Physics 123C
Waves
Lecture 10 (T&M: 15.4)
Waves & Barriers
April 23, 2008 (27 Slides)
John G. Cramer
Professor of Physics
B451 PAB
cramer@phys.washington.edu
Sound Wave Intensity
 E av  12  2 s02V
 E av 1 2
av 
V
 2  s02
 E av  av V  av Avt
Pav 
 E av
t
  E av  av VAv
Pav
1
1 p02
2 2
I
 av v   s0 V 
A
2
2 v
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  (10 dB) log10  I / I0 
I 0  1012 W/m 2
Physics 123C - Lecture 10
 Thresho ld of hearing
0 dB    120 dB
2
Sound Intensities
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Physics 123C - Lecture 10
3
Hearing Response of the Ear
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Physics 123C - Lecture 10
4
Waves in an Open-Open Pipe
n
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Physics 123C - Lecture 10
5
Waves in an Open-Closed Pipe
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6
Pipes and Modes
Open-Open or Closed-Closed
2L

 1 / m 

m
 m  1, 2,3, 4,
v
fm  m
 mf1 

2L
m 
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Open-Closed
4L

 1 / m 

m
 m  1,3,5, 7,
v
fm  m
 mf1 

4L
m 
Physics 123C - Lecture 10
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Example:
The Length of an Organ Pipe
An organ pipe open at both ends sounds its 2nd harmonic at a
frequency of 523 Hz (one octave above middle C).
What is the length of the pipe from sounding hole to end?
v
v
f2  2

2L L
L
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v (343 m/s)

 0.656 m  65.6 cm
f 2 (523 Hz)
Physics 123C - Lecture 10
8
Clicker Question 1
An open-open tube of air supports standing waves of frequencies
of 300 Hz and 400 Hz, with no frequencies between these two.
The second harmonic (m=2) of this tube has frequency:
(a) 100 Hz; (b) 200 Hz;
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(c) 400 Hz;
(d) 600 Hz;
Physics 123C - Lecture 10
(e) 800 Hz.
9
Woodwinds vs. Strings
Many woodwind instruments are effectively an
open-closed pipe. This means they have only odd
harmonics. Their fundamental frequency will be:
The vibrating string of a stringed instrument is
the equivalent of a closed-closed pipe. This means it
will have both odd and even harmonics.
vstring
Its fundamental frequency is:
f1 
2L
vsound
f1 
4L

1 Ts
2L 
Note that for wind instruments, L
is the only adjustable parameter, while for stringed instruments, L,
Ts and  can, in principle, be varied. However, wind instruments can
be played at relatively pure harmonic frequencies, while strings
cannot.
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Physics 123C - Lecture 10
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Example:
The Notes of a Clarinet
A clarinet (an open-closed instrument) is 66 cm long. The speed of
sound in warm air is 350 m/s.
What are the frequencies of the lowest note on a clarinet and of
the next highest harmonic?
f1 
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v
(350 m/s)

 133 Hz
4 L 4(0.66 m)
f3  3 f1  399 Hz
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11
Reflection from a Boundary
When a traveling wave encounters
a “terminating” discontinuity in the
medium (R=), there is a complete
negative reflection at the
discontinuity. All of the wave energy
is reflected as the negative of the
incoming wave.
While the wave is inverted in
displacement direction, its amplitude
is unchanged. At the boundary point
the wave and its reflection always
subtract to produce zero deflection.
The situation can be simulated as
an un-terminated string with positive
and negative amplitude waves moving
in opposite directions and meeting at
the boundary.
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Note that the reflected wave has
the same speed and wavelength (and
energy) as the incident wave.
Physics 123C - Lecture 10
12
Creating Standing Waves
Plucking a Standing Wave
Considering the reflections
at boundaries, it is easy to see
how string vibration occur.
When a string is plucked in
the middle, waves travel in
both directions to the
boundaries, where they are
reflected and propagate back
and forth along the string.
The net result is a
superposition of right and left
moving traveling waves that
produce a standing wave. The
waves so produced must have
nodes at both boundaries.
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Physics 123C - Lecture 10
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Standing Wave Normal Modes
Standing waves have the form:
D(x,t) = (2a sin kx)cos t
The two string boundary conditions are:
D(x=0, t) = 0 and D(x=L, t) = 0.
Therefore, 2a sin kL = 0 , which implies
that kL = mp, where m is an integer. But
k = 2p/, so:
m 
2L
, m  1, 2, 3, 4,
m
The frequency f is related to the
wavelength  by: f = v/, so the allowed waves
on a string of length L will have frequencies:
fm 
v
v
m
, m  1, 2, 3, 4,
2L / m
2L
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Physics 123C - Lecture 10
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About Normal Modes
1. The integer m is the number of antinodes
of the standing wave. The number of nodes
of the wave is m + 1.
2. The fundamental mode, with m = 1, has
wavelength 1 = 2L (not L). Half a
wavelength fits on the string, because the
spacing between nodes is /2.
3. The frequencies of the normal modes of a
string form an arithmetic series: f, 2f, 3f, 4f,
… Therefore, the fundamental frequency f1
can be found as the difference between the
frequencies of any two adjacent modes, i.e.,
f = fm+1  fm = f1.
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Physics 123C - Lecture 10
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Clicker Question 1
A standing wave on a string vibrates as shown.
If the tension is quadrupled while the frequency and distance
between boundaries remain the same, which diagram represents the
new vibration?
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Physics 123C - Lecture 10
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Waves, Power, and Energy
y  A sin  kx  t 
y
  A cos  kx  t 
t
y
 kA cos  kx  t 
x
y y
x t
  FT  kA cos  kx  t     A cos  kx  t  
P  FTy v y   FT tan  v y   FT
 E av  Pav t
 FT k A2 cos 2  kx  t    v 2 A2 cos 2  kx  t 
 12  v 2 A2 t
 12  2 A2 x
Pav  v A
1
2
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2
2
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Reflection and Transmission
Slow to Fast Transition
Fast to Slow Transition
When a traveling wave encounters
a “speed-up” discontinuity in the
medium (L>R), there is a positive
reflection at the discontinuity. Part
of the wave energy is reflected and
part is transmitted.
When a traveling wave encounters
a “slow-down” discontinuity in the
medium (L<R), there is a negative
reflection at the discontinuity. Again,
part of the wave energy is reflected
and part is transmitted.
SlowFast  positive reflection
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Fast Slow  negative reflection
Physics 123C - Lecture 10
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Transmission Coefficients
r  hr / hin
and   ht / hin
v2  v1
r
v2  v1
2v2
and  
v2  v1
Note that if v2  0, r  1 and   0
and if v1  v2 , r  0 and
 1
v2 2
1 r  
v1
2
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Physics 123C - Lecture 10
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Example:
Two Soldered Wires
Two wires with different linear mass densities are soldered end-to-end
and then stretched to a tension FT. The wave speed v1 on the first wire is
twice the wave speed v2 on the second wire
(a) If the incident wave amplitude is A, what are the amplitudes Ar and
At of the reflected and transmitted waves?
(b) What is linear mass density ratio 1/2 of the wires?
(c) What fraction of the incident average power is reflected at the
junction, and what fraction is transmitted?
v  v v  2v2
1
1
r 2 1  2
  so Ar   A
v2  v1 v2  2v2
3
3

2v2
2v2
2
2


so At  A
v2  v1 v2  2v2 3
3
1 FT / v12 v22  2v1 

 2 
4
2
2
2 FT / v2 v1
v1
2
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Pr av
Pin av
Pt av
Pin av
1 2 Ar2v1 12 1 2 ( 13 A)2 v1 1

 1

2 2
2 2
1 A v1
  A v1
9
2 1
1
2
1
2
2 2 At2 v2 12 2 2 ( 23 A) 2 v2

 1
2 2
2 2
1 A v1


A v1
1
2
1
2
1
2
4 2 v2 4 v12 v2 4 v1 8




2
9 1v1 9 v2 v1 9 v2 9
Physics 123C - Lecture 10
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Transparent Optical Media
Rather surprisingly, there are types
of matter, solids, liquids, and gasses,
that are transparent and that transmit
light almost unimpeded. When you
consider that such matter is made of
atoms, electrically charged nuclei
orbited by clouds of electrically charged
electrons, it is quite remarkable that
electromagnetic radiation, the carrier
of electric fields that interact strongly
with these charged particles, is not immediately absorbed.
Instead, within the transparent medium the bound electrons vibrate
together at the frequency of the incoming electric field to “help along” the
incident light without absorbing its energy. This usually reduces its speed
through the material as it is transmitted.
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Physics 123C - Lecture 10
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The Index of Refraction
Light travels through transparent media at
a speed less than its speed c in vacuum.
We define the index of refraction in a
transparent medium as:
n
c
vmedium
Is n always greater than 1?
Almost always. There are a few media in which
the phase velocity of light waves is greater than c.
However, this super-luminal speed cannot be used
to send signals or energy at a speed greater than c.
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Physics 123C - Lecture 10
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Waves vs. Particles
If two
pitching
machines
simultaneously throw
baseballs,
they will
collide and
bounce.
Two
particles
cannot
occupy the same space point at the
same time.
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On the
other hand,
if two loudspeakers
make sound
waves at
the same
time, they
will pass
through
each other
without collision. Two waves can
occupy the same space point at the
same time.
Physics 123C - Lecture 10
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Diffraction
Particles
Waves
/d
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1
Physics 123C - Lecture 10
 / d 1
24