Transform this to an easier integral?

In summary: I substitute J with K and solve for K, I get a frequency of oscillation of 2.7167...which is very different from what i get with J. What am I doing wrong?
  • #1
JohanL
158
0
I have three questions about a problem in mechanics


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

[tex]

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

[/tex]

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

[tex]

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

[/tex]

Under what conditions can action-angle variables be used?

3.

If you have an integral where the integrand is

[tex]

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

[/tex]

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.
 
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  • #2
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 [itex] x^{-3} [/itex] doesn't look good at all,u can't even expand it in harmonics...

Daniel.
 
  • #3
1. Can be solved analytically:
[tex]m\ddot{x}+\frac{ax_{0}^{2}}{x^{3}}=0, x(0)=\hat{x}_{0},\dot{x}(0)=v_{0}[/tex]
Multiply with [tex]\dot{x}[/tex] and integrate:
[tex]\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}}[/tex]
Multiply the equation with [tex]x^{2}=2y(t)[/tex]:
[tex]\frac{m}{2}(\frac{dy}{dt})^{2}=By+C, C=\frac{ax_{0}^{2}}{\hat{x}_{0}^{2}}[/tex]
Or semi-finally:
[tex]\frac{dy}{dt}=\pm\sqrt{Ay+D}[/tex]
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..:wink:
 
  • #4
yeah that's right...i must have been tired yesterday. :smile:

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

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

With

[tex]H = \frac {p^2} {2m} + V = \alpha[/tex]

the constant action variable J is give by

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

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

then

[tex]
\alpha = H = H(J)
[/tex]

and

the frequency of oscillation is

[tex]
\frac {dH} {dJ}
[/tex]

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
 
  • #5
Your integral can be put under the form

[tex]\sqrt{2m}\int\sqrt{\alpha-\frac{a}{\sin^{2}\frac{x}{x_{0}}}} \ dx [/tex]

which by a redefinition of constants can be proportional to

[tex] \int \sqrt{a-\frac{b}{\sin^{2}\frac{x}{c}}} \ dx [/tex]

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

Daniel.
 

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  • #6
Thanks. But i need to solve for [tex]\alpha[/tex] when i have solved the integral.

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

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

then

[tex]
\alpha = H = H(J)
[/tex]

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...
 

1. What does it mean to "transform a integral"?

Transforming an integral refers to the process of changing the form of an integral equation to make it easier to solve or evaluate. This can involve using different techniques such as substitution, integration by parts, or trigonometric identities.

2. Why would someone want to transform an integral?

Sometimes an integral may be too complex to solve in its original form, or it may be difficult to find a closed form solution. Transforming the integral can make it easier to evaluate or solve, allowing for a more efficient and accurate solution.

3. What are some common techniques used to transform integrals?

Some common techniques for transforming integrals include substitution, integration by parts, partial fractions, and trigonometric identities. These techniques can simplify the integral or make it more manageable to solve.

4. Are there any rules or guidelines for transforming integrals?

There are no specific rules for transforming integrals, as it depends on the specific integral and the desired outcome. However, it is important to consider the properties of the integral, such as the limits of integration, to ensure that the transformed integral is still equivalent to the original one.

5. How do I know when I have successfully transformed an integral?

When you have successfully transformed an integral, the resulting integral should be easier to solve or evaluate than the original one. It should also have a simpler form, with fewer terms or a more recognizable function. Additionally, you should be able to use basic integration rules to solve the transformed integral.

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