• Support PF! Buy your school textbooks, materials and every day products Here!

A particle attached to a spring

  • Thread starter nowits
  • Start date
18
0
This is a Hamilton-Jacobi problem.

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.
 

Answers and Replies

Related Threads for: A particle attached to a spring

Replies
1
Views
5K
Replies
13
Views
3K
Replies
1
Views
2K
Replies
3
Views
2K
Replies
3
Views
3K
Replies
1
Views
4K
Replies
2
Views
2K
Top