= Solution
Label the equal <masses> $1,2$ and the mass $2m$ by $3$, in cyclic order. Let their equilibrium angles be $0,2\pi/3,4\pi/3$, with angular displacements $q_1,q_2,q_3$. Because the hoop has unit radius, its three arc lengths are
$$
\ell_1=\frac{2\pi}{3}+q_2-q_1,\quad
\ell_2=\frac{2\pi}{3}+q_3-q_2,\quad
\ell_3=\frac{2\pi}{3}+q_1-q_3.
$$
The <springs> themselves are constrained to the hoop, so these are arc lengths. If their common natural length is $\ell_0$, <Hooke's law> gives $V=\frac k2\sum_i(\ell_i-\ell_0)^2$. Since the arc lengths have fixed sum $2\pi$, all terms linear in the $q_i$ cancel. Dropping the constant,
$$
\boxed{T=\frac m2(\dot q_1^2+\dot q_2^2+2\dot q_3^2),\qquad
V-V_0=\frac k2\big[(q_2-q_1)^2+(q_3-q_2)^2+(q_1-q_3)^2\big]}.
$$
Thus the unspecified common natural length does not affect the motion. The mass and restoring matrices in the <generalized eigenvalue problem> are
$$
M=m\begin{pmatrix}1&0&0\\0&1&0\\0&0&2\end{pmatrix},\qquad
K=k\begin{pmatrix}2&-1&-1\\-1&2&-1\\-1&-1&2\end{pmatrix}.
$$
Exchange of the two equal <masses> separates symmetric and antisymmetric <normal modes>. Direct multiplication gives the complete set
$$
\boxed{\begin{array}{c|c}
\text{mode shape}&\omega^2\\ \hline
(1,1,1)&0\\
(1,-1,0)&3k/m\\
(1,1,-1)&2k/m
\end{array}}.
$$
The first is rigid <rotation> of all three beads: every arc length stays fixed, giving a <zero mode>. In the second, the equal <masses> move oppositely while the heavy bead stays fixed; the net force on that bead cancels. In the third, the equal <masses> move together and the heavy bead moves oppositely with equal angular amplitude. Both nonzero <normal modes> have zero total <angular momentum>. These are the <normal modes of three beads connected by springs on a circular hoop> with $M=2m$.
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