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Chapter 15TRAVELING WAVES

15.1 Simple Wave Motion

Waves in which the disturbance is perpendicular to the direction of propagation are called the transverse waves.

Waves in which the disturbance is parallel to the direction of propagation are called the longitudinal waves.

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Wave pulses

t = 0, y = f(x) t > 0, y = f(x-vt)

For wave pulses on a rope, the speed of waves

/ ,Tv F µ=

where FT is the tension, µ is the linear mass density (mass per unit length).

Example 1. When will the pulse reach the left pole if the mass of the rope is 0.5 kg and the mass of the hanging weight is 10 kg?

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The pulse travels with the speed / .Tv F µ= The tension is

10 9.81 98.1 .TF mg N N= = =i

The linear density of mass is

/ 0.5 / 25 / 0.02 /rope ropem L kg m kg mµ = = =Then the speed of pulse is

/ 98.1/ 0.02 / 70.04 /Tv F m s m sµ= = =

and the propagation time is 20 / 70.04 / 0.29 .t m m s s∆ = =

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For sound waves in liquids and gases, /v B ρ=

where B is the bulk modulus and ρ is the density.

In ideal gases the bulk modulus is proportional to the pressure and (Chapter 17)

/v RT Mγ=where T is the absolute temperature in K, M is the molar mass (mass of 1 mol of the gas), R = 8.314 J/(mol K) is the universal gas constant and γ =

== 1.4 for

diatomic gases ( γ = 1.67 for monatomic gases).

Example 2. Calculate speed of sound in air at 20o C.

For air, the molar mass M = 29x10-3 kg/mol, T = (273 + 20) K = 293 K, and

( )3/ 1.4 8.314 293/ 29 10 / 343 /v RT M m s m sγ −= = ⋅ ⋅ ⋅ =

(at 0o C, v = 331 m/s).

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The Wave Equation2 2

2 22y y

t xv∂ ∂

∂ ∂=

Any function ( ) ( )y y x vt y α= − ≡ satisfies this equation. Indeed,

( ) ( )2 2

2 22

,y dy dyt d t d

y dy dy d ydt d d d tt d

v

v v v

αα α

αα α α α

∂ ∂∂ ∂

∂ ∂ ∂∂ ∂∂

= = −

= − = − =Similarly,

( ) ( )2 2

2 2

,y dy dyx d x d

y dy dy d ydx d d d xx d

αα α

αα α α α

∂ ∂∂ ∂

∂ ∂ ∂∂ ∂∂

= =

= = =Combining, we get the wave equation

2 2

2 22y y

t xv∂ ∂

∂ ∂=

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15. 2 Periodic Waves

Harmonic (sinusoidal) waves

The minimum distance after which the wave repeats is called the wavelength λDuring the period T = 1/f the wave moves a distance of one wavelength:

/ 2Tv fλ λ ωλ π= = =The harmonic (sinusoidal) traveling wave has the shape

( ) ( ) ( ) ( )( ), sin 2 sin sinxy x t A t A kx t A k x vtλπ ω ω= − ≡ − ≡ −where

2 / / 2 /k v f vπ λ ω π= = =is called the wave number.The velocity of a point in the wave is ( )cosy

x tv A kx tω ω∂∂= = − −

and the acceleration of this point is ( )2

22 siny

x ta A kx tω ω∂

∂= = − −

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Example 3. The wave function for a harmonic wave on a string is

( ) ( ) ( ) ( )1 1, 0.21 sin 4.1 9.2y x t m m x s t− − = − Find parameters of the wave.

The amplitude A = 0.21m.

The wave number k = 4.1 m-1 and the wavelength 2 / 1.53 .k mλ π= =

ω = 2 / 0.74 .T sπ ω= =The frequency 9.2 s-1 and

Then the speed of the wave9.24.1 / 2.24 /T kv m s m sλ ω= ≡ = =

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Energy Transfer via Waves

The rate of energy transfer is power,

T tP F v= iThe vector components are

, ,

sinx y

y

T tT T t

T t T t

F F i F j v v j

P F v F vθ

= + =

= ⋅ = − ⋅

At small angles, sin tan yxθ θ ∂∂≈ = In the end, the power

( ) ( )tan

cos cos

y yT t T t T x t

T

P F v F v F

F kA kx t A kx t

θ

ω ω ω

∂ ∂∂ ∂= = − = − ⋅

= − − ⋅ − −

i

/ .Tv F µ=The tension can be expressed via the wave speed v using Then

( )2 2 2cosP v A kx tµ ω ω= −and the average power is

2 212avP v Aµ ω=

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E∆ t∆The energy travels along the string at the wave speed v The average energy

flowing past a point in time

( ) 2 212avav

E P t v A tµ ω∆ = ∆ = ∆x v t∆ = ∆

x∆This energy is distributed over the length so the average energy

isin the length

( ) 2 212av

E A xµω∆ = ∆Example 4. A wave of wavelength 42 cm and amplitude 0.8 cm moves along a 10 m segment of a 75 m string with mass 300 g and tension 12 N. What are the parameters of the wave including the average total energy?

The speed of the wave is given by tension and mass density:

( )/ 12 / 0.3/ 75 / 54.77 /Tv F m s m sµ= = =

The angular frequency is2 / 6.28 54.77 / 0.42 / 818.9 /v rad s rad sω π λ= = ⋅ =

Then the average total energy is( ) ( ) ( )2 22 2 0.31 1

2 2 75 818.9 0.008 10 0.86av

E A x J Jµω∆ = ∆ = =

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Harmonic Sound Waves

The displacement of molecules from their equilibrium positions, ( ) ( )0, sin ,s x t s kx tω= −

is out of phase by 90o degrees with the pressure and density wave,

( ) ( )0, sin / 2 ,p x t p kx tω π= − −

where p is the change in pressure from the equilibrium. The amplitude

0 0p vsρω=

The energy and the energy density of sound waves are

( ) 2 2 2 21 10 02 2, avav

E s V u sρω ρω∆ = ∆ =

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15.3 Waves in Three Dimensions

The motion of wavefronts can be represented by the rays perpendicular to the wavefronts.

The average power in the wave per unit area that is perpendicular to the direction of propagation, is called the intensity,

2/ /av WI P A m= If a point source emits uniformly in all directions, then the wavefronts are spherical, 24 ,A rπ= and the intensity at a distance r from the source is

2/ 4avI P rπ=

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There is a link between intensity and energy density:

In the shell,

2102

,/ ,

/ /

av av av

av av av

av av

E u V u Av tP E t u Av

I P A u v p vρ

∆ = ∆ = ∆= ∆ ∆ =

= = =

where we used 2 210 0 02 , /avu s s p vρω ρω= =

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Example 5. A speaker diaphragm 5 cm in diameter is vibrating at 10 kHz with an amplitude 0.015 mm. Find the parameters of the sound waves immediately in front of the speaker and intensity at a distance 5 m.

The pressure amplitude is

( )( )( )( )3 4 1 5 20 0 1.29 / 2 10 340 / 1.5 10 413.1 /p vs kg m s m s m N mρω π − −= = ⋅ ⋅ =

The intensity at the diaphragm is

( ) ( )22 2 21 102 2/ 413.1 / 1.29 340 / 20.3 /I p v W m W mρ= = ⋅ =

The power of the speaker is the intensity multiplied by the area of the diaphragm,

( )220.3 0.05 / 4 0.04P IA W Wπ= = ⋅ ⋅ =

Then the intensity at a distance 5 m, assuming the uniform radiation in the forward hemisphere, is

( ) ( )2 2 25 / 2 0.04 / 6.28 25 / 0.24 /mI P r W m mW mπ= = ⋅ =

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The intensity level of sound is measured in decibels defined as

010log I

Iβ =where the reference level I0 is usually taken as threshold of hearing,

12 20 10 /I W m−=

Example 6. Find the intensity level of sound at a distance 5 m from the speaker from Example 5.

The intensity 4 2

5 2.4 10 /mI W m−= ⋅ . Then the intensity level

( )( )

0

4 12

8

10log 10log 2.4 10 /10

10log 2.4 10 80 10 0.38 83.8

II db

db db db

β − −= = ⋅

⋅ = + ⋅ =

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15.4 Waves Encountering Barriers

When a wave encounters an obstacle, part of it is reflected and part is transmitted through:

Here the reflected pulse is inverted; the transmitted pulse is not (the second string is heavier)

Here both the reflected and transmitted pulses are not inverted (the second string is lighter)

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Example 7. The incident wave with amplitude A encounters the junction of two wires with the same tension FT. The wave speed in the first wire is three times that in the second, and the amplitude of the reflected wave is a quarter of the amplitude of the incident wave. What is the amplitude of the transmitted wave?

3By energy conservation, the incident power is equal to the sum of powersin reflected and transmitted waves,

2 212

2 2 22 2 21 1 11 1 1 1 2 22 2 2

, ,i r t av

i r t

P P P P v A

v A v A v A

µ ω

µ ω µ ω µ ω

= + =

= +

Since the wave speed / ,Tv F µ= and the tension is the same in both wires,2 22

2 2 2 1 1 21 1 2

2 2 21 1 2

i tT T T rA AF F F Ai r t v v vv v v

v A v A v A= + ⇒ = +

We know that 1 2/ 4, 3 .r iA A v v= = Therefore,

( )22 2 2

1 1 1 1

/ 4 2 2 251/3 16 163ii t iAA A A

t t iv v v v A A A = + = + ⇒ =

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Diffraction

Bending of a wavefront behind an obstacle is called diffraction.

Almost all of the diffraction occurs within several wavelengths form the edge of the obstacle

Flat wavefronts passing through a small aperture bend, spread out, and become spherical.

The larger is the wavelength in comparison to the aperture, the larger are the diffraction effects.

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The approximation that waves propagate in straight lines without diffraction is called the ray approximation (zero wavelength approximation).

3 3cm mλ ÷∼7 74 10 7 10 mλ − −⋅ ÷ ⋅∼

For sound waves,

and for visible light

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15. 5 The Doppler Effect

Change in frequency of the received signal in comparison to the frequency of the emitted signal, which results from a relative motion of the source and receiver, is called the Doppler effect.

If the source and receiver move closer together, the received frequency is greater than the source frequency. If the source and receiver move away from each other, the received frequency is lower than the source frequency.

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Example 1. Source moves to the right with us; receiver is stationary; speed of the waves is v; the time between emission of two consecutive crests is Ts.

Between two consecutive events, the first crest moves vTs while the source moves usTs. Thus, the distance between two crests (the wavelength) is

( ) ( ) /s s s sv u T v u fλ = ± = ±

/s

vr sv uf v fλ ±= =

Then the frequency in the receiver

(+ the receiver behind the source, - in front)Example 2. Source does not move; receiver moves with ur; speed of the waves is v; the time between arrivals of two consecutive crests is Tr.

Between detecting two consecutive crests, the receiver moves by urTr while the crests first crest moves by vTr. The distance between two crests is the wavelength r r rvT u Tλ = ± (+ receiver moves away from the source, - it moves towards the source). Thus the frequency is

( ) ( )/ /1 /r rr s rf u v u vT fv λ= = ± = ±

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Example 3. If both move,r

s

v ur sv uf f±

±=

The proper signs are chosen by remembering that the frequency goes up when the source moves towards the receiver and the receiver – towards the source.

If both ,s ru u v

s

f uf v∆ ≈±

Example 4. The frequency of car horn is 500 Hz. Find the frequency and thewavelength of the received signal if the car moves towards receiver with u = 90 km/h.

The car velocity is 90 km/h =90 (1000 m)/(3600 s) = 25 m/sThe wavelength in front of the car

( ) ( )/ 340 25 / 500 0.63sv u f m mλ = − = − =and the received frequency is

340340 25/ 500 540v

r sv uf v f Hz Hzλ − −= = = =

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Example 5. The same if the honking car is stationary while the receiver moves towards this car with u = 60 km/h

60 km/h = 60(1000)/3600 m/s =16.7 m/sThe received frequency is

( ) ( )/ 500 340 16.7 / 340 525r s rf f v u v Hz Hz= ± = + =Example 6. The car speeds away from the police car. The radar unit emits the electromagnetic waves with frequency f. Find the speed of the car u if the frequency of reflected waves, received by the radar unit, is f f− ∆ .The radar waves strike the speeding car at a frequency

( ) ( ) ( )/ / 1 ur s r s s cf f v u v f c u c f= ± = − = −

and are reflected from (emitted by) the car with the same frequency. The received signal by the police is then

( ) ( ) ( ) ( )11 /' 1 1 1 1 2

s

v c u u u ur r r r r s sv u c u u c c c c cf f f f f f f± + += = = ≈ − = − − ≈ −

Therefore, ' 2 us r scf f f f∆ = − = and

2 s

ffu c∆=

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Doppler Shift and Relativity

In non-relativistic systems, the Doppler effect is determined by the speed of both the source and the receiver relative to the medium which affected the speed of propagation of waves.

For light and other electromagnetic waves the speed of propagation – c – is a universal constant which does not depend on the speed of source or receiver (no “wind”).

Second, the time between the events such as emission of wave crests, Ts, depends on the reference frame and is different in the reference frames of source and receiver.

In the end, the Doppler shift for electromagnetic waves depends only on the relative velocity of the source and receiver u:

c ur sc uf f±= ∓

where sign is such so that the frequency goes up when the source and receiver are approaching each other.

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/Tv F µ=

/v B ρ=

/v RT Mγ=

Chapter 15 review

Waves on a string:

Sound waves in liquids and gases:

Sound waves in dilute ideal gases:

Electromagnetic waves: the speed is a universal constant, c = 3x108 m/s2 2

2 22y y

t xv∂ ∂

∂ ∂=

( ) ( ), sin :y x t A kx tω= ±

2 / / , 2 2 / ,/ 2 /

k v f T kvv f k T

π λ ω ω π πλ ω πωλ λ

= = = = == = = =

Wave equation:

Parameters of harmonic waves

The energy in harmonic waves is proportional to the square of the amplitude

Power for waves on the string: ( ) 2 21/ 2avP v Aµ ω=

Energy density for sound waves:

( ) 220 0 01/ 2 ,avu s p vsρω ρω= =

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/avI P A= avI u v=Intensity for sound waves:

Sound intensity level: ( ) 12 20 010log / , 10 /I I I W mβ −= =

Doppler effectr

s

v ur sv uf f±

±=

Moving source: ( ) /s sv u fλ = ±

Moving receiver: ( ) ( )/ /r r s rf v u f v u vλ= ± = ±

Small speed of source or receiver: / /sf f u v∆ ≈ ±