Lagrangian for Coupled Ocillator problem

[MikeD1]

Homework Statement

|--------------------|
m|----------m-------- |m
|--------------------|

-------> x : positive x-axis

This is a picture of a coupled oscillator in equilibrium. All three masses are equal and the spring constant on the long springs are k and the two short middle springs is κ. the displacement of the three masses are defined as x1, x2, x3.

Homework Equations

L=T-U
T=1/2 m (xdot)^2
U= 1/2kx^2

The Attempt at a Solution

I am having trouble with the lagrangian for this problem, mostly with the sign of the coefficients. I think that the kinetic energy T= 1/2 m((x1dot)^2+(x2dot)^2+(x3dot)^2)
and the potential U= 1/2 [kx1^2 + 2κx2^2 + kx3^2 ] .
is this the correct approach?

 

[BruceW]

That is correct. Just be careful that x1,x2,x3 are the displacements of the masses from their equilibrium positions (not from some shared origin).

Edit: Sorry, its not correct. There are 4 springs in the problem, and you've correctly identified 1/2kx1^2 + 1/2kx3^2 is the potential energy due to the outer springs. But the potential energy due to the inner springs is not κx2^2. If you think about it, the extension of the inner springs depends on the position of the other masses as well (not just the middle one). So you need to calculate the extension of the inner springs (compared to their equilibrium length) and this will tell you what the potential energy must be.

 

[MikeD1]

So the extension of the inner springs due to the movement of the two outer masses would be the displacement relative to the equilibrium positions of the the two masses. So would it be correct if I assume that the displacement on the inner spring is just x3-x1? where x1 and x3 are the displacement of the masses relative to the equilibrium.

 

[BruceW]

no, remember there are two inner springs. You should calculate the potential energy due to each inner spring. To do this, think about the positions of the masses and how that affects the extension of each of the inner springs.