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**1. Homework Statement**

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. Homework Equations**

[tex]p=\sqrt{2(E-V)}\ \ \ \ \ \ I=\frac{1}{\pi}\int p\ dq[/tex]

**3. The Attempt at a Solution**

I have:

[tex]H=\frac{p^2}{2}+\frac{1}{2}\omega^2_0(|q|-d)^2=E[/tex]

and the limits of the integral:

[tex]q^E_\pm=\pm\left(\frac{\sqrt{2E}}{\omega_0}+d\right)[/tex]

Then

[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

\right)[/tex]

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

\right)[/tex]

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.