Lagrangian (chain off spring connected masses)

1. Mar 27, 2009

ballzac

1. The problem statement, all variables and given/known data
______|equillibrium position________
______|__i_____________________
m^^^^^m^^^^^m^^^^^m^^^^^m
____k_|qi|____k______ k________k

A 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.

2. Relevant equations
$$K=\frac{1}{2}mv^2$$
$$V=\frac{1}{2}kx^2$$
$$L=K-V$$

3. 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.

Last edited: Mar 27, 2009
2. Mar 27, 2009

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.

3. Mar 29, 2009

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.