Speed of Waves
Speed of waves in strings (review):
v
F
v
E

  m/ L
F - tension
 - linear density
m - mass
L – length of the string
Speed of longitudinal waves in a solid rod:

E - Young’s modulus
  m /V
Young’s modulus (optional):
 - density
m - mass
V - volume

E

Stress:
Strain:
  F A
  L L0
The Young’s modulus is the
stress divided by the strain.
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Speed of Waves (continued)
Speed of sound waves in a fluid:
v
B
B

P
V V0
B - bulk modulus
 - density
V - volume
P - pressure
Speed of sound waves in an ideal gas:
v
RT
T  t  273 .15
M
 - ratio of heat capacities
T - temperature in Kelvin
t - temperature in Celsius
M - molar mass
R - gas constant
R=8.31451 J/mol K
For air:
  1.4
M  28.9645g / m ol  2.89645 10 2 kg / m ol


v  20.047 T m / s  331.3  0.606t m / s
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Speed of Sound in Some Common Substances
Substance
Speed (m/s)
1. Air (20 oC)
344
2. Helium (20 oC)
999
3. Water (0 oC)
1,402
4. Lead
1,200
5. Human tissue
1,540
6. Aluminum
6,420
7. Iron and steel
5,941
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Reflection
a) Mirror reflection
1   2
normal to
surface
 1 2
in
 1 2
S
S’
out
source
image
wall (mirror)
b) Diffuse reflection
c) Reflection from curved surfaces
•acoustical mirrors (curved mirrors)
•whispering galleries
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Refraction: Snell’s Law
1
sin  2 v2 2


sin 1 v1 1
2
f 2  f1
In optics:
n1 v 2 2


n2 v1 1
n1 sin 1  n2 sin 2
Example:
1
n1
index of refraction:
c
n
v
n1 sin 1  n2 sin 2  n3 sin 3
 2 2
n2
n3
3
 3 is independent fromn2
If n3  n1  3  1
•Sound waves in atmosphere when temperature varies with height
•Sound traveling against the wind
•Mirage
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Doppler Effect
What do you hear when a sound-emitting object (train, car) passes you?
The sound changes in pitch (frequency) as the object goes.
Why does this happen?
v  f
•As the object travels towards you, the distance
between wavefronts is compressed; this makes it
seem like  is smaller and frequency is higher.
•As the object travels away from you, the distance
between wavefronts is extended; this makes it
seem like  is larger and frequency is lower.
Because the speed of a wave in a medium is
constant, change in  will affect change in f.
Velocity of listener (L): vL
Velocity of source (S): vS
Velocity of sound: v
 v  vL
f L  f S 
 v  vS



vL or vS is “+” if in the same
direction as from listener to
source and is “-” otherwise
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Diffraction
1) What it is?
The bending of waves behind obstacles or apertures into the ”shadow region”,
that can be considered as interference of many waves.
2) Haw to observe?
Diffraction is most pronounced when the wavelength of the wave is similar to the
size of the obstacle or aperture.
For example, the diffraction of sound waves is commonly observed because the
wavelength of sound is similar to the size of doors, etc.
The waves
spread out from
the opening!
•Waves will diffract around a single slit or obstacle.
•The resulting pattern of light and dark stripes on a screen is called
a diffraction pattern (fringes).
•This pattern arises because different points along a slit create
wavelets that interfere with each other just as a double slit would.
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3a) Diffraction from a single slit (intensity)
Minima (dark fringes):
D sin  m  m
m  1,2,3,...
Example: In order to obtain a good single slit
diffraction pattern, the slit width could be:
A. /100 ; B. /10; C. ; D. 10; E.100
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Example: Light of wavelength 610 nm is incident on a slit 0.20 mm wide and
the diffraction pattern is produced on a screen that is 1.5 m from the slit.
What is the width of the central maximum?
  610nm
D  0.20m m
x  1.5m
2y  ?
m
yx
D
610 109
3
2 y  2  1.5m

9
.
1510
m  9.2m m
3
0.20  10
Example: Light of wavelength 687 nm is incident on a single slit 0.75 mm wide.
At what distance from the slit should a screen be placed if the second dark fringe
in the diffraction pattern is to be 1.7 mm from the center of the screen?
  687nm
D  0.75m m
y  1.7m m
x?
m
yx
D
yD
x
m
1.7 103 m 0.75103 m 1.7 0.75
x

m  0.93m
9
2 68710 m
20.687





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