# Period of oscillation (potential energy equation given)

## Homework Statement

Let the potential energy of particle depend upon coordinate x as:
U(x) = U0(1-cos(ax)). Where "U0" and "a" are constants. Find the period of small oscillations that particle performs about its equilibrium position.

T = 2∏√(m/a2U0)

## The Attempt at a Solution

It can be seen from the equation that equilibrium will be at x = 0, where forces acting are zero. Also after integrating the equation with dx, it is seen that the motion is relevant to forced oscillations.

ehild
Homework Helper
U(x) is the potential function. Why do you integrate it?
The motion is not forced oscillation. For that, an external time dependent force is needed.

ehild

There a two sufficient conditions to prove something is simple harmonic oscillator
F = -kx
or
Etotal = $\frac{1}{2}$Av2+$\frac{1}{2}$Bq2. Where A and B are some constants and, q is some coordinate (in your case ax). Then the period is T = 2π√(A/B) Also since it says small oscillations, I would use the small angle approximation for cosine.

Etotal = $\frac{1}{2}$Av2+$\frac{1}{2}$Bq2. Where A and B are some constants and, q is some coordinate (in your case ax). Then the period is T = 2π√(A/B) Also since it says small oscillations, I would use the small angle approximation for cosine.

Thanks for that. But this questions has two parts which I didn't mention.
The part-2 of the problem has the equation:
U(x)= a/x2 - b/x
How to deal with this one ?
I mean in above equation x has powers -2 and -1. So how to form its differential equation?

Well assuming the kinetic energy is of the form: $\frac{1}{2}$mv2, the the period for small oscillations is 2π√(k/m) where, k is the second derivative of the potential energy function evaluated at the equilibrium position (you might have to take the limit as it goes to the equilibrium position in this case...)