparticle isy =asinωt ...(1)

The amplitude of the vibrating particle

is defined as its maximum

displacement from the mean position.

Velocity in SHM The rate of change of displacement is

the velocity of the vibrating particle.Differentiating eqn. (1) with respect

to timetdy/dt = d/dt (a sin ωt) ∴ v =a cosωt The velocityv of the particle moving

along the circle can also be obtained

byresolving it into two components.(i)vcos θ in a direction parallel to OY(ii)v sin θ in a direction

perpendicular toOY

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Figure showing velocity of particle executing

SHM

The componentv sinθ has no effect

alongYOY' since it is perpendiculartoOY.∴ velocity =vcosθ=vcosωt We know that,linear velocity = radius×angular

velocity∴v =aω∴ velocity =aωcosωt∴ velocity =aω √(1- sin2ωt)

velocity =aω √(1-y2/a2)@ sinθ = y/a velocity = ω √(a2 – y2)Special cases

(i) When the particle is at mean

position, (i.e)y = 0. velocity is aω and is

maximum.v = ± aω is calledvelocity amplitude.(ii) When the particle is in the

extreme position i.e y = ± a, the

velocity is zero.

Acceleration in SHM

The rate of change of velocity is the

acceleration of the vibrating particle.d2 y/dt2 =d/dt(a ω

cosωt) = -ω2a sin ωt∴ acceleration = d2 y/dt2 = -ω2 y.

The negative sign indicates that the

acceleration is always opposite to the

direction of displacement and is

directed towards the centre.

Special Cases(i) When the particle is at the mean

position (i.e)y = 0, the acceleration is

zero.(ii) When the particle is at the

extreme position i.e, y =+a,

acceleration is±a ω2 which is called

asacceleration amplitude.

The differential equation of simple

harmonic motion is

d2 y/dt2 + ω2 y = 0.

Using the above equations, the

values of displacement, velocity and

acceleration for the SHM are given in

the Table 1.1.

It will be clear from the above, that at

the mean positiony = 0, velocity of

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the particle is maximum but

acceleration is zero.

At extreme position y = ±a, the velocity is zero but the acceleration is

maximum ±a ω2 acting in the

opposite direction.

Table 1.1 - Displacement, Velocity and

Acceleration

Time Phas

e

ωt

y v=

ω√(a2 – y2)

a =

-ω2 y.

t=0

t=T/4

t=T/2

t=3T/

4

t=T

0

π/2

π

3π/2

2π

0

+a

0

-a

0

aω

0

- aω

0

aω

0

-ω2a

0

ω2a

0

Graphical Representation of SHMGraphical representation of

displacement velocity and

acceleration of a particle vibrating

simple harmonically with respect to

time t is shown in Fig. 1.4Displacement graph is a sine curve.

Maximum displacement of the

particle is y= ±a.

Fig. 14.4 Graphical representation.

(ii) The velocity of the vibrating

particle is maximum at the mean

position i.e v =± aω and it is zero

at the extreme position.(iii) The acceleration of the vibrating

particle is zero at the mean position

and maximum at the extremeposition (i.e)±a ω2. The velocity is ahead of displacement

by a phase angle of π/2. The

acceleration is ahead of the velocity

by a phase angle π/2 or by a phase π

ahead of displacement. (i.e) when the

displacement has its greatest positive

value, acceleration has its negativemaximum value or vice-versa.Some Important terms in SHM

(i) Time period

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The time taken by a particle to

complete one oscillation is called time

period T.

In the Fig. 1.1, as the particle Pcompletes one revolution with

angular velocity ω, the foot of the

perpendicular N drawn to the

vertical diameter completes one

vibration. HenceT is the time period.

Then ω = 2π/T or T = 2π/ω.

The displacement of a particleexecuting simple harmonic motion

may be expressed asy(t) =a sin2πt/T ...... (1)

Or y(t) =a cos 2πt/T ....... (2) WhereT represents the time period,

‘a’ represents maximum

displacement

(amplitude).

These functions repeat when t is

replaced by (t +T).y (t +T)=a sin 2π (t+ T)/T ........(3) = a sin( 2πt/T + 2π ) =a sin2πt/T = y(t)In generaly (t +nT) =y (t)

Above functions are examples ofperiodic function with time periodT.

It is clear that the motion repeats

after a timeT = 2π/ω where ω is the

angular frequency of the motion. In

one revolution, the angle covered bya

particle is 2ω in timeT.

(ii) Frequency and angular frequency

The number of oscillations produced

by the body in one second is known

as frequency. It is represented by n.

The time period to complete one

oscillation is 1/n. T =1/nshows the time period is the

reciprocal of the frequency. Its unit ishertz. ω = 2π/n, is called as angular

frequency. It is expressed in rad s-1.

(iii) Phase

The phase of a particle vibrating in

SHM is the state of the particle as

regards to its direction of motion and

position at any instant of time.

In the equationy =a sin (ωt + φo) the

term (ωt +φo) = φ, is known as the

phase of the vibrating particle.

(iv) Epoch

It is the initial phase of the vibrating

particle (i.e) phase at t = 0.

∴φ = φo (@ φ = ωt + φo)

The phase of a vibrating particle

changes with time but the epoch is

phase constant.

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Fig. 14.5 (a) (b)

(i) If the particleP starts from the

position X, the phase of the particle

is zero.(ii) Instead of counting the time from

the instant the particle is at X, it is

counted from the instant when thereference particle is at A(Fig. 1.5a).

Then XOP = ωt -φoHere (ωt -φo) = φ is called the phase

of the vibrating particle. (-φo) is initial

phase or e poch.(iii) If the time is counted from the

instant the particleP is above X (i.e)

atB [Fig. 1.5b] then (ωt + φo) = φ.Here (+φo) is theinitial phase.

Phase difference

If two vibrating particles executing

SHM with same time period, cross

their respective mean positions at the

same time in the same direction, they

are said to be in phase.If the two vibrating particles cross

their respective mean position at the

same time but in the opposite

direction, they are said to be out of

phase (i.e they have a phase

difference of π).If the vibrating motions are

represented by equationsy1 =a sinωt andy2 =a sin (ωt - φ)then the phase difference between

their phase angles is equal to the

phase difference between the two

motions.∴phase difference = ωt- φ – ωt= - φ

negative sign indicates that the

second motion lags behind the first.Ify2 =a sin (ωt + φ), Then phase difference = ωt + φ - ωt =

φHere the second motion leads the

first motion. We have discussed the SHM without

taking into account the cause of the

motion which can be a force (linear

SHM) or a torque (angular SHM).

Example 14.1

A harmonic oscillation is represented

by : y = 0.34 cos (3000t + 0.74) where y and t are in mm and sec

respectively. Deduce (i) its amplitude

(ii) its frequency (iii) its time period

(iv) angular frequency and (v) initial

phase.Solution:

y = 0.34 cos (3000t + 0.74)Comparing with general equation, y = a cos (ωt + φ)

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a = 0.34 mm, ω = 3000 rad/s, φ=

0.74 rad

(i) ω=2πn, frequency n =ω/2π

= 1500/π Hz(ii) ω=2π/T, T=0.0021 sec(iii) ω = 3000 rad/s(iv) φ= 0.74 rad

Example 14.2

A body oscillates with SHM,

according to the equation, y = 5 m

cos [(2π)t + π/4]. At t=1.5 sec,

calculate the (i) displacement (ii)speed and (iii) acceleration of the

body.Solution:

Here, ω = 2π rad/s, T= 2π/ω =1 sec,

t=1.5 sec

(i) Displacement y = 5 m cos [(2π rad/s) 1.5 +

π/4]= 5m cos [3π = π/4] = -5 cos

π/4= -3.535 m

(ii) Velocity v = dx/dt= -5 m 2π sin[(2π rad/s)t +

π/4]= -5 x 2π sin [3π + π/4]

=5x 2π x sin π/4= 22.22 m/s

(iii) Acceleration = dv/dt= -ω2 y= - [4π2][-3.535]= 139.56 m/s2

Energy In Simple Harmonic Motion

The total energy (E) of an oscillating

particle is equal to the sum of its

kinetic energy and potential energy if

conservative force acts on it. The

velocity of a particle executing SHM

at a position where its displacement

isy from its mean position isv =

ω√(a2-y2)Kinetic Energy

Kinetic energy of the particle of mass

misK= ½ m [ω√(a2-y2)]2

K= ½ m ω2(a2-y2) -----

(1) Potential energy

From definition of SHM F = –ky the

work done by the force duringthe

small displacementdy isdW = -F.dy = -(-ky)dy =kydy∴ Total work done for the

displacementyis, W= dW= kydy W= m ω2 y dy @[k =m ω2]

∴ W= ½ m ω2 y2

This work done is stored in the body

as potential energyU =½ m ω2 y2 ......(2)

Total energy E = K + U =½ m ω2(a2-y2) + ½ m ω2 y2

E = ½ m ω2 a2

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Thus we find that the total energy of

a particle executing simple harmonic

motion is E = ½ m ω2 a2

Special cases

(i)When the particle is at the mean

positiony = 0, from eqn (1) it is

known that kinetic energy is

maximum and from eqn. (2) it is

known that potential energy is

zero. Hence the total energy is

wholly kinetic E =Kmax= ½ m ω2 a2

(ii) When the particle is at the

extreme position y =+a, from eqn.

(1) it is known that kinetic energy

is zero and from eqn. (2) it is

known that Potential energy is

maximum. Hence the total energy

is wholly potential.E = Umax= ½ m ω2 a2

Graphical Representation of

Energy

The values ofK andU in terms ofE

for different values ofyare given in

the Table 1.2.

The variation of energy of anoscillating particle with the

displacement can be represented in

a graph as shown in the Fig. 1.6.

Fig. 14.6 Energy-displacement graph.

Example 14.3

A particle executes SHM of amplitude

‘a’. At what distance from the mean

position its KE equals to its PE?Solution:

KE = ½ m ω2(a2-y2)PE = ½ m ω2 y2

As KE =PE½ m ω2(a2-y2) = ½ m ω2 y2

(a2-y2) = y2

Y =a/√2 = 0.71 a

Some Examples of SHM

(i) Horizontal and vertical oscillations

of a loaded spring.(ii) Vertical oscillation of water in a U-

tube(iii) Oscillations of a floating cylinder(iv) Oscillations of a simple pendulum

(v) Vibrations of the prongs of atuning fork.

Dynamics of Harmonic Oscillations

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The oscillations of a physical system

results from two basic properties

namely elasticity and inertia. Let us

consider a body displaced from amean position. The restoring force

brings the body to the mean position.

(i) At extreme position when the

displacement is maximum, velocity is

zero. The acceleration becomes

maximum and directed towards the

mean position.

(ii) Under the influence of restoring

force, the body comes back to the

mean position and overshoots

because of negative velocity gained at

the mean position.

(iii) When the displacement is

negative maximum, the velocity becomes zero and the acceleration is

maximum in the positive direction.

Hence the body moves towards the

mean position. Again when the

displacement is zero in the mean

position velocity becomes positive.

(iv)Due to inertia the body overshootsthe mean position once again. This

process repeats itself periodically.

Hence the system oscillates. The

restoring force is directly proportional

to the displacement and directed

towards the mean position.(i.e)F ∞ - y

F = -ky ............. (1) Wherek is the force constant defined

as the force required to give unit

displacement. It is expressed in N m-

1.From Newton’s second law,

F =ma ..........(2)∴ -k y=maOr a= -(k/m)y .........(3)

From definition of SHM accelerationa = -ω2y

The acceleration is directly

proportional to the negative of the

displacement. The acceleration is directly

proportional to the negative of the

displacement.Comparing the above equations we

get,ω = √(k/m)...(4) Therefore the period of SHM isT =2π/ω = 2π √(m/k)

Horizontal Oscillations Of Spring

Consider a mass(m) attached to anend of a spiral spring (which obeys

Hooke’s law) whose other end is fixed

to a support as shown in Fig. 6.9.

The body is placed on a smooth

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horizontal surface. Let the body be

displaced through a distance x

towards right and released. It will

oscillate about its mean position. Therestoring force acts in the opposite

direction and is proportional to the

displacement.∴ Restoring forceF = -kx.From Newton’s second law,F =ma∴ ma= -k x a=-(k/m)x

Comparing with the equation of SHMa = -ω2 x we get,ω2= k/m or ω = √(k/m)ButT =2π/ω Time periodT= 2π√(m/k)

∴ Frequencyn = 1/T

n= 1

2π √ "

m

Fig 14.7 Linear Harmonic Oscillator

Vertical Oscillations Of A Spring

Fig 1.8a shows a light, elastic spiral

spring suspended vertically from a

rigid support in a relaxed position.

When a mass ‘m’ is attachedto the spring as in Fig. 1.8b, the

spring is extended by a small lengthdl such that the upward force F

exerted by the spring is equal to the

weightmg. The restoring force is F = -kdl =mg ...(1) wherek is spring constant.If we further extend the given spring

by a small distance by applying a

small force by our finger, the springoscillates up and down about its

mean position. Now suppose the

body is at a distance y above the

equilibrium position as in Fig. 1.8c.

The extension of the spring is (dl -y).

The upward force exerted on the body

isk(dl- y) and the resultant forceF

on the body isF =k (dl- y) -mg = -ky ...(2) The resultant force is proportional to

the displacement of the body from its

equilibrium position and the motion

is simple harmonic.If the total extension produced is (dl

+y) as in Fig. 1.8d the restoring force

on the body isk(dl +y) which actsupwards.So, the increase in the upward force

on the spring isk(dl +y) -mg =ky

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Therefore if we produce an extension

downward then the restoring force in

the spring increases byky in the

upward direction. As the force acts inthe opposite direction to that of

Fig. 14.8 Vertical Oscillations of loaded

Spring

displacement, the restoring force isF = -ky

and the motion is SHM. ma =-ky

a =-(k/m) ya = -ω2y ......(3)(expression for SHM)

Comparing the above equations,ω = √(k/m)---(4)

ButT =2π/ω = 2π √(m/k) ......(5)From equation (1)mg =k lm/k = l/g

Therefore time periodT =2π√(l/g) --- (6)Frequencyn = 1/T = [1/2π ]√(g/l)

Case1: When two springs are

connected in Parallel

Two springs of spring factorsk1 and

k2suspended from a rigid support as

shown in Fig.1.9 A load m is attached

to the combination.Let the load be pulled downwards

through a distance y from its

equilibrium position. The increase in

length isy for both the springs but

their restoring forces are different.IfF1andF2are the restoring forces,

then,F1= = -k1y and F2= = -k2y

Fig.14.9 Springs in parallel

∴ Total restoring force =F= (F1+F2) = -( k1+k2 )ySo, time period of the body is given

by T =2π √(m/( k1+k2 )Ifk1 = k2 = k, T =2π √(m/( 2k)

∴ frequency n = 1/T = [1/2π ]

√(2k/m)

Case2: When two springs are

connected in series.

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Two springs are connected in series

in two different ways. This

arrangement is shown in Fig. 1.10a

and 1.10b.

Fig 14.10 (a) (b)

WhereF is the restoring force. Total extension,y =y1 + y2 = -F[1/k1 + 1/k2]

We know thatF = -ky ∴y = -F/k

Oscillations of A Simple Pendulum

A simple pendulum consists of mass-

less and inelastic thread whose oneend is fixed to a rigid support and a

small bob of massm is suspended

from the other end of the thread. Let

l be the length of the pendulum.

When the bob is slightly displaced

and released, it oscillates about its

equilibrium position. Fig.1.11 shows

the displaced position of thependulum.

Suppose the thread makes an angle θ

with the vertical. The distance of the

bob from the equilibrium position A

is AB. AtB, the weightmg acts

vertically downwards. This force is

resolved into two components.(i) The component mg cosθ is

balanced by the tension in the thread

acting along the length towards the

fixed point O.(ii)mg sinθ which is unbalanced, acts

perpendicular to the length of thread.

This force tends to restore the bob to

the mean position. If the amplitude ofoscillation is small, then the path of

the bob is a straight line. ∴F = -mg sin θ ...(1)If the angular displacement is smallsin θ= θ∴F = -mgθ -- (2)

Fig.14.11 Simple Pendulum

But θ = x/l

∴F = - mg x/lComparing this equation with

Newton’s second law,F = ma we get,

accelerationa =gx/l ..........(3)

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(negative sign indicates that the

direction of acceleration is opposite to

the displacement)

Hence the motion of simplependulum is SHM. We know thata = -ω2xComparing this with ..... (3)ω2=g/l , ω = √(g/l) ....... (4)time periodT = 2π/ ω =2π√(l/g) – (5)frequency n = 1/T = 1/2π √(g/l) -- (6)

Oscillation of Liquid Column in a

U – TubeConsider a non viscous liquid column

of lengthl of uniform cross-sectional

area A (Fig. 6.13).Initially the level of

liquid in the limbs is the same. If the

liquid on one side of the tube is

depressed by blowing gently the

levels of the liquid oscillates for a

short time about their initialpositions O and C, before coming to

rest. If the liquid in one of the limbs

is depressed byy, there will be a

difference of 2y in the liquid levels in

the two limbs. At some instant,

suppose the level of the liquid on the

left side of the tube is atD, at a

Fig.1.12 Oscillations of a liquid column in U-tube

heighty above its original positionO,the levelB of the liquid on the other

side is then at a depthy below its

original position C. So the excess

pressure P on the liquid due to the

restoring force isP = excess height× density ×gP = 2yρg

∴ Force on the liquid = pressure ×areaof the cross-section of the tubeF =– 2 yρg × A ..... (1) The negative sign indicates that the

force towards O is opposite to the

displacement measured from O at

that instant. The mass of the liquid column of

lengthlis volume × density ;m =lAρ∴ F =l Aρa .... (2)

From equations (1) and (2)l Aρa = - 2y Aρg

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∴a = –2gy/l .......(3) We know thata =-ω2y(i.e)a = –2gy/l= –ω2y where ω = √(2g/l)

Here, the acceleration is proportional

to the displacement, so the motion is

simple harmonic and the time period

T= 2π/ ω =2π√(l/2g) ..... (3)

Types of oscillations

There are three main types ofoscillations.(i) Free oscillations

When a body vibrates with its own

natural frequency, it is said to execute

free oscillations.

The frequency of oscillations depends

on the inertial factor and spring

factor, which is given by,n = [1/2π ] √(2k/m)Examples

(i) Vibrations of tuning fork(ii) Vibrations in a stretched string(iii) Oscillations of simple pendulum(iv) Air blown gently across the

mouth of a bottle.

(ii) Damped oscillationsMost of the oscillations in air or in

any medium are damped. When an

oscillation occurs, some kind of

damping force may arise due to

friction or air resistance offered by

the medium. So, a part of the energy

is dissipated in overcoming the

resistive force. Consequently, the

amplitude of oscillation decreases with time and finally becomes zero.

Such oscillations are called damped

oscillations (Fig. 1.13).

Examples :

(i) The oscillations of a pendulum(ii) Electromagnetic damping in

galvanometer (oscillations of a coil in

galvanometer)(iii) Electromagnetic oscillations in

tank circuit

Fig. 14.13 Damped Oscillations

(iii) Maintained oscillations

The amplitude of an oscillating

system can be made constant by

feeding some energy to the system.

If an energy is fed to the system tocompensate the energy it has lost,

the amplitude will be a constant.

Such oscillations are called

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maintained / sustained oscillations

(Fig. 14.14).Example :

A swing to which energy is fedcontinuously to maintain amplitude

of oscillation.

Fig. 14.14 Maintained Oscillations.

(iv) Forced oscillations

When a vibrating body is maintained

in the state of vibration by a periodic

force of frequency (n) other than its

natural frequency of the body, the

vibrations are called forced vibrations. The external force isdriverand body

is driven. The body is forced to

vibrate with an external periodic

force. The amplitude of forced

vibration is determined by the

difference between the frequencies of

the driver and the driven. The larger

the frequency difference, smaller will be the amplitude of the forced

oscillations.Examples :

(i) Sound boards of stringed

instruments execute forced vibration,(ii) Press the stem of vibrating tuning

fork, against tabla. The tabla suffersforced vibration.(v) Resonance

In the case of forced vibration, if the

frequency difference is small, the

amplitude will be large.Ultimately when the two frequencies

are same, amplitude becomes

maximum. This is a special case of

forced vibration. If the frequency of

the external periodic force is equal to

the natural frequency of oscillation of

the system, then the amplitude of

oscillation will be large and this is

known as resonance. Advantages

(i) Using resonance, frequency of a

given tuning fork is determined with

a sonometer.(ii) In radio and television, using tank

circuit, required frequency can be

obtained.

Disadvantages

(i) Resonance can cause disaster in

an earthquake, if the naturalfrequency of the building matches

the frequency of the periodic

oscillations present in the Earth. The

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building begins to oscillate with large

amplitude thus leading to a collapse.(ii) A singer maintaining a note at a

resonant frequency of a glass, cancause it to shatter into pieces.

Example 14.4

At what point during the oscillation

of a spring is the force on the mass

greatest?Solution:

Recall thatF = -kx . Thus the force

on the mass will be greatest when thedisplacement of the block is

maximum, or whenx = ±x m .

Example 14.5

What is the period of oscillation of a

mass of 40 kg on a spring with

constantk = 10 N/m?

Solution:

We have derived thatT =2Π . To

find the period of oscillation we

simply plug into this equation:

T = 2Π = 4Π secondsNo matter what initial conditions are

placed on the system, the period of

oscillation will be same. Notice againthat period, frequency and angular

frequency are properties of the

system, not of the conditions placed

on the system.

Example 14.6

A mass of 2 kg is attached to a spring

with constant 18 N/m. It is thendisplaced to the pointx = 2 . How

much time does it take for the block

to travel to the pointx = 1 ?For this problem we use the sin and

cosine equations we derived for

simple harmonic motion.Recall thatx =x mcos(ωt) . We are givenx andx m in the

question, and must calculateω before

we can findt . We know, however,

that no matter the initial

displacement,

ω = = = = 3 . Thus we can plug in our values:

= cosωt = cos3t

3t= cos-

1

t = = 0.35

seconds

This problem was a simple example

of how to use our equations for

simple harmonic motion.

Example 14.7

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A 4 kg mass attached to a spring is

observed to oscillate with a period of

2 seconds. What is the period of

oscillation if a 6 kg mass is attachedto the spring?Solution:

To find the period of oscillation we

need only knowm andk. We are givenm and must findk for

the spring. If a 4 kg mass oscillates

with a period of 2 seconds, we can

calculatek from the followingequation:

T = 2Π Implying that

k = = = 4Π 2

Now that we havek , calculating the

period for a different mass is easy:

T = 2Π = 2Π = = 2.45

seconds A general statement can be made

from this problem: a larger mass

attached to a given spring will

oscillate with a longer period.

Example 14.8:

A mass of 2 kg oscillating on a spring with constant 4 N/m passes through

its equilibrium point with a velocity

of 8 m/s. What is the energy of the

system at this point? From your

answer derive the maximum

displacement,x m of the mass.Solution:

When the mass is at its equilibriumpoint, no potential energy is stored in

the spring. Thus all of the energy of

the system is kinetic, and can be

calculated easily:

K = mv 2 = (2)(8)2 = 64 JoulesSince this is the total energy of the

system, we can use this answer to

calculate the maximum displacementof the mass. When the block is

maximally displaced, it is at rest and

all of the energy of the system is

stored as potential energy in the

spring, given byU = kx m 2 .Since energy is conserved in the

system, we can relate the answer we

got for the energy at one position with the energy at another:

E f =E o

kx m 2= mv 2 = 64

x m = = = 4

meters

We used energy considerations in

this problem in much the same way

we did when we first encountered

conservation of energy- whether the

motion is linear, circular or

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oscillatory, our conservation laws

remain powerful tools.

Exercise


1.A particle executing SHM has

amplitude 0.01 m and frequency60 Hz. The maximum acceleration

of the particle is(i) 144 π2 m/s2

(ii) 120 π2 m/s2

(iii) 80 π2 m/s2

(iv) 60 π2 m/s2

2.A simple pendulum is executing

SHM with a time period T. If thelength of the pendulum is

increased by 21%, the increase in

the time period is(i) 21 %(ii) 10 %(iii) 30 %(iv) 50 %

3.A hollow sphere filled with water

forms the bob of a simplependulum. A small hole at the

bottom of the bob allows the water

to slowly flow out as it is set into

small oscillations and its period of

oscillations is measured. The time

period will(i) Increase

(ii) Decrease(iii) Remain constant(iv) First (i) and then (ii)

4.A simple pendulum hangs from the

ceiling of a car. IF the car

accelerates with uniform

acceleration, the frequency of the

simple pendulum will be(i) Increase

(ii) Decrease(iii) Become infinite(iv) Remains constant

5.The total energy of a body

performing SHM depends on(i) K, a,m

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(ii) K,a, x(iii) K,a(iv) K,x

6.Two springs of constants k1 and k2have equal maximum velocities

when executing SHM. The ratio of

their amplitudes (masses are

equal) will be(i) k1/k2(ii) k2/k1(iii) [k1/k2]1/2

(iv) [k2/k1]1/2

7.Two springs A and B [ka = 3 k b] arestretched by the same force. Then

ratio of work done in stretching is(i) 1 : 2 (iii) 1 : 4(ii) 1 : 3 (iv) 2 : 1

8.A body of mass 5 kg hangs from a

spring and oscillates with a time

period of 2π seconds. If the body

is removed, the length of the

spring will decrease by(i) 2π m (iii) g/k m(ii) g m (iv) k/g m

9.A mass m is suspended from a

spring of negligible mass and the

system oscillates with a frequency

n. What will be the frequency of

oscillation if a mass of 4m is

suspended from the same spring?(i) n/4 (iii) n/2(ii) 2n (iv) 4n

10. If metal bob of a simple

pendulum is replaced by a wooden

bob, then its time period will

(i) Increase(ii) Decrease(iii) Remain the same(iv) First (i) then (ii)

Answers:1.(i) 2. (ii) 3. (iv) 4. (i)5. (iii) 6. (iv) 7. (ii) 8. (ii)9. (iii) 10. (iii)

Questions1.The amplitude of a simple

harmonic oscillator is doubled.

How does this effect (i) the period

(ii) the total energy and (iii) the

maximum velocity of the oscillator?2. A girl is swinging in a swing in

the sitting position. How is the

period of swing affected if she

stands up?3.Show that the time period of SHM

is given by 2π/ω where ω is the

angular frequency of SHM.4.On an average a human heart is

found to beat 75 times in a minute.

Calculate its beat frequency and

period.

5.What will be the time period ofoscillation, if the length of a

second pendulum is one third/6.The bob of a vibrating simple

pendulum is made of ice. How will

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the period of swing will change

when the ice starts melting ?7.The periodic time of a mass

suspended by a spring of forceconstant K is T. If the spring is cut

in three equal parts, what will be

the force constant of each part? If

the same mass be suspended from

one piece, what will be the periodic

time?8.Which of the following functions of

time represent (i) periodic and (ii)non-periodic motion? Give the

period for each motion. (ω is a

positive constant) (i) sin ωt + cosωt

(ii) log ωt (iii) e-ωt

9.A pendulum clock normally shows

correct time. On an extremely cold

day, its length decreases by 0.2%.

Compute the error in time per day.10. A block of mass 1 kg is

fastened to a spring. The spring

has a spring constant 50 N/m.

The block is pulled to a distance y

= 10 cm from its equilibrium

position at y = 0 on a frictionless

surface at t=0. Calculate the

kinetic, potential and total

energies of the block when it is 5

cm away from the mean position.11. The shortest distance travelled

by a particle executing SHM from

mean position in 2 seconds is

equal to (√3/2) times its

amplitude. Determine its time

period.

12. A body is dropped in a holedrilled across a diameter of the

earth. Show that it executes

SHM. Assume the earth to be a

homogenous sphere.

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Introduction:

Waves are everywhere. Whether we

recognize it or

not, we encounter

waves on a daily basis.

Sound waves, visible light waves,

radio waves, microwaves, water

waves, sine waves, cosine waves,

stadium waves, earthquake waves,

waves on a string, and slinky waves

and are just a few of the examples of

our daily encounters with waves.In addition to waves, there are a

variety of phenomena in our physical

world that resemble waves so closely

that we can describe such

phenomenon as being wavelike. The motion of a pendulum, the

motion of a mass suspended by a

spring, the motion of a child on a

swing, and the "Hello, Good

Morning!" wave of the hand can be

thought of as wavelike phenomena. Waves (and wavelike phenomena) are

everywhere!

Before beginning a formal discussion

of the nature of waves, it is often

useful to ponder the various

encounters and exposures that we

have of waves. Where do we see waves or examples

of wavelike motion?

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What experiences do we already have

that will help us in understanding

the physics of waves?

For many people, the first thoughtconcerning waves conjures up a

picture of a wave moving across the

surface of an ocean, lake, pond or

other body of water. The waves are created by some form

of a disturbance, such as a rock

thrown into the water, a duck

shaking its tail in the water or a boatmoving through the water. The water wave hasa crest and a

trough and travels from one location

to another. One crest is often followed

by a second crest that is often

followed by a third crest. Every crest

is separated by a trough to create an

alternating pattern of crests and

troughs. A duck or gull at rest on the surface

of the water is observed to bob up-

and-down at rather regular time

intervals as the wave passes by. The

waves may appear to be plane waves

that travel together as a front in a

straight-line direction, perhaps

towards a sandy shore.Or the waves may be circular waves

that originate from the point where

the disturbances occur; such circular

waves travel across the surface of the

water in all directions.

Another picture of waves involves the

movement of a slinky or similar set of

coils. If a slinky is stretched out from

end to end, a wave can be introduced

into the slinky by either vibrating the

first coil up and down vertically or

back and forth horizontally. A wave will subsequently be seen

travelling from one end of the slinky

to the other. As the wave moves along

the slinky, each individual coil is

seen to move out of place and then

return to its original position. The

coils always move in the same

direction that the first coil was

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vibrated. A continued vibration of the

first coil results in a continued back

and forth motion of the other coils.

If looked at closely, one notices thatthe wave does not stop when it

reaches the end of the slinky; rather

it seems to bounce off the end and

head back from where it started. A

slinky wave provides an excellent

mental picture of a wave and will be

used in discussions and

demonstrations throughout this unit.

Waves are everywhere in nature. Our

understanding of the physical world

is not complete until we understand

the nature, properties and

behaviours of waves. The goal of this

unit is to develop mental models of

waves.

In conclusion, a wave can be

described as a disturbance that

travels through a medium,

transporting energy from one location

(its source) to another location

without transporting matter.

Each individual particle of themedium is temporarily displaced and

then returns to its original

equilibrium positioned.

A pulse is a single disturbance

moving through a medium from one

location to another location.

The repeating and periodic

disturbance that moves through a

medium from one location to another

is referred to as a wave. Amedium is a substance or material

that carries the wave. Waves are said to be an energy

transport phenomenon.

Categories of Waves

A transverse wave is a wave in

which particles of the medium move

in a direction perpendicular to the

direction that the wave moves. Transverse waves are always

characterized by particle motion

being perpendicular to wave motion.

A longitudinal wave is a wave in

which particles of the medium move

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in a direction parallel to the direction

that the wave moves.Longitudinal waves are always

characterized by particle motion being parallel to wave motion.

Anelectromagnetic wave is a wave

that is capable of transmitting itsenergy through a vacuum (i.e., empty

space). Electromagnetic waves are

produced by the vibration of charged

particles. Electromagnetic waves that

are produced on the sun

subsequently travel to Earth through

the vacuum of outer space.

Amechanical wave is a wave that is

not capable of transmitting its energy

through a vacuum. Mechanical waves

require a medium in order to

transport their energy from one

location to another. A sound wave is

an example of a mechanical wave.

The Anatomy of a Wave

Atransverse wave is a wave in which

the particles of the medium are

displaced in a direction

perpendicular to the direction of

energy transport.

A transverse wave can be created in arope if the rope is stretched out

horizontally and the end is vibrated

back-and-forth in a vertical direction.

The dashed line drawn through thecenter of the diagram represents the

equilibrium or rest position of the

string. This is the position that the

string would assume if there were no

disturbance moving through it. Thecrest of a wave is the point on

the medium that exhibits the

maximum amount of positive orupward displacement from the rest

position. Points C and J on the

diagram represent the troughs of this

wave.

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Thetrough of a wave is the point on

the medium that exhibits the

maximum amount of negative or

downward displacement from the restposition. The wave shown above can be

described by a variety of properties.

One such property is amplitude. Theamplitude of a wave refers to the

maximum amount of displacement of

a particle on the medium from its

rest position. In a sense, theamplitude is the distance from rest to

crest. Similarly, the amplitude can be

measured from the rest position to

the trough position.In the diagram above, the amplitude

could be measured as the distance of

a line segment that is perpendicular

to the rest position and extends vertically upward from the rest

position to point A.

The energy transported by a wave is

directly proportional to the square of

the amplitude of the wave. This

energy-amplitude relationship is

sometimes expressed in the following

manner.

The wavelength is another property of

a wave that is portrayed in the

diagram above. The wavelength of a wave is simply

the length of one complete wave

cycle. In the diagram above, the

wavelength is the horizontal distance

from A to E, or the horizontal

distance from B to F, or the

horizontal distance from D to G, or

the horizontal distance from E to H.

Any one of these distance

measurements would suffice in

determining the wavelength of this

wave.

A longitudinal wave is a wave in

which the particles of the mediumare displaced in a direction parallel to

the direction of energy transport.

A longitudinal wave can be created in

a slinky if the slinky is stretched out

horizontally and the end coil is

vibrated back-and-forth in a

horizontal direction.

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A region where the coils are pressed

together in a small amount of space

is known as a compression. A compression is a point on a

medium through which a

longitudinal wave is travelling that

has the maximum density. A region

where the coils are spread apart,

thus maximizing the distance

between coils, is known as a

rarefaction. Ararefaction is a point on a medium

through which a longitudinal wave is

travelling that has the minimum

density. Points A, C and E on the

diagram above represent

compressions and points B, D, and F

represent rarefactions. While a transverse wave has an

alternating pattern of crests and

troughs, a longitudinal wave has an

alternating pattern of compressions

and rarefactions.In the case of a longitudinal wave, a

wavelength measurement is made by

measuring the distance from a

compression to the next compression

or from a rarefaction to the next

rarefaction.

On the diagram above, the distancefrom point A to point C or from point

B to point D would be representative

of the wavelength. Thefrequency of a wave refers to

how often the particles of the

medium vibrate when a wave passes

through the

medium.

Unit for

frequency is theHertz (abbreviated

Hz) where 1 Hz is equivalent to 1

cycle/second. Theperiod of a wave is the time for a

particle on a medium to make one

complete Vibrational cycle. Period,

being a time, is measured in units of

time such as seconds.

In equation form,

Since the symbolf is used for

frequency and the symbol T is used

for period, these equations are alsoexpressed as:

Thespeed of an object refers to how

fast an object is moving and is

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usually expressed as the distance

travelled per time of travel. In the

case of a wave, the speed is the

distance travelled by a given point onthe wave (such as a crest) in a given

interval of time. In equation form,

Speed = Wavelength • Frequency

The above equation is known as the

wave equation. It states the

mathematical relationship between

the speed ( v) of a wave and its

wavelength (λ) and frequency (f).

Using the symbols v, λ, andf, the

equation can be rewritten as v = f • λ

Example 15.1

Two waves on identical strings have

frequencies in a ratio of 2 to 1. If

their wave speeds are the same, then

how do their wavelengths compare?

a. 2:1 b. 1:2 c. 4:1 d. 1:4 Answer:BFrequency and wavelength are

inversely proportional to each other.

The wave with the greatest frequency

has the shortest wavelength. Twice

the frequency means one-half the

wavelength. For this reason, the

wavelength ratio is the inverse of thefrequency ratio.

Sine Waves and Periodic Waves

Functions

We can write the wave function for an

arbitrary disturbance as y(x,t) = f(x - vt) with f(x) describing an arbitrary

function.For periodic waves we can use

sin/costo give functionality to the

wave as they are periodic in 2For a periodic wave, the equation of a

plane progressive simple harmonic

wave function is y(x,t ) = Asin 2 /λ( x – vt)

The term 2 scales the wave to thenatural period of the sin function. The term A gives the amplitude of the

wave which is the maximum

displacement of the wave. A more elegant way of writing the

wave function is y (x,t ) = A sin(kx -wt +φ) The phase of the wave is kx - wt + φ.The phase is always measured in

radians.

The term k = 2 /λ is called the

angular wave number.

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The term ω = 2 v/λ = 2 fand is the

angular frequency. The term φis the initial phase of the

wave. The initial phase of a wave

A harmonic wave is generally

described by the wave function y(x,t) = A sin(kx - wt + φ) To what does φ correspond?Let us set x = 0m and t = 0 s. y(0,0) = A sin(φ)So φ gives the displacement of the

wave at x = 0m and t = 0 s.Hence the name- initial phase.φ does not change the sequence of

the events in a wave it only makes

them happen sooner or later in the

sequence.

Phase and Phase Difference

Phase of the wave is given by the

argument of sine or cosine in the

equation of the wave.Φ(x,t) = 2 /λ(vt – x) + Φ0

Φ0represents initial phase.dΦ/dt = 2 v/λ = 2 /TdΦ/dx = -2 /λNegative sign shows phase lag. It

means that when a particle is atlarger distance from the origin, its

phase lag is greater.Phase difference between any two

particles in a wave determines lack of

harmony in the vibrating state of two

particles, ie how for one particle

leads the other or lags behind the

other.

Particle Velocity and Acceleration

Consider the restoring force in SHM.

How is it related to the displacement

of the oscillating particle?

You will notice that the acceleration

and therefore (by Newton's second

law) the resultant force is always in

the opposite direction to the

displacement of the pendulum. This

is the case forall examples of SHM.For the spring we know from Hooke's

Law that...F = -kx

F = ma

So equating the two and rearranging,

equation for the restoringacceleration of a mass-spring

system...a= - (k/m) x

What was the relationship between

displacement and acceleration of an

oscillating system?

Acceleration is negatively

proportional to displacement...a ∞ -xand we can always re-write equations

of this form...

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a = constant (-x)For SHM the constant is ω2

a = - ω2x

This is the generally defining

equation for all SHM.ω is the angular velocity of an object

performing circular motion, and it is

here that we come back to the close

link between circular motion and

SHM.

Acceleration is the second derivative

of position with respect to time, and

thus we can rewrite equation (1) as...a= d2 x/dt2 = -w2 xNow going back to our graphs of the

spring from the previous lesson...

We can see that if we chose to define

the position of the mass-spring

system at a particular time we could

write an equation of the form...

x = x0sin wt

Taking its derivative with respect to

time gives us the velocity …

dx/dt = x0w cos wt

Taking the derivative again with

respect to time gives us

acceleration...

d

2

x/dt

2

= -x0w

2

sin wt And once again we see that a = -w2 x.NB:Depending on the shape of the

graph you start with you could

equally well usex = x0coswtas your

starting equation and adjust the first

and second derivatives appropriately.

Example 15.2

The wave function for a wave is given

by y (x,t ) = 0.02sin(0.4x - 50t +

0.8)m Find (a)

the amplitude, (b) the wavelength, (c)

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the period, (d) the initial phase and

(e) the wave speed.Solution

Compare the given wave function with the algebraic version. y (x,t ) =Asin(kx -wt +φ) y (x,t ) = 0.02sin(0.4x - 50t + 0.8)m y (x,t ) =Asin(kx -wt +φ)(a) Amplitude A = 0.02 m(b) Wave number k = 0.4 rad m-1. As

k = 2 /λ then λ = 5 m(c) Angular frequency w = 50 rad s-1.

As w = 2 f and f = 1/T then T = 2 /ω.Hence T = /25s.

(d) Initial phase φ = 0.8 rad.(e) Wave speed v = ω /k = 50 rad s-1

or 0.4 rad m-1 = 125 m s-1

Characteristics of Wave Motion

• Wave motion is the propagation

of a disturbance produced in a

medium by the repeated periodic

motion of the particles of the

medium.• It is only the energy of the

disturbance that moves forward

without any mass transfer.• In an elastic medium, the

frequency and the amplitude of

oscillation of each particle is the

same. However, each particle starts

vibrating a little later than its

predecessor and thus, there is a

regular phase difference between

the successive particles of themedium.

• The wave velocity is different

from the particle velocity. While the

wave travels with a uniform

velocity, the velocity of the particle

is different at different positions. It

is maximum at the mean position

and zero at the extreme positionsof oscillation.

• Waves may undergo reflection,

refraction, interference, dispersion

and diffraction.• The energy of a wave is partly

kinetic and partly potential and the

total energy remains constant.

Frequency and Pitch

The human ear is capable of

detecting sound waves with a wide

range of frequencies, ranging between

approximately 20 Hz to 20 000 Hz.

Any sound with a frequency below

the audible range of hearing (i.e., less

than 20 Hz) is known as an

infrasound and any sound with a

frequency above the audible range of

hearing (i.e., more than 20 000 Hz) is

known as anultrasound.

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The sensation of a frequency is

commonly referred to as thepitch of

a sound. A high pitch sound

corresponds to a high frequencysound wave and a low pitch sound

corresponds to a low frequency

sound wave.

Speed of a transverse waves on a

stretched string Assume that the velocity of the wave

v depends upon,(a) the tension in the string (T),

(b) the mass of the string (M) and

(c) the length of the string (L)

Therefore: v = kT xM yLz

Solving this gives x = ½, z = ½, y = -½

The constant k can be shown to be

equal to 1 in this case and we write

m as the mass per unit length where

m = M/L. The formula therefore becomes:

Velocity of waves on a stretchedstring = [T/m]1/2

Velocity = √ /ρ = modulus of rigidity and ρ is

density of material of solid.Since velocity = frequency x

wavelengthFrequency of a vibrating string

n = [T/m]1/2 / λ

Speed of longitudinal waves

Velocity of Sound in Gases (Newton's

Formula)

We know that the properties of a

medium that govern the propagation

of a mechanical wave are:

• a restoring force• an inertial mass

The restoring force acting on the

particles of the medium is intimately

connected to the approximate elastic

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modulus of the medium and the

inertial mass, to its density.In 1676, Newton derived an

expression for the velocity of sound ina homogenous medium. He showed

that

where v is the velocity of sound, E is

the modulus of elasticity and r is the

density of the medium. When the

medium is a gas only the bulk

modulus is to be considered.

Where B is the bulk modulus of

elasticity. Newton assumed that the

temperature remains constant when

sound travels through a gas (air).

Therefore the process is isothermaland Boyle's law can be applied. At a

region of compression, the pressure

increases and volume decreases.Let the initial pressure = PInitial volume = VIncrease in pressure = PDecrease in volume = V Then, final pressure = P + PFinal volume = V - V Applying Boyle's law, (P + P) (V - V)

= PVExpanding the terms, PV - P V + V

P - P.V = PV

Since the changes in pressure and

volume are small, P. V can be

neglected. Then,

Comparing equations 3 and 4 We get B = P Therefore, Newton's formula for

velocity of sound can be written as

At NTP., with pressure of air = P =0.76 x 9.8 x 13.6 x 103

= 1.013 x 105 Pa or Nm-2

Laplace's Correction to Newton's

Formula

Assuming isothermal conditions to

prevail when sound travels through

air, Newton has applied Boyle's law to

the changes in pressure and volume.

In a region of compression, there is a

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slight increase in temperature and in

a region of rarefaction, there is a

slight decrease in temperature. These

changes in pressure occur rapidlyand air is a poor conductor of heat

thus, equalization of temperature

among the different regions was

improbable, according to Laplace.. He

was of the view that the changes in

temperature occur under adiabatic

conditions, i.e., no heat enters the

gas from outside or leaves it frominside. The heat developed in the

compressed layers remains fully

confined to those layers and has no

time to get dissipated into the entire

body of the gas. Similarly, the cold

caused in the rarefied layers cannot

be compensated for, by flow of heat

into it from other layers. Thus Boyle's law does not apply in

this case. The relation between pressure and

volume of a gas under adiabatic

conditions is given by

where i.e., the ratio of the

principle specific heats of the gas at

constant pressure and constant

volume respectively.

Let the pressure change by an

amount DP, producing a change in

volume by DV. Then

Taking out from the 2nd factor in

the above expression,

But from binomial approximation,

Cancelling P on both sides and

neglecting the term containing P.

V because it is too small, we get

But the LHS in the above equation

represents the bulk modulus, B.

Substituting in equation (1-23), we

get

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This is known as Newton-Laplace

formula for the velocity of sound in a

gas.

This is in close agreement with the

experimental value.

Factors affecting velocity of soundEffect of temperature

If we supply heat to a gas, or raise

temperature, there are two

possibilities:

1) If the gas is free to expand, then

on being heated its volume will

increase while pressure will remain

constant. i.e, its density woulddecrease and pressure will remain

constant.

i.e, the value of ratio P/d will

increase.

2) If the gas is closed in a container,

then on being heated its pressure will

increase while the density will not

change. i.e, again the value of P/d will increase.

Thus in both situations,the speed

of sound will increase with

increase in temperature.

Effect of density

If we are considering a single gas

system, then the density is aconstant. Otherwise, velocity is

inversely proportional to square root

of density.

The velocity of sound in hydrogen is

greater than that in oxygen because

hydrogen is a lighter gas.

Effect of wind

The velocity of sound is also affected

by the wind velocity. Suppose S is a

source of sound, O is an observer.

The sound travels along SO with

velocity v. If wind blows with velocity

u in the direction SQ making an

angle θ with SO, then component of

wind velocity along SO is u cosθ. This

component is added to velocity of

sound, and sound travels in the

direction SO with a velocity,

V = v + u cosθ

Thus, sound will travel faster, if θ is

acute and slower, if θ is obtuse.The

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wind will have no effect on velocity on

velocity of sound, if θ = 90.

Effect of Frequency There is no effect of frequency on the

speed of sound.Sound waves of

different frequencies travel with same

speed in air although their

wavelengths in air are different. In

case, the speed of sound through air

depended on frequency, then we

could not have enjoyed orchestra.

Effect of amplitude

If amplitude of sound waves is very

large, the compressions and

rarefactions may result in large

temperature variations and this may

affect the velocity of sound.

Reflection of Waves

Boundary Behavior

The behaviour of a wave (or pulse)

upon reaching the end of a medium

is referred to as boundary behaviour. When one medium ends,

another medium begins; the interface

of the two media is referred to as the

boundary and the behaviour of a

wave at that boundary is described

as its boundary behaviour.

Fixed End Reflection

First consider an elastic rope

stretched from end to end. One end

will be securely attached to a pole on

a lab bench while the other end will

be held in the hand in order to

introduce pulses into the medium.

Because the right end of the rope isattached to a pole (which is attached

to a lab bench) (which is attached to

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will be unable to move when a

disturbance reaches it. This end of

the rope

is

referred to as afixed end.

If a pulse is introduced at the left end

of the rope, it will travel through the

rope towards the right end of the

medium. This pulse is called the

incident pulse since it is incident

towards (i.e., approaching) the

boundary with the pole. When the

incident pulse reaches the boundary,

two things occur:

• A portion of the energy carried

by the pulse is reflected and returns

towards the left end of the rope. The

disturbance that returns to the left

after bouncing off the pole is known

as thereflected pulse.• A portion of the energy carried

by the pulse istransmitted to the

pole, causing the pole to vibrate.

Because the vibrations of the pole are

not visibly obvious, the energy

transmitted to it is not typically

discussed. The focus of thediscussion will be on the reflected

pulse. What characteristics and

properties could describe its motion? When one observes the reflected

pulse off the fixed end, there are

several notable observations. First

the reflected pulse isinverted.

That is, if an upward displaced pulseis incident towards a fixed end

boundary, it will reflect and return as

a downward displaced pulse.Similarly, if a downward displaced

pulse is incident towards a fixed end

boundary, it will reflect and return as

an upward displaced pulse.

The inversion of the reflected pulse

can be explained by returning to our

conceptions of the nature of amechanical wave. When a crest reaches the end of a

medium ("medium A"), the last

particle of the medium A receives an

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upward displacement. This particle is

attached to the first particle of the

other medium ("medium B") on the

other side of the boundary. As the last particle of medium A pulls

upwards on the first particle of

medium B, the first particle of

medium B pulls downwards on the

last particle of medium A. This is merelyNewton's third law of

action-reaction. For every action,

there is an equal and oppositereaction. The upward pull on the first

particle of medium B has little effect

upon this particle due to the large

mass of the pole and the lab bench to

which it is attached. The effect of the downward pull on

the last particle of medium A (a pull

that is in turn transmitted to the

other particles) results in causing the

upward displacement to become a

downward displacement. The upward displaced incident pulse

thus returns as a downward

displaced reflected pulse. It is

important to note that it is the

heaviness of the pole and the lab

bench relative to the rope that causes

the rope to become inverted upon

interacting with the wall. When two media interact by exerting

pushes and pulls upon each other,

the most massive medium wins the

interaction. Just like in arm wrestling,

the medium that loses receives a

change in its state of motion.Other notable characteristics of the

reflected pulse include:

• The speed of the reflected pulse

is the same as the speed of the

incident pulse.• The wavelength of the reflected

pulse

is the

same

as the

wavelength of the incident pulse.• The amplitude of the reflected

pulse is less than the amplitude

of the incident pulse.

Of course, it is not surprising that

the speed of the incident and

reflected pulse are identical since the

two pulses are travelling in the same

medium. Since the speed of a wave

(or pulse) is dependent upon the

medium through which it travels,two pulses in the same medium will

have the same speed. A similar line of

reasoning explains why the incident

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and reflected pulses have the same

wavelength.Every particle within the rope will

have the same frequency. Beingconnected to one another, they must

vibrate at the same frequency.Since the wavelength of a wave

depends upon the frequency and the

speed, two waves having the same

frequency and the same speed must

also have the same wavelength.Finally, the amplitude of the reflected

pulse is less than the amplitude of

the incident pulse since some of the

energy of the pulse was transmitted

into the pole at the boundary. The reflected pulse is carrying less

energy away from the boundary

compared to the energy that the

incident pulse carried towards the

boundary.Since the amplitude of a pulse is

indicative of the energy carried by the

pulse, the reflected pulse has a

smaller amplitude than the incident

pulse.

Free End Reflection

Now consider what would happen if

the end of the rope were free to move.

Instead of being securely attached toa lab pole, suppose it is attached to a

ring that is loosely fit around the

pole.Because the right end of the rope is

no longer secured to the pole, the last

particle of the rope will be able to

move when a disturbance reaches it. This end of the rope is referred to as

afree end.Once more if a pulse is introduced at

the left end of the rope, it will travel

through the rope towards the right

end of the medium. When the incident pulse reaches the

end of the medium, the last particle

of the rope can no longer interact

with the first particle of the pole.

Since the rope and pole are no longer

attached and interconnected, they

will slide past each other.So when a crest reaches the end of

the rope, the last particle of the rope

receives the same upward

displacement; only now there is no

adjoining particle to pull downwardupon the last particle of the rope to

cause it to be inverted. The result is that the reflected pulse

is not inverted. When an upward

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displaced pulse is incident upon a

free end, it returns as an upward

displaced pulse after reflection.

And when a downward displacedpulse is incident upon a free end, it

returns as a downward displaced

pulse after reflection. Inversion is not

observed in free end reflection.

Transmission of a Pulse across a

Boundary from Less to More Dense

Let's consider a thin rope attached to

a thick rope, with each rope held at

opposite ends by people. Andsuppose that a pulse is introduced by

the person holding the end of the

thin rope.If this is the case, there will be an

incident pulse travelling in the less

dense medium (the thin rope)

towards the boundary with a more

dense medium (the thick rope).

Upon reaching the boundary, the

usual two behaviors will occur.

•

A portion of the energy carried by the incident pulse is reflected and

returns towards the left end of the

thin rope. The disturbance that

returns to the left after bouncing off

the boundary is known as the

reflected pulse.• A portion of the energy carried

by the incident pulse is transmittedinto the thick rope. The disturbance

that continues moving to the right is

known as thetransmitted pulse.

The reflected pulse will be found to

be inverted in situations such as

this. During the interaction between

the two media at the boundary, the

first particle of the more dense

medium overpowers the smaller mass

of the last particle of the less dense

medium. This causes an upward displaced

pulse to become a downward

displaced pulse. The more dense

medium on the other hand was atrest prior to the interaction. The first

particle of this medium receives an

upward pull when the incident pulse

reaches the boundary. Since the

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more dense medium was originally at

rest, an upward pull can do nothing

but cause an upward displacement.

For this reason, the transmittedpulse is not inverted. In fact,

transmitted pulses can never be

inverted. Since the particles in this

medium are originally at rest, any

change in their state of motion would

be in the same direction as the

displacement of the particles of the

incident pulse. TheBefore and After snapshots of the

two media are shown in the diagram

below.

Comparisons can also be made

between the characteristics of thetransmitted pulse and those of the

reflected pulse. Once more there are

several noteworthy characteristics.

• The transmitted pulse (in the

denser medium) is travelling

slower than the reflected pulse (in

the less dense medium).• The transmitted pulse (in the

denser medium) has a smaller

wavelength than the reflected

pulse (in the less dense medium).• The speed and the wavelength

of the reflected pulse are the same

as the speed and the wavelength

of the incident pulse.

The speed of a wave is dependent

upon the properties of the medium.

In this case, the transmitted and

reflected pulses are travelling in two

distinctly different media. Waves always travel fastest in the

least dense medium. Thus, the

reflected pulse will be travelling faster

than the transmitted pulse. Second,

particles in the denser medium will

be vibrating with the same frequency

as particles in the less dense

medium.Since the transmitted pulse was

introduced into the denser medium

by the vibrations of particles in theless dense medium, they must be

vibrating at the same frequency. So

the reflected and transmitted pulses

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have the different speeds but the

same frequency.Since the wavelength of a wave

depends upon the frequency and thespeed, the wave with the greatest

speed must also have the greatest

wavelength.Finally, the incident and the reflected

pulse share the same medium. Since

the two pulses are in the same

medium, they will have the same

speed.Since the reflected pulse was created

by the vibrations of the incident

pulse, they will have the same

frequency. And two waves with the same speed

and the same frequency must also

have the same wavelength.

Transmission of a Pulse across a

Boundary from More to Less Dense

Finally, let's consider a thick rope

attached to a thin rope, with the

incident pulse originating in the

thick rope.If this is the case, there will be an

incident pulse travelling in thedenser medium (thick rope) towards

the boundary with a less dense

medium (thin rope).

Once again there will be partial

reflection and partial transmission at

the boundary.

The reflected pulse in this situation

will not be inverted. Similarly, the

transmitted pulse is not inverted (as

is always the case).Since the incident pulse is in a

heavier medium, when it reaches the

boundary, the first particle of the less

dense medium does not have

sufficient mass to overpower the last

particle of the denser medium.

The result is that an upward

displaced pulse incident towards the

boundary will reflect as an upward

displaced pulse.For the same reasons, a downward

displaced pulse incident towards the boundary will reflect as a downward

displaced pulse. TheBefore and After snapshots of the

two media are shown in the diagram

below.

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Comparisons between thecharacteristics of the transmitted

pulse and the reflected pulse lead to

the following observations.

• The transmitted pulse (in the

less dense medium) is traveling

faster than the reflected pulse (in

the more dense medium).• The transmitted pulse (in the

less dense medium) has a larger

wavelength than the reflected

pulse (in the more dense

medium).• The speed and the wavelength

of the reflected pulse are the

same as the speed and the

wavelength of the incident pulse.

The boundary behavior of waves

in ropes can be summarized by

the following principles:

• The wave speed is always

greatest in the least dense rope.• The wavelength is always

greatest in the least dense rope.• The frequency of a wave is not

altered by crossing a boundary.• The reflected pulse becomes

inverted when a wave in a less

dense rope is heading towards a

boundary with a more dense

rope.• The amplitude of the incident

pulse is always greater than the

amplitude of the reflected pulse.

The Principle of Superposition

The task of determining the shape of

the resultant demands that the

principle of superposition is applied. Theprinciple of superposition is

sometimes stated as follows:

When two waves interfere, the

resulting displacement of the

medium at any location is thealgebraic sum of the

displacements of the individual

waves at that same location.

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Analytical Treatment

Consider two waves which are out of

phase by a constant angle φ, then

Y1 = rsin(ωt –kx) Y2 = rsin(ωt –kx+φ) According to superposition principle: y = y1 + y2(Simplified using trigonometric

relation sinA + sin B = 2cos(A-

B)/2sin(A+B)/2)

y = 2rcosφ/2 sin(ωt –kx + φ/2)

Amplitude = 2rcosφ/2

Wave interference is the

phenomenon that occurs when two

waves meet while travelling along the

same medium.

if two upward displaced pulses

having the same shape meet up withone another while travelling in

opposite directions along a medium,

the medium will take on the shape of

an upward displaced pulse with twice

the amplitude of the two interfering

pulses. This type of interference is

known as constructive

interference.If an upward displaced pulse and a

downward displaced pulse having the

same shape meet up with one

another while travelling in opposite

directions along a medium, the two

pulses will cancel each other's effect

upon the displacement of the

medium and the medium will assumethe equilibrium position. This type of

interference is known asdestructive

interference. The diagrams below show two waves -

one is blue and the other is red -

interfering in such a way to produce

a resultant shape in a medium; the

resultant is shown in green.In two cases (on the left and in the

middle), constructive interference

occurs and in the third case (on the

far right, destructive interference

occurs.

Traveling Waves vs. Standing Waves

A mechanical wave is a disturbance

that is created by a vibrating object

and subsequently travels through a

medium from one location to another,

transporting energy as it moves. The mechanism by which a

mechanical wave propagates itself

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through a medium involves particle

interaction;

one particle applies a push or pull on

its adjacent neighbour, causing a

displacement of that neighbour fromthe equilibrium or rest position. As a wave is observed travelling

through a medium, a crest is seen

moving along from particle to

particle. This crest is followed by a

trough that is in turn followed by the

next crest.In fact, one would observe a distinct

wave pattern (in the form of a sine

wave) travelling through the medium.

This sine wave pattern continues to

move in uninterrupted fashion until

it encountersanother wave along

the medium or until it encountersa

boundary with another medium. This type of wave pattern that is seen

travelling through a medium is

sometimes referred to as atravelling

wave. Travelling waves are observed when a

wave is not confined to a given space

along the medium. The most

commonly observed travelling wave is

an ocean wave.

What is a Standing Wave Pattern?

It is however possible to have a wave

confined to a given space in amedium and still produce a regular

wave pattern that is readily

discernible amidst the motion of the

medium.For instance, if an elastic rope is held

end-to-end and vibrated at just the

right frequency, a wave pattern would

be produced that assumes the shapeof a sine wave and is seen to change

over time. The wave pattern is only

produced when one end of the rope is

vibrated at just the right frequency.

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When the proper frequency is used,

the interference of the incident wave

and the reflected wave occur in such

a manner that there are specificpoints along the medium that appear

to be standing still. Because the

observed wave pattern is

characterized by points that appear

to be standing still, the pattern is

often called a standing wave

pattern.

There are other points along themedium whose displacement changes

over time, but in a regular manner.

These points vibrate back and forth

from a positive displacement to a

negative displacement; the vibrations

occur at regular time intervals such

that the motion of the medium is

regular and repeating. A pattern isreadily observable.

Note that point A on the medium

moves from a maximum positive to a

maximum negative displacement over

time. The diagram only shows one-

half cycle of the motion of thestanding wave pattern. The motion would continue and

persist, with point A returning to the

same maximum positive

displacement and then continuing its

back-and-forth vibration between the

up to the down position.Note that point B on the medium is a

point that never moves. Point B is apoint of no displacement. Such

points are known asnodes and will

be discussed in more detaillater in

this lesson. The standing wave pattern that is

shown at the right is just one of

many different patterns that could be

produced within the rope. Otherpatterns will be discussedlater in the

lesson.

How is a Standing Wave Formed?

Let’s consider a snakey stretched

across the room, approximately 4-

meters from end to end. (A "snakey"is a slinky-like device that consists of

a large concentration of small-

diameter metal coils.)If an upward displaced pulse is

introduced at the left end of the

snakey, it will travel rightward across

the snakey until it reaches the fixed

end on the right side of the snakey.Upon reaching the fixed end, the

single pulse will reflect and undergo

inversion. That is, the upward

displaced pulse will become a

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downward displaced pulse. Now

suppose that a second upward

displaced pulse is introduced into the

snakey at the precise moment thatthe first crest undergoes itsfixed end

reflection.If this is done with perfect timing, a

rightward moving, upward displaced

pulse will meet up with a leftward

moving, downward displaced pulse in

the exact middle of the snakey.

As the two pulses pass through eachother, they will undergodestructive

interference. Thus, a point of no

displacement in the exact middle of

the snakey will be produced.

A standing wave pattern is an

interference phenomenon. It is

formed as the result of the perfectly

timed interference of two waves

passing through the same medium. A standing wave pattern is not

actually a wave; rather it is the

pattern resulting from the presence

of two waves (sometimes more) of the

same frequency with different

directions of travel within the same

medium. The physics of musical instruments

has a basis in the conceptual and

mathematical aspects of standing

waves.

What are Nodes and Antinodes?

One characteristic of every standing

wave pattern is that there are pointsalong the medium that appear to be

standing still. These points,

sometimes described as points of no

displacement, are referred to as

nodes. There are other points along

the medium that undergo vibrations

between a large positive and large

negative displacement. These are the points that undergo

the maximum displacement during

each Vibrational cycle of the standing

wave. In a sense, these points are the

opposite of nodes, and so they are

calledantinodes. A standing wave pattern always

consists of an alternating pattern ofnodes and antinodes.

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Harmonics

Standing wave patterns can only beproduced within the medium when it

is vibrated at certain frequencies.

There are several frequencies with

which the snakey can be vibrated to

produce the patterns.Each frequency is associated with a

different standing wave pattern.

These frequencies and their

associated wave patterns are referred

to asharmonics. The simplest standing wave pattern

that could be produced within a

snakey is one that has points of no

displacement (nodes) at the two ends

of the snakey and one point of

maximum displacement (antinode) in

the middle.First Harmonic Standing Wave Pattern

The above standing wave pattern is

known as thefirst harmonic. It isthe simplest wave pattern produced

within the snakey. The standing wave pattern

characterized by nodes on the two

ends of the snakey and an additional

node in the exact center of the

snakey is shown in the figure. As in all standing wave patterns,

every node is separated by anantinode. This pattern with three

nodes and two antinodes is referred

to as thesecond harmonic.

Second Harmonic Standing Wave

Pattern

The standing wave pattern for the

third harmonic has an additional

node and anti-node between the ends

of the snakey. The pattern is depicted

in the figure shown below.

Third Harmonic Standing Wave

Pattern

Analyzing the First Harmonic

Pattern

One complete wave in a standing

wave pattern consists of twoloops.

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Thus, one loop is equivalent to one-

half of a wavelength.

In comparing the standing wave

pattern for the first harmonic with its

single loop to the diagram of a

complete wave, it is evident that there

is only one-half of a wave stretching

across the length of the string. That is, the length of the string is

equal to one-half the length of a

wave. Put in the form of an equation:

Frequency (f1) = 1/2L[T/m]1/2

Analyzing the Second Harmonic

Pattern

Now consider the string being vibrated with a frequency that

establishes the standing wave pattern

for the second harmonic.

The second harmonic pattern

consists of two anti-nodes. Thus,

there are two loops within the lengthof the string.Since each loop is equivalent to one-

half a wavelength, the length of the

string is equal to two-halves of a

wavelength.

Put in the form of an equation:

Frequency (f2) = 1/L[T/m]1/2

F2 = 2f1

Analyzing the Third Harmonic

Pattern

The same reasoning pattern can be

applied to the case of the string being

vibrated with a frequency that

establishes the standing wave pattern

for the third harmonic.

The third harmonic pattern consists

of three anti-nodes. Thus, there are

three loops within the length of the

string.Since each loop is equivalent to one-

half a wavelength, the length of the

string is equal to three-halves of a wavelength. Put in the form of an

equation:

Frequency (f3) = 3/2L [T/m]1/2

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f3 = 3f1

When inspecting the standing wave

patterns and the length-wavelength

relationships for the first three

harmonics, a clear pattern emerges.

The number of antinodes in the

pattern is equal to the harmonic

number of that pattern. The first harmonic has one antinode;

the second harmonic has two

antinodes; and the third harmonic

has three antinodes. Thus, it can be generalized that the

nth harmonic hasn antinodes where

n is an integer representing the

harmonic number.Furthermore, one notices that there

are n halves wavelengths present within the length of the string. Put in

the form of an equation:

Frequency fn = n f1

Closed Organ Pipe

The organ pipe in which one end is

opened and another end is closed is

called organ pipe. Bottle, whistle, etc.

are examples of closed organ pipe. The different mode of the vibration in

the closed organ pipe is discussed as below.

A. First Mode Of Vibration

In the first mode of vibration in the

closed organ pipe, an antinode is

formed at the open end and a node is

formed at the close end. If ‘l’ be the

length of pipe and be the

wavelength of wave emitted in this

mode of vibration. Then,

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if ‘v’ be the velocity of sound and to

be the frequency of wave in this mode

of vibration.

This is fundamental frequency of 1st

harmonics

B. Second Mode Of Vibration:

In this mode of vibration two

antinodes and two nodes are formed

as in fig. If ‘λ’ be the length of pipeand be the wavelength of wave

emitted in this mode of vibration.

Then,

if v be the velocity of wave and f1 be

the frequency of wave, in this mode of

vibration

This is the frequency of 1st overtone

of third harmonics.

C. Third Mode Of Vibration:

In this mode of vibration three

antinodes and three nodes are

formed as in fig. If ‘l’ be the length of

the pipe and be the wavelength of

wave admitted in this mode of

vibration. Then,

If v be the velocity of wave and f2 be

the frequency of wave. If this mode of vibration,f2 = 5f0 This is the frequency of 2nd overtone

& firth harmonics.In this way, for the 4th, 5th…. modes

of vibration the frequency of wave

emitted are 7f0, 9f0……. which are

called 7th , 9th …… harmonics i.e.in

the closed organ pipe only odd

harmonic are present.

Open Organ Pipes

The pipe in which the both of its ends

are open is called open organ pipe.

Flute is the example of organ pipe.

Different modes of vibration in

open pipe

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A. First Mode Of Vibration:

In the mode of vibration in the organ

pipe, two antinodes are formed at twoopen ends and one node is formed in

between them. If length of the pipe be

‘l’ and be the wavelength of wave

emitted in this mode of vibration.

Then,

if ‘v’ be the velocity of sound and to be the frequency of wave in this mode

of vibration. Then,

This is the fundamental frequency of

1st overtone or 2 harmonious.

B. Second Mode Of Vibration: In the second mode of vibration in

the open organ pipe, there antinodes

are formed at two ends and two

nodes between them. If ‘l’ be the length of the pipe and λ1 be the wavelength of wave emitted in

this mode of vibration. Then,

If v be the velocity of sound and to

be the frequency of wave in this mode

of vibration, then,

This is the fundamental frequency of2nd overtone or harmonics.

C. Third Mode Of Vibration:

In the third mode of vibration in the

open organ pipe, four antinodes areformed and three nodes between

them. If ‘l’ be the length of the pipe

and λ2 be the wavelength of wave

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emitted in this mode of vibration

then

if v be the velocity of sound and f2 be

the frequency of wavelength on this

mode of vibration, then,

This is the fundamental of 2nd

overtone & 3rd harmonics. In this way, for the 4th, 5th

…………….. modes of vibration, the

frequency of wave are f3= 4f0; f4=

5f0…………….are the frequency of 4th,

5th harmonics.

Hence in open organ pipe odd and

even both harmonic are present.

Then the sound heard by open organ

is so sonorous.

Beats

Beats are caused by the interference

of two waves at the same point in

space. The phenomenon of alternate

variation in the intensity of sound

with time a a particular position,

when two sound waves of nearly

same frequencies and amplitudessuperimpose on each other is called

beats. The time interval between two

successive beats is called beat

period. The number of beats

produced per second is called the

beat frequency.

For the formation of beats,

frequencies of two sources of sound

should be nearly equal ie difference

in frequencies of two sources must be

small, say less than 10. This plot of the variation of resultant

amplitude with time shows the

periodic increase and decrease for

two sine waves.

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Analytical treatment

Two waves of equal amplitude are

travelling in the same direction. The

two waves have different frequencies

and wavelengths, but they both travel with the same wave speed.Using the principle of superposition,

the resulting particle displacement

may be written as:

y(x,t)=ym sin(k1 x - ω1t) + ymsin(k2 x +

ω2t)= 2 ymcos[( k1 - k2)2x - (ω1 -ω2)2t] x

sin [(k1 + k2)2x - (ω1 +ω2)2t]

This resulting particle motion is the

product oftwo travelling waves.One part is a sine wave which

oscillates with the average frequency

f = ½( f1 + f2). This is the frequency which is perceived by a listener. The other part is a cosine wave which

oscillates with the difference

frequency f = ½( f1 - f2).

This term controls the amplitude

"envelope" of the wave and causes the

perception of "beats".

The beat frequency is actually twicethe difference frequency, f beat = ( f1 - f2).In the figure given below, at left two

waves with slightly different

frequencies are travelling to the right.

Since the two waves are travelling in

the same medium, they travel with

the same speed. The resulting superposition sum

wave travels in the same direction

and with the same speed as the two

component waves, but its local

amplitude depends on whether the

two individual waves have the same

or opposite phase. The "beat" wave oscillates with the

average frequency, and its amplitude

envelope varies according to the

difference in frequency.

Applications of Beats

RADAR speed detectors bounce

microwave radiation off of moving

vehicles and detect the reflected waves. These waves are shifted in

frequency by theDoppler effect, and

the beat frequency between the

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directed and reflected waves provides

a measure of the vehicle speed.

Doppler Effect

The Doppler Effect is observed

whenever the source of waves is

moving with respect to an observer.

TheDoppler Effectcan be described

as the effect produced by a moving

source of waves in which there is an

apparent upward shift in frequency

for observers towards whom the

source is approaching and an

apparent downward shift in

frequency for observers from whom

the source is receding. It is important

to note that the effect does not result

because of an actual change in the

frequency of the source. You recall an instance in which a

police car or emergency vehicle was

travelling towards you on the

highway. As the car approached with

its siren blasting, the pitch of the

siren sound (a measure of the siren's

frequency) was high; and then

suddenly after the car passed by, the

pitch of the siren sound was low.

That was the Doppler effect - an

apparent shift in frequency for a

sound wave produced by a moving

source.

The Apparent change in the

frequency of sound when the source,the observer and the medium are in

a relative motion is calledDoppler

Effect. For the waves which require a

medium for their propagation, the

apparent frequency depends on three

factors :1. Velocity of the source2. Velocity of the observer3. Velocity of the medium or wind.

Doppler Shift Formula

Doppler Shift was given by Christian

Johann Doppler in 1842. This

apparent change in the frequency of

sound as a result of relative motion

between the source and the observer

is the Doppler Effect.

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Expression for apparent frequency

Let S be a source of sound and L, the

listener of sound. Both are initially at

rest. Let f be the actual frequency of

sound emitted by the source and λ bethe actual wavelength of the sound

emitted.

λ = v/fLet the distance between the source

and listener be v, so that ν waves

from the source reach the listener in

one second. Therefore, frequency of

sound heard by the listener is also f.Let us consider a situation, in which

the medium, source and listener

move in the same direction from S to

L with velocities vm, vs and vLrespectively. As a result of the motion of the

medium, the sound waves in one

second, travel a distance

= (v +vm). This is the resultant velocity of sound along SL.Further, the distance moved by the

source in one second = SS’ = vs, along

SL.

Relative velocity of sound with

respect to the source = [(v +vm) – vs].

This is the distance covered by the

sound waves in one second, relativeto the source. As the frequency of sound waves

remains unchanged by the motion of

the source or the medium, therefore,

v waves emitted in one second occupy

the distance [(v +vm) – vs]. Therefore,

the apparent wavelength λ’ of the

sound waves would become λ’ = [(v +vm) – vs]/ f These waves of wavelength λ’ travel

towards the listener, who himself

moves through a distance LL’ = vL in

one second.Relative velocity of sound waves with

respect to listener = (v +vm) – vL

This is the distance available in onesecond to waves of wavelength λ’. Apparent frequency of sound waves

heard by the listener isf’ =[ (v + vm) – vL]/λ’Substituting for λ’ = [(v +vm) – vs]/ f f’ = [(ν + νm) – νL] f [(ν + νm) – νs]

All velocities along the direction S toL are taken as positive and all

velocities along the direction L to S

are taken as negative.

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In case the medium is moving in the

opposite direction (from listener to

source), vm is negative. Now the

effective velocity of sound waves = (v –vm). Hence, observed frequency f’ is

given byf’ = [(ν - νm) – νL]/f

[(ν - νm) – νs]Special cases

There are eight Doppler Effect

Formulas for frequency depending on

cases:(i) When the source is moving

towards a observer at rest

(ii) When the source is moving away

from the observer at rest

(iii) When observer is moving towards

the stationary source

(iv) When observer moving away from

a stationary source

(vii) When the Source is approaching

the Stationary observer and observer

moving away from observer

(viii) When the Observer is

approaching the Stationary source

Where vs = velocity of the Source,

vo = velocity of the Observer, v = velocity of sound in

medium,

f = Real frequency,

f' = Apparent frequency.

Doppler effect formula is used to find

the apparent frequency and

wavelength for the source moving

towards the observer and away fromthe observer or observer moving

towards the source or away from the

source.Numerical

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