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Chapter 16

SUPERPOSITION & STANDING WAVES

16.1 Superposition of waves

Principle of superposition: When two or more waves overlap, the resultant wave is the algebraic sum of the individual waves.

Illustration: when there are two pulses on a string, the resulting wave function is a sum of two individual ones.

In a special case when these two pulses are identical but inverted, there is a moment in time when the sum is exactly zero. At this instant, the string is straight - not stationary - and the total energy is nonzero.

The principle of superposition is valid because the wave function is governed by the linear differential equation (the overall solution is a sum of individual solutions).

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Interference of harmonic waves

The superposition of two or more waves with almost the same frequency, which produces an observable pattern in intensity, is called interference.

Example 1. Interference of two waves,

( ) ( )( ) ( )

1

2

, sin ,

, sin ,

y x t A kx t

y x t A kx t

ω

ω δ

= −

= − +that are almost identical except for a phase.

( ) ( )( ) ( )

1 2 si

2 cos / 2 si

s

2

n n

/

i

n

y

A kx

y y A kx A kx t

t

tω ω

δ ω δ

δ= + = −

−

+ −

+

+

= i

( )( ) ( )( )1 11 2 1 2 1 22 2sin sin 2cos sinθ θ θ θ θ θ+ = − +

The resultant wave is the sum

where we used the trigonometric identity

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( )2 cos / 2A δ

The result of superposition of two waves of equal amplitudes, wave numbers and frequency is the harmonic wave with the same wave number andfrequency and amplitude

If the two waves are in phase, 0, cos0 1δ = =the resultant amplitude is 2A (constructiveinterference).

, cos / 2 0δ π π= =If the two waves are 180o out of phase,

the resultant amplitude is 0 (destructive interference).

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Beats – interference of two (sound) waves with slightly different frequencies. At a fixed point (ear), the spatial difference of the wave contributes just a phase constant and the waves at this point are the harmonic time oscillations

1 0 1 2 0 2sin , sin ,p p t p p tω ω= =

and the superposition yields

( )( ) ( )( )1 11 2 11 20 22 22 cos sinp p p p t t ωω ω ω= −= + +

( )11 22 ω ω−

( )11 22 ω ω+

( )11 22 ω ω−

The “outside” sinusoid has a low frequency , the “inside” one has a high

. Thus the amplitude

and

frequencyoscillates with the frequency

the intensity, being the square of the

1 2ω ω− , - the so-called beat frequency.

amplitude, with the double frequency,

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( )11 22 ω ω+ 1 2ω ω−

What one hears is a pulsating tone at the average frequency of two waves,

, with the frequency of pulses

highest when the amplitude function is either at a maximum or a minimum).

(the sound intensity is the

Example 2. When a 540 Hz tuning fork is struck simultaneously with a musical instrument string, 4 beats per second are heard. As the string is tightened, the frequency of the beats goes up. What is the initial frequency of the string?

1 2 4f f Hz− =

1 2f f− 2f

Since , the initial frequency is either 544 Hz or 536 Hz. When the string is tightened, its frequency goes up. Since the beat frequency and, therefore, also go up, this means that was initially 544 Hz.

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Phase difference due to path difference

A difference in path length between two sources and the observation (interference) point is a common cause of phase difference.

Example 3. Two sources, S1 and S2, oscillate in phase, but are at different distances, x1 and x2, from the observationpoint,

( ) ( )1 0 1 2 0 2sin , sinp p kx t p p kx tω ω= − = −

Then the phase difference in the observation point is

( ) ( ) ( )1 2 1 2

2 /kx t kx t k

kx

x xxω ωδ

π λ≡ ∆ ≡

− − −

∆

= − =

x∆ x Nλ∆ =

x∆ / 2x Nλ∆ =

When is some integer number N of wavelengths, the waves come to the observation point in phase and the amplitudes add up (constructive interference). When

is a semi-integer number N/2 of wavelengths, the waves come to the observation point out of phase by 180o

and the waves cancel each other (destructive interference).

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Example 4. Two sound sources oscillate in phase with the frequency 2 kHz and amplitude p0. Find the phase difference and the amplitude of the resultant wave at a point 4.2 m from one source and 4.4 m from the other.

The wavelength and the wave vector of the sound waves are

1 1

/ 340 / 2000 0.17 ,2 / 2 / 6.28 2000 / 340 36.94v f m m

k f v m mλ

π λ π − −

= = =

= = = ⋅ =The phase difference is then

( )36.94 0.2 7.388k x rad radδ = ∆ = ⋅ =

The resultant wave,( ) ( )

( )( ) ( )( )1 2 0 1 0 2

0 1 2 1 2

sin sin

2 cos / 2 sin / 2

p p p p kx t p kx t

p k x x k x x t

ω ω

ω

= + = − + −

= − + −i

has the amplitude

( )( )( )

1 10 1 2 0 02 2

0 0

2 cos / 2 2 cos 2 cos

2 cos 36.94 0.2 / 2 1.702

p k x x p p k x

p p

δ− = = ∆

= ⋅ = −

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Example 5. Two loudspeakers broadcast identical waves at 1.2 kHz at a distance 1.8 m from each other. Locate the points between the speakers where the sound intensity is zero.

The amplitude of the signal at a point x1 from the first speaker and x2 = 1.8 - x1from the second is

( )( ) ( )( )( )( ) ( )( )

210 1 2 0 0 12

6.28 12000 1 0 1340

2 cos / 2 2 cos 2 cos 2 1.8 / 2

2 cos 0.9 2 cos 22.16 0.9

fvp k x x p k x p x

p x p x

π

⋅

− = ∆ = ⋅ −

= ⋅ − = ⋅ −

This amplitude is zero when

( )1 222.16 0.9x Nπ⋅ − = ±with odd integer N or

1 2 22.160.9 0.9 0.071x N Nπ⋅= ± = ±

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Two sources do not have to be in phase to produce interference patterns. Similar interference patterns can be produced by any two sources whose phase difference remainsconstant.

Two sources that remain in phase or maintain a constant phase difference are called coherent. Wave sources whose phase difference is not constant but varies randomly are called incoherent.

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16.2 Standing Waves

If we fix both ends of a string, we can excite only with certain resonance frequencies and in certain modes of vibration (resonance modes). These patterns are called the standing waves (the nodes and the positions of maximum amplitude) do not move.

nf1 1, .nf f n f= ⋅

The lowest resonance frequency (the mode with the largest wavelength) is called the fundamental mode or the first (lowest) harmonic. The higher harmonics (the second, third, etc.) which are equal to

The set of all resonant frequencies is called the frequency spectrum. Sometimes, the higher resonance frequencies are called the overtones.

the fundamental frequency have the frequencies

the harmonic number n times

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The points on the standing wave where there is no motion are called the nodes. Half-way between the nodes are the points of maximum amplitude – the antinodes.

The wavelength and frequency of the n-th harmonic on the string of length Lwith fixed ends are

2

1

, 2 / ,/ / 2

nn

n n

L n L nf v nv L nf

λ λλ

= =

= ≡ =(the standing wave condition).To produce a standing wave with the help of a vibrator, the frequency of the vibrator should coincide (be at resonance) with one of the natural (resonance) frequencies of the string .nfThe resonance of standing waves is similar to the resonance of a harmonic oscillator with a harmonic driving force.

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Example 6. Two adjacent resonance frequencies of the string are 354 and 442 Hz. The tension in the string is 500 N, the density is 0.032 kg/m. What are the fundamental frequency and the length of the string.

Since the adjacent resonance frequencies of the string are 354 and 442 Hz, the fundamental frequency is

1 / 2 (442 354) 88f v L Hz Hz≡ = − =The wave velocity

/ 500 / 0.032 / 158.1 /Tv F m s m sµ= = =

Therefore, the string length is

( )1/ 2 158.1/ 2 88 0.898L v f m m= = ⋅ =

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String free at one end, fixed at the other

Since a finite vertical force at the free end (ring) would give the massless ring an infinite acceleration, to keep the acceleration finite the slope of the string at the free end should be horizontal – the free end of the string is an antinode.It is self-explanatory that in each of vibration there is an odd number of quarter-wavelengths in length L:

1 1

/ 4, 1;3;5;7... 4 / ,/ / 4 , / 4n n

n n

L n n L nf v nv L nf f v L

λ λλ

= = == = ≡ =

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Wave function for the standing wave

In n-th mode, each point on the string moves with simple harmonic motion,

( ) ( ) ( ), cos 2n n ny x t A x f tπ δ= +where the amplitude is

( ) sinn n nA x A k x=and ( ) ( ) ( ), sin cos 2n n n ny x t A k x f tπ δ= +

Example 7. Show that the superposition of two identical waves traveling in the opposite directions, ( )1,2 0 siny y kx tω= ∓ is a standing wave.

The superposition,( ) ( ) ( ) ( )1 2 0 0 0sin sin 2 sin cosy y y y kx t y kx t y kx tω ω ω= + = − + + =

decouples into the product of ( )sin kx and ( )cos tω and is, therefore, astanding wave.

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Example 8. What are the allowed frequencies in an unstopped (open ad bothends) organ pipe with (effective) length 1.5 m?

The open end of the organ pipe is displacement antinode (and pressure node). Therefore, the pipe length is equal to an integer number of half-wavelengths,

/ 2, / 2 , / / 2340 / 2 1.5 113.33n n n nL n n L f v nv L

n Hz n Hzλ λ λ= = = =

= ⋅ ⋅ = ⋅

Example 9. The sound wave resonances at frequency 488.6 Hz in the open part of the tube are found at Lequal to 125 and 90 cm. What is the speed of sound at this temperature and atmospheric pressure?

The wavelength is twice the distance between successive water levels,

( ) ( )12 2 1.25 0.90 0.7 .n nL L m mλ += − = − =

Then the sound velocity is

0.7 488.6 / 342 /v f m s m sλ= = ⋅ =

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In general, the motion of vibrating system consists of a mixture of individual harmonics,

( ) ( ) ( ), sin cos 2n n n nn

y x t A k x f tπ δ= +∑

,n nk f,n nA δ

where are the wave numbers and frequencies of individualharmonics, and the amplitudes and phases are determined byinitial position and velocity of the string.

Analysis of the wavefronts in terms of individual harmonics that comprise them is called the harmonic or spectral or Fourier analysis. The inverse of harmonic analysis – the construction of a signal from harmonic components – is harmonic synthesis.

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Review of Chapter 16

Superposition & interference of two identical, except for the phase, harmonic waves: ( ) ( )

( ) ( )1 2 sin sin

2 cos / 2 sin / 2

y y y A kx t A kx t

A kx t

ω ω δ

δ ω δ

= + = − + − +

= − +i

Constructive interference: waves are in phase or differ in phase by an integer number times 2π

.

Destructive interference: waves differ in phase by an odd integer number times π .

f∆ f∆Beats: result of interference of two waves with a slight difference in frequencies

; the beat frequency is .

Phase difference due to path difference 2 /k x xδ π λ= ∆ = ∆

.

The distance between a node and an adjacent antinode in a standing wave is a quarter-wavelength.

String fixed at both ends has nodes at the ends while the standing wave condition is / 2, 1, 2,3...L n nλ= = which is equivalent to 1/ / 2n nf v nv L nfλ= = =

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For a string with one end fixed and the other one free, there is a node at the fixed end and the antinode at the free end,

1/ 4, / / 4 , 1,3,5,7...n n nL n f v nv L nf nλ λ= = = = =

Standing sound waves in a pipe open at both ends have pressure nodes (displacement antinodes) at both ends; the standing wave condition is the same as for a string with two fixed ends.

( ) ( ) ( ), sin cos 2n n n ny x t A k x f tπ δ= +

Wave function for a standing wave is