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Sinusoidal Diff EQ

  1. Jun 21, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\ddot{\Theta}=C \sin{\Theta}[/tex] where [tex]\Theta[/tex] is a function of time, and C is a constant.

    I ran into this on a Lagrangian Equation problem, and though the problem doesn't ask for the solution, I'm wondering how one would solve this Diff EQ. I'm afraid my intro to Diff EQ class was a couple years back and I'm coming up short with any ideas.

  2. jcsd
  3. Jun 21, 2008 #2
    what function do you differentiate twice to get a sin with a constant in front?
  4. Jun 21, 2008 #3


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    You can solve this by noting that the second order derivative of theta is purely a function of theta alone. This means that [tex] \theta'' = f(\theta)[/tex]

    By the chain rule:

    [tex]\frac{d^2 \theta}{dt^2} = \frac{d}{d\theta} \left( \frac{{\theta'}^2}{2} \right) = C \sin \theta[/tex]

    This is separable and solvable. Once you have solved for [itex]\theta ' [/itex], you can use solve for theta again using separation of variables
  5. Jul 5, 2008 #4
    Alright, so:

    [tex]\frac{d^2 \theta}{dt^2} = \frac{d}{d\theta} \left( \frac{{\theta'}^2}{2} \right) = C \sin \theta[/tex]

    [tex]\left( \frac{{\theta'}^2}{2} \right) = (C_{1} \cos \theta) + C_{2}[/tex]

    [tex] \theta = \int \sqrt{(C_{1} \cos \theta)+C_{2}} dt [/tex]

    How would I use separation of variables here?
  6. Jul 5, 2008 #5


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    By separating the variables before you integrate!
    [tex]\theta '= \sqrt{2C cos(\theta)+ C_2}[/tex]
    Where I have incorporated the "2C2[/sup]" into the unknown constant C2 (but not C1= 2C; C is a given number, not an unknown).

    [tex]d\theta= \sqrt{2C cos(\theta)+ C_2}dt[/tex]
    [tex]\int \frac{d\theta}{\sqrt{2C cos(\theta)+ C2}}= \int dt[/itex]

    That integral on the left looks to me like an "elliptic integral" which cannot be integrated in terms of elementary functions.
  7. Jul 5, 2008 #6


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    That's not what you should get. You should have [tex] \frac{1}{\sqrt{c_1 - 2 \cos \theta}} d\theta = dt[/tex]

    But I see a problem here. I don't know of a way to integrate the LHS.
  8. Jul 5, 2008 #7
    Thanks all. That helps. It's coming back, just slowly at times.
  9. Jul 5, 2008 #8


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    Of course, the traditional way to solve it is to make a small angle approximation, and forget about the cases where that isn't valid. :p
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