1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Frequency of small oscillations about equilibrium point.

  1. May 14, 2009 #1
    A particle of mass m moves in one dimension subject to the potential:
    V(x)=(-12/x)+(x^-12)
    Find the equilibrium point and the frequency of small oscillations about that point.



    I think I've found the equilibrium point 'a', but using the formula V'(a)=0, and i got the answer a=1.


    However, im completely stuck on finding the frequency, I've found the Lagrangian and hence the equation of motion, but then dont know what to do.

    any help much appreciated!! :)
     
  2. jcsd
  3. May 14, 2009 #2
    Ok, you don't need the equations of motion to find the frequency of small oscillations.

    Here's what you need to do:

    At x=1, there's a stable equilibrium. For very small oscillations, it will approximate a SHO. So all you need to do is fit a parabola to that point by:

    1. Putting the equilibrium point of the parabola at the (x=1, y=???) point.
    2. Making the first derivative of the parabola equal the derivative at the point (of course, this has already been done for you by using a parabola).
    3. Making the second derivative match the second derivative of your potential function there.

    So what's the equation of a parabola?

    y = ax^2 + bx + c
    y' = 2ax + b
    y'' = 2a

    So all you need to do is to make sure that y(1) = V(1), y'(1) = V'(1), and y''(1) = V''(1).

    Then simply realize that F = -kx by Hooke's law, and that F = -y'. So whatever the coefficient on the 'x' term in your y' happens to be, that equals k. From that and the mass m, it should a straightforward exercise to find everything else you could want.
     
  4. May 14, 2009 #3
    ive managed to work out what you've said but i dont understand why?
     
  5. May 14, 2009 #4
    sorry,
    i dont understand why you need to fit the parabola at that point, and i dont really get how you do it, do you equate the V's and y's to find the values of a, b and c? and how does this help?
    and i dont see how to use the info to find the oscillations :(
     
  6. May 14, 2009 #5
    forget it!! ive figured it out! temporary blank.

    thanks very much for your help.
     
  7. May 14, 2009 #6
    The idea is that for small displacements from a stable equilibrium, all oscillations look like simple harmonic oscillations.

    Simple harmonic oscillations are described by a parabolic potential well.

    So by finding the parabola which best approximates your potential function V at the point of stable equilibrium, you will find the SHO potential which best approximates the oscillatory behavior near the equilibrium.

    To find the best-fit parabola at your point, you have to make it match your V in its zeroth, first, and second derivatives; you have to find the 2nd order Taylor polynomial for the potential function V expanded around the point of equilibrium.

    The system of equations I suggested fits the parabola.

    Once you have the parabola for the SHO, it's easy to use Hooke's law to find k, and once you have k and m, you can easily find everything else.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Frequency of small oscillations about equilibrium point.
  1. Equilibrium points (Replies: 0)

  2. Equilibrium points (Replies: 1)

Loading...