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,
y1 ( x, t ) = A sin ( kx − ω t ) ,
y2 ( x, t ) = A sin ( kx − ω t + δ ) ,
that are almost identical except for a
phase.
The resultant wave is the sum
y = y1 + y2 = A sin ( kx − ωt ) + A sin ( kx − ωt + δ )
= 2 A cos (δ / 2 )isin ( kx − ωt + δ / 2 )
where we used the trigonometric identity
sin θ1 + sin θ 2 = 2 cos ( 12 (θ1 − θ 2 ) ) sin ( 12 (θ1 + θ 2 ) )
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The result of superposition of two waves of equal amplitudes, wave numbers
and frequency is the harmonic wave with the same wave number and
frequency and amplitude 2 A cos (δ / 2 )
If the two waves are in phase, δ = 0,
cos 0 = 1
the resultant amplitude is 2A (constructive
interference).
If the two waves are 180o out of phase,
δ = π , cos π / 2 = 0
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
p1 = p0 sin ω1t ,
p2 = p0 sin ω 2t ,
and the superposition yields
p = p1 + p2 = 2 p0 cos ( 12 t (ω1 − ω2 ) ) sin ( 12 t (ω1 + ω2 ) )
The “outside” sinusoid has a low frequency
1
, the “inside” one has a high
2 (ω1 − ω 2 )
frequency 12 (ω1 + ω 2 ) . Thus the amplitude
oscillates with the frequency 12 (ω1 − ω 2 )
and the intensity, being the square of the
amplitude, with the double frequency,
ω1 − ω 2 , - the so-called beat frequency.
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What one hears is a pulsating tone at the average frequency of two waves,
1
2
(ω1 + ω 2 ) , with the frequency of pulses ω1 − ω 2 (the sound intensity is the
highest when the amplitude function is either at a maximum or a minimum).
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?
Since f1 − f 2 = 4 Hz , 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, f1 − f 2 also go up, this means that f 2 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 observation
point,
p1 = p0 sin ( kx1 − ω t ) , p2 = p0 sin ( kx2 − ω t )
Then the phase difference in the observation point is
δ = ( kx1 − ωt ) − ( kx2 − ωt ) = k ( x1 − x2 )
≡ k ∆x ≡ 2π∆x / λ
When ∆x is some integer number N of wavelengths, ∆x = N λ
the waves come to the observation point in phase and the
amplitudes add up (constructive interference). When
∆x is a semi-integer number N/2 of wavelengths, ∆x = N λ / 2
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
λ = v / f = 340 / 2000 m = 0.17 m,
k = 2π / λ = 2π f / v = 6.28 ⋅ 2000 / 340 m −1 = 36.94 m −1
The phase difference is then
δ = k ∆x = ( 36.94 ⋅ 0.2 ) rad = 7.388 rad
The resultant wave,
p = p1 + p2 = p0 sin ( kx1 − ω t ) + p0 sin ( kx2 − ω t )
= 2 p0 cos ( k ( x1 − x2 ) / 2 )isin ( k ( x1 + x2 ) / 2 − ω t )
has the amplitude
2 p0 cos ( k ( x1 − x2 ) / 2 ) = 2 p0 cos 12 δ = 2 p0 cos 12 k ∆x
= 2 p0 cos ( 36.94 ⋅ 0.2 / 2 ) = −1.702 p0
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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 - x1
from the second is
2 p0 cos ( k ( x1 − x2 ) / 2 ) = 2 p0 cos 12 k ∆x = 2 p0 cos
(
2π f
v
⋅ ( 2 x1 − 1.8 ) / 2
)
⋅1200
= 2 p0 cos ( 6.28340
⋅ ( x1 − 0.9 ) ) = 2 p0 cos ( 22.16 ⋅ ( x1 − 0.9 ) )
This amplitude is zero when
22.16 ⋅ ( x1 − 0.9 ) = ± π2 N
with odd integer N or
π
x1 = 0.9 ± 2⋅22.16
N = 0.9 ± 0.071N
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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 remains
constant.
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.
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.) have the frequencies f n which are equal to
the harmonic number n times the fundamental frequency f1 , f n = n ⋅ f1.
The set of all resonant frequencies is called the frequency spectrum.
Sometimes, the higher resonance frequencies are called the overtones.
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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 L
with fixed ends are
λ
L = n 2n , λn = 2 L / n,
f n = v / λn ≡ nv / 2 L = nf1
(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 f n .
The 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
f1 ≡ v / 2 L = (442 − 354) Hz = 88 Hz
The wave velocity
v = FT / µ = 500 / 0.032 m / s = 158.1 m / s
Therefore, the string length is
L = v / 2 f1 = (158.1/ 2 ⋅ 88 ) m = 0.898 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:
L = nλn / 4, n = 1;3;5;7... λn = 4 L / n,
f n = v / λn = nv / 4 L ≡ nf1 ,
f1 = v / 4 L
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Wave function for the standing wave
In n-th mode, each point on the string moves with simple harmonic motion,
y ( x, t ) = An ( x ) cos ( 2π f nt + δ n )
where the amplitude is
and
An ( x ) = An sin kn x
y ( x, t ) = An sin ( kn x ) cos ( 2π f nt + δ n )
Example 7. Show that the superposition of two identical waves traveling in the
opposite directions, y1,2 = y0 sin ( kx ∓ ω t ) is a standing wave.
The superposition,
y = y1 + y2 = y0 sin ( kx − ωt ) + y0 sin ( kx + ωt ) = 2 y0 sin ( kx ) cos (ωt )
decouples into the product of sin ( kx ) and cos (ω t ) and is, therefore, a
standing wave.
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Example 8. What are the allowed frequencies in an unstopped (open ad both
ends) 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,
L = nλn / 2, λn = 2 L / n, f n = v / λn = nv / 2 L
= n ⋅ 340 / 2 ⋅1.5 Hz = n ⋅113.33 Hz
Example 9. The sound wave resonances at frequency
488.6 Hz in the open part of the tube are found at L
equal 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,
λ = 2 ( Ln +1 − Ln ) = 2 (1.25 − 0.90 ) m = 0.7 m.
Then the sound velocity is
v = λ f = 0.7 ⋅ 488.6 m / s = 342 m / s
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In general, the motion of vibrating system consists of a mixture of individual
harmonics,
y ( x, t ) = ∑ An sin ( kn x ) cos ( 2π f nt + δ n )
n
where
kn , f n
are the wave numbers and frequencies of individual
harmonics, and the amplitudes and phases
initial position and velocity of the string.
An , δ n
are determined by
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:
y = y1 + y2 = A sin ( kx − ωt ) + A sin ( kx − ωt + δ )
= 2 A cos (δ / 2 )isin ( kx − ωt + δ / 2 )
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 .
Beats: result of interference of two waves with a slight difference in frequencies
. is ∆f .
∆f ; the beat frequency
Phase difference due to path difference δ = k ∆x = 2π∆x / 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 L = nλ / 2, n = 1, 2,3... which is equivalent to f n = v / λn = nv / 2 L = nf1
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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,
L = nλn / 4,
f n = v / λn = nv / 4 L = nf1 , n = 1,3,5, 7...
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.
Wave function for a standing wave is
y ( x, t ) = An sin ( kn x ) cos ( 2π f nt + δ n )
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