What Is the Period of This Particle's Motion in the Given Oscillator Potential?

[Mentz114]

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 ?

 

[dextercioby]

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

Daniel.

 

[Mentz114]

Thanks, Dextercioby.

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

$$t + C =\frac{1}{2}\sqrt{ 2ak}( x\sqrt{a^2 - x^2} + arcsin( x) )$$

for the relation between x and t. Doing some numerical work this looks
like a straight line, and not an oscillation.

 

[StatMechGuy]

I would suggest that you try your numerical calculation again. The arcsin(s) term would indicate that there should definitely be oscillations.

 

[Mentz114]

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.

 

[Mentz114]

First post now Latexed

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

$$V(x)=\frac{-k}{a-x}-\frac{k}{a+x}$$
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 2a.

I start with the force equation

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

I can integrate this to get

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

This ought to be textbook example, can anyone point me to an authoritative solution or come up with x ?

 

[Mentz114]

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

This equation is correct ( thanks Dan )

$$\frac{dx}{dt}=\sqrt{\frac{4ak}{m}}\frac{1}{\sqrt{a ^2-x^2}}$$

and has solution

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

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

$$C = -\frac{a^2}{2}\left(\frac{m}{4ak}\right)^\frac{1}{2}$$

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