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

Solution strange, differential equation

  1. May 28, 2008 #1
    Hi!
    I have big problem with solve this equation:
    [tex]m\frac{d^{2}x}{dt^{2}}+ksinx=0[/tex]
    I can't go ahead, because I don't know how solve this
    [tex]\frac{dx}{\sqrt{cosx}}=\sqrt{\frac{2k}{m}}dt[/tex]
    Phizyk
     
  2. jcsd
  3. May 28, 2008 #2

    HallsofIvy

    User Avatar
    Science Advisor

    Since the independent variable, t, does not appear explicitely in the equation, that is a candidate for "quadrature".

    Let v= dx/dt. Then d2x/dt2= dv/dt. But by the chain rule, dv/dt= (dv/dx)(dx/dt). And dx/dt= dv/dt, of course. That is d2x/dt2= (dx/dt)(dv/dt)= vdv/dx.

    Your differential equation can be reduced to vdv/dx= -ksin(x) which is "separable":
    mvdv= - k sin(x)dx. Integrating both sides, (m/2)v2= k cos(x)+ C. (That square is the reason for the name "quadrature".) Then v2= (2k/m) cos(x)+ C' or
    v= dx/dt= sqrt((2k/m) cos(x)+ C').

    I assume that, in order to get rid of that constant of integration, C', you must have some initial condition on dx/dt.
     
  4. May 28, 2008 #3
    Great. Thanks Hallsoflvy.
     
  5. May 28, 2008 #4
    But this equation [tex]\frac{dx}{dt}=\sqrt{\frac{2k}{m}cosx+C^{'}}[/tex] can I solve? Can I obtain x(t)? For t=0 x=0. It's a equation of motion.
     
    Last edited: May 28, 2008
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook