1. ## Conservative systems

A charged particle moves along a line joining two charged particles which are fixed at the locations $\displaystyle x = 0$ and $\displaystyle x = a > 0$. 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
$\displaystyle x'' = -\frac{k_1}{(x - a)^2} + \frac{k_2}{x^2}$
for some constants $\displaystyle k_1 > 0$ and $\displaystyle k_2 > 0$ models the motion of the trajectory $\displaystyle x(t)$ 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 $\displaystyle x(0) = a/2$ and $\displaystyle x'(0) = 0$ and suppose $\displaystyle k_2 > k_1$. 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:
$\displaystyle x_e = \frac{a}{1 - (\frac{k_1}{k_2})^{1/2}}$
Also that the potential $\displaystyle V(x) = \frac{k_1}{x - a} - \frac{k_2}{x}$

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!

2. The DE is of the type $\displaystyle x^{''}= f(x)$ and can be reduced to a first order DE as...

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

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. Originally Posted by chisigma
The DE is of the type $\displaystyle x^{''}= f(x)$ and can be reduced to a first order DE as...

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

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

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

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

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

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

Kind regards

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

Secondly I'm still not sure how to do problem 1 from this.

4. Originally Posted by mukmar
How did you get the initial condition of $\displaystyle x^{'} (\frac{a}{2})=0$ from $\displaystyle x(0) = a/2$ and $\displaystyle x'(0) = 0$?
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...

5. 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?

6. To get the frequency w you must get the equation

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

To get this make substitution

$\displaystyle y=x-x_e$

and assume

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