Finding the frequency of small oscillations given potential energy U

[spacetimedude]

Homework Statement

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.

Homework Equations

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

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.

 

[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 that's 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.

 

[spacetimedude]

Hello. Thanks for the reply.

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.

 

[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).

 

[spacetimedude]

Thank you so much!