Chapter 14: Wave Motion - Stony Brook...

Chapter 14: Wave Motion

Types of mechanical wavesMechanical waves

• are disturbances that travel through some material or substancecalled medium for the waves.

• travel through the medium by displacing particles in the medium

• travel in the perpendicular to or along the movement of theparticles or in a combination of both

transverse waves:waves in a string etc.

longitudinal waves:sound waves etc.

waves in water etc.

Types of mechanical waves (cont’d)

Longitudinal and transverse waves

sound wave = longitudinal waveC = compressionR = rarefaction

air compressedair rarefied


Longitudinal-transverse waves


Periodic waves• When particles of the medium in a wave undergo periodicmotion as the wave propagates, the wave is called periodic.

x=0 x

λA

t=0

t=T/4

t=T

period

amplitudewavelength

Mathematical description of a wave

Wave function• The wave function describes the displacement of particles

in a wave as a function of time and their positions:

txytxyy ,atntdisplaceme is;),(=• A sinusoidal wave is described by the wave function:

)//(2cos)/(2cos

)]/(cos[)]/(cos[),(

TtxAtvxfA

tvxAvxtAtxy

−=−=

−=−=

λππωω sinusoidal wave moving in

+x directionangular frequency

fπω 2=velocity of wave, NOT ofparticles of the medium

wavelength periodvf =λ

Tf /1=

)]/(cos[),( xvtAtxy += ω sinusoidal wave moving in-x direction v->-vphase velocity

Mathematical description of a wave (cont’d)

Wave function (cont’d)

t=0

x=0 x

λ

t=T/4

t=T period

wavelength)//(2cos),( TtxAtxy −= λπ

),(),(

Ttxytxy

+=+= λ


Wave number and phase velocityλπ /2=kwave number:

)cos(),( tkxAtxy ω−=phase

The speed of wave is the speed with which we have tomove along a point of a given phase. So for a fixed phase,

.consttkx =−ω

vkdtdx == // ω phase velocity

)](cos[)cos(),( vtxkAtkxAtxy −=−= ω


Particle velocity and acceleration in a sinusoidal wave

)cos(),( tkxAtxy ω−=

),(

)cos(/),(),(

)sin(/),(),(

2

222

txy

tkxAttxytxa

tkxAttxytxv

y

y

ω

ωω

ωω

−=

−−=∂∂=

−=∂∂=

acceleration

u in textbookvelocity

Also ),()cos(/),( 2222 txyktkxAkxtxy −=−−=∂∂ ω

222

222222

/),(/),()/(/),(

tvtxyttxykxtxy

∂∂=

∂∂=∂∂ ωwave equation


General solution to the wave equation

22

2

2

2

2

2

2

2 ),(),(),(tv

txyt

txykx

txy∂

∂=

∂∂

=∂

∂ω

wave equation

)(),( vtxftxy ±= )cos( tkx ω−Solutions: such as

The most general form of the solution:

)()(),( vtxgvtxftxy ++−=

Speed of a transverse waveWave speed on a string

x∆

2FyF2

FF x =2yF1

FF x =1

1F

•Consider a small segment of string whoselength in the equilibrium position is .x∆

•The mass of the segment is .xm ∆= µ• The x component of the force (tension) at bothends have equal in magnitude and opposite indirection because this is a transverse wave.

x• xxyxy xyFFxyFF ∆+∂∂=∂∂−= )/(/,)/(/ 21

• The total y component of the forces is:

xx ∆+

)/(

])/()/[(22

21

tyx

xyxyFFFF xxxyyy

∂∂∆=

∂∂−∂∂=+= ∆+

µNewton’s 2nd law

massacceleration

Speed of a transverse wave (cont’d)Wave speed on a string (cont’d)

x∆

2FyF2

F

yF1

F

x

• The total y component of the forces is:

)/(

])/()/[(22

21

tyx

xyxyFFFF xxxyyy

∂∂∆=

∂∂−∂∂=+= ∆+

µ

)/)(/(

/])/()/[(22 tyF

xxyxy xxx

∂∂

=∆∂∂−∂∂ ∆+

µ0→∆x

)/)(/(/ 2222 tyFxy ∂∂=∂∂ µ

1F

wave eq.

)/()(

/

inertiaforcerestoring

Fv == µ

Energy in wave motionTotal energy of a short string segment of mass

x∆

2FyF2

F

yF1

F

1F

x

dxdm µ=• At point a, the force

a

yF1 does work on the string segment right of point a.

• Power is the rate of work done :

)/),()(/),((

)/),()(,(),( 10

ttxyxtxyF

ttxytxFtxP yt

∂∂∂∂−=

∆∆=→∆

)sin()/()sin()/(

)cos(),(

tkxAtytkxkAxy

tkxAtxy

ωωω

ω

−=∂∂−−=∂∂

−=

)(sin

)(sin)(sin/),(

222

222

22

tkxAF

tkxAvtkxAFkdtdEtxP

ωωµ

ωωµ

ωω

−=

−=

−==

µω /, 2 Fvvk ==

work done

Pmax

Energy in wave motion (cont’d)Maximum power of a sinusoidal wave on a string:

22max AFP ωµ=

Average power of a sinusoidal wave on a string

21sin

21 2

0

2 =∫π

θθπ

d

• The average of )(sin2 tkx ω− over a period:

22)2/1( AFPave ωµ=• The average power:

Wave intensityWave intensity for a three dimensional wave from a pointsource:

22 W/mofunitsin

4 rPIπ

= power/unit area

1r

2r

22

212

1 44 IrIr ππ =

21

22

2

1

rr

II=

Wave interference, boundary condition, and superposition

The principle of superposition

• When two waves overlap, the actual displacement of anypoint at any time is obtained by adding the displacementthe point would have if only the first wave were present andthe displacement it would have if only the second wave werepresent:

),(),(),( 21 txytxytxy +=

Wave interference, boundary condition, and superposition (cont’d)

Interference

• Constructive interference (positive-positive or negative-negative)

• Destructive interference (positive-negative)


Reflection incident wave reflected wave

• Free end )cos()cos(),( txkBtxkAtxy ωω ++−=

Bx

For x<xB

At x=xB ABxtxyBxx +=→=∂∂ = 0)/),((

Vertical component of the forceat the boundary is zero.


Reflection (cont’d)

)cos()cos(),( txkBtxkAtxy ωω ++−=• Fixed end

For x<xB

ABtxyBxx −=→== 0),(At x=xB

Displacement at the boundary is zero.



• At high/low density



• At low/high density

Standing waves on a stringSuperposition of two waves moving in the same direction

Superposition of two waves moving in the opposite direction

Standing waves on a string (cont’d)

Superposition of two waves moving in the opposite directioncreates a standing wave when two waves have the samespeed and wavelength.

))(sin(sin2)cos()cos(

),(),(),( 21

txkAtkxAtkxA

txytxytxy

ωωω

=−−−=

+=

incident reflected

,..)2,1,0(2//or

when0sin

===

==

nnknx

nkxkxλπ

π

N=node, AN=antinode

Normal modes of a string

There are infinite numbers of modes of standing waves

fixed end fixed endL

,...)3,2,1(2

== nnL λ

µ

λ

FL

fLvnf

nL

n

n

21

2

/2

1 ===

=

2/1λ

2λ

2/3 3λ

42λ

firstovertone

secondovertone

thirdovertone

funda-mental

Sound wavesSound

• Sound is a longitudinal wave in a medium

• The simplest sound waves are sinusoidal waves whichhave definite frequency, amplitude and wavelength.

• The audible range of frequency is between 20 and 20,000 Hz.

Sound waves (cont’d)Sound wave (sinusoidal wave)

)cos(),( tkxAtxy ω−=Sinusoidal sound wave function:

x x+∆x∆x

S

),(1 txyy = ),(2 txxyy ∆+=undisturbedcyl. of air

disturbedcyl. of air

)],(),([)( 12

txytxxySyySV

−∆+=−=∆

Change of volume:

)(/),(/ SdxVxtxyVdV =∂∂= Q

)//(),( VdVtxpB −=bulk modulus pressure

)sin()/),((),( tkxBkAxtxyBtxp ω−=∂∂−=

Pressure:

Pressure amplitude and earPressure amplitude for a sinusoidal sound wave

)sin(),( tkxBkAtxp ω−=• Pressure:

BkAp =max• Pressure amplitude:

Ear

Perception of sound wavesFourier’s theorem and frequency spectrum

• Fourier’s theorem:Any periodic function of period T can be written as

∑ +=n nnnn tfBtfAty )]2cos()2sin([)( ππ

,...)3,2,1(,/1 11 === nnffTf n

fundamental freq.

where

• Implication of Fourier’s theorem:

Perception of sound wavesTimbre or tone color or tone quality

Frequency spectrum

music

noise

piano

piano

Speed of sound waves (ref. only)The speed of sound waves in a fluid in a pipe

movable piston

pA pA

pAApp )( ∆+

yvyv

yv yv

tvy

vt

fluid inequilibrium

fluid in motion

yvvtA)(ρlongitudinal momentum carriedby the fluid in motion

tAvy−

Avtoriginal volume of the fluid inmotion

bulk modulus B:-pressure change/frac. vol. change

change in volume of the fluid in motion

)/()( AvttAvp

y−∆−

change in pressure in the fluid in motion v

vBp y=∆

fluid at restboundary moves at speed of wave

velocity of wave

velocityof fluid

Speed of sound waves (ref. only) (cont’d)

The speed of sound waves in a fluid in a pipe (cont’d)

longitudinal impulse = change in momentum

yy vtAvAt

vv

BpAt ρ==∆

ρBv =speed of a longitudinal

wave in a fluid

The speed of sound waves in a solid bar/rod

== YYv ,ρ

Young’s modulus

Speed of sound waves (cont’d)The speed of sound waves in gases

===

==

MTR

pMRTv

ργγ 0

==

=

0

0

p

pBγγ

bulk modulus of a gas ratio of heat capacities

equilibrium pressure of gas

gas constant 8.314472 J/(mol K)

temperature in Kelvin

molar mass

- P in textbook(backgroundpressure).

- ρ density

In textbook

speed of a longitudinalwave in a fluid

Sound level (Decibel scale)Decibel scale

As the sensitivity of the ear covers a broad range of intensities,it is best to use logarithmic scale:

2120

0

W/m10,log)dB10( −== IIIβDefinition of sound intensity:

( unit decibel or dB)

Intensity (W/m2)Sound intensity in dB

Military jet plane at 30 m 140 102

Threshold of pain 120 1

Whisper 20 10-10

Hearing thres. (100Hz) 0 10-12

Standing sound wavesSound wave in a pipe with two open ends

Standing sound wavesStanding sound wave in a pipe with two open ends

Standing sound wavesSound wave in a pipe with one closed and one open end

Standing sound wavesStanding wave in a pipe with two closed ends

Displacement

Normal modesNormal modes in a pipe with two open ends

2nd normal mode

,...)3,2,1(2

== nLvnfn,...)3,2,1(2

2=== n

nLornL n

n λλ

Normal modesNormal modes in a pipe with an open and a closed end

(stopped pipe)

,...)5,3,1(44

=== nnLornL n

n λλ,...)5,3,1(

4== n

Lvnfn

ResonanceResonance

• When we apply a periodically varying force to a system that canoscillate, the system is forced to oscillate with a frequency equalto the frequency of the applied force (driving frequency): forcedoscillation. When the applied frequency is close to a characteristicfrequency of the system, a phenomenon called resonance occurs.

• Resonance also occurs when aperiodically varying force is appliedto a system with normal modes. When the frequency of the appliedforce is close to one of normalmodes of the system, resonanceoccurs.

Interference of wavesTwo sound waves interfere each other

constructivedestructive

d1 d2

,....2,1,0)()2/1()(21

=+=

=−

nedestructivnveconstructindd

λλ

BeatsTwo interfering sound waves can make beat

Two waves with differentfrequency create a beatbecause of interferencebetween them. The beatfrequency is the differenceof the two frequencies.

Beats (cont’d)Two interfering sound waves can make beat (cont’d)

Suppose the two waves have frequencies af .bfFor simplicity, consider two sinusoidal waves of equal intensity:

and

tfAty bb π2sin)( −=;2sin)( tfAty aa π=

Then the resulting combined wave will be:

]))(2(21cos[]))(2(

21sin[2)()( tfftffAtyty bababa +−=+ ππ

))(21cos)(

21sin2sinsin( bababa +−=−Q

As human ears does not distinguish negative and positive amplitude,they hear two max. or min. intensity per cycle, so 2 x (1/2)|fa-fb|=|fa-fb| is the beat frequency fbeat.

Doppler effectMoving listener

Source at restListener moving right

Source at restListener moving left

Doppler effect (cont’d)Moving listener (cont’d)

•The wavelength of the sound wave does not change whetherthe listener is moving or not.

• The time that two subsequent wave crests pass the listenerchanges when the listener is moving, which effectively changesthe velocity of sound.

Lf

sffreq. listener hears

s

LLL fv

vvvvf/±

=±

=λ

freq. source generates

vvelocity of sound at source

Lv - for a listener moving away from + for a listener moving towards

the source.

velocity of listener

Doppler effect (cont’d)Moving source

When the source moves

Doppler effect (cont’d)Moving source (cont’d)• The wave velocity relative to the wave medium does notchange even when the source is moving.

• The wavelength, however, changes when the source is moving.This is because, when the source generates the next crest, thethe distance between the previous and next crest i.e. the wave-length changed by the speed of the source.

s

s

s

s

s fvv

fv

fv ±

=±=λs

s

fv

=λ

The source at rest When the source is moving

+ for a receding source- for a approaching source

Doppler effect (cont’d)Moving source and listener

+ for a receding source- for a approaching source

ss

LLL f

vvvvvvf

±±

=±

=λ

- for a listener moving away from + for a listener moving towards

the source.

Effect of change of source speed

The signs of vL and vS are measuredin the direction from the listener L to thesource S.

svv = svv <

Doppler effect (cont’d)Example 1• A police siren emits a sinusoidal wave with frequency fs=300 Hz.The speed of sound is 340 m/s. a) Find the wavelength of the wavesif the siren is at rest in the air, b) if the siren is moving at 30 m/s, findthe wavelengths of the waves ahead of and behind the source.

a)

b) In front of the siren:

Behind the siren:

m. 1.13 Hz /300m/s 340/ === sfvλ

m 1.03 Hz m/s)/300 30 - m/s 340(/)( ==−= ss fvvλ

m 1.23 Hz m/s)/300 30 m/s 340(/)( =+=+= ss fvvλ

Doppler effect (cont’d)Example 2• If a listener l is at rest and the siren in Example 1 is moving awayfrom L at 30 m/s, what frequency does the listener hear?

Hz. 276Hz) 300(m/s 30 m/s 340

m/s 340=

+=

+= s

sL f

vvvf

Example 3• If the siren is at rest and the listener is moving toward the left at 30m/s, what frequency does the listener hear?

Hz. 274Hz) 300( m/s 340m/s) 30-m/s 340

==−

= sL

L fvvvf

Doppler effect (cont’d)Example 4• If the siren is moving away from the listener with a speed of 45 m/srelative to the air and the listener is moving toward the siren with aspeed of 15 m/s relative to the air, what frequency does the listenerhear?

Hz. 277Hz) 300(m/s 45 m/s 340m/s 15m/s 340

=++

=++

= ss

LL f

vvvvf

Example 5• The police car with its 300-MHz siren is moving toward a warehouseat 30 m/s, intending to crash through the door. What frequency doesthe driver of the police car hear reflected from the warehouse?

Hz. 329Hz) 300( m/s 30m/s 340

m/s 340=

−=

−= s

sW f

vvvf

Hz. 358Hz) 329( m/s 340m/s 30m/s 340

=+

=+

= WL

L fvvvf

Freq. reachingthe warehouse

Freq. heard bythe driver

ExercisesProblem 1

A transverse wave on a rope is given by:])250()400.0[(cos)750.0(),( 11 tsxcmcmtxy −− += π

(a) Find the amplitude, period, frequency, wavelength, and speed ofpropagation. (b) Sketch the shape of the rope at the following valuesof t : 0.0005 s, and 0.0010 s. (c) Is the wave traveling in the +x or –xdirection? (d) The mass per unit length of the rope is 0.0500 kg/m.Find the tension. (e) Find the average power of this wave. Solution

(a)A=0.75 cm, λ=2/0.400 = 5.00 cm, f=125 Hz, T=1/f=0.00800 s andv=λf=6.25 m/s.

(b) Homework(c) To stay with a wave front as t increases, x decreases. Therefore the

wave is moving in –x direction.(d) , the tension is(e)

)//(2cos),( TtxAtxy += λπ

)/( µFv = .6.19)/25.6)(/050.0( 22 NsmmkgvF === µ

.2.54)2/1( 22 WAFPav == ωµ

ExercisesProblem 2

A triangular wave pulse on a taut string travels in the positive +x directionwith speed v. The tension in the string is F and the linear mass density ofthe string is µ. At t=0 the shape of the pulse I given by

LxforLxforLxLh

xLforLxLhLxfor

><<−<<−+

−<

00/)(

0/)(0

=)0,(xy

(a)Draw the pulse at t=0. (b) Determine the wave function y(x,t) at alltimes t. (c) Find the instantaneous power in the wave. Show that thepower is zero except for –L < (x-vt) < L and that in this interval thepower is constant. Find the value of this constant.

Solution y(a)

hL-L

x

ExercisesProblem 2 (cont’d)

Solution

(b) The wave moves in the +x direction with speed v, so in the experessionfor y(x,0) replace x with –vt:

LxforLxforLvtxLh

xLforLvtxLhLxfor

><<+−<<−−+

−<

00/)(

0/)(0

=),( txy

(c)

LxforFLxforLhFvLhvLhF

xLforLhFvLhvLhFLxforF

>=−<<=−−

<<−=−−

−<=−

0)0)(0(0)/()/)(/(

0)/()/)(/(00)0(

2

2

=∂∂

∂∂

−=ty

xyFtxP ),(

Thus the instantaneous power is zero except for –L < (x-vt) < L whereIt has the constant value Fv(h/L)2.

ExercisesProblem 3

The sound from a trumpet radiates uniformly in all directions in air. At adistance of 5.00 m from the trumpet the sound intensity level is 52.0 dB.At what distance is the sound intensity level 30.0 dB?

Solution

The distance is proportional to the reciprocal of the square root of theintensity and hence to 10 raised to half of the sound intensity levelsdivided by 10:

.9.6210)00.5( 2/)00.320.5( mm =−

210/00

2 ,10)/log(10,4/ −∝=→== dIIIIIdPI ββπ

22/)10/10/(

12

1210/10/

212121 10)/(10/ ddddII =→== −− ββββ

ExercisesProblem 4An organ pipe has two successive harmonics with frequencies 1,372 and1,764 Hz. (a) Is this an open or stopped pipe? (b) What two harmonics arethese? (c) What is the length of the pipe?

Solution

(a)For an open pipe, the difference between successive frequencies isthe fundamental, in this case 392 Hz, and all frequencies are integermultiples of this frequency. If this is not the case, the pipe cannot bean open pipe. For a stopped pipe, the difference between the successivefrequencies is twice the fundamental, and each frequency is an oddinteger multiple of the fundamental. In this case, f1 = 196 Hz, and1372 Hz = 7f1 , 1764 Hz = 9f1 . So this is a stopped pipe.

(b) n=7 for 1,372 Hz, n=9 for 1,764 Hz.(c) so),4/(1 Lvf = .439.0)784/()/344()4/( 1 mHzsmfvL ===

ExercisesProblem 5

Two identical loudspeakers are located atpoints A and B, 2.00 m apart. The loud-speakers are driven by the same amplifierand produce sound waves with a frequencyof 784 Hz. Take the speed of sound in air tobe 344 m/s. A small microphone is moved outfrom Point B along a line perpendicular to theline connecting A and B. (a) At what distancesfrom B will there be destructive interference?(b) At what distances from B will there beconstructive interference? (c) If the frequencyis made low enough, there will be no positionsalong the line BC at which destructiveinterference occurs. How low must thefrequency be for this to be the case?

A

B2.00 m

C

x

ExercisesProblem 5

Solution(a) If the separation of the speakers is denoted by h, the condition for

destructive interference iswhere is an odd multiple of one-half. Adding x to both sides, squaring,canceling the x2 term from both sides and solving for x gives:

Using and h from the given data yields:

,22 βλ=−+ xhxβ

].2/)2/([ 2 βλβλ −= hx fv /=λ

,2/101.9 =βform ,2/371.2 =βform ,2/527.1 =βform ,2/753.0 =βform .2/9026.0 =βform

(b) Repeating the above argument for integral values for , constructiveinterference occurs at 4.34 m, 1.84 m, 0.86 m, 0.26 m.

(c) If , there will be destructive interference at speaker B.If , the path difference can never be as large as . The minimum frequency is then v/(2h)=(344 m/s)/(4.0 m)=86 Hz.

β

2/λ=h2/λ<h 2/λ

ExercisesProblem 6

A 2.00 MHz sound wave travels through a pregnant woman’s abdomenand is reflected from fetal heart wall of her unborn baby. The heart wall ismoving toward the sound receiver as the heart beats. The reflected soundis then mixed with the transmitted sound, and 5 beats per second aredetected. The speed of sound in body tissue is 1,500 m/s. Calculate thespeed of the fetal heart wall at the instance this measurement is made. Solution

Let f0=2.00 MHz be the frequency of the generated wave. The frequencywith which the heart wall receives this wave is fH=[(v+vH)/v]f0, and this isalso the frequency with which the heart wall re-emits the wave. The detectedfrequency of this reflected wave is f’=[v/(v-vH )]fH, with the minus sign indicatingthat the heart wall, acting now as a source of waves, is moving toward thereceiver. Now combining f’=[(v+vH)/(v-vH)]f0, and the beat frequency is:

Solving for vH ,)./(2)]/()[(' 000 HHHHbeat vvfvfvvvvfff −=−+=−=

./1019.3)}85)1000.2(2/[85){/1500()]2/([ 260 smHzHzHzsmfffvv beatbeatH

−×=+×=+=

Chapter 14: Wave Motion - Stony Brook...

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