SHM / phase

  • Thread starter tigger88
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  • #1
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I've got a review question on classical waves (SHM) where I'm given the frequency as f = 20 (sec^-1) and velocity v = 80m/s.
The question is: How far apart are two points whose displacements are 30 degrees apart in phase?

My own reasoning doesn't seem to agree with the solution I was provided.

Here is my thinking:
Phase = wt +/- kx
For the first point, let x = x1 and t = t.
For the second point, let x = x2 (and t = t)
So then the phase difference would be (wt-kx1) - (wt-kx2) = k(x2-x1).
Then, given v and f, lambda = v/f = 4 m, so k = 2pi/lambda = pi/2.
So phase difference, which is given to be 30 degrees, or pi/6 rad, becomes:
pi/6 = (pi/2)(x2-x1) -where solving for x2-x1 gives an answer of 1/3.

The solution I was given says that the phase difference = 2pi(x2-x1) = pi/6, which gives x2-x1 = 1/12.

I don't understand how the 2pi comes into it!
Could someone tell me where my thinking went wrong?
 

Answers and Replies

  • #2
alphysicist
Homework Helper
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Hi tigger88,

Your answer looks right to me. Another way of thinking about it would be to say that two points with displacement 360 degrees apart in phase would be one wavelength apart in distance, and since 30 degrees is 1/12 of 360 degrees, the distance you're looking for is 1/12 of a wavelength.
 

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