# Oscillations of beat in tube

1. Oct 12, 2013

### Numeriprimi

1. The problem statement, all variables and given/known data
Bead (mass point - mass m and charge q) is free to move in a horizontal tube (without friction). The tube is between two spheres (separated by a distance 2a) with charges Q = − q. What is the frequency of small oscillations around the equilibrium point of the bead?

2. Relevant equations
Coulomb's law

3. The attempt at a solution
Hmm... Coulomb's law defines electrostatic interaction between two spheres. The spheres have same charges - repel, but the bead have opposite charge, so the spheres and repel attract. It causes oscillation? Then we may use an equation for the mechanical oscillation... I don't know how combine mechanical oscillation and electrostatic interaction in one equation.
Can you help me?

Thank you very much and sorry for my bad English.

2. Oct 12, 2013

### voko

Obviously, midway between the spheres the forces from both spheres cancel each other out, so at this point the bead is in equilibrium. Let x be the coordinate of the bead, and let x = 0 at the equilibrium. What is the net force on the bead at an arbitrary x?

3. Oct 12, 2013

### nasu

Are you sure that the charges at the ends of the tube have sign opposite to the charge in the middle?

Can you put the actual text of the problem? Or diagram.
The way I understand it, there will be no oscillations.
The middle point is indeed an equilibrium position but when the bead is moved from equilibrium the net force will be directed away from equilibrium.

4. Oct 12, 2013

### Numeriprimi

Ok, I may write it bad, so I'm going to write it better and I take you a picture...

A small bead of mass m and charge q is free to move in a horizontal tube. The tube is placed in between two spheres with charges Q = − q. The spheres are separated by a distance 2a. What is the frequency of small oscillations around the equilibrium point of the bead? You can neglect any friction in the tube.

Hint: When the bead is only slightly displaced, the force acting on it changes negligibly

So, it is better? Do you understand me?

5. Oct 12, 2013

### voko

We understand you. Now answer the question in #2.

6. Oct 12, 2013

### nasu

Yes, like this it works. I thought that the line connecting the two sphere goes along the tube.
Thank you for the clarification.

7. Oct 12, 2013

### Numeriprimi

So, the net force is the vector sum of forces from spheres... Ok? And if the bead isn't between the speheres, vector sum isn't 0, so beat have to oscillate?

8. Oct 12, 2013

### voko

OK.

It will feel some force that depends on x. Find the force first, then you can think about the effect of the force.