Transform this to an easier integral?

[JohanL]

I have three questions about a problem in mechanics


1. If you have found the equation of motion for a system

$$m\ddot{x} + \frac {2ax_0^2} {x^3} = 0$$

where a and x0 are constants.
and you want to find the frequency of oscillations which ansatz should you make. You can't use x = A*exp(iwt)...i think.

2. If a particle of mass m moves in one dimension subject to the potential

$$V = \frac {a} {[sin(x/x_0)]^2}$$

Under what conditions can action-angle variables be used?

3.

If you have an integral where the integrand is

$$\sqrt{2m*(b - \frac{a}{[sin(x/x_0)]^2})}$$

how could you transform this to an easier integral?
When i integrate over a complete period ,0 pi, with MATLAB i get an infinite answer, of course. I guess that question number 2 could help me with this...but I am not sure.

Any ideas?

Thank you.

 

[dextercioby]

For the first,your equation is awfully nonlinear,therefore u can't use solutions which span the space of solution for a linear one...That $x^{-3}$ doesn't look good at all,u can't even expand it in harmonics...

Daniel.

 

[arildno]

1. Can be solved analytically:
$$m\ddot{x}+\frac{ax_{0}^{2}}{x^{3}}=0, x(0)=\hat{x}_{0},\dot{x}(0)=v_{0}$$
Multiply with $$\dot{x}$$ and integrate:
$$\frac{m}{2}\dot{x}^{2}-\frac{ax_{0}^{2}}{x^{2}}=\frac{B}{2},\frac{B}{2}=\frac{m}{2}v_{0}^{2}-\frac{ax_{0}^{2}}{\hat{x}_{0}^{2}}$$
Multiply the equation with $$x^{2}=2y(t)$$:
$$\frac{m}{2}(\frac{dy}{dt})^{2}=By+C, C=\frac{ax_{0}^{2}}{\hat{x}_{0}^{2}}$$
Or semi-finally:
$$\frac{dy}{dt}=\pm\sqrt{Ay+D}$$
for appropriate constants A,D.
This can be worked with more, if you like, but I don't think solving the damn thing was the question..

 

[JohanL]

yeah that's right...i must have been tired yesterday.

But question 1 was really only to check the answer for small oscillations.

The problem is to find the frequency of oscillations with the action-angle variables method for a particle of mass m moving in the potential

$$V = \frac {a} {[sin(x/x_0)]^2}$$

With

$$H = \frac {p^2} {2m} + V = \alpha$$

the constant action variable J is give by

$$J = \int p dq = \int \sqrt{2m*(\alpha - \frac{a}{[sin(x/x_0)]^2})} dx$$

where the integration is to be carried over a complete period.

then

$$\alpha = H = H(J)$$

and

the frequency of oscillation is

$$\frac {dH} {dJ}$$

So as soon i know that integral the problem is solved.

the problem has a hint:
The integral for J can be evaluated by manipulating the integrand so that the square root appears in the denominator.

But i don't understand how you should do this

 

[dextercioby]

Your integral can be put under the form

$$\sqrt{2m}\int\sqrt{\alpha-\frac{a}{\sin^{2}\frac{x}{x_{0}}}} \ dx$$

which by a redefinition of constants can be proportional to

$$\int \sqrt{a-\frac{b}{\sin^{2}\frac{x}{c}}} \ dx$$

which is evaluated by Mathematica to be (see attached thumbnail).

Daniel.

 

[JohanL]

Thanks. But i need to solve for $$\alpha$$ when i have solved the integral.

$$J = \int p dq = \int \sqrt{2m*(\alpha - \frac{a}{[sin(x/x_0)]^2})} dx$$

where the integration is to be carried over a complete period.

then

$$\alpha = H = H(J)$$

and that will not be easy with the expression mathematica gives.
There must be some clever substitution.

the problem has a hint:
The integral for J can be evaluated by manipulating the integrand so that the square root appears in the denominator.

I only get strange results tho...