= Solution
Put $\theta_i=\theta_i^0+\eta_i$. Differentiating the spring length gives
$$
\delta\ell_{ij}=\frac{\sin(\theta_i^0-\theta_j^0)}{\sqrt3}(\eta_i-\eta_j),\qquad
(\delta\ell_{ij})^2=\frac14(\eta_i-\eta_j)^2.
$$
The springs are unextended at <equilibrium>, so the quadratic <potential energy> contains only the squares of these first-order extensions:
$$
U_2=\frac k8\left[(\eta_1-\eta_2)^2+(\eta_2-\eta_3)^2+(\eta_3-\eta_1)^2\right].
$$
The linearized <Euler-Lagrange equations> are therefore
$$
m\ddot{\boldsymbol\eta}+\frac k4
\begin{pmatrix}2&-1&-1\\-1&2&-1\\-1&-1&2\end{pmatrix}\boldsymbol\eta=0.
$$
The <stiffness matrix> is $(k/4)(3I-\mathbf 1\mathbf 1^T)$. Its <eigenvector> $\mathbf 1=(1,1,1)^T$ has zero <eigenvalue>, and every vector whose components sum to zero has <eigenvalue> $3k/4$. One convenient orthonormal set of <normal modes> is
$$
\boxed{\begin{aligned}
\mathbf v_0&=\frac1{\sqrt3}(1,1,1)^T,&\omega_0&=0,\\
\mathbf v_1&=\frac1{\sqrt2}(1,-1,0)^T,&\omega_1&=\sqrt{\frac{3k}{4m}},\\
\mathbf v_2&=\frac1{\sqrt6}(1,1,-2)^T,&\omega_2&=\sqrt{\frac{3k}{4m}}.
\end{aligned}}
$$
Any orthonormal basis of the sum-zero plane is equally valid for the two degenerate <normal modes>. The general small motion is
$$
\boldsymbol\eta(t)=(A_0+B_0t)\mathbf v_0+\sum_{j=1}^2\left[A_j\cos(\omega_1t)+B_j\sin(\omega_1t)\right]\mathbf v_j.
$$
The linear drift in the <zero mode> describes a slow uniform rotation. These are <angular frequencies>; the ordinary <frequencies> of the two vibrational modes are $\omega_1/(2\pi)$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2-q7-normal-modes.png]
{title=Tangential displacement patterns of the three bead normal modes}
{description=Arrows show the signs and relative sizes of the angular displacements in one choice of orthonormal <normal modes>. The equal-arrow pattern is a rigid rotation, while the two sum-zero patterns have equal nonzero <angular frequency>.}
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