# Lagrange ' s equations for spring problem

1. Apr 6, 2004

### student1938

Ok, there are two objects of mass m on a frictionless table. The 2 masses are connected to the other by a spring of spring constant k. One mass is connected to a wall with a spring of the same constant k. Solve for the motion using Lagrange' s equations.

I used generalized coordinates x starting from the centre of the mass closer to the wall and y starting from teh center of the mass farthest from the wall. I get T = (1/2)mxdot ^ 2 + (1/2)mydot ^ 2

V = (1/2)kx ^ 2 +(1/2)k(x-y) ^ 2

I know that this contains cross terms but uptill this stage , is it correct?

2. Apr 7, 2004

### jamesrc

That's fine, except you didn't include the rest lenghts of the springs. If we stick with your notation, let's use xo as the rest length of the 1st spring and yo as the rest length of the second spring. Kinetic coenergy is the same:
$$T^* = \frac 1 2 m\dot{x}^2 + \frac 1 2 m\dot{y}^2$$
Potential energy looks more like this:
$$V = \frac 1 2 k\left(x-x_o\right)^2 + \frac 1 2 k\left(y-x-y_o\right)^2$$

Don't worry about simplfying any algebra at this point; just right L = T*-V and start differentiating away (or, as one of my buddies used to say, Lagrangiate). Have fun.

3. Apr 7, 2004

### student1938

Couldn' t I take x0, y0 to be zero?

4. Apr 7, 2004

### jamesrc

Yes, provided that you define your coordinates properly. I thought you had defined the distances from the wall. I just re-read your post and I think I'm wrong about that, but I'm still not quite sure.

To avoid any ambiguity:
Let x = the distance of the first mass (the one closer to the wall) measured from the original position of that mass (its position when the system is in equilibrium).
and similarly:
Let y = the distance of the second mass (the one farthest from the wall) measured from the original position of that mass (its position when the system is in equilibrium).

That may have been what you originally meant, but I didn't read it that way. Sorry for any confusion.