Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]

    Now,

    [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.

    So,

    [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
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook