# Homework Help: Time period of osciilation

1. Jul 13, 2013

### Saitama

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
I have tried to attack this problem with two methods but no luck. I always end up with something gibberish. :/

Method 1:
Considering the forces on any one charge, the net force on a charge is $$-\frac{kq^2(3\sqrt{2}-4)}{4\sqrt{2}r^2}$$
hence,
$$ma=-\frac{kq^2(3\sqrt{2}-4)}{4\sqrt{2}r^2}$$
where $a=\frac{d^2r}{dt^2}$, this leads to a differential equation of the form
$$\frac{d^2r}{dt^2}=-\frac{k}{r^2}$$
where k is some constant. Plugging this in wolfram alpha gives me nothing. :(

Method 2:
Throughout the motion, energy and angular momentum is conserved. The potential energy of charges at any instant is
$$-\frac{kq^2(3\sqrt{2}-4)}{\sqrt{2}r}$$
Let the velocity of charge along the radius vector be $v_r$ and perpendicular to radius vector is $v_p$. Hence, the total kinetic energy is
$$4\left(\frac{1}{2}m(v_r^2+v_p^2)\right)$$
The total energy of the system at any time t is
$$E=-\frac{kq^2(3\sqrt{2}-4)}{\sqrt{2}r}+2m(v_r^2+v_p^2)$$
I don't really know what to do with this. Usually, in SHM problems, I would differentiate the expression for energy wrt time and set the derivative to zero but doing so here doesn't seem helpful. I can make one from equation from the fact that $dL/dt=0$ where L is the angular momentum of the system. L at any time is $4mv_pr$. Even if I differentiate the expression for L, I will end up with an equation which doesn't look useful.

Any help is appreciated. Thanks!

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2. Jul 13, 2013

### Staff: Mentor

How did you get your equation for the force?

The motion in a 1/r^2-potential is called Kepler problem (he studied it for gravity), I guess you can use the formulas for that.

It is amazing that such a system can exist (even if it is unstable).

3. Jul 13, 2013

### rude man

My inclination is to solve the problem assuming L = constant = L0. The fact that the oscillations in L are not at all defined in terms of L(t) leads me to suspect that the net centripetal force on each +q charge is independent of L. That problem is pretty straight-forward.

However, I'm too lazy to prove that centripetal force is not a function of L.

4. Jul 13, 2013

### Saitama

That's the net force in the radial direction on a single charge. Do I have to consider the centrifugal force too? :uhh:

5. Jul 13, 2013

### WannabeNewton

What happens to the radial velocity at the instants of maximum and minimum elongation of the sides of the square?

6. Jul 13, 2013

### Saitama

It becomes zero.

7. Jul 13, 2013

### ehild

dr/dt = 0 when the sides are Lo or 1/4 Lo. Lo is given in the problem. Use conservation or energy and angular momentum for these configurations.

ehild

8. Jul 13, 2013

### WannabeNewton

Ok now use conservation law(s) for those instants. See where that takes you.

EDIT: darn it ehild. >.>

9. Jul 13, 2013

### ehild

You were not present yet when I started to solve the problem and post. You were quicker, do it further.

ehild

10. Jul 13, 2013

### WannabeNewton

It was a joke xD <3

Last edited: Jul 13, 2013
11. Jul 13, 2013

### Saitama

$$E_i=-\frac{kq^2(3\sqrt{2}-4)}{L_0/4}+2mv_{pi}^2$$
$$E_f=-\frac{kq^2(3\sqrt{2}-4)}{L_0}+2mv_{pf}^2$$
Equating them
$$\frac{3kq^2(3\sqrt{2}-4)}{L_0}=2m(v_{pi}^2-v_{pf}^2)$$
Using conservation of angular momentum at these two instants,
$$4mv_{pi}\frac{L_0}{4\sqrt{2}}=4mv_{pf}\frac{L_0}{\sqrt{2}} \Rightarrow v_{pi}=4v_{pf}$$

Substituting the relation in energy in energy equation and solving for $v_{pf}^2$,
$$v_{pf}^2=\frac{kq^2(3\sqrt{2}-4)}{10mL_0}$$

What should I do with this?

12. Jul 14, 2013

### ehild

How are vpi and vpf related to the angular speed and Lo?

ehild

Last edited: Jul 14, 2013
13. Jul 14, 2013

### Saitama

I still don't get it. The problem asks the time period of oscillation but what I have is the square of final velocity.

$v_p$ is the component of velocity perpendicular to the radius vector. $v_{pi}$ is the initial and $v_{pf}$ is the final.

$v_{pf}=\omega L_0/\sqrt{2}$?

Last edited: Jul 14, 2013
14. Jul 14, 2013

### ehild

Yes, vpffL0/√2. So you know the maximum and minimum ω, but I have to think about the time period...

ehild

Last edited: Jul 14, 2013
15. Jul 14, 2013

### TSny

I agree with your expression for the force.

In polar coordinates, the radial acceleration is not $\ddot{r}$ but rather $\ddot{r}-r\dot{\theta}^2$.

The net force on a particle is an inverse square force. So, each particle travels an ellipse. I would follow mfb here. One of Kepler's laws should get you the answer straight away.

16. Jul 14, 2013

### WannabeNewton

TSny mentioned Kepler's 3rd law, so you could roll with that and get the answer in a direct fashion analogous to the case for gravitational interactions. Since you already got the final angular velocity $\omega_{f}$, you could also use $r^{2}_{f}\omega_{f} = \frac{2\pi}{P}ab$ where $P$ is the period and $a,b$ are the semimajor and semiminor axes of the elliptical orbit (which you are given in the problem statement) of any one of the charges respectively if you want: http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion#Second_law

17. Jul 14, 2013

### Saitama

Is there no other way?
Kepler's third law?

18. Jul 14, 2013

### Saitama

How do I find a and b here?

Are they simply $L_0/\sqrt{2}$ and $L_0/4\sqrt{2}$?

19. Jul 14, 2013

### WannabeNewton

The semimajor axis is the arithmetic average of the min and max position and the semiminor is the geometric average. I think it would be much easier to just go with what TSny said and just use Kepler's 3rd law. It's (Kepler's 3rd law) is usually written down in terms of the gravitational attraction but the potential here is essentially the same, just with different constants, so you would just have to adjust that part.

EDIT: By the way, by arithmetic average I meant $a = (r_f + r_i) /2$ and by geometric average I meant $b = (r_f r_i)^{1/2}$

Last edited: Jul 14, 2013
20. Jul 14, 2013

### Saitama

What is $r_f$ then? Is it equal to $a$?

21. Jul 14, 2013

### TSny

No, those are not the lengths of the semi-major and semi-minor axes. They are the greatest and least distances to the central charge.

22. Jul 14, 2013

### Saitama

I am still confused. Why the path of the charges is an ellipse? Shouldn't they travel in a circle with a changing radius? This is how I analysed the situation before posting the problem here.

23. Jul 14, 2013

### ehild

What can be a circle with periodically changing radius? :)

24. Jul 14, 2013

### TSny

A circle with a changing radius???

The motion of a particle acted on by an inverse square attractive force must be a conic section.

25. Jul 14, 2013

### WannabeNewton

To put it another way, there is a maximum distance from the stationary charge that each of the orbiting charges reach and there is a minimum distance from the stationary charge that they reach. For the given potential, we can have an elliptic (which includes circular), parabolic, or hyperbolic trajectory. Here we have a periodic orbit with each charge having a maximum distance from the central chatge (apoapsis) and minimum distance from the central charge (periapsis) so we have an elliptical orbit.

Last edited: Jul 14, 2013