waves_02
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SUPERPOSITION OF WAVES
• Two waves passing through the same region will
superimpose - e.g. the displacements simply add
• Two pulses travelling in opposite directions will pass
through each other unaffected
• While passing, the displacement is simply the sum of the
individual displacements
Animations courtesy of Dr. Dan Russell, Kettering University
CP 514
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waves_02: MINDMAP SUMMARY – SUPERPOSITION PRINCIPLE
Travelling waves, superposition principle, interference,
constructive interference, destructive interference, intermediate
interference, nodes, antinodes, phase, phase difference, in phase,
out of phase, path difference, two point interference, standing
waves on strings, standing waves in air columns, thin film
interference
Superposition (at time to)
Wave 1 + Wave 2
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y ( x, to )  y1 ( x, to )  y2 ( x, to )
Constructive interference:
 = m
 = m (2)
Destructive interference:
 = (m + 1/2) 
 = (m + ½) (2)
m = 0, 1, 2, 3, ...
SUPERPOSITION
 INTERFERENCE
3
28
24
20
16
12
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In phase  constructive interference 4
Out of phase 
destructive interference
0
0
10
20
30
40
50
60
70
80
90
CP 510
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Problem 1
Two sine waves travelling in the
same direction  Constructive
and Destructive Interference
Two sine waves travelling in
opposite directions  standing
wave
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Interference of two overlapping travelling waves depends on:
* relative phases of the two waves
* relative amplitudes of the two waves
fully constructive interference: if each wave reaches a max at the same time,
waves are in phase (phase difference between waves two waves  = 0)
greatest possible amplitude ( ymax1 + ymax2)
fully destructive interference: one wave reaches a max and the other a min at the
same time, waves are out phase (phase difference between two waves  = 
rad), lowest possible amplitude |ymax1 - ymax2|
0 < phase difference  <  rad
or  < phase difference  < 2
intermediate interference:

SUPERPOSITION  INTERFERENCE
A phase difference of 2 rad corresponds to a shift of one
wavelength between two waves.
For m = 0, 1, 2, 3
fully constructive interference  phase difference = m 
fully destructive interference  phase difference = (m + ½) 
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SUPERPOSITION  INTERFERENCE
60
60
50
50
40
40
30
30
20
20
10
10
0
0
0
100
200
300
400
500
600
700
800
0
100
200
300
400
position
500
600
700
800
position
A
B
60
50
40
30
Which graph corresponds to
constructive, destructive and
intermediate interference ?
20
10
0
0
100
200
300
400
500
position
C
600
700
800
SUPERPOSITION  INTERFERENCE
What do these pictures tell you ?
16 17 18 19 20 21
8
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CONSTRUCTIVE INTERFERENCE
path length difference = 2 
phase difference = 2 (2 ) = 4 
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DESTRUCTIVE INTERFERENCE
path length difference = 3 ( / 2)
phase difference = 3 
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PARTIAL INTERFERENCE
Problem solving strategy: I S E E
Identity:
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What is the question asking (target variables) ?
What type of problem, relevant concepts, approach ?
Set up:
Diagrams
Equations
Data (units)
Physical principals
PRACTICE ONLY
MAKES PERMANENT
Execute: Answer question
Rearrange equations then substitute numbers
Evaluate:
Check your answer – look at limiting cases
sensible ?
units ?
significant figures ?
SUPERPOSITION  INTERFERENCE
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In phase

Audio
oscillator
s2
s1
Out of phase

Path difference  = |s2 - s1|
Phase difference  = 2 ( / )
CP 523
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Problem 2
Two small loudspeakers emit pure sinusoidal waves that are in
phase.
(a) What frequencies does a loud sound occur at a point P?
(b) What frequencies will the sound be very soft?
(vsound = 344 m.s-1).
2.00 m
3.50 m
P
2.50 m
CP 523
Solution 2
s2  2.52  3.52  4.30 m
s1  2.02  3.52  4.03 m
Construction interference
v
= s2 -s1  m   m
f
mv
f 

m  0,1,2,...
= s2 -s1  4.30  4.03 m=0.27 m
 344 
f 
 m  1274 m
0.27


 1.27 kHz, 2.55 kHz, 3.82 kHz, … , 19.1 kHz
Destructive interference
(m  2 )
v
1
1
= s2 -s1  (m  2)   (m  2)  f 
f

1
 344 
1
1
f 
  m  2   1274  m  2 
 0.27 
m  0,1,2,...
 0.63 kHz, 1.91 kHz, 3.19 kHz,… , 19.7 kHz
Problem 3
Two speakers placed 3.00 m apart are driven by the same oscillator.
A listener is originally at Point O, which is located 8.00 m from the
center of the line connecting the two speakers. The listener then walks
to point P, which is a perpendicular distance 0.350 m from O, before
reaching the first minimum in sound intensity. What is the frequency
of the oscillator? Take speed of sound in air to be 343 m.s-1.
r1  (8.00) 2  (1.15) 2 m  8.08 m
r2  (8.00)2  (1.85) 2 m  8.21 m
1
r2  r1  (n  ) and n  0
2
  2(r2  r1 )  0.26 m
f 
v


343
Hz  1.3 103 Hz
0.26
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FOURIER ANALYSIS
• A sinusoidal sound wave of frequency f is a pure tone
• A note played by an instrument is not a pure tone - its
wavefunction is not of sinusoidal form
• The wavefunction is a superposition (sum) of a sinusoidal
wavefunction at f (fundamental or 1st harmonic), plus one
at 2f (second harmonic or 1st overtone) plus one at 3f (third
harmonic or second overtone) etc, with progressively
decreasing amplitudes
• The harmonic waves with different frequencies which sum
to the final wave are called a Fourier series. Breaking up
the original wave into its sinusoidal components is called
Fourier analysis.
CP 521
Superimpose  resultant (add)
waveform
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Waveform
Fundamental 1st harmonic
1st overtone 2nd harmonic
2nd overtone 3rd harmonic
CP 521
FOURIER ANALYSIS  any wave pattern can be decomposed
into a superposition of appropriate sinusoidal waves.
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FOURIER SYNTHESIS  any wave pattern can be constructed
as a superposition of appropriate sinusoidal waves
An = A1 / n
fn = n f1
Electronic Music ???
n
y   An sin(2  f n )
n 1
CP 521
Quality of Sound
Timbre or tone color or tone quality
http://paws.kettering.edu/~drussell/Demos/
superposition/superposition.html
Frequency spectrum
noise
music
piano
Harmonics
piano
Harmonics
Some of the animations are from the web site
http://paws.kettering.edu/~drussell/demos.html
The Physics Teacher Vol 33, Feb 1995
INTERFERENCE PATTERNS AND
LANDING AIRCRAFT
Aircraft are guided in landing with the aid of the
interference pattern from two aerials A1 and A2
about 40 m apart. The aerials emit coherent waves
at 30 MHz. The wavelength is  = c / f = 10 m . The
lines of maximum signal strength are shown in the
diagram.
The number of lines emanating between the aerials depends upon the wavelength and the distance
between the aerials such that centre line corresponds to the central maximum. The plane should fly
along the line of the central maximum. If a plane flies along an adjacent (weaker) line of maximum
signal strength, the planes position can be in error by about 500 m. Aerials are also placed so a
vertical interference pattern is set up so that the height of the plane can be controlled to fly along this
central maximum.
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