# Lecture 30 ```Interference of Sound
Waves
Interference
2 waves, of the same frequency; out of phase.
Eg.
Then
y1=A0sin (kx - wt)
y2=A0sin (kx - wt +f)
yR=ARsin(kx-wt+f /2),
and the resultant amplitude is
AR=2A0cos(&frac12;f ).
Identical waves which travel different distances will
arrive out of phase and will interfere, so that the
resultant amplitude varies with location.
Quiz
You are located at position y, where you can hear a loud
sound - the first maximum in intensity from two speakers.
The speakers are then connected ‘out of phase’ (difference
of π). What will you hear?
A) no change – same loud sound
B) no sound
C) something between ‘no sound’ and ‘loud sound’
y
Example:
Two sources, in phase; waves arrive at P by
paths of different lengths:
S1
P
detector
x1
x2
S2
At P:
y1  A0 sin( kx1  wt )
y2  A0 sin( kx2  wt )

Phase difference :
 = (kx  wt )  (kx
1
2
 wt )  k ( x1  x2 )  k x
Define x to be the path difference
  k x 
2

Then (using trig), at detector:

x1  x2

yR  (2 A0 cos )sin(k (
)  wt  )
2
2
2
kx terms don’t cancel!
Example
A pair of speakers is separated by 3.0m and driven by the
same oscillator. The listener walks perpendicular from a
point on the centerline 8m away to a distance of 0.35m
before reaching the first minimum in sound intenstiy.
What is the frequency of the oscillator?
3m
8.0m
0.35m
Solution
Intensity I
I = Power per unit area
Units: W / m2
(the area is measured perpendicular to the wave velocity)
Intensity is proportional to (resultant amplitude)2 ,
since I=P/A and power is proportional to A2
Two sources, each with amplitude Ao, intensity Io ,
phase difference f:

resultant amplitude AR  2 Ao cos( )
2


I R  4 I o cos ( )  I max cos ( )
2
2
2
2
Notes:
1) Maximum IR is 4 x IO
2) Maxima when f = 0, 2π, 4π, 6π , …
x = 0, &plusmn; λ, &plusmn; 2λ,…
3) Minima (zero intensity) when
f = π, 3π, 5π , …
x = &plusmn; λ/2, &plusmn; 3λ/2, &plusmn; 5λ/2,…
Note: The sources are in phase !!!
Example
6m
I think I
hear
something!
The detector
x
2 speakers, same intensity, in phase; f = 170 Hz
(so  = 2.0 m when speed of sound is 340 m/s)
At the position x=9 m, find the intensity in terms of the
intensity of a single speaker
Solution
```