Lagrangian (chain off spring connected masses)

[ballzac]

Homework Statement

______|equillibrium position________
______|__i_____________________
m^^^^^m^^^^^m^^^^^m^^^^^m
____k_|qi|____k______ k________kA collection of particles each of mass m separated by springs with spring constant k. The displacement of the ith mass from its equilibrium position is q_i=q_i(t). Write the Langrangian for this one dimensional chain of masses.

Homework Equations

K=\frac{1}{2}mv^2
V=\frac{1}{2}kx^2
L=K-V

The Attempt at a Solution

So we have K=\frac{1}{2}mv^2=\frac{1}{2}m(\frac{dx}{dt})^2
Can I now say
K=\frac{1}{2}m\frac{dq_i}{dt}^2
?
(I think I have to because I then have to differentiate wrt q_i
Also, the extension of the springs to begin with must matter because if they are taught then the potential energy must be higher. Am I right? Does it matter that there is a spring on either side? I guess it does, but as the mass is just moving between its equilibrium point maybe they kind of cancel out. There are N particles, so am I right in thinking that the energy required is the sum of all functions of q_i for every i\leq N?

I haven't been given very long for this, and I've never come across this stuff before, so I'm having issues figuring it out. Thanks for the help.

EDIT: Yeah, so I'm thinking V=\frac{1}{2}kq_i ^2 or V=kq_i ^2 but then I'm still not sure if the original taughtness of the spring comes into it.

 

[Winzer]

I think I would make a different coordinate for each mass. So mass 1-x1, mass 2-x2 and so on{at equilibrium position}. Get the potentials for each mass in terms of those different coordinates. Get kinetic energy for each mass, then do L=K-V. I think the typical convention though is L=T-U.

 

[ballzac]

Okay, thank you I will try that. I'm just using the notation that was given in the handouts. Not sure if there is any significance.