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Math Help - Conservative systems

  1. #1
    Newbie mukmar's Avatar
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    Conservative systems

    A charged particle moves along a line joining two charged particles which are fixed at the locations x = 0 and x = a > 0. Assume the interactions between the moving charged particle and each of the fixed charges is repulsive and obeys an inverse square law. The differential equation
     x'' = -\frac{k_1}{(x - a)^2} + \frac{k_2}{x^2}
    for some constants k_1 > 0 and k_2 > 0 models the motion of the trajectory x(t) of the moving particle.

    Problem:
    1)Calculate the frequency of the small periodic oscillations near the equilibrium point.
    2) Suppose that the initial condition is x(0) = a/2 and x'(0) = 0 and suppose k_2 > k_1. Find a formula for the velocity of the particle as it passes through the equilibrium position.

    Attempt at solution:
    I know that the equilibrium point is:
    x_e = \frac{a}{1 - (\frac{k_1}{k_2})^{1/2}}
    Also that the potential V(x) = \frac{k_1}{x - a} - \frac{k_2}{x}

    I'm not really sure how to proceed for part 1 and for part 2, the only way I can imagine is to find an analytic solution to the given differential equation and plug in the initial conditions. I don't think that's the correct method to obtain the solution.

    Edit:
    I was able to solve part 2. I only need assistance with part 1 at the moment.


    Any hints or suggestions would be greatly appreciated!
    Last edited by mukmar; March 10th 2011 at 11:42 PM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    The DE is of the type x^{''}= f(x) and can be reduced to a first order DE as...

    \displaystyle x^{'}\ d x^{'}= f(x)\ dx = \{\frac{k_{2}}{x^{2}} - \frac{k_{1}}{(x-a)^{2}} \}\ dx (1)

    ... where the variables x^{'} and x are separated. Integrating (1) in standard fashion You obtain...

    \displaystyle x^{'}= \pm \sqrt{\frac{2 k_{1}}{x-a} - \frac{2 k_{2}}{x} + c_{1}} (2)

    Now the 'initial conditions' say that is x^{'} (\frac{a}{2})=0 so that is...

    \displaystyle c_{1}= \frac{4 (k_{1}+ k_{2})}{a} (3)

    ... and x^{'} as function of x is found. Now are You able to proceed?...

    Kind regards

    \chi \sigma
    Last edited by chisigma; March 11th 2011 at 05:24 AM. Reason: trivial error in (1)... sorry!...
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  3. #3
    Newbie mukmar's Avatar
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    Quote Originally Posted by chisigma View Post
    The DE is of the type x^{''}= f(x) and can be reduced to a first order DE as...

    \displaystyle x^{'}\ d x^{'}= f(x)\ dx = \{\frac{k_{2}}{x} - \frac{k_{1}}{(x-a)^{2}} \}\ dx (1)

    ... where the variables x^{'} and x are separated. Integrating (1) in standard fashion You obtain...

    \displaystyle x^{'}= \pm \sqrt{\frac{2 k_{1}}{x-a} - \frac{2 k_{2}}{x} + c_{1}} (2)

    Now the 'initial conditions' say that is x^{'} (\frac{a}{2})=0 so that is...

    \displaystyle c_{1}= \frac{4 (k_{1}+ k_{2})}{a} (3)

    ... and x^{'} as function of x is found. Now are You able to proceed?...

    Kind regards

    \chi \sigma
    How did you get the initial condition of x^{'} (\frac{a}{2})=0 from x(0) = a/2 and  x'(0) = 0?

    Secondly I'm still not sure how to do problem 1 from this.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by mukmar View Post
    How did you get the initial condition of x^{'} (\frac{a}{2})=0 from x(0) = a/2 and  x'(0) = 0?
    The first step of the solution find x' ad function of x. Now the 'initial conditions' say that when t=0 x'=0 and x=a/2, so that when x=a/2 is x'=0...
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  5. #5
    Newbie mukmar's Avatar
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    Quote Originally Posted by chisigma View Post
    The first step of the solution find x' ad function of x. Now the 'initial conditions' say that when t=0 x'=0 and x=a/2, so that when x=a/2 is x'=0...
    Oh okay, so with the notation x'(a/2) = 0, you mean that when x = a/2 that x' = 0. I was just confused because I thought you meant that when t = a/2, that x' = 0.

    Alright, I understand that, however I still don't understand how this helps me find the frequency of the small periodic oscillations near the equilibrium point?
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  6. #6
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    To get the frequency w you must get the equation

    <br />
y''=-w^2 y \ .<br />

    To get this make substitution

    <br />
y=x-x_e<br />

    and assume

    <br />
y<<a, \ y<<x_e \ . <br />
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