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Homework Help: SHM and phase difference

  1. Apr 22, 2008 #1
    Two particles are executing SHM with same amplitude and frequency (but with a phase difference). The maximum separation between them was found to be equal to the amplitude. What is the phase difference. This is a question that I came across.

    And I solved it the following way.

    Assume one is y1 = A sin (wt) and the other one y2 = A sin (wt + phi)
    max(y1-y2) = A (given)
    Solve for phi.

    Since both y1 and y2 are sinusoids of w frequency, their difference will also be a sinusoid with a different amplitude but with same frequency and a different phase.

    Since we are just interested in the max value of the difference, I calculated the amplitude of the sinusoid using phasors A e^j(phi) - A. It turned out to be 2A sin(phi/2). From that I calculated phi as pi/3 but the answer I have is as pi/2. I think the book is correct but I could not find the mistake in my calculation.

    Can some one point out if I am wrong and if yes, where?
     
  2. jcsd
  3. Apr 22, 2008 #2

    Shooting Star

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    The correct answer is pi/3. The answer in the book is wrong.
     
  4. Apr 26, 2008 #3
    I think I am wrong because the difference between A sin(wt) and A sin (wt+pi/2) is A at t=0 itself. I think the difference between A sin(wt) and A sin (wt+pi/3) is never A ( which is what is required). I am not able to find a mistake in my method though. Can you tell at what 't' the the difference between A sin(wt) and A sin (wt+pi/3) is A? Thanks for your time and effort.
     
  5. Apr 26, 2008 #4

    alphysicist

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    Hi manjuvenamma,

    You can't find a specific time in seconds until you specify [itex]\omega[/itex], but here is one time that works:

    [tex]t = \frac{1}{\omega} (2.61799\ \mbox{radians})[/tex]

    (So if you just use [itex]\omega t=2.61799[/itex] radians as a whole into your functions you will get the right answer.)

    You can find this by solving (after cancelling out the A's) for [itex]t[/itex]:

    [tex]
    \sin(\omega t +\pi/3) -\sin(\omega t) = \pm 1
    [/tex]
     
  6. Apr 26, 2008 #5

    Shooting Star

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    Yes, at first glance it may seem so. When particle 1 is at x=0, and particle 2 is at x=A, then their separation is A. But the problem states that the maximum separation is A, which is not true here..

    Let us analyze this situation stated above.

    Case 1:

    2 is at x=A, and 1 is going away from x=0 in the -ve direction. Since the speed of 2 at the extreme point is very small compared to 1, without any calculation we can say that their separation will be more than A for some time after this.

    Case 2:

    2 is at x=A, and 1 is going toward 2 from x=0, that is, in the +ve direction. We know that when 1 will reach the extreme position, 2 would have come to the mean position x=0. After that, the same argument as in case 1 applies.

    As we see, in both of these cases, the maximum separation is more than A.

    What you have written in the first post is correct, and the answer is pi/3.
     
  7. Apr 28, 2008 #6
    Thanks so much, Shooting Star and alphysicist!
    I am now convinced that I am right and I know the correct method too!
     
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