Conservative systems

**mukmar** · March 10th 2011, 09:19 PM

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!

**chisigma** · March 11th 2011, 01:00 AM

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

$\chi$ $\sigma$

**mukmar** · March 11th 2011, 02:08 AM

Originally Posted by chisigma

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.

**chisigma** · March 11th 2011, 02:21 AM

Originally Posted by mukmar

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...

**mukmar** · March 11th 2011, 02:26 AM

Originally Posted by chisigma

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?

**zzzoak** · March 11th 2011, 10:34 AM

To get the frequency w you must get the equation

$ y''=-w^2 y \ . $

To get this make substitution

$ y=x-x_e $

and assume

$ y<<a, \ y<<x_e \ . $

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