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SHM Proof

  1. Oct 8, 2005 #1
    Question: A particle undergoing simple harmonic motion has a velocity v1 when the displacement is x1 and a velocity v2 when the displacement is x2. Find the angular frequency and the amplitutde of the motion in terms of the given quantities. The answers given in the back of the book are: angular frequency = [(v2^2 - v1^2)/(x1^2 - x2^2)]^(1/2) and amplitude = [(x1^2*v2^2 - x2^2*v1^2)/(v2^2 - v1^2)]^(1/2).

    I've tried approaching this several ways, but can't seem to get my answers to agree. Should I treat the solution as a sum of two separate solutions with two different amplittudes since this is a linear differential equation, such that x(t) = x1(t) + x2(t) = A*cos(w*t + phi) + B*cos(w*t + phi) and similarly with v(t) = v1(t) + v2(t) = -w*A*sin(w*t + phi) - w*B*sin(w*t + phi). Or is there an easier way? I'm not sure what is appropriate to generalize and what is not. For example, I've taken the phase angle phi to be zero which I believe is acceptable. Any other suggestions? Thanks.
  2. jcsd
  3. Oct 8, 2005 #2


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    You only need two arbitrary constants in your general solution. You can use either

    [tex]A \cos(\omega t + \phi)[/tex]
    [tex]A \cos \omega t + B \sin \omega t[/tex].
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