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Second order differential equation

  1. Dec 18, 2009 #1
    I've created a second order homogeneous equation from my orginal data
    m(d^2x/dt^2) + kx = 0
    how can I turn it into a expression of displacement relevant to time?
  2. jcsd
  3. Dec 18, 2009 #2
    [tex]\frac{d^2x}{dt^2} = \frac{-kx}{m}[/tex]


    [tex]\frac{d^2x}{dt^2} = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx} = \frac{d(\frac{1}{2}v^2)}{dv}\frac{dv}{dx} = \frac{d(\frac{1}{2}v^2)}{dx}=\frac{d}{dx}\left(\frac{1}{2}\left(\frac{dx}{dt}\right)^2\right)[/tex]

    By applying the chain rule twice.


    [tex]\frac{d}{dx}\left(\frac{1}{2}\left(\frac{dx}{dt}\right)^2\right) = \frac{-kx}{m}[/tex]

    Hopefully you can now solve this as a differential equation
  4. Dec 18, 2009 #3
    Isn't it simpler to solve the first one, it being standard? [tex]x=ACos(\sqrt{\frac{k}{m}}t+\phi_{0})[/tex]
  5. Dec 18, 2009 #4
    Haha yeah, I guess so. But since I haven't learnt the theory for second order differential equations, this is a way using highschool calculus.

    If you go through it, it comes out to be the same answer you have
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