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 coordinatescoordinate $x$ and $y$ to be the distance between one pair of neighbouring masses. The potential energy is then $\frac{k}{2} (2x^2 + 2y^2)$$\frac{k}{2} (2x^2 + 2(\pi r - x)^2) + \dots$ while the kinetic energy is of the form $\frac{m}{2} (2\dot x^2 + 2\dot y^2 + \dots)$$\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$$\frac{k}{2} r^2(\theta_1)^2 + \frac{k}{2}r^2(\theta_1 - \pi)^2 + \dots$, so the leading term is $\sim \frac{2k}{2} x^2$)