•Sound Waves
•Speed of sound in fluids in fluids, gases and solids
•Doppler Effect
•Beats

Sound waves
•
Unlike waves on a string (transverse waves), sound waves are longitudinal waves
•Particles undergo SHM about equilibrium positions PARALLEL to direction of wave propagation
•Compression/rarefaction of the medium through which the sound wave travels………………………………………………
•Wave travelling from left to right/right to left still described by: y ( x , t )

A cos( kx

/
  t )
But this time, y is in same direction as x……

(Ear, for example is sensitive to pressure, not dislacement
Fluctuations)
Sound waves

In a fluid:
In a solid rod:
Speed of sound v

B

Bulk modulus density v

Y

Young’s modulus

Speed of sound

The Doppler Effect
•
If we stand still near a source of sound, and the source is also not moving we hear sound of a certain frequency f
0
•If either, or both, the listener and the source are moving relative to one another, then a different frequency f
L is heard by the listener.
•In general, if the listener and the source are moving towards each other, f
L
> f
0
, ie a higher pitch is heard.
•If the listener and the source are moving away from each other, f
L
<f
0
, ie a lower pitch is heard.
•This is known as the Doppler Effect
•Ambulance sirens, car horns, etc. etc.

Explanation of the Doppler Effect v
 f

Listener moving towards stationary source with velocity v
L

Explanation of the Doppler Effect
Effective velocity of sound heard by listener = v
L
+v: v
 f
 
( v
 v
L
)
 f
 f


( v

 v
L
)
 v f
0 f
 f
0
( v
 v
L
) v
Moving listener, stationary source

Moving Source
It takes a time T for one complete cycle of the wave to be emitted.
(T = 1/f
0
)
In front of source
During this time, the wavefront has moved a distance vT, and the source has moved a distance v s
T
The wavelength is:
 
T ( v
 v s
)

( v
 v s
) f
0
 
T ( v
 v s
)

( v
 v s
) f
0
Behind source

Moving listener approaching moving source f

( v

 v
L
)
(Stationary source, earlier slide) f
 f
0
( v
( v

 v
L v s
)
)
(Modified λ, due to moving source)

And now the good news…………………… f
 f
0
( v
( v

 v
L v s
)
)
Works for ALL relative motions of source and listener in acoustic Doppler shift calculations, PROVIDED you DEFINE the positive direction as the listener to source direction, and set the signs of v
L and v s accordingly……………………………………………………………………………
………

Example calculations
Two train whistles A and B each have a frequency of 392 Hz. A is stationary, while B is moving to the right, away from A, at a speed of
35ms -1 . A listener is between the 2 trains and is also moving to the right, with a speed of 15ms -1 . What frequencies does the listener hear from whistles A and B?
(speed of sound in air = 340ms -1 )

Whistle A
Positive direction f
 f
0
( v
( v

 v
L v s
)
) f
 f
0
( v
( v

 v
L
0 )
)

Whistle B
Positive direction f
 f
0
( v
( v

 v
L v s
)
) f
 f
0
( 340

15 )
( 340

35 )

A police car (whose driver has clearly been watching too much
Starsky and Hutch), with a 300Hz siren is approaching a warehouse at 30ms -1 with the intention of crashing through the door. What frequency does the driver hear reflected from the warehouse?
(speed of sound in air = 340ms -1 )
Two part question………………………..

In the first part, treat the warehouse door as the “listener”…..
f
 f
0
( v
( v

 v
L v s
)
)
Sound at this frequency is reflected back towards the police car. This becomes the “new” f
0 for the 2 nd part of the question…………………..

Now, the warehouse door is the stationary source of the reflected sound……..
f
 f
0

( v
( v

 v
L
) v s
) f


 f
0
(
( v v

0 )

 v s
)   v v

 v
L
0

Beats
These occur from the superposition of 2 waves of close, but different frequency: f beat
 f a
 f b