Interference with sound
Superposition works exactly as it did for transversal waves.
Additional complication: 3D waves! (next lecture)
Reflection of sound waves against a surface
Consider a sound pulse (air moves to the right and back to initial
position) traveling along a pipe toward a closed end:
A closed end is
a “fixed end”
Wave must be inverted
(s becomes –s)
Reflection of sound at an open end
Pulse travels out into open air
Oscillation back from a larger slice
moves more air into pipe…
…and increases pressure…
…and causes a wave to propagate
back in (a reflection!)
(and another wave is
Beyond the open end of the pipe, variations in the pressure must be much
smaller than pressure variations (gauge pressure) in pipe.
Just beyond the open end,
Boundary conditions for sound
• gauge pressure = 0
• maximum (absolute) air
• air displacement = 0
• maximum (absolute) gauge
Standing sound waves in pipe open at both
A harmonic wave and its reflection on an open end:
s1 (x ,t ) = smax sin(kx − ωt )
s2 (x ,t ) = smax sin(kx + ωt )
sall (x ,t ) = smax sin(kx − ωt ) + sin(kx + ωt )
sin ( a ) + sin ( b ) = 2cos
sall (x ,t ) = 2smax sin ( kx ) cos ( ωt )
At the openings:
Standing wave within pipe: does not travel, bounces back and forth.
Amplitude will decrease as energy is transported out of the pipe
ACT: Pipe open at both ends
This is the air displacement for a standing wave inside this tube.
Sketch the gauge pressure vs
position for this wave.
Compare with your neighbor and
p = 0 at open ends
Maximum/minimum p at node
Each harmonic is a standing wave.
n = 1,2,3...
n = 1,2,3...
λ gets shorter, frequency increases
Visualize them: http://www.walter-fendt.de/ph11e/stlwaves.htm
ACT: Pipe closed at one end
Max s, p = 0
s = 0, max p
First harmonic or fundamental
λ1 = 4L
What is the standing wave for the next harmonic?
s = 0 at an open end? (No!)
And s max at a closed end? (No!)
L = nodd
nodd = 1,3...
A tube with both ends open has a fundamental frequency f. What is
the fundamental frequency of the same tube if one end is closed?
None of the above
Close end = node
Open end = antinode
λ = 2L
λ ′ = 4L = 2λ
λ ′ 2λ 2
A little music
When you blow air into a pipe, all the harmonics are present.
Example: Blow into a tube of length 19.2 cm open at one end
λ1 = 2L
= 890 Hz
2L 2 ( 0.192 m )
Approx. A5 (La)
To produce a wave, we need to apply an external force (driving
force). This driving force can be periodic with frequency fD.
The amplitude of the perturbation is maximum when the
frequency of the driving force is equal to one of the natural
(or harmonic, or normal) frequencies of the system.
Pendulum: resonance occurs when fD = 2π
(A pendulum has only one normal frequency)
String fixed at both ends: when fD = fn =
Pipe closed at one end: when fD = fn = v sound
(L = length of string)
for n = 1,2,...
for nodd = 1,3,...
(see lecture 6)
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