= Solution
To find the <normal modes of two arc-spring-coupled beads on a hoop>, take $x_i=a_i e^{i\Omega t}$ for a <normal mode>. The resulting <generalized eigenvalue problem> is
$$
\begin{pmatrix}
2k-m_1\Omega^2&-2k\\
-2k&2k-qm_1\Omega^2
\end{pmatrix}
\begin{pmatrix}a_1\\a_2\end{pmatrix}=0.
$$
Its <determinant> is
$$
(2k-m_1\Omega^2)(2k-qm_1\Omega^2)-4k^2
=m_1\Omega^2\left(qm_1\Omega^2-2k(1+q)\right).
$$
Thus the <angular frequencies> are
$$
\boxed{\Omega_0=0,\qquad
\omega=\sqrt{\frac{2k(1+q)}{qm_1}}.}
$$
The ordinary nonzero <frequency> is $\omega/(2\pi)$. The <zero mode> has <mode shape> $(1,1)$ and represents a common <rotation> with no change in <spring> length. The oscillatory <mode shape> is $(1,-1/q)$.
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