 Solution

Varying the Lagrangian gives, with the cyclic angle conventions,All successive angular gaps are $2\pi /N$ in equilibrium, so their two contributions cancel. Because the potential depends only on gaps, a uniform change of all angles costs no energy. Its Hessian therefore annihilates $(1,\dots ,1)$, giving a zero-frequency normal mode. The corresponding motion is rigid rotation, ${\theta }_i(t)={\theta }_i^{(0)}+A+Bt$, including a constant common angular shift.

For two particles write ${\theta }_2-{\theta }_1=\pi +{\delta }_2-{\delta }_1$. Both cyclic gaps contribute, so the quadratic potential is $k({\delta }_2-{\delta }_1{)}^2$. With the specified masses, the mass matrix is $km\mathrm{diag}(1,2)$ and the stiffness matrix is $2k\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$. The generalized eigenvalue problem givesThe finite-frequency normal mode keeps the mass-weighted mean angle fixed and changes the angular separation sinusoidally. Together with the two rigid-rotation constants, it spans all small motions about equilibrium.