• Support PF! Buy your school textbooks, materials and every day products Here!

Frequency of small oscillations about equilibrium point.

  • Thread starter bjw1311
  • Start date
  • #1
6
0
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!! :)
 

Answers and Replies

  • #2
492
0
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.
 
  • #3
6
0
ive managed to work out what you've said but i dont understand why?
 
  • #4
6
0
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 :(
 
  • #5
6
0
forget it!! ive figured it out! temporary blank.

thanks very much for your help.
 
  • #6
492
0
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.
 

Related Threads for: Frequency of small oscillations about equilibrium point.

  • Last Post
Replies
2
Views
917
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
1
Views
931
  • Last Post
Replies
1
Views
976
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
1K
Replies
4
Views
651
Replies
3
Views
2K
Top