Chapter 04

OscillatiOns and Waves

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introduction

What we learn in this chapter about oscillations in mechanical systems and of the waves that

oscillating systems may set up, forms the basis for gaining an understanding of many other areas of physics.

A study of oscillations is important for many reasons, not least safety in design. For example oscillations may be set up in a bridge as traffic passes over it and these oscillations can lead to structural damage in the bridge. Many types of machines (lathes, car engines etc) are also subject to oscillations and again, these oscillations can produce damage.

Of course oscillating systems may also be very useful. The oscillations of a simple pendulum may be used as an accurate timing device and the oscillations set up in a quartz crystal may be used as an even more accurate timing device.

If an oscillating body causes other particles with which it is in contact to oscillate, then the energy of the oscillating body may be propagated as a wave. An oscillating tuning fork, vibrating string and vibrating reed cause the air molecules with which they are in contact to oscillate thereby giving rise to a sound wave that we may hear as a musical note. As we shall see, oscillating systems and waves are intimately connected.

On a fundamental level, all atoms and molecules are in effect oscillating systems. An understanding of these oscillations is crucial to understanding both the microscopic and macroscopic properties of a substance. For example, the dependence of specific heat capacity on temperature, (a topic well beyond the scope of an IB Physics Course) arises from studying atomic oscillations. Also, by analysing the oscillations of atoms and molecules, we gain an understanding of the interaction between matter and radiation. For example, we shall see in Chapter 8, that the Greenhouse Effect is essentially due to the interaction of infrared radiation with gases such as carbon dioxide.

It must also be mentioned that the oscillations of electrically charged particles give rise to electromagnetic waves (light, radio waves X-rays etc). This is examined in more detail in the Option on Electromagnetic waves (Option G). However, the concept of electromagnetic waves occurs in several other places in both the Core and AHL material.

We look first at the oscillations in mechanical systems.

4. 1 Kinematics of Simple Harmonic Motion (2H)

4.2 Energy changes during Simple Harmonic Motion (SHM)

4.3 Forced Oscillations and Resonance

4.4 Wave Characteristics

4.5 Wave Properties

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4.1.1 Describe examples of oscillations.

4.1.2 Define the terms displacement,amplitude, frequency,period,phasedifference.

4.1.3 Define simpleharmonicmotion(SHM) and state the defining equation as a= - ω2x.

4.1.4 Solve problems using the defining equation for simple harmonic motion.

4.1.5 Apply the equations v= -v0sin ωt,v= v

0

cosω t,v= ±ω√(x0 2 - x2),x= x

0cosωt,

x= x0sinωt,as solutions to the defining

equation for simple harmonic motion.

4.1.6 Solve problems, both graphically and by calculation, for acceleration, velocity and displacement during simple harmonic motion

© IBO 2007

4.1.1 ExamplEsOfOsCillating

systEms

In the introduction, we mentioned several oscillating systems. Here are a few more:

• a boat at anchor at sea• the human vocal chords• an oscillating cantilever• the Earth’s atmosphere after a large explosion.

4.1.2 DEfinitiOnsOftERms

assOCiatEDwithOsCillating

systEms

In order to understand the terms associated with oscillating systems, let us look at the oscillation of a so-called simple pendulum. A simple pendulum consists of a weight (called the bob) that is suspended vertically by a light string or thread as shown in Figure 401. The bob X is suspended by the thread from the point P.

X

BCE D

P

θ0

Figure401 Asimpleoscillatingsystem

To set the pendulum oscillating, the bob is pulled up to a position such as B where the angle XPB is θ0. It is then released. The bob will now oscillate between the positions B and C.

displacement (x, θ)

This refers to the distance that an oscillating system is from its equilibrium position at any particular instant during the oscillation. In the case of the simple pendulum the displacement is best measured as an angular displacement. For example, in Figure 401 when the bob is at the position D, the displacement is the angle XPD = θ and when at E, is the angle XPE = - θ. The displacement when the bob is at X is θ = 0.

amplitude (x0, θ

0)

This is the maximum displacement of an oscillating system from its equilibrium position. For the simple pendulum in Figure 401 this is clearly θ0.

Period (T)

This is the time it takes an oscillating system to make one complete oscillation. For the simple pendulum in Figure 401, this is the time it takes to go from X to B, B to C and then back to X.

4. 1 Kinematics of simPle Harmonic motion



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frequency (f)

This is the number of complete oscillations made by the system in one second.

relation between frequency and

period

The time for one complete oscillation is the period T. Therefore the number of oscillations made in one second is f 1T---= . The number of oscillations made in one second is

also defined as the frequency f, hence:

f 1

T---=

Equation 4.1

Phase difference

Suppose we have for instance, two identical pendulums oscillating next to each other. If the displacements of the pendulums are the same at all instances of time, then we say that they are oscillating in phase. If on the other hand the maximum displacement of one of them is θ0 when the maximum displacement of the other is -θ0, then we say that they are oscillating in anti-phase or that the phase difference between is 180°. The reason for the specification in terms of angle will become clear in section 4.1.5. In general, the phase difference between two identical systems oscillating with the same frequency can have any value between 0 and 360° (or 0 to 2π radians- see Figure 402). We shall see that the concept of phase difference is very important when discussing certain aspects of wave motion.

radian measure

When dealing with angular displacements, it is often useful to measure the displacement in radians rather than in degrees. In Figure 402, the angle θ measured in radians is defined as the arc length AB (l) divided by the radius r of the circle i.e

(rad) lr

θ =

0(180 ) rr

θ = = (rad)

ω = 2πf

2T

ω =

2Tω

= or 2T

ω =

Equation 4.2

r

A

B

l

θ

Figure402 Theradian

If θ = 180° then l is equal to half the circumference of the circle i.e. l = πr. Hence from equation 4.2, we have that

(rad) lr

θ =

0(180 ) rr

θ = = (rad)

ω = 2πf

2T

ω =

2Tω

= or 2T

ω =

Hence 1 radian (rad) is equal to 57.3°.

4.1.3 DEfinitiOnOfsimplE

haRmOniCmOtiOn(shm)

description of sHm

Suppose we were to attach a fine marking pen to the bottom of the bob of a simple pendulum and arrange for this pen to be in contact with a long sheet of white paper as shown in Figure 403.

Motion of pendulum

Direction of paper

Paper

Figure403 ArrangementfordemonstratingSHM


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As the pendulum oscillates, the paper is pulled at a constant speed in the direction shown.

Figure 404 shows a particular example of what is traced on the paper by the marker pen.

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0

0.1

0.2

0.3

0.4

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5time/s

disp

lacemen

t/cm

Figure404 Asampletrace

The displacement is measured directly from the trace and the time is calculated from the speed with which the paper is pulled.

There are several things to notice about the trace.

1. One complete oscillation is similar to a sine or cosine graph.

2. The period stays constant at about 0.8 s. (Oscillations in which the period is constant are called isochronous.)

3. The amplitude is decaying with time. This is because the pendulum is losing energy to the surroundings due to friction at the point of suspension and to air resistance.

Based on this sort of time trace, we can define a special type of oscillatory motion. Oscillators that are perfectly isochronous and whose amplitude does not change with time are called simple harmonic oscillators and their motion is referred to as simple harmonic motion (SHM). Clearly SHM does not exist in the real world as the oscillations of any vibrating system will eventually die out. Interestingly enough, the simple pendulum does not perform SHM for yet another reason, namely that the period is actually dependant on the amplitude. However, this only becomes noticeable as θ0 exceeds about 40°.

Although SHM does not exist in the real world, many oscillatory systems approximate to this motion. Furthermore, as part of the scientific method, it makes good sense to analyse a simple situation before moving onto more complex situations.

angular frequency (ω)

A very useful quantity associated with oscillatory motion is angular frequency, ω. This is defined in terms of the linear frequency as

(rad) lr

θ =

0(180 ) rr

θ = = (rad)

ω = 2πf

2T

ω =

2Tω

= or 2T

ω =

Equation 4.3

Using equation 4.1 we also have that

(rad) lr

θ =

0(180 ) rr

θ = = (rad)

ω = 2πf

2T

ω =

2Tω

= or 2T

ω =

Equation 4.4

There is a connection between angular frequency and the angular speed of a particle moving in a circle with constant speed. The angular speed of the particle is defined as the number of radians through which the particle moves in one second. If the time for one complete revolution of the circle is T , then from equation 4.2 we have that

(rad) lr

θ =

0(180 ) rr

θ = = (rad)

ω = 2πf

2T

ω =

2Tω

= or 2T

ω =

There is actually a physical connection between angular speed and SHM in the respect that it can be shown that the projection of the particle onto any diameter moves with SHM. See Figure 405.

P

N

As the particle P movesround the circle, itsprojection N ontoadiameter moves backwardsand forwards along thediameter with SHM.

Figure405 Circularandharmonicmotion

definition of sHm

If it were possible to remove all frictional forces acting on an oscillating pendulum, then the displacement –time graph for the motion would look like that in Figure 406. The amplitude does not decay with time. This therefore is a displacement-time graph for SHM



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-0.3

-0.2

-0.1

0

0.1

0.2

0.3

0.4

1 2 3 4 5time/s

disp

lacemen

t/cm

Figure406 Adisplacement–timegraph

It turns out that if the acceleration a of a system is directly proportional to its displacement x from its equilibrium position and is directed towards the equilibrium position, then the system will execute SHM. This is the formal definition of SHM.

We can express this definition mathematically as

a = -const x Equation 4.5

The negative sign indicates that the acceleration is directed towards equilibrium. Mathematical analysis shows that the constant is in fact equal to ω2 where ω is the angular frequency (defined above) of the system. Hence equation 4.5 becomes

a = -ω2 x Equation 4.6 This equation is the mathematical definition of SHM.

If a system is performing SHM, then to produce the acceleration, a force must be acting on the system in the direction of the acceleration. From our definition of SHM, the magnitude of the force F is given by

F = -kx Equation 4.7

where k is a constant and the negative sign indicates that the force is directed towards the equilibrium position of the system. (Do not confuse this constant k with the spring constant. However, when dealing with the oscillations of a mass on the end of a spring, k will be the spring constant.)

a matHematical PersPective of sHm

SHM is a very good example in which to apply the Newtonian method discussed in Chapter 2 i.e. if the forces that act on a system are known, then the future behaviour of the system can be predicted. Here we have a situation in which the force is given by –kx. From Newton’s second law therefore –kx = ma, where m is the mass of the system and a

is the acceleration of the system. However, the acceleration is not constant. For those of you who have a mathematical bent, the relation between the force and the acceleration is written as

2

2

ddxkx mt

− = . This is what is called a “second order differential equation”. The solution of the equation gives x as a function of t. The actual solution is of the SHM equation is

x = Pcosωt + Qsinωt where P and Q are constants and ω is the angular frequency of the system and is equal to . Whether a particular solution involves the sine function or the cosine function, depends on the so-called ‘boundary conditions’. If for example x = x0 (the amplitude) when t = 0, then the solution is x = x0cosωt.

The beauty of this mathematical approach is that, once the general equation has been solved, the solution for all systems executing SHM is known. All that has to be shown to know if a system will execute SHM, is that the acceleration of the system is given by Equation 4.5 or the force is given by equation 4.7. The physical quantities that ω will depend on is determined by the particular system. For example, for a weight of mass m oscillating at the end of a vertically supported spring whose spring constant is k, then ω = or , from equation 4.4. For a simple pendulum, ω = where l is the length of the pendulum and g is the acceleration of free fall such that .

4.1.4 sOlvingpROblEmsinvOlving

a = -ω2xTo understand how the equation a = -ω2x is used in particular situations involving SHM’ let us look at an example.

toK a mathematical Perspective

Galileo stated that ‘the book of nature is written in mathematical terms’. In this respect, it is impossible to explain to somebody why the period of oscillation of a simple pendulum depends on the square root of its length without recourse to solving a second order differential equation.


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Example

1. A cylindrical piece of wood floats upright in water as shown in Figure 407.

pushdownwards

water

wood

l

Figure407 SHMofafloatingpieceofwood

The wood is pushed downwards and then released. The subsequent acceleration a of the wood is given by the expression

where ρ = density of water, σ = density of the wood, l = length of wood, g = acceleration of free fall and x = displacement of the wood from its equilibrium position.

(a) Explain why the wood executes SHM.Answer

The equation shows that the acceleration of the wood is proportional to its displacement from equilibrium and directed towards the equilibrium position.

(b) The length of the wood is 52 cm and it is pushed downwards a distance of 24 cm. Calculate the maximum acceleration of the wood. (ρ = 1.0 × 103 kg m-3, σ = 8.4 × 102 kg m-3, g = 9.8 m s-2).

Answer

Maximum acceleration is when x = A (the amplitude)

= 2.4 × 10-2 m

therefore maximum acceleration

= 3 2

2 2

1.0 10 9.8 24 108.4 10 52 10

−

−

× × × ×× × ×

= 5.4 m s-2

(c) Calculate the period T of oscillation of the wood.

Answer

ω2 =

rad2

ω = 4.8 rad

= 1.3 s

(d) State and explain in terms of the period T of oscillation of the wood, the first two instances when the acceleration is a maximum.

Answer

The amplitude will be a maximum at T = 0 and again when T =

so acceleration is a maximum at T = 0 and T = 0.65 s

4.1.5 sOlutiOnsOfthEshm

EquatiOn

To understand the solutions of the SHM equation, let us consider the oscillations of a mass suspended from a vertically supported spring. We shall consider the mass of the spring to be negligible and for the extension x to obey the rule F = kx for all values of x. F is the force causing the extension. Figure 408(a) shows the spring and a suspended weight of mass m in equilibrium. In Figure 408(b), the weight has been pulled down a further extension x0.

spring

weight of mass mx0

x0

Px

unbalancedforce F onweight =-kx

a b

equilibriumx = 0

unstretched length

eke

mg

Figure408 SHMofamasssuspendedbyaspring

In Figure 408 (a), the equilibrium extension of the spring is e and the net force on the weight is mg - ke = 0.



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In Figure 408 (b), if the weight is held in position then released, then when the weight is at position P say, distance x from the equilibrium position x = 0, then the net force on the weight is mg - ke – kx. Clearly, then the unbalanced force on the weight is –kx. When the weight reaches a point distance x above the equilibrium position, the compression force in the spring provides the unbalanced force towards the equilibrium position of the weight.

The acceleration of the weight is given by Newton’s second law;

F = -kx = ma

i.e. Equation 4.8

This is of the form a = -ω2 x where ω = , that is the weight will execute SHM with a frequency

.

The displacement of the weight x, the solution of the SHM equation, is given by

ω Equation 4.9

This is the particular solution of the SHM equation for the oscillation of a weight on the end of a spring. This system is often referred to as a harmonic oscillator.

The velocity v of the weight at any instant can be found by finding the gradient of the displacement-time graph. From equation 4.9, the displacement graph is a cosine function and the gradient of a cosine function is a negative sine function. The gradient of

x = x0cosωt is in fact ω ω so

ω ω ω Equation 4.10

where v0 is the maximum and minimum velocity equal in magnitude to ωx0.

Students familiar with calculus will recognise the velocity v as

( )0d d cos sind d oxv x t x tt t

ω ω ω= = = − . Similarly,

( ) 2 20 0

d d sin cosd dva x t x t xt t

ω ω ω ω ω= = − = − = −

which of course is just the defining equation of SHM.

However, we have to bear in mind that ωt varies between 0 and 2π such that the cosωt is negative for ωt in the

and sinωt is negative for ωt in the range π to

2π. This effectively means when the displacement from equilibrium is positive, the velocity is negative and so directed towards equilibrium. When the displacement from equilibrium is negative, the velocity is positive and so directed away from equilibrium

The sketch graph in Figure 409 shows the variation with time t of the displacement x and the corresponding variation with time t of the velocity v. This clearly demonstrates the relation between the sign of the velocity and sign of the displacement.

T 2T0,0

0,0

velo

city

dis

pla

cem

ent

t

Figure409 Displacement-timeandvelocity-timegraphs

We can also see how the velocity v changes with displacement x.

From equation 4.10, we have that

However we can express sinωt in terms of cosωt using the trigonometric relation

2 2sin cos 1θ θ+ =

From which it can be seen that

Replacing θ with ωt we have

20 1 cosv x tω ω= − −

Remembering that and putting x0 inside the square root gives


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Bearing in mind that v can be positive or negative, we must write

Equation 4.11

The velocity is zero when the displacement is a maximum and is a maximum when the displacement is zero.

The graph in Figure 410 shows the variation with x of the velocity v for a system oscillating with a period of 1 sec and with an amplitude of 5 cm. The graph shows the variation over a time of any one period of oscillation.

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-4

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-2

-1

1

2

3

4

5

6

7

-5 -4 -3 -2 -1 1 2 3 4 5

v/

cms

-1

x / cm

Figure410 Velocity-displacementgraph

Boundary conditions

The two solutions to the general SHM equation are

and . Which solution applies to a particular system depends, as mentioned above, on the boundary conditions for that system. For systems such as the harmonic oscillator and the simple pendulum, the boundary condition that gives the solution is that the displacement x = x0 when t = 0. For some other systems it might turn out that x = 0 when t =0. This will lead to the solution . From a practical point of view, the two solutions are essentially the same; for example when timing the oscillations of a simple pendulum, you might decide to start the timing when the pendulum bob passes through the equilibrium position. In effect, the two solutions differ in phase by .

The table in Figure 411 summarises the solutions we have for SHM.

x = x 0 cosωt x = x 0 sinωtν = − ν 0 sinωt ν = ν 0 cosωt

ν = – ωx 0 sinωt ν = – ωx 0 cosωtν = ± ω √

_______ ( x 0 – x 2 ) ν = ± ω √

_______ ( x 0 – x 2 )

Figure411 Commonequations

We should mention that since the general solution to the SHM equation is there are in fact three solutions to the equation. This demonstrates a fundamental property of second order differential equations; that one of the solutions to the equation is the sum of all the other solutions. This is the mathematical basis of the so-called principle of superposition.

4.1.6 sOlvingshmpROblEms

In this section we look at an example of a typical SHM problem and its solution.

Examples

The graph in Figure 412 shows the variation with time t of the displacement x of a system executing SHM.

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-8

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-4

-2

0

2

4

6

8

10

0.5 1 1.5 2 2.5 3 3.5t /s

x/cm

Figure412 Displacement–timegraphforSHM

Use the graph to determine the

(i) period of oscillation(ii) amplitude of oscillation(iii) maximum speed(iv) the speed at t = 1.3 s(v) maximum acceleration



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Solutions

(a) (i) 2.0 s(ii) 8.0 cm(iii) Using gives

(remember that ) = 25 cm s-1

(iv) v = −v0sinωt = −25sin (1.3π). To find the value of the sine function, we have to convert the 1.3π into degrees (remember ω and hence ωt, is measured in radians)

1 deg therefore 1.3π = 1.3 × 180

= 2340 therefore v1 = −25sin (234) = +20 cm s-1.

Or we can solve using

from the graph at t = 1.3 s, x = -4.8 cm

therefore v = π x = 20 cm s-1

(v) Using = π2 × 8.0 = 79 m s-2

Exercise

Answer the same questions (a)(i) to (a)(iv) in the above example for the system oscillating with SHM as described by the graph in Figure 413. Also state two values of t at which the magnitude of the velocity is a maximum and two values of t at which the magnitude of the acceleration is a maximum.

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-1

0

1

2

3

4

5

6

7

0.5 1 1.5 2 2.5 3 3.5t /s

x/cm

4.2 energy cHanges during simPle Harmonic motion (sHm)

4.2.1 Describe the interchange between kinetic energy and potential energy during SHM 4.2.2 Apply the expression E

K= ½mω2(x02–x2)for

the kinetic energy of a particle undergoing simple harmonic motion, ET= ½mω2x02for the total energy and E

P= ½mω2x2for the potential

energy.

4.2.3 Solve problems, both graphically and by calculation, involving energy changes during simple harmonic motion.

© IBO 2007

4.2.1 KinEtiCanDpOtEntial

EnERgyChangEs

We must now look at the energy changes involved in SHM. To do so, we will again concentrate on the harmonic oscillator. The mass is stationary at x = +x0 (maximum extension) and also at x = -x0 (maximum compression). At these two positions the energy of the system is all potential energy and is in fact the elastic potential energy stored in the spring. This is the total energy of the system ET and clearly

Equation 4.12

That is, that for any system performing SHM, the energy of the system is proportional to the square of the amplitude. This is an important result and one that we shall return to when we discuss wave motion.

At x = 0 the spring is at its equilibrium extension and the velocity v of the oscillating mass is a maximum v0. The energy is all kinetic and again is equal to ET. We can see that this is indeed the case as the expression for the maximum kinetic energy Emax in terms of v0 is

Equation 4.13


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Clearly ET and Emax are equal such that

=

From which

Therefore

which ties in with the velocity being equal to the gradient of the displacement –time graph. (See 4.1.5) As the system oscillates there is a continual interchange between kinetic energy and potential energy such that the loss in kinetic energy equals the gain in potential energy and ET = EK + EP.

4.2.2 thEshmEnERgyEquatiOns

Remembering that , we have that

Equation 4.14

Clearly, the potential energy EP at any displacement x is given by

Equation 4.15

At any displacement x, the kinetic energy EK is Hence remembering that

, we have

Equation 4.16

Although we have derived these equations for a harmonic oscillator, they are valid for any system oscillating with SHM. The sketch graph in Figure 414 shows the variation with displacement x of EK and EP for one period.

displacement

energy

potential

kinetic

Figure414 Energyanddisplacement

4.2.3 sOlvingshmEnERgy

pROblEms

Let us now look at solving a problem involving energy in SHM.

Example

The amplitude of oscillation of a mass suspended by a vertical spring is 8.0 cm. The spring constant of the spring is 74 N m-1. Determine

(a) total energy of the oscillator

(b) the potential and the kinetic energy of the oscillator at a displacement of 4.8 cm from equilibrium.

Solution

(a) Spring constant k = 74 N m-1 and

x0 = 8.0 × 10-2 m

ET = ½ kx02

= ½ × 74 × 64 × 10-4 = 0.24 J

(b) at x = 4.8 cm

therefore EP = ½ × 74 × (4.8 × 10-2)2

= 0.085 × 10-4 J

EK = ET − EP = 0.24 − 0.085 = 0.16 J

Exercise

In a simple atomic model of a solid, the atoms vibrate with a frequency of 2.0 × 1011 Hz. The amplitude of vibration of the atoms is 5.5 × 10-10 m and the mass of each atom is 4.8 × 10-26 kg. Calculate the total energy of the oscillations of an atom.



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4.3.1 State what is meant by damping.

4.3.2 Describe examples of damped oscillations.

4.3.3 State what is meant by naturalfrequencyof vibrationand forcedoscillations.

4.3.4 Describe graphically the variation with forced frequency of the amplitude of vibration of an object close to its natural frequency of vibration.

4.3.5 State what is meant by resonance.

4.3.6 Describe examples of resonance where the effect is useful and where it should be avoided.

© IBO 2007

4.3.1 Damping

In this section, we look at oscillations of real systems. In section 4.1.3, we described an arrangement by which the oscillations of a pendulum could be transcribed onto paper. Refer to Figure 415.

time

disp

lace

men

t

Figure415 Damping

The amplitude of the oscillations gradually decreases with time whereas for SHM, the amplitude stays at the same value for ever and ever. Clearly, the pendulum is losing energy as it oscillates. The reason for this is that dissipative forces are acting that oppose the motion of the pendulum. As mentioned earlier, these forces arise from air resistance and though friction at the support. Oscillations, the amplitude of which decrease with time, are called damped oscillations.

4.3.2 ExamplEsOfDampED

OsCillatiOns

All oscillating systems are subject to damping as it is impossible to completely remove friction. Because of this, oscillating systems are often classified by the degree of damping. The oscillations shown in Figure 415 are said to be lightly damped. The decay in amplitude is relatively slow and the pendulum will make quite a few oscillations before finally coming to rest. Whereas the amplitude of the oscillations shown in Fig 416 decay very rapidly and the system quickly comes to rest. Such oscillations are said to be heavily damped.

time

amplitud

e

Figure416 Heavilydampedoscillations

Consider a harmonic oscillator in which the mass is pulled down and when released, and the mass comes to rest at its equilibrium position without oscillating. The friction forces acting are such that they prevent oscillations. However, suppose a very small reduction in the friction forces would result in heavily damped oscillation of the oscillator, then the oscillator is said to be critically damped.

The graph in Figure 417 shows this special case of damping known as critical damping.

4.3 forced oscillations and resonance


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0

0.2

0.4

0.6

0.8

1

1.2

0 0.2 0.4 0.6 0.8 1 1.2

time / s

disp

lace

men

t / m

Figure417 Criticaldamping

Although not in the IB syllabus, a useful way of classifying oscillating systems, is by a quantity known as the quality factor or Q-factor. The Q-factor does have a formal definition but it is approximately equal in value to the number of oscillations that occur before all the energy of the oscillator is dissipated. For example, a simple pendulum has a Q-factor of about 1000.

As mentioned in the introduction to this chapter, the oscillations (vibrations) made by certain oscillatory systems can produce undesirable and sometimes, dangerous effects. Critical damping plays an important role in these situations. For example, when a ball strikes the strings of a tennis racquet, it sets the racquet vibrating and these vibrations will cause the player to lose some control over his or her shot. For this reason, some players fix a “damper” to the springs. If placed on the strings in the correct position, this has the effect of producing critically damped oscillations and as a result the struck tennis racquet moves smoothly back to equilibrium. The same effect can be achieved by making sure that the ball strikes the strings at a point known as the ‘sweet spot’ of which there are two, one of which is know as the ‘centre of percussion (COP)’. Cricket and baseball bats likewise have two sweet spots.

Another example is one that involves vibrations that may be set up in buildings when there is an earthquake. For this reason, in regions prone to earthquakes, the foundations of some buildings are fitted with damping mechanisms. These mechanisms insure any oscillations set up in the building are critically damped.

Exercise

Identify which of the following oscillatory systems are likely to be lightly damped and which are likely to be heavily damped.

1. Atoms in a solid.2. Car suspension3. Guitar string4. Harmonic oscillator under water.5. Quartz crystal.6. A cantilever that is not firmly clamped.7. Oil in a U-tube8. Water in a U-tube

4.3.3 natuRalfREquEnCyanD

fORCEDOsCillatiOns

Consider a small child sitting on a swing. If you give the swing a single push, the swing will oscillate. With no further pushes, that is energy input, the oscillations of the swing will die out and the swing will eventually come to rest. This is an example of damped harmonic motion. The frequency of oscillation of the swing under these conditions is called the natural frequency of oscillation (vibration). So far in this topic, all the systems we have looked at have been systems oscillating at their natural frequency.

Suppose now when each time the swing returns to you, you give it another push. See Figure 418

Figure418 Aforcedoscillation

The amplitude of the swing will get larger and larger and if you are not careful your little brother or sister, or who ever the small child might be, will end up looping the loop.

The frequency with which you push the swing is exactly equal to the natural frequency of oscillation of the swing and importantly, is also in phase with the oscillations of the spring. Since you are actually forcing the swing to oscillate,



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the swing is said to be undergoing forced oscillations. In this situation the frequency of the so-called driver (in this case, you) is equal to the natural frequency of oscillation of the system that is being driven (in this case, the swing). If you just push the swing occasionally when it returns to you, then the swing is being forced at a different frequency to its natural frequency. In general, the variation of the amplitude of the oscillations of a driven system with time will depend on the

• frequency of the driving force• frequency of natural oscillations• amplitude of the driving force • phase difference between driving force frequency

and natural frequency• amount of damping on the system

(There are many very good computer simulations available that enable you to explore the relation between forced and natural oscillations in detail.)

The driving force and system are in phase if, when the amplitude of system is a maximum, it receives maximum energy input from the driver. Clearly this is when the amplitude of the driver is a maximum.

What is of particular interest is when the forced frequency is close to and when it equals the natural frequency. This we look at in the next two sections.

4.3.4 fORCEDfREquEnCyanD

amplituDE

We now look to see how the amplitude of an oscillating system varies with the frequency of the driving force.

The graph in Figure 419 shows the variation with frequency f of the driving force of the amplitude A of three different systems to which the force is applied.

0

5

10

0 5 10 15 20 25 30f/Hz

A /cm

heavy damping

medium damping

light damping

Figure419 Forcedfrequency

Each system has the same frequency of natural oscillation, f0 = 15 Hz. The thing that is different about the systems is that they each have a different degree of damping: heavy (low Q), medium (medium Q), light (large Q).

For the heavily damped system we see that the amplitude stays very small but starts to increase as the frequency approaches f0 and reaches a maximum at f = f0; it then starts to fall away again with increasing frequency.

For the medium damped system, we see that as f approaches f0, the amplitude again starts to increase but at a greater rate than for the heavily damped system. The amplitude is again a maximum at f = f0 and is greater than that of the maximum of the heavily damped system.

For the lightly damped system, again the amplitude starts to increase as f approaches f0, but at a very much greater rate than for the other two systems; the maximum value is also considerably larger and much more well-defined i.e. it is much easier to see that the maximum value is in fact at f = f0.

If there were such a thing as a system that performs SHM, then if this system were driven at a frequency equal to its natural frequency, its amplitude would be infinite. Figure 420 shows how the amplitude A for a driven system with very little damping and whose natural frequency of oscillation f0 = 15 Hz , varies with the frequency f of the driving force.

0

50

100

150

0 5 10 15 20 25 30f /Hz

A /cm

Figure420 Amplitude-frequencygraphforalightlydampedoscillator

We see that the maximum amplitude is now very large and also very sharply defined. Also, either side of f0, the amplitude drops off very rapidly.


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4.3.5 REsOnanCE

(Note: As well as the availability of a large number of computer simulations that demonstrate resonance, there are also many laboratory demonstrations and experiments that can be done to demonstrate it).

We have seen that when an oscillatory system is driven at a frequency equal to its natural frequency, the amplitude of oscillation is a maximum. This phenomenon is known as resonance. The frequency at which resonance occurs is often referred to as the resonant frequency.

4.3.6 ExamplEsOfREsOnanCE

In the introduction to this Topic, we referred to resonance phenomena without actually mentioning the term resonance. For example we can now understand why oscillations in machinery can be destructive A piece of machinery will have a natural frequency of oscillation. If moving parts in the machine act as a driver of forced oscillations and have a frequency of oscillation equal to the natural frequency of oscillation of the machine, then the amplitude of vibration set up in the machine could be sufficient to cause damage.

Similarly, if a car is driven along a bumpy road, it is possible that the frequency with which the bumps are crossed by the car, will just equal the natural frequency of oscillation of the chassis of the car, If this is the case then the result can be very uncomfortable.

We now also see why it is important that systems such as machines, car suspensions, suspension bridges and tall buildings are critically or heavily damped.

Resonance can also be very useful. For example, the current in a particular type of electrical circuit, oscillates. However, the oscillating current quickly dies out because of resistance in the circuit. Such circuits have a resonant frequency and if driven by an alternating current supply, the amplitude of the current may become very large, particularly if the resistance of the circuit is small. Circuits such as this are referred to as resonant circuits. Television and radios have resonant circuits that can be tuned to oscillate electrically at different frequencies. In this way they can respond to the different frequencies of electromagnetic waves that are sent by the transmitting station as these waves now act as the driving frequency. This is discussed in more detail in Topic F.1

In the introduction we mentioned the use of quartz crystal as timing devices. If a crystal is set oscillating at its natural frequency, electric charge constantly builds up and dies away on it surface in time with the vibration of the crystal (This is known as the piezoelectric effect.). This makes it easy to maintain the oscillations using an alternating voltage supply as the driving frequency. The vibrations of the crystal are then used to maintain the frequency of oscillation in a resonant circuit. It is the oscillations in the resonant circuit that control the hands of an analogue watch or the display of a digital watch. (The concept of analogue and digital signals is discussed in Topics 14.1 and C.1).

It is left as an exercise to you to think of other situations in which resonance can be useful or can be harmful.

4.4 wave cHaracteristics

4.4. 1 Describe a wave pulse and a continuous progressive (travelling) wave. (Students should be able to distinguish between oscillations and wave motion and appreciate that in many examples, the oscillations of the particles are simple harmonic).

4.4.2 State that progressive (travelling) waves transfer energy. (Students should understand that there is no net motion of the medium through which the wave travels).

4.4.3 Describe and give examples of transverse and of longitudinal waves.

4.4.1 Wave pulses and continuous travelling waves© IBO 2007

Perhaps one of the most familiar types of wave motion is a water wave. However, we can also set up waves in strings very easily. A simple demonstration is to take a length of rubber tubing. Hold one end of it and shake that end up and down. A wave will travel down the tube. If we give the end of the tube just one shake then we observe a pulse to travel down the tube. By this we can see that we can have either a continuous travelling wave or a travelling pulse. This is illustrated in Figure 421 in which we have taken an instantaneous snap shot of the tube.



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pulse

handmovement

(a) (b)

Figure421 Simplewaves

We can also set up another type of wave by using a slinky spring. In this demonstration we lay the spring along the floor. Hold one end of it and move our hand backwards and forwards in the direction of the spring. In this way we see a wave travelling down the spring as a series of compressions and expansions of the spring as illustrated in Figure 422.

We can also set up a pulse in the spring by moving our hand backwards and forwards just once in the direction of the spring.

Of course we can set up a wave in the spring that is similar to the one we set up in the rubber tube. We shake the spring in a direction that is at right angles to the spring as shown in Figure 422.

compressions

expansionscompress

expand

compress expand compress

2(a) 2(b)

Figure422 Slinkysprings

A very important property associated with all waves is their so-called periodicity. Waves in fact are periodic both in time and space and this sometimes makes it difficult to appreciate what actually is going on in wave motion. For example, in our demonstration of a wave in a rubber tube we actually drew a diagram that froze time- an instantaneous snapshot of the whole string. Hence Figure 421 shows the periodicity of the wave in space.

The diagram is repeated as a sketch graph in Figure 423. The y-axis shows the displacement of the tube from its equilibrium position. The graph is a displacement space graph.

We now look at one particle of the tube labelled P and “unfreeze” time. The diagram in Figure 424 shows how the position of P varies with time. This illustrates the

periodicity of the wave in time. We recognise that the point P is actually oscillating with SHM.

Equilibrium

displacemen

toftube

distance along tube

P position

Figure423 Displacement-spacegraph

The y-axis now shows the displacement of the point P from equilibrium. The graph is a displacement-time graph.

The space diagram and the time diagram are both identical in shape and if we mentally combine them we have the whole wave moving both in space and time.

displacemen

tofp

articleP

from

equilibriu

mposition

time

Figure424 Displacement-Timegraph

For the longitudinal wave in the slinky spring, the displacement-space graph actually shows the displacement of the individual turns of the spring from their equilibrium position as a function of distance along the spring. However, it could equally show how the density of turns of the spring varies with length along the spring. The displacement–time graph shows the displacement of one turn of the spring from its equilibrium positions as a function of time.

4.4.2 whatisawavE?

We have shown that waves fall into two distinct classes but as yet we haven’t actually said what a wave is. Quite simply a wave is a means by which energy is transferred between two points in a medium without any net transfer of the medium itself. By a medium, we mean the substance or object in which the wave is travelling. The important thing is that when a wave travels in a medium, parts of the medium do not end up at different places. The energy of the source of the wave is carried to different parts of the medium by the wave. When for example you see a field of corn waving in the wind you actually see a wave carry across the field but none of the corn moves to another field. Similarly with the


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demonstrations above, the tube and the spring do not end up in a different part of the laboratory. Water waves however, can be a bit disconcerting. Waves at sea do not transport water but the tides do. Similarly, a wave on a lake does not transport water but water can actually be blown along by the wind. However, if you set up a ripple tank you will see that water is not transported by the wave set up by the vibrating dipper.

4.4.3 tRansvERsEanD

lOngituDinalwavEs

transverse waves

In these types of wave, the source that produces the wave vibrates at right angles to the direction of travel of the wave i.e. the direction in which the energy carried by the wave is propagated. It also means that the particles of the medium through which the wave travels vibrate at right angle to the direction of travel of the wave (direction of energy propagation).

Figure 421 illustrates an example of a transverse wave. Light is another example of a transverse wave although this a very special kind of wave. Light waves are discussed in more detail in Topic G.1.

An important property of transverse waves is that they cannot propagate through fluids (liquid or gases). This is one reason why light wave are special; they are transverse and yet can propagate through fluids and through a vacuum.

longitudinal waves

In these types of wave, the source that produces the wave vibrates in the same direction as the direction of travel of the wave i.e. the direction in which the energy carried by the wave is propagated. It also means that the particles of the medium through which the wave travels vibrate in the same direction of travel of the wave (direction of energy propagation). The wave in the slinky spring in Figure 422 is a longitudinal wave as is sound (see Section 4.4.5).

4.4.4 Describe waves in two dimensions, including the concepts of wavefronts and rays.

4.4.5 Describe the terms crest,trough, compressionand rarefaction.

4.4.6 Define the terms displacement,amplitude, frequency,period,wavelength,wavespeed and intensity.

© IBO 2007

4.4.4 wavEfROntsanDRays

Place a piece of white card beneath a ripple tank and shine a light on the ripples from above. The image that you get on the card will look something like that shown in Figure 425.

vibratingdipper

lamp

white paper

ripple tank

Figure425 Arippletank

Again this is a snapshot. However, by using a stroboscope as the source of illumination it is possible to “freeze” the waves.

Each bright area of illumination represents a trough or crest. (Light incident on the top or bottom of a crest will be transmitted through the water and not reflected by the surface).

Each bright line representing a crest can be thought of as a “wavefront” and this is a very good way of representing a travelling wave as shown in Figure 426.



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directionof travel

λ λλλ

Figure426 Awavefront

If the wave is a light wave then the arrow that shows the direction of travel of the wave is none other than what we call a light ray.

4.4.5 qualitativEtERms

assOCiatEDwithwavEs

Crest This is a term coined from water waves and is used in connection with transverse waves. It refers to the maximum height of the wave i.e. the point at which the particles in the medium in which the waves travels have maximum displacement.

Trough Again, a term coined from water waves and used in connection with transverse waves. It refers to the lowest point of the wave i.e. the point at which the particles in the medium in which the waves travels have minimum displacement (strictly speaking, maximum negative displacement).

Compression This is a term used in connection with longitudinal wave and refers to the region where the particles of the medium are “bunched up”.

Rarefaction Again, a term used in connection with longitudinal waves referring to the regions where the particles are “stretched out”.

In a sound wave, the compression is the region in which the molecules of the air are pushed together (region of greatest pressure) and the rarefaction is the region where the molecules are well separated ( region of least pressure). In a physiological sense, sound is the ear’s response to the changes in pressure in a longitudinal wave in air. If the frequency (see below) of the wave is too low (≈10 Hz) then the ear will not detect the pressure changes nor will it if the frequency is too high (≈15 kHz).

A longitudinal wave in a slinky spring is analogous to a sound wave in that each turn of the spring represents an air molecule.

4.4.6 DEfinitiOnsassOCiatEDwith

wavEs

amplitude (A,a)

This is the maximum displacement of a particle from its equilibrium position. (It is also equal to the maximum displacement of the source that produces the wave).

The energy that a wave transports per unit time across unit area of the medium through which it is travelling is called the intensity (I). From our knowledge of SHM we know that the energy of the oscillating system is proportional to the square of the amplitude (equation 4.12). Hence for a wave of amplitude A, we have that

Period (T)

This is the time that it takes a particle to make one complete oscillation. (It is also equal to the time for the source of the wave to make one complete oscillation).

frequency (f)

This is the number of oscillations made per second by a particle. (It is also equal to the number of oscillations made per second by the source of the wave). The SI unit of frequency is the hertz-Hz. Clearly then,

wavelength (λ)

This is the distance along the medium between two successive particles that have the same displacement

wave speed (v,c)

This is the speed with which energy is carried in the medium by the wave. A very important fact is that wave speed depends only on the nature and properties of the medium.

You can demonstrate this by sending pulses along different types of rubber tubes or by sending pulses along a slinky in which you alter the distance between successive turns by stretching it.


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(In some circumstances the wave speed is a function of wavelength, a phenomenon known as dispersion)

Figure 427 shows how the different terms and definitions associated with waves relate to both transverse and longitudinal waves. From these diagrams, we also see that the wavelength of a transverse wave is equal to the distance between successive crests and also between successive troughs. For a longitudinal wave the wavelength is equal to the distance between successive points of maximum compression and also between successive points of maximum rarefaction.

dis

pla

cem

ent o

f eq

uilib

rium

pos

itio

n equilibrium positionof tube

amplitude, A

wavelengthcrest crest

trough trough

amplitude wavelength

distance along tube

d

isp

lace

men

t of p

arti

cle

period, T

time

compression

rarefaction maximumcompression

wavelength

tube from

TRANSVERSE

TRANSVERSEAND

LONGITUDINAL

LONGITUDINAL

period, T

Figure427 Howthedefinitionsapply

4.4.7 Draw and explain displacement-timeand displacement-positiongraphs for transverse and for longitudinal waves

4.4.8 Derive and apply the relationship between wave speed, wavelength and frequency

4.4.9 State that all electromagnetic waves travel with the same speed in free space and recall the orders of magnitude of the wavelengths of the principal radiations in the electromagnetic spectrum.

© IBO 2007

4.4.7 DisplaCEmEnt-timEanD

DisplaCEmEnt-pOsitiOngRaphs

4.4.8 thERElatiOnshipbEtwEEn

wavEspEED,wavElEngth

anDfREquEnCy

Figure 428 shows an instantaneous snapshot of a medium through which a wave is travelling. A particle of the medium is labelled ‘P’.

displacemen

tofm

edium

distance alongmedium

P At time t = 0

Figure428 Instantaneoussnapshotofdisplacementofmedium

If we take another photograph half a period later then the particle ‘P’ will be in the position shown in Figure 429.

displacement

ofmed

ium

distance along medium

At time tT

2---=

P

Figure429 Particle‘P’halfaperiodlater

In this time the wave will have moved forward a distance of half a wavelength

, but,

λ2---

dis cetantime

---------------------- λT---= f 1

T---=

v fλ=

.

We have therefore that the speed v of the wave

= , but,

λ2---

dis cetantime

---------------------- λT---= f 1

T---=

v fλ=

,

Hence we have, but,

λ2---

dis cetantime

---------------------- λT---= f 1

T---=

v fλ=



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Example

Water waves of wavelength 5.0 cm are travelling with a speed of 1.0 ms-1.

Calculate the frequency of the source producing the waves?

The waves travel into deeper water where their speed is now 2.0 m s –1. Calculate the new wavelength λnew of the waves.

Solution

Using v = fλ we have that 1.0 = f × .05

Such that f = 20 Hz.

In the deeper water using v = fλ we now have that 2.0

= 20 × λnew

Such that λnew

= 10 cm.

4.4.9 ElECtROmagnEtiCwavEs

We mentioned above that light is a transverse wave but in fact light is just one example of a most important class of waves known as electromagnetic waves. There are called electromagnetic waves (EM waves) because they

actually consist of an electric field and a magnetic field oscillating at right angles to each other. There existence was first predicted by Clerk Maxwell in 1864 and verified some 20 years later by Heinrich Hertz (hence Hz for the unit of frequency). This was one of the great unifications in Physics.

Before Maxwell’s prediction there were the separate studies of Optics (light) and Electricity and Magnetism and that after the verification of his prediction, there was just the study of Electromagnetism. Maxwell’s theory also predicted that all em waves would have the same speed in free space (vacuum) of very nearly 3 × 108 m s-1. This prediction had great implications for the development of Physics as will be seen in Topics D.1 and H.1.

The source of all EM waves is essentially the accelerated motion of electric charge. If the charge is oscillating then the frequency with which the charge oscillates determines the frequency of the em wave. The so-called spectrum of em waves is vast. Suppose in a thought experiment, we were to have a charged metal sphere and were able to set it oscillating at different frequencies. When vibrating with a frequency of 103 Hz, the oscillating charged sphere would be a source of long wave radio signals, at 109 Hz a source of television signals, at 1015 Hz a source of visible light and at 1018 Hz, it would emit X-rays. Of course this is just a thought experiment and we should identify the actual sources of the different regions of the em spectrum. These are shown in Figure 430.

Our thought experiment might be somewhat absurd, but it does emphasise the point that the origin of all em waves is the accelerated motion of electric charge.

103 106

Radio waves

104 10181016101410121010108

106 102104 10–1010–810–610–410–2100

f / Hz

Source Electrons moving in conductors Hot Objects

MicrowavesInfra-red

Ultra-violet

Gasdischarge

X-rays

γ-rays

Electronsstriking targets

Radioactivedecay

Visi

ble

(ver

y ho

t ob

ject

s)

λ / m

Figure430 Regionsoftheelectromagneticspectrum


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4.5.1 Describe the reflection and transmission of waves at a boundary between two media.

4.5.2 State and apply Snell’s law.© IBO 2007

4.5.1 REflECtiOnanDREfRaCtiOn

reflection of a single pulse

We now look to see what happens to a wave when it is incident on the boundary between two media. First of all, we shall look at a single pulse travelling along a string.

Figure 431 shows a pulse travelling in a string that is fixed to a rigid support, it is essentially the positive part of a sine curve.

Figure431 Theincidentpulse

When the pulse reaches the end of the string it is reflected Figure 432 shows the reflected pulse, this pulse is the negative part of a sine curve. Some of the energy of the pulse will actually be absorbed at the support and as such, the amplitude of the reflected pulse will be less than that of the incident pulse.

Figure432 Thereflectedpulse

The main thing to note is that the pulse keeps its shape except that now it is inverted i.e. the pulse has undergone a 180° (π) phase change.

The reason for this is a little tricky to understand but it is essentially because the end of string that is fixed cannot move. As any part of the forward pulse reaches the fixed end the associated point on the string is moving upward and so if the fixed point is not to move, the point on the string that is moving upward must be cancelled out by a point moving downwards. This is an example of the so-called principle of superposition that we look at in Section 4.4.5

If the string is not attached to a support then a pulse is still reflected from the end of the string but this time there is no phase change and so the reflected pulse is not inverted.

reflection of wavefronts

reflection

We can use the idea of wavefronts to see what happens when a wave strikes a barrier. This is demonstrated with a ripple tank.

Figure 433 shows the incident wavefronts and the reflected wavefronts.

incident waves re ected waves

i.e.i r

i r

Figure433 Incidentandreflectedwavefronts

By constructing the associated rays (see Figure 433), we see that the angle at which the waves are reflected from thethe angle at which the waves are reflected from the barrier is equal to the angle at which they are incident on the barrier (the angles are measured to the normal to the barrier). That is ∠i = ∠r . All waves, including light, sound, water obey this rule, the so-called law of reflection.

We can use the idea of reflection to make a very simple measurement of the speed of sound. If you stand about 100 m away from a tall wall and clap your hands once, a short time later you will hear an echo of the clap. The sound pulse produced by your hand clap travels to the

4.5 wave ProPerties



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wall and is reflected back. The trick now is to clap your hands continuously until each clap is synchronous with the echo. You should then get a friend to time the number of claps that you make in say 30 seconds. Let us suppose that this is 50. The number of claps per second is therefore 0.6. This means that it takes the sound 0.6 seconds to travel to the wall and back. The speed of sound can then be approximated as 330 m s–1. If you should get this result then you have done very well indeed since this is very nearly the value of the speed of sound in air at standard temperature and pressure. However, in the true spirit of experimental physics, you should repeat the experiment several times and at different distances and you should also assess the quantitative error in your result.

refraction

We now look to see what happens to a wave when it is incident on the boundary between two media and passes from one medium to the other (transmission). As for the single pulse, some energy will be absorbed at the boundary. Also, as well as energy being transmitted by the wave, some energy will be reflected at the boundary. Here we will concentrate on the transmitted energy.

We have discussed previously the idea that the speed of a wave depends only on the nature and properties of the medium through which the wave travels. This gives rise to the phenomenon of refraction. That is the change in direction of travel of a wave resulting from a change in speed of the wave. This is easily demonstrated with a ripple tank by arranging two regions of different depth. To achieve this a piece of flat perspex or glass is placed a short distance from the source of the waves as shown in Figure 434.

source of plane waves ripple tank

perspex sheet water

Figure434 Rippletanksetuptodemonstraterefraction

Figure 435, shows the result of a continuous plane wave going from deep to shallow water.

deep water shallow water deep water

λ d λ s

L I G H T S O U R C E

NB: λd > λ

s

perspex sheet

Figure435 Wavefrontsinwaterofdifferentdepth

In this diagram the wave fronts are parallel to the boundary between the two regions. The frequency of the waves does not alter so, as we have mentioned before, the wavelength in the shallow water will be smaller. If the speed of the waves in the deep water is vd and the speed in the shallow water is vs,

then, λ λ

such that

λλ

In Figure 436 the wavefronts are now incident at an angle to the boundary between the deep and shallow water.

shallow deep

A

BC

barrier

Figure436 Wavesincidentatanangle

As well as the wavelength being smaller in the shallow water the direction of travel of the wavefronts also alters. We can understand this by looking at the wavefront drawn in bold. By the time that part A of this wavefront reaches the barrier at B the refracted wave originated from the barrier will have only reached C since it is travelling more slowly.


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4.5.2 snEll’slaw

In 1621 the Dutch physicist Willebrord Snell discovered and published a very important rule in connection with the refraction of light.

Figure 437 shows a light ray travelling from one medium to another. The line AB is a line constructed such that it is at right angles to the surface between the two media (called the normal). It is used as a reference to measure the angle (the angle of incidence) and the angle (the angle of refraction). Snell discovered that for any two media

θ

θ

normal

medium 1

medium 2

θ1

θ2

A

B

Figure437 Refractionoflightrays

This is known as Snell’s law. In fact it enables us to define a property of a given optical medium by measuring the angles when medium 1 is a vacuum. (In the school laboratory air will suffice.). The constant is then a property of medium 2 alone called its refractive index n.

We usually write

When Snell published his law it was essentially an empirical law and the argument as to the nature of light was still debatable.

Although we have described Snell’s law for light rays, we must remember that a ray is a line that is perpendicular to the wavefronts of a wave. In this respect Snell’s law is true for all waves. Example

In Figure 438 the wavefronts in medium 1 travel with speed v1 and in medium 2 with speed v2 .

A

θ1

θ2

YX

θ1

θ2

medium 1

medium 2

boundary

B

Figure438 Snell’slaw

In the time that it takes point A on the wavefront XA to reach Y, point X will have travelled to point B. If we let this time be ∆t then we have

AY = v1∆t and XB = v2∆t

However from the geometry of the situation we have that

AY = XYsinθ1 and XB = XYsinθ2

From which

θθ

ΔΔtt

that is

θ

θ

That is, the constant in Snell’s law is the ratio of the speed of light in medium 1 to that in medium 2. The result will of course be valid for all types of waves.

For light we have that the refractive index of a material is the ratio of the speed of light in vacuo (c) to that in the material (v). We can write this as

As mentioned above this result has been confirmed for light by a direct measurement of the speed of light in water and subsequently in other materials.

Example

The refractive index of a certain type of glass is 1.5. The speed of light in free space is 3.0 × 108 m s-1.

Calculate the speed of light in glass.



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Solution

From Snell’s law above, the speed of light in glass can be found using c = 3 × 108 m s-1.

So, with n = 1.5 and , we have that v = 2 .0 × 108 m s–1.

4.5.3 Explain and discuss qualitatively the diffraction of waves at apertures and obstacles.

4.5.4 Describe examples of diffraction. © IBO 2007

4.5.3,4 DiffRaCtiOn

When waves pass through a slit or any aperture, or pass the edge of a barrier, they always spread out to some extent into the region that is not directly in the path of the waves. This phenomenon is called diffraction. This is clearly demonstrated in a ripple tank.

Figure 439 (a) shows plane waves incident on a barrier in which there is a narrow slit, the width of which is similar in size to the wavelength of the incident waves. Figure 439 (b) shows plane waves passing the edge of a barrier.

a b

Figure439(a)Diffractionataslit(b)Diffractionatanedge

To understand this, we use an idea put forward by Christiaan Huygens (1629-1695). Huygens suggested that each point on any wavefront acts as a source of a secondary wave that produce waves with a wavelength equal to the wavelength associated with the wavefront. In Figure 440 we can see how this works in the case of plane wavefronts.

new wavefront

secondarywave

wavefront

Figure440 Planarwaves

Each point on the wavefront is the source of a secondary wave and the new wavefront is found by linking together the effect of all the secondary waves. Since there are, in this case in infinite number of them, we end up with a wavefront parallel to the first wavefront. In the case of the plane waves travelling through the slit in Figure 438 (b), it is as if the slit becomes a secondary point source.

If we look at the effect of plane waves incident on a slit whose width is much larger than the incident wavelength as shown in Figure 441 then we see that diffraction effects are minimal. We can understand this from the fact that each point on the slit acts as a secondary source and we now have a situation where the waves from the secondary sources results in a wavefront that is nearly planar (see Figure 440).

Figure441 Diffractionatalargeaperture

Diffraction effects at edges can also be understood on the basis of Huygens’ suggestion. By sketching the wavefronts of secondary sources at points on a wavefront close to the edge, it easy to see that diffraction effects become more pronounced as the wavelength of the incident wave is increased.

4.5.4 ExamplEsOfDiffRaCtiOn

We have seen that the ripple tank can be used to demonstrate the phenomenon of diffraction, i.e. the spreading out of a wave has it goes through an aperture or encounters an object.

Diffraction of sound waves can be demonstrated using the set up shown Figure 442.


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The barrier would be expected to prevent the waves reaching points such as X. However, because of diffraction at the slit, the waves spread out in the region beyond the slit and the microphone will detect sound at points such as X.

CROspeaker

signalgenerator

microphoneX

Figure442 Diffractionofsoundwaves

You can also demonstrate the diffraction of light by shining laser light through a single narrow slit such that after passing through the slit, the light is incident on a screen. The effect of diffraction at the slit produces a pattern on the screen that consists of areas of illumination (bright fringes) separated by dark areas (dark fringes). In this situation, each point on the slit is acting as a secondary source and the pattern of light and dark fringes is a result of the interference (see next section) of the waves from these sources. (The diffraction of light is considered in more detail in Topic 11.3).

As mentioned above, diffraction effects at a slit really only becomes noticeable when the slit width is comparable to the wavelength. In this respect, if the laser is replaced by a point source, then as the slit is made wider, the diffraction pattern tends to disappear and the illumination on the screen becomes more like what one would expect if light consisted of rays rather than waves. Historically, the diffraction of light was strong evidence for believing that light did indeed consist of waves.

The diffraction of light can also be demonstrated by looking directly at a point source through a narrow slit. Unless the point source is monochromatic – a monochromatic source is one that emits light of a single colour (i.e. wavelength) and you will see a series of different coloured fringes interspersed by dark fringes.

Exercise

1. (a) Calculate the wavelengths of

i) FM radio waves of frequency 96 MHz

ii) long wave radio waves of frequency 200 kHz.(Speed of light in free space c = 3.0 × 108 m s-1)

(b) Use your answers to (i) and (ii) to explain why if your car is tuned to FM, it cuts out when you enter a tunnel but doesn’t if you are tuned to long wave reception.

2. Suggest one reason why ships at sea use a very low frequency sound for their foghorn.

4.5.5 State the principleofsuperpositionand explain what is meant by constructive interferenceand by destructiveinterference.

4.5.6 State and apply the conditions for constructive and for destructive interference in terms of path difference and phase difference.

4.5.7 Apply the principle of superposition to determine the resultant of two waves

© IBO 2007

4.5.5 thEpRinCiplEOf

supERpOsitiOn

We now look at what happens when two waves overlap. For example in Figure 443 what will happen when the two pulses on a string travelling in opposite directions cross each other?

pulse1 pulse 2

Figure443 Twopulsesapproachingeachother

There is a very important principle in physics that applies not only to waves but to other situations as well. This is the principle of superposition. What it effectively tells us is that if you want to find out the effect of two separate causes then you need to add the effects of each separate



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cause. In the example of the two pulses, to find the net displacement at any point we add the displacement of pulse 1 at that point to the displacement of pulse 2 at the point. We show two examples of this in Figure 444 below as the pulses move across each other.

pulse 1 pulse 2

pulse 1pulse2

a

2a

Figure444 (a)Constructiveinterference

Figure444

pulse 1 pulse 2

pulse 1pulse 2

pulse 1

pulse 2

authordot the pulses in the last diag pls

(b) Destructiveinterference

1. In the first diagram the pulses do not fully overlap and in the second diagram they do. In the second diagram we have we what call full constructive interference. The two pulses add to give a single pulse of twice the amplitude of each separate pulse.

2. We now consider two pulses as shown in the diagram below.

When these two pulses completely overlap the net displacement of the string will be zero. We now have what we call complete destructive interference.

4.5.6 pathDiffEREnCEanDphasE

DiffEREnCE

The schematic diagram in Figure 445 shows part of the interference pattern that we get from two point sources, S1 and S2. The left hand side is a reflection of the pattern on the right hand side. The dashed line shows where the crests from S1 meet the crests from S2, creating a double crest, i.e., constructive interference.

S1 S2

The dashed line shows where the crests from Smeet the crests from Screating a double crestconstructive interference.

constructive interference

destructive interference(crest meets trough)

(crest meets crest)

(antinodes)

(nodes)

1

2,, i.e.,

Figure445 Twosourceinterference

The sources are two dippers connected to a bar that is vibrated by an electric motor. The dippers just touch the surface of the water in a ripple tank. The sources therefore have the same frequency. They also are in phase. By this we mean that when a crest is created by one dipper a crest is created by the other dipper. Sources that have the same frequency and that are in phase are called coherent sources.

The bold lines in the diagram show the places where a trough of one wave meets the crest of the other wave producing complete destructive interference. The water at these points will not be displaced and such points are called nodal points or nodes. Points of maximum constructive interference are called antinodes. The bold lines are therefore called nodal lines.

The overall pattern produced by the interfering waves is called an interference pattern.

To emphasise the idea of phase the diagrams in Figure 446 show snapshots of two waves in phase and two waves that are π out of phase.


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Two waves in phase Twowaves¼ out of phase.

wave A

wave B

wave A

wave B

Diagram 1 Diagram 2

Figure446 Phasedifference

The idea of π out of phase comes from the idea that if the space displacement of wave A in diagram 2 is represented by y = Asinθ , then the space displacement of wave B is y = Asin (θ + π). At the instant shown the waves in diagram 1 will reinforce and produce constructive interference whereas the waves in diagram 2 will produce destructive interference.

Let us now look at the interference between wave sources from two points in a little more detail.

In Figure 447 we want to know what will be the condition for there to be a point of maximum or minimum interference at P.

S1

S2 X

P

Figure447 Interference

S1 and S2 are two coherent point sources. For there to be a maximum at P, a trough must meet a trough or a crest must meet a crest. The waves from the sources will have travelled a different distance to P and therefore will not be necessarily in phase when they reach P. However, if a crest meets a crest (or trough a trough) at P then they will be in phase. They will be in phase only if the difference in the distances travelled by the two waves is an integral number of wavelengths.

This means that path difference

= = = =S2P S1P S2X nλ n, 0 1 2 …, , ,

S2P S1P S2X n 12---+ λ n, 0 1 2 …, , ,= = = =

The waves will be out of phase if the difference in the distance travelled, is an odd number of half-wavelengths.

So for a minimum we have path difference =

4.5.7 applyingthEpRinCiplEOf

supERpOsitiOn

It will help you understand the principle of superposition to do the following exercise.

Exercise

Sophia sounds two tuning forks A and B together and places them close to her ear.

The graph in Figure 448 shows the variation with time t of the air pressure close to her ear over a short period of time.

Figur

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

t

pressure/arbita

ryunits

tuning fork A

tuning fork B

e448 Tuningforkgraph

On a separate piece of graph paper, use the principle of superposition to construct a graph that shows the variation with time t of the resultant pressure close to Sophia’s ear.

Use your graph to suggest the nature of the sound that Sophia will hear up until the time that the vibrations of the tuning forks die out.

= = = =S2P S1P S2X nλ n, 0 1 2 …, , ,

S2P S1P S2X n 12---+ λ n, 0 1 2 …, , ,= = = =


Chapter 04

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