Approximation to simple harmonic motion.

[Angelos]

[SOLVED] Approximation to simple harmonic motion.

Homework Statement

A small mass $$m$$, which carries a charge $$q$$, is constrained to move vertically inside a narrow, frictionless cylinder. At the bottom of the cylinder is a point mass of charge $$Q$$ having the same sign as $$q$$. Show that if the mass $$m$$ is displaced by a small amount from its equilibrium position and released, it will exhibit simple harmonic motion with angular frequency
$$\omega = \sqrt{\frac{2g}{y_0}}$$
where $$y_0$$ is the height, when the mass is in equilibrium.

Homework Equations

Simple harmonic motion:
$$F=-kx$$

Coulomb's Law(one dimension, same sign of charges):
$$F= \frac{kq_1q_2}{y^2}$$

Newton's Second Law:
$$F=ma$$

The Attempt at a Solution

Force acting on the mass $$m$$ at a height $$y$$ is:
$$F=\frac{kqQ}{y^2} - mg$$

I found the potential energy by:
$$U=-\int F dy= \frac{kqQ}{y} + mgy$$

I found the equilibrium point:
$$mg = \frac{kqQ}{y_0^2}$$
$$y_0 = \sqrt{\frac{kqQ}{mg}}$$

Near (stable) equilibrium I can approximate the potential energy by some parabola with equation:
$$U_n = U(y_0) + B(y - y_0)^2 = 2\sqrt{kqQmg} + B(y - \sqrt{\frac{kQq}{mg}})^2$$
where $$B$$ is some (positive) constant.

Than I can find the approximate force by taking the negative derivative of $$U_n$$. That is
$$F = -2B(y - y_0)$$

That is the same as simple harmonic motion and so we can say that
$$\omega = \sqrt{\frac{2B}{m}}$$

Well my problem is that I have no idea how to find the $$B$$ or how to show that the $$B = \frac{mg}{y_0}$$. The thing is I don't know how to mathematically describe that I want the parabola of the approximate potential energy function to be the most similar to the original function near equilibrium.

I will really appreciate any help from you.
Thanks.

 

[CompuChip]

You said
$$U(y) = \frac{kqQ}{y} + mgy$$
(which I didn't check, but I assume it is correct).

I would consider making a Taylor expansion around the equilibrium point, i.e.
$$U(y) = U(y_0) + U'(y_0)(y - y_0) + \frac12 U''(y_0) (y - y_0)^2 + \mathcal{O}((y - y_0)^3).$$
Since it's an equilibrium point, the linear term will vanish. Then approximate the potential by throwing away the (y - y0)^3 terms and you will have a parabola, from which you can easily read off the coefficient.

I quickly tried this and it will give you the B you stated, so it should work. But there might be another way.

 

[Angelos]

Hi...I had never heard about Taylor's polynomials, but I looked it up, when I read your reply and it's a great stuff. Exactly what I needed and how you said - it works. So thanks a lot. I really appreciate it.