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Simple Harmonic Motion Differential Equation

  1. Oct 1, 2008 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m moves in one dimension under the action of a force given by -kx where x is the displacement of the body at time t, and k is a positive constant. Using F=ma write down a differential equation for x, and verify that its solution is x=Acos([tex]\omega[/tex]t+[tex]\phi[/tex]), where [tex]\omega[/tex]2=k/m (omega squared, that is). If the body starts from rest at the point x=A at time t=0, find an expression for x at later times.

    2. Relevant equations



    3. The attempt at a solution
    I think the differential equation they're looking for is,
    a=-kx/m

    As a=d2x/dt2

    But from here I can't see where to go; integration of course leads to the wrong formula.
     
  2. jcsd
  3. Oct 1, 2008 #2

    Kurdt

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    You have the differential equation so just substitute the solution in and show that both sides are equal.
     
  4. Oct 1, 2008 #3
    So differentiate the solution given to us to get it in terms of acceleration and then just compare that with the a=-kx/m?
     
  5. Oct 1, 2008 #4

    Kurdt

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    Thats correct.
     
  6. Oct 1, 2008 #5
    I'm sorry I'm still sightly confused; I now have:

    [tex]a=-A\omega^{2}cos(\omega t+\phi)=\frac{-kx}{m}[/tex]

    Which implies that [tex]x=Acos(\omega t+\phi)[/tex] but doesn't really show why? Is this what you intended or am I missing something?
     
  7. Oct 1, 2008 #6

    Kurdt

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    You're also told what omega is in the question, and you haven't substituted for x on the right hand side.
     
  8. Oct 1, 2008 #7
    Oh of course it is! We're told what x is!
     
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