Finding the frequency of small oscillations given potential energy U

1. May 8, 2015

spacetimedude

1. The problem statement, all variables and given/known data
The potential energy of a particle of mass m near the position of equilibrium is given by U=U0sin2(αx) where U0 and α are constants. Find the frequency of the small oscillations about the position of equilibrium.

2. Relevant equations
Work energy equation (1/2)kx12+(1/2)mv12=(1/2)kx22+(1/2)mv22

3. The attempt at a solution
(1/2)kx2=U0sin2(αx)
Differentiating twice and rearranging:
k=2U0α2cos(2αx)

I'm confused from here. Am I supposed to use the work energy relation?
I vaguely remember learning about sin(θ)≈θ when θ is small and also about Taylor expansion.
Any hints will be appreciated.

2. May 8, 2015

Brage

Hiya I think I might be able to help you out a bit here. I must admit I don't understand what the "work energy equation" is supposed to represent. I assume x1 and x2 are independent degrees of freedom, in which case you have the total energy of one particle (work done + kinetic energy) equals the total energy of another particle?
Regardless, you should be able to solve the problem without it. In your attempt at a solution you seem to be solving for k but thats not what helping, as k is a spring constant (if I understood your equation correctly). Basically you wish to find an equation for position x and then find the frequency (inverse of the period) of the system. Remember Newton's equation for the unbalanced force on a system $m\ddot{x}=-\frac{\partial U}{\partial x}$ where $\ddot{x}$ denotes the second derivative of position with respect to time (i.e. acceleration). Solving this equation should yield the answer.

3. May 8, 2015

spacetimedude

I understand. So I got to a=(U0α/m)sin(2αx) where 'a' denotes x double dot.
Since the angle is small, we can think of the system as simple harmonic motion. In that case, the general form should be d2x/dt2+w2x=0.
We already have dx2/dt2 but how do find w2x? Or am I not going the right direction?

Thank you.

4. May 8, 2015

Brage

Yes you are on the right track, remember if $\theta$ is small then $sin(\theta)=\theta$, as we may taylor expand the sin function about zero to find $sin(\theta) = \theta + \frac{\theta^3}{3!} + \frac{\theta^5}{5!} +...$ where $\theta<<1 \Rightarrow \theta^3 \approx 0$. This means $sin(2\alpha x) \approx 2\alpha x$ for small x. Also you have missed out a negative in your equation (check your calculations).

5. May 9, 2015

spacetimedude

Thank you so much!