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Solving a Second-Order Differential Equation

  1. Nov 17, 2006 #1
    I've been trying to derive the period of a physical pendulum, and I ended up with a differential equation that boils down to this:

    [tex]
    \frac{d^2 \theta}{dt^2} = C \sin{\theta}
    [/tex]

    where C is some constant.

    With no experience in differential equations, I have no idea how to solve this. I can't find anything about how to solve this type of differential equation, so if someone could point me in the right direction, that would be helpful.

    Thanks!
     
  2. jcsd
  3. Nov 17, 2006 #2

    StatusX

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    It can't be solved nicely, which is why we usually assume the angle is small and make the approximation [itex] \sin \theta \approx \theta[/itex]. If you want, there's a way to reduce it to finding the integral of something like [itex]1/\sqrt{a-\cos(\theta)}[/tex], but that's not a nice integral to do (ie, you need special functions).
     
  4. Nov 17, 2006 #3

    HallsofIvy

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    That is an extremly non-linear equation and cannot be solved in any simple form. As StatusX said, for smal values of the angle, you can make the approximation [itex]sin(\theta)= \theta[/itex] so the equation becomes [itex]d^2\theta/dt^2= C\theta[/itex].

    Another technique is called "quadrature". Since t does not appear explictly in the equation, let [itex]\omega= d\theta/dt[/itex]. Then apply the chain rule: [itex]d^2\theta/dt^2= d\omega/dt= d\omega/d\theta d\theta/dt= \omega d\omega/dt= Csin \theta[/itex]. That's a separable first order differential equation which can be written as [itex]\omega d\omega= Csin(\theta)d\theta[/itex] and can be integrated directly:
    [itex](1/2) \omega^2= -Ccos(\theta)+ D[/itex]. The rub is when you replace [itex]\omega[/itex] by [itex]d/t\eta/dt[/itex] and solve for [itex]d\theta/dt[/itex]: [itex] d\theta/dt= \sqrt{D -2C cos\theta}[/itex] so
    [itex]\frac{d\theta}{\sqrt{D- 2C cos(\theta}}= dt[/itex] and the left side cannot be integrated in terms of any elementary function. (That's called an "elliptic" integral. I've seen whole book cases of tables of values.)
     
  5. Nov 18, 2006 #4
    Does this mean that it is impossible to generate a theoretical formulation for pendulums that includes large-angle values?

    I suspected that I would have to make a small-angle assumption. If I did have [itex]d^2\theta/dt^2= C\theta[/itex], how would I solve that? (Sorry for my lack of experience - once again, I just need to be pointed in the right diretcion.)
     
  6. Nov 18, 2006 #5

    HallsofIvy

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    For small angles, yes. Of course, in order to get periodic solutions, C must be negative.
     
  7. Nov 18, 2006 #6

    daniel_i_l

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    what function equals itself times a constant when differentiating twice?
    (HINT: whats the derivative of e^x ?)
     
  8. Nov 18, 2006 #7

    robphy

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  9. Nov 18, 2006 #8

    arildno

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    You can get an approximate solution to this, to whichever degree of accuracy you desire, by perturbation theory.

    Alternatively, you may use numerics
     
  10. Nov 18, 2006 #9
    I don't understand - how does that help me solve the differential equation?

    Thanks, everyone! This page answers all of my questions.
     
  11. Nov 18, 2006 #10

    HallsofIvy

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    I can think of 4 such (independent) functions. That's why I said " Of course, in order to get periodic solutions, C must be negative."
     
  12. Nov 18, 2006 #11
    just a curious question, has there ever been a series solution to the exact motion of undamped pendulum?
     
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