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Interesting oscillator potential

  1. Jan 22, 2007 #1

    Mentz114

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    Hi Everyone,

    I think I've solved the equations of motion for a particle in this one-dimensional potential -

    V(x) = -k/(a-x) - k/(a+x) with |x| < a, a is real +ve, x is real.
    K is a constant of suitable dimension.
    It's a charge between two like charges with separation 2*a.

    I start with the force equation

    m*x'' = -4*a*k*x / (a^2 - x^2)^2
    (Using ' to indicate differentiation wrt time)

    I can integrate this to get

    m*x' = sqrt(2*k*a)/sqrt(a^2 - x^2)

    This ought to be textbook example, can anyone point me to an authoritative solution or come up with x ?
     
    Last edited: Jan 22, 2007
  2. jcsd
  3. Jan 23, 2007 #2

    dextercioby

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    I get the separable ODE
    [tex] \frac{dx}{dt}=\sqrt{\frac{4ak}{m}}\frac{1}{\sqrt{a^{2}-x^{2}}} [/tex]

    Daniel.
     
  4. Jan 23, 2007 #3

    Mentz114

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    Thanks, Dextercioby.

    We agree to a factor of sqrt(2). With some help I got

    [tex]t + C =\frac{1}{2}\sqrt{ 2ak}( x\sqrt{a^2 - x^2} + arcsin( x) )[/tex]

    for the relation between x and t. Doing some numerical work this looks
    like a straight line, and not an oscillation.
     
    Last edited: Jan 23, 2007
  5. Jan 23, 2007 #4
    I would suggest that you try your numerical calculation again. The arcsin(s) term would indicate that there should definitely be oscillations.
     
  6. Jan 23, 2007 #5

    Mentz114

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    I agree. Obviously, without the x*sqrt(a^2-x^2) term it's a sine wave.
    I start with x=a/2 at t=0 which gives me k. The calculation gives the correct
    x at t=0, then x falls linearly as t increases. Could be something wrong with some scaling.
     
  7. Jan 23, 2007 #6

    Mentz114

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    First post now Latexed

    I think I've solved the equations of motion for a particle in this one-dimensional potential -

    [tex]V(x)=\frac{-k}{a-x}-\frac{k}{a+x}[/tex]
    with |x| < a, a is real +ve, x is real.
    K is a constant of suitable dimension.
    It's a charge between two like charges with separation 2a.

    I start with the force equation

    [tex]m\frac{d^2x}{dt^2} = -4akx / (a^2 - x^2)^2}[/tex]

    I can integrate this to get

    [tex]m\frac{dx}{dt} = \sqrt{2ka}/\sqrt{a^2 - x^2}[/tex]

    This ought to be textbook example, can anyone point me to an authoritative solution or come up with x ?
     
    Last edited: Jan 23, 2007
  8. Aug 6, 2008 #7

    Mentz114

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    Sorry to necro-post, I made are errors in the above which I can now correct.

    This equation is correct ( thanks Dan )

    [tex] \frac{dx}{dt}=\sqrt{\frac{4ak}{m}}\frac{1}{\sqrt{a ^2-x^2}} [/tex]

    and has solution

    [tex]t + C = \frac{1}{2}\left(\frac{m}{4ak}\right)^\frac{1}{2}\left[x(a^2-x^2)^\frac{1}{2} - a^2cosh^{-1}(\frac{x}{a})\right][/tex]

    If we choose coords so when x=0, t=0 then

    [tex]C = -\frac{a^2}{2}\left(\frac{m}{4ak}\right)^\frac{1}{2}[/tex]

    Not simple harmonic motion. How do I calculate ( guess ?) the period ?
     
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