Wave - III
Sound Resonances
Consider a pipe of length L, open at one end, closed
at the other end.
At resonance, a displacement antinode at the open
end, and a displacement node at the closed end.
The longest wavelength to satisfy this condition is
1
v
v
1  L or f1 

4
1 4 L
Harmonics:
3v1
5v
f2 
 3 f1 and f 3 
 5 f1
4L
4L
Fundamental
resonant
frequency
v
f n  2n  1
4L
Pipe open at both ends: displacement antinodes at
both ends. open end closed at the other end.
Pipe closed at both ends: displacement nodes at
both ends.
In both cases:
v
fn  n
 nf1
2L
The same expression as in string with both ends
fixed.
cos   cos   2 cos
Beats
1
1
   cos    
2
2
Two sound waves with different but close
frequencies give rise to BEATS
Consider
s1 x,t   sm cos 1t
s2  x,t   sm cos 2 t
1   2
s  s1  s2  2sm cos t cos t
Very
small
1
1
 1   2     1   2 
2
2
≈1≈2
s  2sm cos t cost
1
 1   2 
2
On top of the almost same frequency, the
amplitude takes maximum twice in a cycle:
cos’t = 1 and -1: Beats
Beat frequency fbeat:
fbeat  f1  f2
The Doppler Effect
The Doppler Effect: the frequency change
related to the motions of the source or/and
detector
In the following, the speed is measured with
respect to the air, through which the sound
wave travels
Detector Moving, Source Stationary
The detector stationary:
Distance the sound
travels in time t
Periods in unit
time: frequency
vt
Divided by  to get
the number of periods
in time t
v

f

t

The detector moving toward the source:
more periods reaches detector. Equivalently:
vt  vD t 
v  vD 


f 

t

f 
v  vD 

f
v


The detector moving toward the source:
v  vD 

f 
f
vD is the SPEED,
always positive
v
In general:
v  vD 

f 
f
v
+ : toward S
-: away from S
Source Moving, Detector Stationary
The source stationary:
Distance between two
wavefronts period T
apart
  vT
f
v

The source moving toward the detector :
waves are squeezed. Equivalently:
v
f 
 vT  vS T


f  f

  vT
 vT  vS T

f  
f

The source moving toward the detector :
v
f 
f
v  vS 
vS is the SPEED,
always positive
In general:
v
f 
f
v vS 
-: toward D
+: away fromD
In General
+ : toward S
-: away from S
+: away from D
-: toward D
v  vD 

f 
f
v vS 
All speeds are measured with respect
to the medium of propagation: the air
At Low Speed
+ : toward
 u 
f  f 1 
 v 
each other
-: away from
each other
Relative speed:
u  vS  v D
Supersonic Speed
When vS>v, the equation no longer applicable:
Supersonic speed
The wavefronts form a Mach Cone
A Shock Wave is generated: abrupt change of
air pressure
The source moving toward the detector :
v
f 
f
v  vS 
HRW 51E (5th ed.). The water level in a vertical glass tube 1.00 m
long can be adjusted to any position in the tube. A tuning fork
vibrating at 686 Hz is held just over the open top end of the tube. At
what positions of the water level will there be a resonance?
Let L be the length of the air column. Then the
condition for resonance is:
f n  2n  1

v
2n  1v
f n  2n  1
or Ln 
4L
4f
343
1 3 5 7
Ln  2n  1
 , , , m
4  686 8 8 8 8
L water  1.0  Ln  0.875,0.625,0.375,0.125m
v
4L
HRW 61E (5th ed.). A tuning fork of unknown frequency makes three beats per
second with a standard fork of frequency 384 Hz. The beat frequency decreases
when a small piece of wax in put on a prong of the first fork. What is the
frequency of this fork?
fbeat  f1  f2
fbeat = 3 Hz  f1 = 381 or 387 Hz
Mass increases  f1 decreases
fbeat decreases  f1 becomes closer to 384 Hz
Therefore, f1 = 387 Hz
Resonant frequency
n 
f
2L 
HRW 68E (5th ed.). The 16,000 Hz whine of the turbines in the jet
engines of an aircraft moving with speed 200 m/s is heard at what
frequency by the pilot of a second aircraft trying to overtake the first
at a speed of 250 m/s?
v  vD 

f 
f
v vS 
The detector moves toward the source: take the plus sign for vD.
The source moves away from the detector : take the plus sign for vS.
v  vD 

343 m/s
f 
f
v  vS  343 m/s
+ 250 m/s
 17,500 Hz
+ 200 m/s
HRW 80P (5th ed.). A person on a railroad car blows a trumpet note at 440 Hz.
The car is moving toward a wall at 20.0 m/s. Calculate (a) the frequency of the
sound as received at the wall and (b) the frequency of the reflected sound arriving
back at the source.
(a) The source moving toward the detector :
v
343 m/s


f
f 
(440 Hz) = 467 Hz
v  vS   343 m/s - 20.0 m/s 
(b) The person (detector) moves toward the source at the wall with
f’ = 467 Hz:
v  vD
343 m/s + 20.0 m/s 

fr 
f
(467 Hz) = 494 Hz
v
343 m/s

