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Integrating a second derivative-involving solution for Simple Harmonic Motion

  1. Aug 27, 2005 #1
    Hello, i am now in the process of integrating m(d^2x/dt^2)=-kx which i know i will have to do twice in order to obtain the general solution to simple harmonic motion, x= Acos(wt+c) c=phi

    but i'm just having problems with the second derivative of acceleration (d^2*x/dt^2) when it comes to integrating, I tried separating them into dv/dt and then dv/dx and dx/dt, so i obtain v, etc... but this complicates things even more for me!! It would be so great if someone could give me a hint as to where i can start in integrating with this second derivative....
    Thank you very much for your time.
  2. jcsd
  3. Aug 27, 2005 #2


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    The 2nd differential is often rearranged like this,

    [tex]\frac{d^2x}{dt^2}= v\frac{dv}{dx}[/tex]

    and you get that this way,

    [tex]\frac{d^2x}{dt^2}=\frac{d}{dt}(dx/dt)=\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt}= \frac{dv}{dx}v = v\frac{dv}{dx}[/tex]

    substitute for [tex]\frac{d^2x}{dt^2}[/tex] to [tex]v\frac{dv}{dx}[/tex] then do the integration.
  4. Aug 28, 2005 #3
    Yes, i tried doing that, but then eventually i get stuck with
    (v^2)/2= (-k/m)(x^2)/2 + C
    Would i have to integrate again to obtain the general solution of
    x=Acos(wt+c)? What can i do with the v^2 and the x^2, and the negative k/m?
  5. Aug 28, 2005 #4


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    Your halfway there.
    From the initial conditions of the problem, work out a value for C.
    rewrite the eqn as v = whatever ...
    convert v to dx/dt and integrate agan.
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