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A particle attached to a spring

  1. Apr 21, 2008 #1
    This is a Hamilton-Jacobi problem.

    1. The problem statement, all variables and given/known data
    A particle (mass m=1) moving on a line is attached to a light spring (length d, another end attached to the origin). The potential is:
    [tex]V(q)=\left\lbrace \begin{array}{cc} \frac{1}{2}\omega^2 _0(|q|-d)^2 & when |q|\geq d \\ 0 & when |q|<d \end{array}[/tex]

    Solve the energy E as a function of the action variable I.
    2. Relevant equations
    [tex]p=\sqrt{2(E-V)}\ \ \ \ \ \ I=\frac{1}{\pi}\int p\ dq[/tex]

    3. The attempt at a solution
    I have:
    and the limits of the integral:
    [tex]I=\frac{1}{\pi}\int^{q^E_+}_{q^E_-} \sqrt{2(E-V)}\ dq=\frac{1}{\pi}\int^{q^E_+}_{q^E_-} \sqrt{2E-\omega^2 _0(|q|-d)^2}\ dq[/tex]

    As V=0 between -d and d, and as |q|=-q when q<0 and |q|=q when q>0 it transforms to:
    [tex]I=\frac{1}{\pi}\left ( \int^{-d}_{q^E_-} \sqrt{2E-\omega^2 _0(-q-d)^2}\ dq
    + \int^{d}_{-d} \sqrt{2E}\ dq +
    \int^{q^E_+}_{d} \sqrt{2E-\omega^2 _0(q-d)^2}\ dq

    The first and the third integral are the same, so I get
    [tex]I=\frac{2}{\pi}\left ( \int^{d}_{0} \sqrt{2E}\ dq +
    \int^{q^E_+}_{d} \sqrt{2E-\omega^2 _0(q-d)^2}\ dq

    My main problem is I can't figure out how to transform that latter integral into anything sensible. I've tried some tricks, but it always turns out to be zero, and the variable I consist only of what the first integral gives out. But that can't be, as I know I is required to have omega in it.
  2. jcsd
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