# Period of oscillation (potential energy equation given)

• sodaboy7
In summary, the potential energy of a particle is given by U(x) = U0(1-cos(ax)), where U0 and a are constants. The period of small oscillations performed by the particle about its equilibrium position can be found using the equation T = 2π√(m/a2U0). The motion is relevant to forced oscillations, as an external time-dependent force is needed for forced oscillation. In addition, there are two sufficient conditions to prove something is a simple harmonic oscillator: F = -kx or Etotal = \frac{1}{2}Av2+\frac{1}{2}Bq2, where A and B are constants and q is a coordinate. For the second

## 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.

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.

therealnihl said:
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...)