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Finding the frequency of small oscillations given potential energy U

  1. May 8, 2015 #1
    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. jcsd
  3. May 8, 2015 #2
    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.
     
  4. May 8, 2015 #3
    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.
     
  5. May 8, 2015 #4
    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).
     
  6. May 9, 2015 #5
    Thank you so much!
     
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