PH 221-3A Fall 2010

Waves - I

Lectures 25-26

Chapter 16 (Halliday/Resnick/Walker, Fundamentals of Physics 8th edition)

1

Chapter 16 Waves I

In this chapter we will start the discussion on wave pphenomena. We will study the following topics:

Types of wavesAmplitude, phase, frequency, period, propagation speed of a wave Mechanical waves propagating along a stretched string Wave equation Principle of superposition of wavesPrinciple of superposition of waves Wave interference Standing waves, resonance

2

Waves

A wave is a vibrational, trembling motion in an elastic, deformable body.

•The wave is initiated by some external force that acts on some parts of the body andThe wave is initiated by some external force that acts on some parts of the body and deforms it.• The elastic restoring force communicates this initial disturbance from one part of the body to the next, adjacent part.y , j p• The disturbance gradually spreads outward through the entire elastic body.

The elastic body in which the wave propagate is called the medium.

When a wave propagates through a medium, the particles in the medium vibrate back and forth, but the medium as a whole does not perform translational motion.

3

Transverse Wave Motion

1 up

If we shake one end of the tightly stretched elastic string up and down

p

2nd up

with a flick of the wrist a disturbance travels along the string.

1 down

String is regarded as a row of particles connected by small, massless springs.

2 down

1 particle up => it will pull the 2nd

particle up.

Transverse wave motion – the direction of the disturbance propagation isof the disturbance propagation is perpendicular to the back and forth vibration of the particles. 4

Longitudinal Wave Motion

Disturbance generated by pushingDisturbance generated by pushing the 1st particle towards the 2nd

L i di l di i fLongitudinal wave - direction of disturbance motion is parallel to the vibration of the particles.p

The wave is a propagation of the disturbance through the medium without any net displacement of the medium.Consider a horizontal string connected to the mass executing simple harmonic motion in the vertical direction.

Each particle transmits the motion to the next particle along the entire length of the string. The resulting wave propagates in the h i l di i i h l ihorizontal direction with a velocity “v”, while any one particle of the string executes simple harmonic motion in the vertical directiondirection.

The particle of the string is moving perpendicular to the direction of wave propagation and is not moving in the direction of the wave.

A transverse wave is a wave in which the particles of the medium execute simple harmonic motion in a direction perpendicular to itsexecute simple harmonic motion in a direction perpendicular to its direction of propagation.

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A wave is defined as a disturbance that is self-sustained and propagates in space with a constant speed

Waves can be classified in the following three categories:

1. Mechanical waves. These involve motions that are governed by Newton’s laws and can exist only within a material medium such as air, water, rock, etc. Common examples are: sound waves, seismic waves etcwaves, etc.2. Electromagnetic waves. These waves involve propagating disturbances in the electric and magnetic field governed by Maxwell’s

ti Th d t i t i l di i hi h tequations. They do not require a material medium in which to propagate but they travel through vacuum. Common examples are: radio waves of all types, visible, infra-red, and ultra-violet light, x-rays, gamma rays. All electromagnetic waves propagate in vacuum with the same speed c = 300,000 km/s3. Matter waves. All microscopic particles such as electrons,3. Matter waves. All microscopic particles such as electrons, protons, neutrons, atoms etc have a wave associated with them governed by Schroedinger’s equation. 7

Waves can be divided into the following two categoriesTransverse and Longitudinal waves

depending on the orientation of the disturbance with respect to the wave propagation velocity . v

If the disturbance associated with a particular wave isperpendicular to the wave propagation velocity, this

i ll d l i iwave is called " ". An example is given in the upper figure which depicts a mechanical wave that

transverse

propagates along a string. The movement of each p p g g gparticle on the string is along the -axis; the wave itselfpropagates along the -axis.

yx

A wave in which the associated disturbance is parallel to the wave propagationvelocity is known as a " ". An example of such a wave is given in the lower figure. It is produced by a

longitudinal wavepiston oscillating in a tube filled withg p y p g

air. The resulting wave involves movement of the air molecules along the axis of the tube which is also the direction of the wave propagation velocity .v8

Consider the transverse wave propagating along the string as shown in the figure. The position of

i h i b d ib d bany point on the string can be described by a function ( , ). Further along the chapterwe shall s

y h x tee that function has to have a specifich

form to describe a wave. Once such suitable function is: ( , ) sin - S h

my x t y kx t

hi h i d ib d b iSuch a wave which is described by a sine (or a cosine) function is known as" ".harmonic waveThe various terms that appear in the expression for a harmonic wave are identified in the lower figureFunction (y ) depends on and There are twox t x tFunction (y , ) depends on and . There are two ways to visualize it. The first is to "freeze" time (i.e. set ). This is like taking a snapshot of theo

x t x t

t t

wave at . , The second is to set

. o o

o

t t y y x t

x x

In this case , .oy y x t 9

The is the absolute value of themaximum displacement from the equilibrium

( , ) sinm

myy x t y kx t

amplitude

maximum displacement from the equilibriumposition. The is defined as the argument kx tphaseof the sine function.The is the wavelength shortest distance between two repetitions of the wave atbetween two repetitions of the wave at a fixed time.

We fix at 0 We have the condition: ( 0) ( 0)t t y x y x

1 1

1 1 1

We fix at 0. We have the condition: ( ,0) ( ,0)

sin sin sin

Since the sine function is periodic with period 2 2 2m m m

t t y x y x

y kx y k x y kx k

k k

Since the sine function is periodic with period 2 2

A is the time it takes (with ed fix

k

T

k

period ) the sine function to complete one oscillation We take 0 (0 ) (0 )

xx y t y t T

one oscillation. We take 0 (0, ) (0, )

sin sin sin 2 2m m m

x y t y t T

y t y t TT

y t T T

In the figure we show two snapshots of a harmonic waveThe speed of a traveling wave v

k

taken at times and . During the time interval the wave has traveled a distance . The wave speed

t t t tx

x

. Onxvt

e method of finding is to imagine that

we move with the same speed along the -axis. In thish ill h i d h

v

xcase the wave will seem to us that it does not change.

Since ( , ) sin this means that the argument of the sine functionmy x t y kx t

is constant. constant. We take the derivative with respect to .

0 The speed

kx t tdx dx dxk vdt dt k dt k

A harmonicdt dt k dt k

wave that propagates along the -axis is described by the equation:

( , ) sin . The function ( , ) describes a general wave that m

xy x t y kx t y x t h kx t

negative

propagates along the positive -axis. A genex

ral wave that propagates along the

-axis is described by the equatio ( , )n: y x t hx kx t negative 11

A particular wave is given by y = (0.200m)sin[(0.500m-1)x – (8.20rad/s)t]Find:a) A e) f h) y at x 10 0m and t 0 500sa) A e) f h) y at x = 10.0m and t = 0.500sb) K f) tc) λ g) v

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Problem 5. If ( , ) (6.0 )sin( (600 / ) ) describes a wave travelingalong a string, how much time does any given point on the string take to move between

y x t mm kx rad s t

displacements y=+2.0 mm and y=-2.00mm?

Let y1 = 2.0 mm (corresponding to time t1) and y2 = –2.0 mm (corresponding to time t2). Then we find

kx + 600t1 + = sin1(2.0/6.0) and

kx + 600t + = sin1( 2 0/6 0)kx + 600t2 + = sin (–2.0/6.0) .

Subtracting equations gives

600( ) i 1(2 0/6 0) i 1( 2 0/6 0)600(t1 – t2) = sin1(2.0/6.0) – sin1(–2.0/6.0). Thus we find t1 – t2 = 0.011 s (or 1.1 ms).

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Problem 9. A transverse sinusoidal wave is moving along a string in the positive directionof an x axis with a speed of 80 m/s. At t=0, the string particle at x=0 has a tranverse displacement of 4.0 cm from its equilibrium position and is not moving. The maximum transverse speed of the string particle at x=0 is 16 m/s (a) What is the frequency of the wave?transverse speed of the string particle at x 0 is 16 m/s. (a) What is the frequency of the wave?(b) What is the wavelength of the wave? If

m

the wave equation is of the form ( , ) sin( )what are (c) y ,( ) , ( ) , (f) , and (g) the correct choice of sign in front of ?

my x t y kx td k e

(a) Recalling from Ch. 12 the simple harmonic motion relation um = ym, we have(a) Recalling from Ch. 12 the simple harmonic motion relation um ym, we have

1 6 4 0 0 r a d / s .0 .0 4 0

Since = 2f, we obtain f = 64 Hz. (b) Using v = f, we find = 80/64 = 1.26 m 1 .3 m . (c) The amplitude of the transverse displacement is 24 .0 c m 4 .0 1 0 m .my (d) The wave number is k = 2/ = 5.0 rad/m.(d) The wave number is k 2/ 5.0 rad/m. (e) The angular frequency, as obtained in part (a), is 21 6 / 0 .0 4 0 4 .0 1 0 r a d / s . (f) The function describing the wave can be written as

0 0 4 0 i 5 4 0 0 0 .0 4 0 s i n 5 4 0 0y x t where distances are in meters and time is in seconds. We adjust the phase constant to satisfy the condition y = 0.040 at x = t = 0. Therefore, sin = 1, for which the “simplest”root is = /2. Consequently, the answer is

0 .0 4 0 s i n 5 4 0 0 .2

y x t

(g) The sign in front of is minus.

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1 1 0.5

Problem 29. Use the wave equation to find the speed of a wave given by ( , ) (2 00 )[(20 ) (4 0 ) ]y x t mm m x s t ( , ) (2.00 )[(20 ) (4.0 ) ] .y x t mm m x s t

The wave 1 1 1/2( , ) (2.00 mm)[(20 m ) (4.0 s ) ]y x t x t is of the form ( )h kx t with angular wave number 120 mk and angular frequency 4.0 rad/s . Thus, the speed of the a e isthe wave is

1/ (4.0 rad/s)/(20 m ) 0.20 m/s.v k

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Below we will determine the speed of a wave thatWave speed on a stretched stringv

ppropagates along a string whose linear mass densityis . The tension on the string is equal to . C id ll

ti f th t i f l th Consider a small section of the string of length .

The shape of the element can be approximated to be an arc of a circle of radius Rwhose center is at O. The net force in the direction of O is 2 sin . F

Here we assume that 1 sin 2

FR R

2 2

The force is also given by Newton's second law: v vF mR R

(eqs.1)

= (eqs.2)

2

If we compare equations 1 and 2 we get: v

R RvR R

The speed depends on the tension and the mass densiv Note : ty but not on the wave frequency .f

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Problem : The drawing shows a 15.0 kg ball being whirled in a circular path on the end of a string. The motion occurs on a frictionless horizontal table. The angular speed of the ball is ω = 12.0 rad/s. The string has a mass of 0.0230 kg. How much time does it take for a wave on the string to travel from the center of the circle to the ball?

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The Speed of a Wave on a StringProblem To measure the acceleration due to gravity on a distant planet, an astronaut hangs a 0.086 kg ball from the end of a wire. The wire has length of 1.5 m and a linear density of 3.1 x 10-4 kg/m. g g y gUsing electronic equipment, the astronaut measures the time for a transverse pulse to travel the length of the wire and obtains a value of 0.083 s. The mass of the wire is negligible compared to the mass of the ball. Determine the acceleration due to gravity.

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The Speed of a Wave on a StringProblem A horizontal wire is under a tension of 315 N and has a mass per unit length of 6.50 x 10-3

kg/m. A transverse wave with an amplitude of 2.50 mm and a frequency of 585 Hz is traveling on g p q y gthis wire. As the wave passes, a particle of the wire moves up and down in simple harmonic motion. Obtain:a) The speed of the waveb) The maximum speed with which the particle moves up and down) p p p

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Problem 23. A 100 g wire is held under a tension of 250 N with one end at x=0 and the other at x=10.0m. At time t=0, pulse 1 is sent along the wire from the end at x=10.0 m.

i l i l h i f h d h i iAt time t=30.0 ms, pulse 2 is sent along the wire from the end at x=0. At what position xdo the pulses begin to meet?

The pulses have the same speed v. Suppose one pulse starts from the left end of the wirep p pp pat time t = 0. Its coordinate at time t is x1 = vt. The other pulse starts from the right end, atx = L, where L is the length of the wire, at time t = 30 ms. If this time is denoted by t0then the coordinate of this wave at time t is x2 = L – v(t – t0). They meet when x1 = x2, or, what is the same when vt = L – v(t – t0) We solve for the time they meet: t = (L + vt0)/2vwhat is the same, when vt L v(t t0). We solve for the time they meet: t (L + vt0)/2vand the coordinate of the meeting point is x = vt = (L + vt0)/2. Now, we calculate the wavespeed:

(250 N) (10.0 m) 158m/s. Lv

0.100 kgm Here is the tension in the wire and m/L is the linear mass density of the wire. Thecoordinate of the meeting point iscoordinate of the meeting point is

310.0 m (158m/s) (30.0 10 s) 7.37 m.2

x

This is the distance from the left end of the wire. The distance from the right end is L – x= (10.0 m – 7.37 m ) = 2.63 m. 20

Consider a transverse wave propagating along a string which is described by the equation:

Rate of energy transmission

which is described by the equation: ( , ) sin - . The transverse velocity

cos At point a both

my x t y kx tyu y kx t

- cos - At point a bothmu y kx tt

and are equal to zero. At point b both and have maximau K u Khave maxima.

2

2

1In general the kinetic energy of an element of mass is given by: 2

1

dm dK dmv

2

2

1 - cos - The rate at which kinetic energy propagates 2

1along the string is equal to

mdK dx y kx t

dK v y

2 2cos - The average ratekx talong the string is equal to 2

v ydt

2 2 2 2 2

cos The average rate

1 1cos - As in the case of the oscillating 2 4

m

m mavgavg

kx t

dK v y kx t v ydt

spring-mass system

g

avgavg avg avg av

dU dK dU dKPdt dt dt dt

2 212 m

g

v y

Problem 26. A string along which waves can travel is 2.70 m long and has a mass of 260 g. The tension in the string is 36.0 N. What must be the frequency of traveling waves of amplitude 7 70 mm for the average power to be 85 0 W?7.70 mm for the average power to be 85.0 W?

2 21

2avg myP v v

Using Eq. 16–33 for the average power and Eq. 16–26 for the speed of the wave, we solve for f = /2:

avg3

21 1 2(85.0 W) 198 Hz.2 2 (7.70 10 m)/ (36.0 N) (0.260 kg / 2.70 m )m

Pf

y

22

Consider a string of mass density and tension A transverse wave propagates along the string

The wave equation2 2

2 2 2

1y yx v t

A transverse wave propagates along the string. The transverse motion is described by ( , ) Consider an element of length and mas

y x tdx s dm dx

1 2The forces the net force along the y-axis is given by the equation:

sin sin sin sin Here we

F F

F F F

2 2 1 1 2 1

1 2 1 11

sin sin sin sin Here we

assume that 1 and 1 sin tan

ynetF F F

yx

θ2

2 2and sin tan yx

2 2 1

ynety yFx x

2

2

22 1

From Newton's second law we have: ynet yy y yF dma dx

t x x

y y

2 2 2 2

2 12 2 2 2 2

2 1

1y y

y y y y y yx xdxx x t dx x t v t

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The Superposition of Waves

A superposition principle: when two or more waves arrive at any given point simultaneously, the resultant instantaneous deformation is the sum of the individual instantaneous deformationsthe sum of the individual instantaneous deformations.

• The waves do not interact, they have no effect on one anotherE h h h h h• Each wave propagates as though the others were not present

• The net displacement of the medium is the vector sum of the individual displacementsp

The superposition principle is very well satisfied for waves of low amplitudeamplitude.

For waves of very large amplitude the superposition principle fails, b th fi t lt th ti f th di dbecause the first wave alters the properties of the medium and therefore affects the behavior of a second wave propagating on the same string, air, etc. (medium)

2 2 21y y y

The principle of superposition for waves1 2( , ) ( , ) ( , ) y x t y x t y x t

2 2 2 2

1The wave equation even though

it was derived for a transverse wave propagating alongt i d t i i t f ll t f

y y yx t v t

a string under tension, is true for all types of wa

1

2

ves.This equation is "linear" which means that if and are solutions of the wave equation, the function

yy

1 1 2 2 1 2 is also a solution. Here and are constants.The principle of superpositionc y c y c c

is a direct consequenceof the linearity of the wave equation This principleof the linearity of the wave equation. This principle can be expressed as follows:

1 2

Consider two waves of the same type that overlap at some point P in space.Assume that the functions ( , ) and ( , ) describe the displacementsif the wave arrived at P alone. The displacement at P

y x t y x t when both wavesp

1 2are present is given by: Overlapping waves do not in any way alter the travel of each oth

( , ) ( , ) ( , )er

y x t y x t y x t Note : 25

Consider two harmonic waves of the same amplitude and frequency which propagate along the x axis The

Interference of waves

and frequency which propagate along the x-axis. The two waves have a phase difference . Wewill combine these waves using the

principle of

superposition. The phenomenon of combing waves is knwon as and the two waves are said to The displacement of the two waves

interferenceinterfere

1

said to . The displacement of the two waves are given by the functions: ( , ) my x t y

interfere

2 1 2

sin

and ( , ) sin . m

kx t

y x t y kx t y y y

, sin sin

, 2 cos sin2 2

m

m

m

y x t y kx t

y x t y kx t y tkx

The resulting wave has the same frequency asthe original waves, and its amplitude

2 2

2 cm my y Its phase is equalos 2

to 2

26

C t ti i t fThe amplitude of two interefering waves is given by:

2 cos It has its maximum value if 0y y

Constructive interference

2 cos It has its maximum value if 02

In this case The displacement of the resul

2ting wave is:

m m

m my y

y y

The displacement of the resulting wave is:

, 2 smy x t y in2

kx t

This phenomenon is known as fully constructive interference

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The amplitude of two interefering waves is given by:

Destructive interference

2 cos It has its minimum value if 2

In this case 0m

m my y

y

The displacement of the resulting wave is:

, 0Thisy x t

phenomenon is known asThis phenomenon is known as fully destructive interference

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The amplitude of two interefering waves is given by:Intermediate interference

p g g y

2 cos When interference is neither fully2

constructive nor fully destructive it is called

m my y

constructive nor fully destructive it is called

An

intermediate interference2example is given in the figure for An example is given in the figure for 3

In this case The displacement of the resulting wave is:

m my y

p g

, sin3

S ti th h diff i

my x t y kx t

N t Sometimes the phase difference is expressed as a difference iNote :

n wavelength In this case remembre that:

2 radians 1 29

Pr oblem 32. What phase difference between two identical traveling waves moving in the samedirection along a stretched string, results in the combined wave having an amplitude 1.50 timesthat of the common amplitude of the two combining waves? Express your answer inthat of the common amplitude of the two combining waves? Express your answer in (a) degrees, (b) radians, (c) wavelengths?

2 cos2m my y

(a) Let the phase difference be . Then from Eq. 16–52, 2ym cos(/2) = 1.50ym, which gives

1 1 50y 1 1.502cos 82.8 .2

m

m

yy

(b) Converting to radians, we have = 1.45 rad.( ) g , (c) In terms of wavelength (the length of each cycle, where each cycle corresponds to 2rad), this is equivalent to 1.45 rad/2 = 0.230 wavelength.

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A phasor is a method for representing a wave whose Phasors

1 1displacement is: , sinThe phasor is defined as a vector with the following

ti

my x t y kx t

properties:Its magnitude is equal to the wave's amp1. 1litude The phasor has its tail at the origin O and rotates in

my2.the clockwise direction about an axis through Owith angular speed .Thus defined the projection of the phasor on the axisy

Thus defined, the projection of the phasor on the -axis (i.e. i

y

1ts y-component) is equal to sinA phasor diagram can be used to represent more than one

my kx t

2 2

waves. (see fig.b). The displacement of the second wave is: , sin The phasor of the my x t y kx t second

f l ith th h f th fi twave forms an angle with the phasor of the first wave indicating that it lags behind wave 1 by a phase angle .

31

Consider two waves that have the same frequencyWave addition using phasorsConsider two waves that have the same frequencybut different amplitudes. They also have a phase difference . The displacements of the two waves

1 1are: , sinmy x t y kx 2 2

and

, sin . The superposition of the twowaves yields a wave that has the same angular frequency

m

t

y x t y kx t

waves yields a wave that has the same angular frequency and is described by: sin Here is thewave amplitude and is the pha

m my y kx t y

se angle. To determine and we add the two phasors representing the waves as vectors (see fig c)

my

(see fig.c). Note: The phasor method can be used to add vectors that have different amplitudes.

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m1 2 1 2

Problem 35. Two sinusoidal waves of the same frequency travel in the same direction along a string. If y 3.0 , 4.0 cm, 0, and / 2 , what is the amlitude of the resultant wave?mcm y rad

The phasor diagram is shown below: y1m and y2m represent the original waves and ymrepresents the resultant wave. The phasors corresponding to the two constituent waves

k l f 90° ith h th th t i l i i ht t i l Th P thmake an angle of 90° with each other, so the triangle is a right triangle. The Pythagoreantheorem gives

2 2 2 2 2 21 2 (3.0cm) (4.0cm) (25cm)m m my y y .

Thus ym = 5.0 cm.

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, 2 sin cosmy x t y kx t

Consider the superposition of two waves that have the sameStanding Waves :

1 2

frequency and amplitude but travel in opposite directions. The displacements of two waves are: , sin , ,my x t y kx t y x t y

sin

The displacement of the resulting wavem kx t

y x t y x t y x t

1 2The displacement of the resulting wave , , ,

, sin sin 2 sin cosThis is not a traveling wave but an oscillation that has a position

m m m

y x t y x t y x t

y x t y kx t y kx t y kx t

g pdependent amplitude. It is known as a standing wave.

34

The displacement of a standing wave is given by the equation

, 2 sin cos:

my x t y kx t

The position dependant amplitude is equal to

These are defined as positions

2 si

where the standing

nmy kx

Nodes :

a a

These are defined as positions where the standing wave aNodes :

mplitude vanishes. They occur when 0,1, 2,2 0,1,2,...

kx n n

x n n nx

n n nn

0,1, 2,...

These are defined as positions where the standing

2nx n n nx

Antinodes : a wave amplitude is maximum.

1They occur when 2

kx n

0,1,2,...n 2

2 1 0,1, 2,...2

12 2nx n x nn

The distance between ajacent nodes and antinodes is /2 The distance between a node and an ajacent antinode is /4

Note 1 : Note 2 :

Consider a string under tension on which is clamped Standing waves and resonance

A B at points A and B separated by a distance . We senda harmonic wave traveling to the right. The wave isreflected at poi

L

nt B and the reflected wave travels to

A B

pthe left. The left going wave reflects back at point Aand creates a thrird wave traveling to the right. Thus

h l b f l i h lf

A B

we have a large number of overlapping waves half of which travel to the right and the rest to the left.A B

For certain frequencies the interference produces a standing wave. Such astanding wave is said to be . The frequencis at which the standingwave occurs are known as the

at resonanceresonant frequencie of the system. s

36

A B

Resonances occur when the resulting standing wavesatisfies the boundary condition of the problem. These are that the Amplitude must be zero at point AA B These are that the Amplitude must be zero at point Aand point B and arise from the fact that the string isclamped at both points and therefore cannot move.

A B1

The first resonance is shown in fig.a. The standing

wave has two nodes at points A and B. Thus 2

L

A B1

2. The second standing wave is sh2 own L

in fig.b. It has three nodes (two of them at A and B)

2In this case 2 2

L L

The third standing wave is shown in fig c It has four nodes (two of them at A and B)

3

The third standing wave is shown in fig.c. It has four nodes (two of them at A and B)

In this case 3 The general expression for the resonant2

2 3

LL

2 wavelengths is: 1, 2,3,..nL nn

. the resonant frequencies 2n

n

v vf nL

Problem 43. A string fixed at both ends is 8.40 m long and has a mass of 0.120 kg. It is subjected to a tension of 96.0 N and set oscillating. (a) What is the speed of the waves on the string? (b) What is the longest possible wavelength for a standing wave?waves on the string? (b) What is the longest possible wavelength for a standing wave? (c) Find the frequency of the wave.

(a) The wave speed is given by ,v where is the tension in the string and is the linear mass density of the string Since the mass density is the mass per unit length =linear mass density of the string. Since the mass density is the mass per unit length, M/L, where M is the mass of the string and L is its length. Thus

(96.0 N) (8.40 m) 82 0 m/sLv 82.0 m/s.0.120 kg

vM

(b) The longest possible wavelength for a standing wave is related to the length of thestring by L = /2, so = 2L = 2(8.40 m) = 16.8 m. (c) The frequency is f = v/ = (82.0 m/s)/(16.8 m) = 4.88 Hz.

38

Problem 50. A rope, under a tension of 200N and fixed at both ends, oscillates in a second harmonic standing wave pattern. The displacement of the rope is given by (0.10 )(sin / 2)sin12 , where x=0

y m x t at one end of the rope, x is in meters, and t is in seconds. What are (a) the length of the rope,

(b) the speed of the wave on the rope, and (c) the mass of the rope? (d) If the rope oscillates in ( ) p p , ( ) p ( ) pa third harmonic standing wave pattern, what will be the period of oscillation?

Since the rope is fixed at both ends, then the phrase “second-harmonic standing wave pattern” describes the oscillation shown in Figure 16–23(b), where (see Eq. 16–65)

a n d = vL fL

.

(a) Comparing the given function with Eq. 16-60, we obtain k = /2 and = 12 rad/s. Since k = 2/ then 2 sin cosy x t y kx t

2 4 .0 m 4 .0 m .2

L

(b) Since = 2f then 2 1 2 r a d / s ,f which yields

, 2 sin cosmy x t y kx t

6 .0 H z 2 4 m / s .f v f

(c) Using Eq. 16–26, we have

2 0 0 N2 4 m /s 2 4 m /s

/ ( 4 .0 m )v

m

which leads to m = 1.4 kg. (d) With ( )

3 3 ( 2 4 m /s ) 9 .0 H z2 2 ( 4 .0 m )

vfL

The period is T = 1/f = 0.11 s.

39