Wave Interference: Solving Homework

[planauts]

Homework Statement

[PLAIN]http://img18.imageshack.us/img18/6175/questionn.gif

Homework Equations

The Attempt at a Solution

At M the waves are in phase and at P the waves have exactly (pi) phase difference. M and P are 2.00 m apart. So it would mean that the largest possible wavelength of sound emitted by the loudspeakers would be 2.00*2 = 4.00 m.


But the answer key says 8.00 m.

Thanks,

 

[vela]

Can you explain why 2*2.00 m is the relevant quantity, i.e. what does it represent?

 

[planauts]

At M, the two waves are in phase. At P, the waves have a pi phase difference. The period is 2pi. So to hear the sound at max again, he would have to travel 2 more meters. So, from M to P is only have the wavelength (from crest to trough). To get the full wavelength, I multiplied by 2.

 

[vela]

OK, that's not quite right. Interference is caused by differences in path length. What's the difference between the distance from one speaker to point P and the distance from the other speaker to P?

 

[planauts]

It doesn't tell you the from one speaker to point P and the distance from the other speaker to P. It just says the distance from M (2m). And at M, there is constructive interference and at P, there is destructive interference.

There is a formula in my book.

[PLAIN]http://img151.imageshack.us/img151/3535/hintg.jpg

I'm not exactly sure what to do...

 

[vela]

Say the distance from the speakers to point M is equal to x. In terms of x, what are the distance between point P and the speaker to the left and the distance between point P and the speaker to the right?

 

[planauts]

Say the distance from the speakers to point M is equal to x. In terms of x, what are the distance between point P and the speaker to the left and the distance between point P and the speaker to the right?

Let x = distance from M to the speakers

From P to Left Speaker
[tex]d_{1} = x + 2[/tex]

From P to Right Speaker
[tex]d_{2} = x - 2[/tex]

[tex]d_{1} - d_{2} = (x+2) - (x-2)[/tex]
[tex]d_{1} - d_{2} = 0x + 4 = 4[/tex]
[tex]4 = (n+\frac{1}{2}) \lambda [/tex]

Let n = 0 to get largest wavelength
[tex]4 = \frac{1}{2} \lambda [/tex]
[tex]8 = \lambda [/tex]

Therefore, 8.00 m is the largest wavelength.

Wow, that makes more sense now. Thanks a lot!

 

[vela]

Perfect! Good job.