Lagrangians and Masses with springs

[ThereIam]

Okay, so two equal masses are connected by spring with spring constant k. The kinetic energy is obviously 1/2*m*x1dot^2 +1/2*m*x2dot^2. Please excuse my notation. x1 and x2 are the positions, x1dot and x2dot are the velocities. L is the length of the spring when not stretched.

So anyway, the potential energy ought to be 1/2*k*(x2-x1-L)^2, I would figure, because when x2-x1 = L, the spring would be unstretched and would store no potential energy. However, my book does not include the -L, and just gives 1/2*k*(x2-x1)^2 as the potential energy. Can anybody explain this?

 

[AlephZero]

Often you don't measure every position from one common reference point, you measure them from the initial or undisplaced position of the system.

The book is measuring x1 from the initial position of one end of the spring, and x2 from the initial position of the other end.

Measuring both x1 and x2 from the same point will give you the same results (except for a constant offset of L) but the math will be messier. Do the problem both ways, to see how it works out.