= Solution
Varying the <Lagrangian> gives, with the cyclic angle conventions,
$$
\boxed{m_i a^2\ddot\theta_i=k(\theta_{i+1}-2\theta_i+\theta_{i-1}).}
$$
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 $(1,\ldots,1)$, 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 $km\operatorname{diag}(1,2)$ and the stiffness matrix is $2k\left(\begin{smallmatrix}1&-1\\-1&1\end{smallmatrix}\right)$. The <generalized eigenvalue problem> gives
$$
\boxed{\omega_0=0,\quad(\delta_1,\delta_2)\propto(1,1);\qquad \omega_1=\sqrt{3/m},\quad(\delta_1,\delta_2)\propto(2,-1).}
$$
The 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.
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