A particle attached to a spring

[nowits]

This is a Hamilton-Jacobi problem.

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:
$$V(q)=\left\lbrace \begin{array}{cc} \frac{1}{2}\omega^2 _0(|q|-d)^2 & when |q|\geq d \\ 0 & when |q|<d \end{array}$$

Solve the energy E as a function of the action variable I.

Homework Equations

$$p=\sqrt{2(E-V)}\ \ \ \ \ \ I=\frac{1}{\pi}\int p\ dq$$

The Attempt at a Solution

I have:
$$H=\frac{p^2}{2}+\frac{1}{2}\omega^2_0(|q|-d)^2=E$$
and the limits of the integral:
$$q^E_\pm=\pm\left(\frac{\sqrt{2E}}{\omega_0}+d\right)$$
Then
$$I=\frac{1}{\pi}\int^{q^E_+}_{q^E_-} \sqrt{2(E-V)}\ dq=\frac{1}{\pi}\int^{q^E_+}_{q^E_-} \sqrt{2E-\omega^2 _0(|q|-d)^2}\ dq$$

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

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

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.

 

[Jhero]

Thank you for posting your Hamilton-Jacobi problem. It seems like you have made good progress in solving it so far. I can offer some suggestions to help you continue towards finding a solution.

Firstly, it is helpful to note that the potential V(q) is piecewise-defined. This means that we can divide the integral into three parts, corresponding to the three regions where V is different. These regions are q < -d, -d < q < d, and q > d. We can then integrate each part separately and combine the results.

Secondly, when integrating over the region -d < q < d, it is important to note that the potential is zero in this region. This means that the integral in this region is simply the integral of a constant, which is just the constant times the width of the region. In this case, the width of the region is 2d, so the integral is 2d times the constant.

Finally, when integrating over the region q > d, we can make a substitution to simplify the integral. Let u = q - d, so that du = dq. This transforms the integral into one that is easier to solve.

Using these suggestions, you should be able to continue towards finding a solution for I in terms of the energy E. I hope this helps, and good luck with your problem!