chapter 14: wave motion types of mechanical waves mechanical waves are disturbances that travel...

Chapter 14: Wave Motion

Types of mechanical waves Mechanical waves

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

• travel through the medium by displacing particles in the medium

• travel in the perpendicular to or along the movement of the particles 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 wave

C = compressionR = rarefaction

air compressedair rarefied


Longitudinal-transverse waves


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

x=0 x

t=0

A

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[),(

TtxA

tvxfA

tvxA

vxtAtxy

sinusoidal wave moving in

+x directionangular frequency

f 2velocity of wave, NOT ofparticles of the medium

wavelengthperiod

vf Tf /1

)]/(cos[),( xvtAtxy sinusoidal wave moving in-x direction v->-v

phase velocity

Mathematical description of a wave (cont’d)

Wave function (cont’d)

x=0 x

t=0

t=T/4

t=T period

wavelength

)//(2cos),( TtxAtxy ),(

),(

Ttxy

txy


Wave number and phase velocity

)cos(),( tkxAtxy

vkdtdx //

wave number: /2k

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

phase

.consttkx

phase velocity

)](cos[)cos(),( vtxkAtkxAtxy


Particle velocity and acceleration in a sinusoidal wave

)cos(),( tkxAtxy

),(

)cos(/),(),(

)sin(/),(),(

2

222

txy

tkxAttxytxa

tkxAttxytxv

y

y

velocity

acceleration

Also ),()cos(/),( 2222 txyktkxAkxtxy

222

222222

/),(

/),()/(/),(

tvtxy

ttxykxtxy

wave equation

u in textbook


General solution to the wave equation

Solutions: )(),( vtxftxy such as )cos( tkx

The most general form of the solution:

)()(),( vtxgvtxftxy

wave equation22

2

2

2

2

2

2

2 ),(),(),(

tv

txy

t

txyk

x

txy

Speed of a transverse wave Wave speed on a string

x

2FyF2

FF x 2yF1

FF x 1

1F

•Consider a small segment of string whose length in the equilibrium position is .x•The mass of the segment is .xm • The x component of the force (tension) at both ends have equal in magnitude and opposite in direction because this is a transverse wave.

x• xxyxy xyFFxyFF )/(/,)/(/ 21

• The total y component of the forces is:

)/(

])/()/[(22

21

tyx

xyxyFFFF xxxyyy

Newton’s 2nd law

massacceleration

xx

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

x

2FyF2

F

yF1

F

1F

x

• The total y component of the forces is:

)/(

])/()/[(22

21

tyx

xyxyFFFF xxxyyy

)/)(/(

/])/()/[(22 tyF

xxyxy xxx

0x

)/)(/(/ 2222 tyFxy wave eq.

)/()(

/

inertiaforcerestoring

Fv

Energy in wave motion Total energy of a short string segment of mass

x

2FyF2

F

yF1

F

1F

x

dxdm • 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(),(

tkxAty

tkxkAxy

tkxAtxy

)(sin

)(sin

)(sin/),(

222

222

22

tkxAF

tkxAv

tkxAFkdtdEtxP

/, 2 Fvvk

work done

Pmax

Energy in wave motion (cont’d)

Average power of a sinusoidal wave on a string

2

1sin

2

1 2

0

2

d

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

• The average power: 22)2/1( AFPave

Maximum power of a sinusoidal wave on a string:

22max AFP

Wave intensity

Wave intensity for a three dimensional wave from a point source:

22

W/mofunitsin4 r

PI

1r

2r

22

212

1 44 IrIr

21

22

2

1

r

r

I

I

power/unit area

Wave interference, boundary condition, and superposition

The principle of superposition

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

),(),(),( 21 txytxytxy

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

Interference

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

• Destructive interference (positive-negative)


Reflection

• Free end )cos()cos(),( txkBtxkAtxy

Bx

For x<xB

At x=xB ABxtxyBxx 0)/),((

incident wave reflected wave

Vertical component of the forceat the boundary is zero.


Reflection (cont’d)

• Fixed end )cos()cos(),( txkBtxkAtxy 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 string

Superposition 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 direction creates a standing wave when two waves have the same speed and wavelength.

N=node, AN=antinode

))(sin(sin2

)cos()cos(

),(),(),( 21

txkA

tkxAtkxA

txytxytxy

incident reflected

,..)2,1,0(

2//or

when0sin

n

nknx

nkxkx

Normal modes of a string

There are infinite numbers of modes of standing waves

fixed end fixed endL

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

nnL

F

Lf

L

vnf

nL

n

n

2

1

2

/2

1

2/1

2

2/3 3

42

firstovertone

secondovertone

thirdovertone

funda-mental

Sound waves Sound

• Sound is a longitudinal wave in a medium

• The simplest sound waves are sinusoidal waves which have 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 Sinusoidal sound wave function:

x x+xx

S

),(1 txyy ),(2 txxyy undisturbedcyl. of air

disturbedcyl. of air

)],(),([

)( 12

txytxxyS

yySV

Change of volume:

)(/),(/ SdxVxtxyVdV

)//(),( VdVtxpB bulk modulus pressure

)sin()/),((),( tkxBkAxtxyBtxp

Pressure:

Pressure amplitude and ear

Pressure amplitude for a sinusoidal sound wave

• Pressure: )sin(),( tkxBkAtxp

• Pressure amplitude: BkAp max

Ear

Perception of sound waves

Fourier’s theorem and frequency spectrum

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

n nnnn tfBtfAty )]2cos()2sin([)(

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

• Implication of Fourier’s theorem:

fundamental freq.

Perception of sound waves

Timbre or tone color or tone quality

piano

piano

music

noise

Frequency spectrum

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)(longitudinal momentum carried by the fluid in motion

tAv y

Avtoriginal volume of the fluid in motion

change in volume of the fluid in motion

bulk modulus B: -pressure change/frac. vol. change )/()( AvttAv

p

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 vtAvAtv

vBpAt

B

v speed of a longitudinal wave in a fluid

The speed of sound waves in a solid bar/rod

YY

v ,

Young’s modulus

Speed of sound waves (cont’d)

The speed of sound waves in gases

M

T

R

p

M

RTv

0

speed of a longitudinal wave in a fluid

0

0

p

pB

bulk modulus of a gasratio of heat capacities

equilibrium pressure of gas

gas constant 8.314472 J/(mol K)

temperature in Kelvin

molar mass

- P in textbook (background pressure).- density

In textbook

Sound level (Decibel scale)

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

2120

0

W/m10,log)dB10( II

IDefinition of sound intensity: ( unit decibel or dB)

Military jet plane at 30 m 140 102

Threshold of pain 120 1

Whisper 20 10-10

Hearing thres. (100Hz) 0 10-12

Sound intensity in dB Intensity (W/m2)

Standing sound waves

Sound wave in a pipe with two open ends


Standing sound wave in a pipe with two open ends


Sound wave in a pipe with one closed and one open end


Standing wave in a pipe with two closed ends

Displacement

Normal modes

Normal modes in a pipe with two open ends

2nd normal mode

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

2 n

n

LornL n

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

nL

vnfn

Normal modes

Normal modes in a pipe with an open and a closed end (stopped pipe)

,...)5,3,1(4

4 n

n

LornL n

n ,...)5,3,1(

4 n

L

vnfn

Resonance

Resonance

• When we apply a periodically varying force to a system that can oscillate, the system is forced to oscillate with a frequency equal to the frequency of the applied force (driving frequency): forced oscillation. When the applied frequency is close to a characteristic frequency of the system, a phenomenon called resonance occurs.

• Resonance also occurs when a periodically varying force is applied to a system with normal modes. When the frequency of the applied force is close to one of normal modes of the system, resonance occurs.

Interference of waves

Two sound waves interfere each other

constructivedestructive

d1 d2

,....2,1,0

)()2/1(

)(21

n

edestructivn

veconstructindd

Beats

Two 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 and .bfFor simplicity, consider two sinusoidal waves of equal intensity:

;2sin)( tfAty aa tfAty bb 2sin)(

Then the resulting combined wave will be:

]))(2(2

1cos[]))(2(

2

1sin[2)()( tfftffAtyty bababa

))(2

1cos)(

2

1sin2sinsin( bababa

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 effect

Moving 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 whether the listener is moving or not.

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

s

LLL fv

vvvvf

/

freq. listener hears Lf

freq. source generates sf

velocity of sound at source

velocity of listener

v

Lv - for a listener moving away from + for a listener moving towards the source.


Moving source

When the source moves


Moving source (cont’d)

• The wave velocity relative to the wave medium does not change 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, the the distance between the previous and next crest i.e. the wavelength changed by the speed of the source.

s

s

s

s

s f

vv

f

v

f

v

s

s

f

v

The source at rest When the source is moving

+ for a receding source- for a approaching source


Moving source and listener

+ for a receding source- for a approaching source

ss

LLL f

vv

vvvvf

- for a listener moving away from + for a listener moving towards the source.

Effect of change of source speed

svv svv

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


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 waves if the siren is at rest in the air, b) if the siren is moving at 30 m/s, find the 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/ sfv

m 1.03 Hz m/s)/300 30 - m/s 340(/)( ss fvv

m 1.23 Hz m/s)/300 30 m/s 340(/)( ss fvv


Example 2

• If a listener l is at rest and the siren in Example 1 is moving away from 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

vv

vf

Example 3

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

Hz. 274Hz) 300( m/s 340

m/s) 30-m/s 340

s

LL f

v

vvf


Example 4

• If the siren is moving away from the listener with a speed of 45 m/s relative to the air and the listener is moving toward the siren with a speed of 15 m/s relative to the air, what frequency does the listener hear?

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

m/s 15m/s 340

ss

LL f

vv

vvf

Example 5

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

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

m/s 340

s

sW f

vv

vf

Hz. 358Hz) 329( m/s 340

m/s 30m/s 340

W

LL f

v

vvf

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 (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, =2/0.400 = 5.00 cm, f=125 Hz, T=1/f=0.00800 s and v=f=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

)/( Fv .6.19)/25.6)(/050.0( 22 NsmmkgvF

.2.54)2/1( 22 WAFPav

ExercisesProblem 2

Solution

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 . At t=0 the shape of the pulse I given by

)0,(xy

Lxfor

LxforLxLh

xLforLxLh

Lxfor

0

0/)(

0/)(

0

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

(a)y

x

h-L L

ExercisesProblem 2 (cont’d)

Solution

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

),( txy

Lxfor

LxforLvtxLh

xLforLvtxLh

Lxfor

0

0/)(

0/)(

0

(c)

t

y

x

yFtxP ),(

LxforF

LxforLhFvLhvLhF

xLforLhFvLhvLhF

LxforF

0)0)(0(

0)/()/)(/(

0)/()/)(/(

00)0(

2

2

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 the intensity and hence to 10 raised to half of the sound intensity levels divided by 10:

.9.6210)00.5( 2/)00.320.5( mm

210/00

2 ,10)/log(10,4/ dIIIIIdPI

22/)10/10/(

12

1210/10/

212121 10)/(10/ ddddII

ExercisesProblem 4

Solution

An 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?

(a)For an open pipe, the difference between successive frequencies is the fundamental, in this case 392 Hz, and all frequencies are integer multiples of this frequency. If this is not the case, the pipe cannot be

an open pipe. For a stopped pipe, the difference between the successive frequencies is twice the fundamental, and each frequency is an odd integer multiple of the fundamental. In this case, f1 = 196 Hz, and 1372 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

BC

x

2.00 m

ExercisesProblem 5

Solution

(a) If the separation of the speakers is denoted by h, the condition for destructive interference is where 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 xhx

].2/)2/([ 2 hx fv /

,2/101.9 form ,2/371.2 form ,2/527.1 form ,2/753.0 form .2/9026.0 form

(b) Repeating the above argument for integral values for , constructive interference 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.

2/h2/h 2/

ExercisesProblem 6

Solution

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.

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 types of mechanical waves mechanical waves are disturbances that travel...

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