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Setting up SHM force equation

  1. Sep 11, 2006 #1
    Say you had a spring hanging vertically-- spring constant k. A mass m is attached to it and it's at its equilibrium position, b.
    Say, I decide to extend it (downward) such that its total extension distance from how it would be normally (without the mass) is x. Say I take the upward direction as positive. The moment I let go of it, the sum of the forces should be:
    [tex]kx - mg = m \dfrac{d^{2}x}{dt^{2}}[/tex]
    Now, at some point I'd be asked a question (on a test) that says "Prove SHM".
    To do so I have to get my force equation to the form [tex]\dfrac{d^{2}x}{dt^{2}} = - \omega^{2} (x-b)[/tex], yet I can only do this if I had gotten:
    [tex]mg - kx = m \dfrac{d^{2}x}{dt^{2}}[/tex]
    As my force equation.
    Can someone tell me what I did wrong in forming my original force equation? I'd like to clear this up before my exam tomorrow.
    Thanks alot.
     
  2. jcsd
  3. Sep 11, 2006 #2

    Doc Al

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    Staff: Mentor

    Even though you took upward as positive, you still defined your "x" as being downward. You have the wrong sign for [tex] \dfrac{d^{2}x}{dt^{2}}[/tex]. An upward acceleration would be [tex]- \dfrac{d^{2}x}{dt^{2}}[/tex].
     
  4. Sep 11, 2006 #3
    Exactly why is [tex]\dfrac{d^{2}x}{dt^{2}}[/tex] negative? Can you elaborate?
    Thanks for the reply.
     
  5. Sep 11, 2006 #4

    Doc Al

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    Staff: Mentor

    When [tex]\dfrac{d^{2}x}{dt^{2}}[/tex] is positive it means that the acceleration is in the +x direction. You defined x to be + downward. (Otherwise the restoring force would be -kx, not kx.)

    You can stick with your definition of x, and call down to be postive. In which case the forces would be -kx & mg; set that sum equal to [tex]m \dfrac{d^{2}x}{dt^{2}}[/tex].
     
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