How is the Lagrangian Derived for a Two-Mass Spring System?

[don_anon25]

The system examined in the problem is depicted below:
^^^^^(m1)^^^^^(m2)
m1 and m2 are connected by a spring and m1 is connected to the wall by a spring. The spring constant is k.

T = m/2 [ x1'^2 +x2'^2 ] kinetic energy of system (x1' is velocity of m1, x2' is velocity of m2)
U = 1/2 m k^2 (x1 - b)^2 + 1/2 m k^2 (x2-x1-b)^2 potential energy of system (x1 is position of m1, x2 is position of m2, b is the unstretched length of the spring)
Is m the reduced mass?
Also, could someone explain how the equation for U is derived? Why is it k^2 and not just k (i.e. potential energy for spring = 1/2 kx^2)? Also, why is there a mass term in the potential energy? Or is this the wrong expression for potential energy altogether?

I known then that the Lagrangian for the system is L = T - U. I can then take derivatives and substitute into the Euler-Lagrange equation. I should have two E.L. equations, correct? But what should I solve for -- x1 and x2?

Any guidance/hints greatly appreciated!

 

[Physics Monkey]

Hi don,

The general form of the potential energy for a linear spring is \frac{1}{2} k (x - d)^2 where k is the spring constant,x is the distance between endpoints of the spring, and d is the unstreched length. Why did you put m k^2 in front of the displacement squared? Perhaps you were thinking of k = m \omega^2, the definition of the frequency for a single spring system.

Also, the differential equations you obtain are two coupled linear second order equations. There are several ways to proceed, one would be to see if you can make a useful separation into the relative and center of mass motion. Alternatively, are you familiar with the theory of coupled oscillations? You can obtain the normal modes (there are two of them) for your system without too much trouble.