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waves_02. 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 - PowerPoint PPT Presentation
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1
• 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
CP 514
SUPERPOSITION OF WAVESwaves_02
Animations courtesy of Dr. Dan Russell, Kettering University
2waves_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
1 2( , ) ( , ) ( , )o o oy x t y x t y x t Superposition (at time to)Wave 1 + Wave 2
Constructive interference: = m m = 0, 1, 2, 3, ... = m (2)
Destructive interference: = (m + 1/2) = (m + ½) (2)
14 15
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0 10 20 30 40 50 60 70 80 90
CP 510
SUPERPOSITION INTERFERENCE
In phase constructive interference
Out of phase destructive interference
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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|
intermediate interference: 0 < phase difference < rad or < phase difference < 2
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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)
SUPERPOSITION INTERFERENCE
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0 100 200 300 400 500 600 700 800position
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0 100 200 300 400 500 600 700 800position
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0 100 200 300 400 500 600 700 800position
Which graph corresponds to constructive, destructive and intermediate interference ?
A B
C
SUPERPOSITION INTERFERENCE
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What do these pictures tell you ?
SUPERPOSITION INTERFERENCE
16 17 18 19 20 21
9CONSTRUCTIVE INTERFERENCE
path length difference = 2
phase difference = 2 (2 ) = 4
10DESTRUCTIVE INTERFERENCE
path length difference = 3 ( / 2)
phase difference = 3
11PARTIAL INTERFERENCE
Problem solving strategy: I S E E
Identity: What is the question asking (target variables) ? What type of problem, relevant concepts, approach ?
Set up: Diagrams Equations Data (units) Physical principals
Execute: Answer question Rearrange equations then substitute numbers
Evaluate: Check your answer – look at limiting cases sensible ? units ? significant figures ?
PRACTICE ONLY MAKES PERMANENT
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Audio oscillator
s1
s2
Path difference = |s2-s1|Phase difference = 2(/ )
In phase
Out of phase
CP 523
SUPERPOSITION INTERFERENCE
14Problem 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).
P3.50 m
2.50 m
2.00 m
CP 523
2 21 2.0 3.5 4.03 ms 2 2
2 2.5 3.5 4.30 ms
2 1= s -s 0,1,2,...v m vm m f mf
121 12 1 2 2
( )= s -s ( ) ( ) 0,1,2,...
mvm m f mf
2 1= s -s 4.30 4.03 m = 0.27 m
344 12740.27
f m m
1 12 2
344 12740.27
f m m
Solution 2
Construction interference
Destructive interference
1.27 kHz, 2.55 kHz, 3.82 kHz, … , 19.1 kHz
0.63 kHz, 1.91 kHz, 3.19 kHz,… , 19.7 kHz
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 thecenter of the line connecting the two speakers. The listener then walksto point P, which is a perpendicular distance 0.350 m from O, beforereaching the first minimum in sound intensity. What is the frequencyof the oscillator? Take speed of sound in air to be 343 m.s-1.
2 21 (8.00) (1.15) m 8.08 mr
2 22 (8.00) (1.85) m 8.21 mr
0 and )21(12 nnrr
m 26.0)(2 12 rr 3343 Hz 1.3 10 Hz0.26
vf
Problem 3
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• 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
FOURIER ANALYSIS
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Waveform
Fundamental
1st overtone2nd overtone
1st harmonic
2nd harmonic3rd harmonic
Supe
rim
pose
r
esul
tant
(add
)
wav
efor
m
CP 521
19
1sin(2 )
nn n
ny A f
FOURIER ANALYSIS any wave pattern can be decomposed into a superposition of appropriate sinusoidal waves.
FOURIER SYNTHESIS any wave pattern can be constructed as a superposition of appropriate sinusoidal waves An = A1 / n fn = n f1 Electronic Music ???
CP 521
Quality of Sound Timbre or tone color or tone quality
piano
piano
music
noise
Frequency spectrum
Harmonics
Harmonics
http://paws.kettering.edu/~drussell/Demos/superposition/superposition.html
Some of the animations are from the web site
http://paws.kettering.edu/~drussell/demos.html
22The 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.