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Math Help - Quartic Oscillators

  1. #1
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    Quartic Oscillators

    Hi, here is a practice question. I'm not sure where the equation of motion comes in to play. Any help would be much appreciated. Thanks!

    The equation of motion for a particle of mass 1 in a quartic oscillator V(x)=(1/4)x^4 is ẍ+x^3=0. Suppose that the maximum amplitude of the oscillator is x(max). Find an expression for the time T that it takes to go from x=0 to x=x(max) and show that this time is inversely proportional to x(max).
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  2. #2
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    Using conservation of energy (let maximum of x be x_0) you must have:

    V(x_0) = \frac{1}{2}m v^2 + V(x)

    but since m=1 we have

    v^2 =2 \left( V(x_0) -  V(x)\right)

    \iff v^2 = \frac{1}{2} \left(x_0^4 - x^4 \right)

    so

    \frac{dx}{dt} = \sqrt{\frac{1}{2} \left(x_0^4 - x^4 \right)}

    which is of separable variable form so

    \int_{0}^{T} \frac{1}{\sqrt{2}} \, \mathrm{d}t = \int_{0}^{x_0} \frac{1}{\sqrt{x_0^4 - x^4}} \, \mathrm{d}x.

    Let x= x_0 u then \mathrm{d}x = x_0 \, \mathrm{d}u so this becomes

    \frac{T}{\sqrt{2}} = \int_{0}^{1} \frac{1}{x_0^2 \sqrt{1 - u^4}} x_0 \, \mathrm{d}u

    \iff T = \frac{\sqrt{2}}{x_0} \int_{0}^{1} \frac{1}{\sqrt{1 - u^4}} \, \mathrm{d}u.

    At this point we have answered the question as the integral has no dependence on x_0 so is simply a numerical constant and so we've shown that the time period is inversely proportional to the amplitude x_0.

    If you must know what the integral evaluates to:

    \int_{0}^{1} \frac{1}{\sqrt{1 - u^4}} \, \mathrm{d}u = \frac{1}{4} \beta \left(\frac{1}{4},\frac{1}{2} \right) = \frac{4 \left[\left(\frac{1}{4}\right)! \right]^2}{\sqrt{2 \pi}},

    so not anything nice in terms of elementary functions!
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  3. #3
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    then x'' + x^3 = 0 doesn't have any meaning here?
    x'' =v*dv/dx = - x ^3 then v dv = - x^3 dx
    so we can get the v= dx/dt
    is this right?
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  4. #4
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    k, maybe it's the same in the end
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  5. #5
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    Quote Originally Posted by zorop View Post
    then x'' + x^3 = 0 doesn't have any meaning here?
    x'' =v*dv/dx = - x ^3 then v dv = - x^3 dx
    so we can get the v= dx/dt
    is this right?
    The solution to \ddot{x} + x^3 = 0 gives the Conservation of Energy equation that the_doc used as his/her starting point:

    \ddot{x} + x^3 = 0

    \Rightarrow \frac{d^2 x}{dt^2} + x^3 = 0

    \Rightarrow \frac{d}{dx} \left[ \frac{v^2}{2}\right] + x^3 = 0

    \Rightarrow \frac{v^2}{2} + \frac{x^4}{4} = C.
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