Period of oscillation (potential energy equation given)

[sodaboy7]

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.

2. Given Answer

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]

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

 

[therealnihl]

There a two sufficient conditions to prove something is simple harmonic oscillator
F = -kx
or
E_total = \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.

 

[sodaboy7]

E_total = \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?

 

[therealnihl]

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...)