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.

 

[paranormal]

Yes, your approach is correct. The Lagrangian for the coupled oscillator problem can be written as:

L = T - U = 1/2 m((x1dot)^2+(x2dot)^2+(x3dot)^2) - 1/2 [kx1^2 + 2κx2^2 + kx3^2]

where T represents the kinetic energy and U represents the potential energy. The sign of the coefficients is correct, as the potential energy is always written with a negative sign in the Lagrangian. This is because the potential energy is a conservative force, meaning that it can be expressed as the gradient of a scalar potential function, and the negative sign ensures that the potential energy decreases in the direction of the force.

In this problem, the masses are equal and the displacement of each mass is defined as x1, x2, and x3 respectively. The spring constants for the long springs are k, and the spring constant for the two shorter middle springs is κ. Therefore, the potential energy can be expressed as 1/2 [kx1^2 + 2κx2^2 + kx3^2], as you correctly stated.

Overall, your approach to finding the Lagrangian for the coupled oscillator problem is correct. Keep in mind that for more complex systems, it may be necessary to use generalized coordinates instead of the individual masses' displacements. But in this case, your approach is sufficient.