1. Interference of Sound Waves

2. Interference 2 waves, of the same frequency; out of phase.
Eg. y1=A0sin (kx - wt) y2=A0sin (kx - wt +f)
Then
yR=ARsin(kx-wt+f /2),
and the resultant amplitude is AR=2A0cos(½f ).
Identical waves which travel different distances will arrive out of phase and will interfere, so that the resultant amplitude varies with location.

3. 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

4. Example:
Two sources, in phase; waves arrive at P by paths of different lengths: S1 P x1
detector x2
At P: S2

5. Phase difference : =
Define x to be the path difference
Then (using trig), at detector:
kx terms don’t cancel!

6. 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?
8.0m
0.35m 3m

7. Solution

8. 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:

9. Notes:
Maximum IR is 4 x IO
Maxima when f = 0, 2π, 4π, 6π , …
Minima (zero intensity) when
f = π, 3π, 5π , …
x = 0, ± λ, ± 2λ,…
 x = ± λ/2, ± 3λ/2, ± 5λ/2,…
Note: The sources are in phase !!!

10. I think I hear something!
Example
6 m x
The detector
I think I hear something!
2 speakers, same intensity, in phase; f = 170 Hz
(so l = 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

11. Solution