# Potential of coupled oscillators

1. Mar 12, 2008

### ehrenfest

[SOLVED] potential of coupled oscillators

1. The problem statement, all variables and given/known data
http://cache.eb.com/eb/image?id=2480&rendTypeId=4
How do you calculate the potential energy of the coupled oscillators in the picture with spring constant k_1,k_2,k_3 as the spring constants from left to write?

2. Relevant equations

3. The attempt at a solution
The force on the left mass is $-k_1 x_1+(x_2-x_1) k_2$ and the force on the right mass is $(x_2 -x_1)k_2 +k_3 x_2$ where x_1,2 is the displacement to the right. We integrate w.r.t x_1 and x_2 (and then reverse the sign) to get $1/2 k_1 x_1^2+1/2 (x_2-x_1)^2 k_2+C$ and $+1/2(x_2-x_1)^2k_2 +1/2k_3x_2^2+C'$. We compare these two find that $C = 1/2k_3x_2^2$ so $V(x_1,x_2)= 1/2 k_1 x_1^2+1/2 (x_2-x_1)^2 k_2+1/2k_3x_2^2$.

My question is how can you do that faster? That is, you should be able to directly calculate the potential without integrating anything. How would you do that? Can you add the "potential of each mass" or "the potential of each spring" or something? That would make it much easier especially when you have many more oscillators!

EDIT: also can someone tell me the thought process that goes into finding the force, say on the left mass? I just found it by kind of guessing and then checking certain cases...

Last edited: Mar 12, 2008
2. Mar 17, 2008

### ehrenfest

anyone?

3. Mar 17, 2008

### kdv

Imagine that the first mass is diplaced by a distance x_1 to the right (measured from its equilibrium position) and the second mass is displaced x_2 to the right from its equilibrium position.

Then clearly the first spring is stretched by a a value equal to x_1 so its strores a potential energy 1/2k x_1^2.

The second spring is stretched by (x_2-x_1) (if this value is negative, it simply means that the spring is compressed) so it stores an energy 1/2 k(x_2-x_1)^2.

The third spring is compressed by a value equal to x_2 so it stores a potential energy equal to 1/2 k x_2^2

Obviously if you have n masses, the potential energy in each will be

$$\frac{1}{2} k x_1^2, \frac{1}{2} k (x_2-x_1)^2, \frac{1}{2} k (x_3-x_2)^2, \ldots \frac{1}{2} k x_n^2$$