Eigenfrequencies of Normal Modes

Question

I understand the whole deal with coupled oscillators and how to solve for normal modes and eigenfrequencies and such. But what is tripping me up is what these eigenfrequencies correspond to. If I have 4 masses along a ring connected to springs, I find that the frequencies are $$\omega^2=\frac km,\frac {2k}{m},0$$ But I do not know how to explain this. I assume 0 means they aren't moving. Does $k/m$ imply the system acts as a simple pendulum? What does this imply about $2k/m$? Thanks!

Answer 1

The best thing to do is to look at the eigenvectors. This will tell you all you need to know.

Anyway, $0$ corresponds to all the masses rotating in unison.

Another mode corresponds to neighbouring masses moving closer and farther away from each other, symmetrically. Here you can choose a generalized coordinate $x$ to be the distance between one pair of neighbouring masses. The potential energy is then $\frac{k}{2} (2x^2 + 2(\pi r - x)^2) + \dots$ while the kinetic energy is of the form $\frac{m}{2} 4(\frac{\dot x}{2})^2 $.

I believe a final mode involves two opposite masses staying fixed, while the other masses in between them oscillate. The oscillating masses move in opposite directions, so that the forces on the stationary masses remain balanced. In this last case, you can replace the fixed masses with immovable points (symmetry). Then each of the remaining masses has two springs attached to it, giving an effective spring constant of $2k$. (In generalized coordinates, the potential for one mass is $\frac{k}{2} r^2(\theta_1)^2 + \frac{k}{2}r^2(\theta_1 - \pi)^2 + \dots$, so the leading term is $\sim \frac{2k}{2} x^2$)