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Homework Help: 2 coupled EDOs

  1. Jan 17, 2007 #1

    quasar987

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    I must solve the following two coupled EDOs in the context of a Lagrangian mechanics problem (a rigid pendulum of lenght l attached to a mass sliding w/o friction on the x axis). The problem statement does not mention that we can make small angle approximation. It says "find the equations of motion and solve them for the following initial conditions:...". Is this feasable?

    [tex](m_1+m_2)\ddot{x}+m_2l\ddot{\theta}\cos(\theta)-m_2l\dot{\theta}^2\sin(\theta)=0[/tex]

    [tex]l\ddot{\theta}+\ddot{x}\cos(\theta)+g\sin(\theta)=0[/tex]

    They can be uncoupled but there remains a second order non-linear ODE to solve.

    Is this doable analytically?

    And an annexed question (perhaps this one is more of a physical nature): why can we say that [itex]\dot{\theta}\approx 0[/itex] in the small angle approximation? The angle can be small and nevertheless vary furiously fast. What indicates that if theta is small, the so is its derivative?
     
    Last edited: Jan 17, 2007
  2. jcsd
  3. Jan 17, 2007 #2

    quasar987

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    Same question but with

    [tex]\ddot{y}+ay+b\cos(y)=0[/tex]

    Solvable? (a,b are constants)
     
  4. Jan 18, 2007 #3

    dextercioby

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    The one in the second post is doable:

    [tex] t+C_{2}=\int \frac{dy}{\sqrt{2C_{1}-ay^{2}-2b\sin y}} [/tex]

    It remains to compute the integral.

    Mathematica returns

    [tex] \int \frac{dy}{\sqrt{2C_{1}-ay^{2}-2b\sin y}} =-\frac{2}{\sqrt{2C_{1}-ay^{2}-2b\sin y}} F\left[\frac{1}{4}\left(\pi -2y\right), \frac{4b}{-2C_{1}+ay^2 +2b}\right]\sqrt{\frac{2C_{1}-ay^{2}-2b\sin y}{2C_{1}-ay^{2}-2b}} [/tex]

    Daniel.
     
    Last edited: Jan 18, 2007
  5. Jan 18, 2007 #4

    cva

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    So how do you inverse the function t?
     
  6. Jan 18, 2007 #5

    dextercioby

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    By an appropriate use of Jacobi's elliptic functions. In this case i think sine amplitudinis.

    Daniel.
     
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