Interference of Sound Interference of Sound WavesWaves

InterferenceInterference

2 waves, of the same frequency; out of phase.

Eg. y1=A0sin (kx - t) y2=A0sin (kx - t +)

Then yR=ARsin(kx-t+/2),

and the resultant amplitude is AR=2A0cos(½.

Identical waves which travel different distances will arrive out of phase and will interfere, so that the resultant amplitude varies with location.

Quiz

y

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 soundB) no soundC) something between ‘no sound’ and ‘loud sound’

Example:

Two sources, in phase; waves arrive at P by paths of different lengths:

At P:

1 0 1

2 0 2

sin( )sin( )

y A kx ty A kx t

detectorS1

S2

Px1

x2

Phase difference :

Then (using trig), at detector:

1 2 1 2( ) ( ) ( )kx t kx t k x x k x

2 radians

k x x

1 2

0(2 cos )sin( ( ) )

2 2 2R

x xy A k t

Define x to be the path difference

=

kx terms don’t cancel!

ExampleExample

A pair of speakers is separated by 3.0m and driven by thesame 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 0.35m3m

Solution

Intensity IIntensity 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

otant amplitude 2 cos( )

2Rresul A A

2 2

o max4 cos ( ) cos ( )

2 2RI I I

Notes:

1) Maximum IR is 4 x IO

2) Maxima when = 0, 2π, 4π, 6π , …

3) Minima (zero intensity) when = π, 3π, 5π , …

x = ± λ/2, ± 3λ/2, ± 5λ/2,…

x = 0, ± λ, ± 2λ,…

Note: The sources are in phase !!!

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

6 m

x

The detector

Example

I think I hear

something!

Solution