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Phase planes of Conservative systems (2nd order ODEs) watch

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    I'm unsure whether my phase plane diagram is correct or not for this question. Could someone verify?

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    Phase diagram for pendulum eq. from lec. notes:

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    My attempt:

    Determine potential energy \displaystyle U(x)= - \int -\omega^2 (x- \frac{1}{6}x^3) .dx =\frac{\omega^2}{24} x^2 (12-x^2) (up to a const.)

    Total energy eq. is hence \frac{1}{2}y^2 + \frac{\omega^2}{24}x^2(12-x^2) = h for some h = \text{const.} (this is the energy level) and y=\dot{x}

    Equilibrium points given by U'(x) = 0 \Rightarrow x-\frac{1}{6}x^3 = 0 thus x=-\sqrt{6}, 0, \sqrt{6}

    U''(x) = \omega^2(1-\frac{1}{2}x^2) which is >0 for x=0 and <0 for x= \pm \sqrt{6} so the former is a local min point, and the latter are the two max points.

    Sketching U(x) against x, and y against x below it, gives:

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    From my understanding you're just drawing the family of curves for different values of h. If we fix w, then you have the right idea for 0<h<(1/24)w^2(6)(12-(6)) = U(sqrt(6)), i.e ellipses at the centre.

    But for h>U(sqrt(6)), your reasoning is a bit off. For x-> +- infinity from the U(x) graph you can see that U-> -infinity, so as you have (1/2) y^2 + U = h this implies (1/2) y^2 = h - U > 0 (clear as we took h greater than max value of U). So as x-> +- infinity, U-> -infinity, so y^2 -> infinity. Of course this makes sense as we expect energy to be conserved so as U decreases KE increases.

    For h<0, there are some more considerations to make, but you can try reasoning like above. Also for 0<h<U(sqrt(6)), you should think about what happens when x as it gets bigger.

    You can try plotting these graphs for varying h.
    Hope that helps.
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    (Original post by MagneticFlux)
    From my understanding you're just drawing the family of curves for different values of h. If we fix w, then you have the right idea for 0<h<(1/24)w^2(6)(12-(6)) = U(sqrt(6)), i.e ellipses at the centre.

    But for h>U(sqrt(6)), your reasoning is a bit off. For x-> +- infinity from the U(x) graph you can see that U-> -infinity, so as you have (1/2) y^2 + U = h this implies (1/2) y^2 = h - U > 0 (clear as we took h greater than max value of U). So as x-> +- infinity, U-> -infinity, so y^2 -> infinity. Of course this makes sense as we expect energy to be conserved so as U decreases KE increases.

    For h<0, there are some more considerations to make, but you can try reasoning like above. Also for 0<h<U(sqrt(6)), you should think about what happens when x as it gets bigger.

    You can try plotting these graphs for varying h.
    Hope that helps.
    Thanks, that helped. Does this look about right now?

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    (Original post by RDKGames)
    Thanks, that helped. Does this look about right now?

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    Yeah that's what I had in mind, nicely done.
 
 
 
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