2.
Interference with sound
Superposition works exactly as it did for transversal waves.
Additional complication: 3D waves! (next lecture)
3.
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:
No displacement:
s=0
A closed end is
a “fixed end”
Wave must be inverted
(s becomes –s)
Reflected pulse
v
Incoming pulse
v
s
s
s
s
v
x
x
v
4.
Reflection of sound at an open end
v
Pulse travels out into open air
s
Oscillation back from a larger slice
moves more air into pipe…
v
…and increases pressure…
v
s
v
…and causes a wave to propagate
back in (a reflection!)
(and another wave is
transmitted outside)
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,
p ≈0
5.
Boundary conditions for sound
Open end
• gauge pressure = 0
• maximum (absolute) air
displacement
Closed end
• air displacement = 0
• maximum (absolute) gauge
pressure
6.
Standing sound waves in pipe open at both
ends
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 )
a +b
sin ( a ) + sin ( b ) = 2cos
2
a −b
÷sin
÷
2
sall (x ,t ) = 2smax sin ( kx ) cos ( ωt )
At the openings:
p~0
Large displacements
Standing wave within pipe: does not travel, bounces back and forth.
Amplitude will decrease as energy is transported out of the pipe
7.
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
discuss.
p
x
p = 0 at open ends
Maximum/minimum p at node
8.
Higher harmonics
DEMO:
Organ pipes
Each harmonic is a standing wave.
λ
2
n = 1,2,3...
2L
n
n = 1,2,3...
L =n
λn =
λ gets shorter, frequency increases
2
Visualize them: http://www.walter-fendt.de/ph11e/stlwaves.htm
9.
ACT: Pipe closed at one end
Open end:
Max s, p = 0
Closed end:
s = 0, max p
s
First harmonic or fundamental
frequency:
λ1 = 4L
What is the standing wave for the next harmonic?
A
s = 0 at an open end? (No!)
B
And s max at a closed end? (No!)
C
λ2 =
4L
3
In general,
λ
L = nodd
nodd = 1,3...
4
10.
In-class example
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?
A.
B.
C.
D.
E.
4f
2f
f
f/4
None of the above
Close end = node
Open end = antinode
λ = 2L
f′=
λ ′ = 4L = 2λ
v
v
f
=
=
λ ′ 2λ 2
11.
A little music
DEMO:
Xylophone
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
f1 =
v
343 m/s
=
= 890 Hz
2L 2 ( 0.192 m )
Approx. A5 (La)
12.
Resonance
DEMO:
Resonant slabs
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.
Examples:
Pendulum: resonance occurs when fD = 2π
(A pendulum has only one normal frequency)
g
L
String fixed at both ends: when fD = fn =
F n
µ 2L
Pipe closed at one end: when fD = fn = v sound
nodd
4L
(L = length of string)
for n = 1,2,...
for nodd = 1,3,...
(see lecture 6)
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