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Transform this to an easier integral?

  1. May 5, 2005 #1
    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 cant 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 infinte answer, of course. I guess that question number 2 could help me with this...but im not sure.

    Any ideas?

    Thank you.
     
  2. jcsd
  3. May 5, 2005 #2

    dextercioby

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    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.
     
  4. May 5, 2005 #3

    arildno

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    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:
     
  5. May 6, 2005 #4
    yeah thats 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 dont understand how you should do this
     
  6. May 6, 2005 #5

    dextercioby

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    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.
     

    Attached Files:

  7. May 14, 2005 #6
    Thanks. But i need to solve for [tex]\alpha[/tex] when i have solved the integral.

    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...
     
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