= Solution
\b[Normal-mode idea.] Near a stable equilibrium, expand the <potential energy> to quadratic order and the <kinetic energy> as a quadratic form in generalized velocities. The <Euler-Lagrange equations> then take the linear form $\mathsf M\ddot{\mathbf x}+\mathsf K\mathbf x=0$. A <normal mode> has fixed displacement ratios, $\mathbf x=\mathbf q\cos(\omega t+\varphi)$, with $\mathsf K\mathbf q=\omega^2\mathsf M\mathbf q$. Real systems admit this description for sufficiently small motion, before higher-order nonlinearities or strong damping matter. <Superposition> permits a general motion to be assembled from the modes. A free symmetry direction may instead be a <zero mode>, with constant or linearly changing coordinate.
\b[Lagrangian and bead equations.] Take each $x_i$ as an arc-length displacement positive in the same sense around the fixed circular wire. Spring length changes are $x_2-x_1$, $x_3-x_2$ and $x_1-x_3$. Equal equilibrium spring tensions give no net tangential <force>; their linear energy terms sum to zero. Dropping an irrelevant constant, the <Lagrangian> is
$$
\boxed{\mathcal L=\frac m2(\dot x_1^2+\dot x_2^2+\dot x_3^2)
-\frac k2\left[(x_2-x_1)^2+(x_3-x_2)^2+(x_1-x_3)^2\right].}
$$
For example, differentiating the potential with respect to $x_1$ gives $k(2x_1-x_2-x_3)$. The three <Euler-Lagrange equations> therefore are
$$
\boxed{\begin{aligned}
m\ddot x_1&=-k(2x_1-x_2-x_3),\\
m\ddot x_2&=-k(2x_2-x_3-x_1),\\
m\ddot x_3&=-k(2x_3-x_1-x_2).
\end{aligned}}
$$
The wire supplies normal constraint forces but does no tangential work. These are the equal-mass case of the <normal modes of three beads connected by springs on a circular hoop>.
\b[Frequencies.] The stiffness matrix divided by $k$ is
$$
\mathsf K/k=\begin{pmatrix}2&-1&-1\\-1&2&-1\\-1&-1&2\end{pmatrix}
=3\mathsf I-\mathbf 1\mathbf 1^{\mathsf T},\qquad
\mathbf 1=(1,1,1)^{\mathsf T}.
$$
It annihilates $\mathbf 1$. On the two-dimensional subspace $q_1+q_2+q_3=0$, its action is multiplication by three. Thus the <angular frequencies> of the <normal modes> are
$$
\boxed{\omega_0=0,\qquad\omega_1=\omega_2=\sqrt{\frac{3k}{m}}.}
$$
The repeated <frequency> is a <degeneracy of normal modes>, while the zero <frequency> expresses free rigid <rotation> of all beads along the wire.
\b[General solution.] Two independent vectors spanning the vibrating subspace are $\mathbf q_A=(1,-1,0)^{\mathsf T}$ and $\mathbf q_B=(1,0,-1)^{\mathsf T}$. Each amplitude satisfies $\ddot Q+\omega^2Q=0$, with $\omega=\sqrt{3k/m}$. The uniform coordinate satisfies $\ddot Q_0=0$. Hence
$$
\boxed{\mathbf x(t)=\mathbf q_A A\cos(\omega t+\phi)
+\mathbf q_B B\cos(\omega t+\psi)
+\mathbf 1(Ct+D).}
$$
The six freely specifiable real constants $A,\phi,B,\psi,C,D$ give the six initial data; $\omega$ is fixed by $k,m$ and is not an additional independent integration constant. The linear zero-mode term must be included to allow nonzero total tangential momentum.
\b[Shapes and uniqueness.] In the first term beads 1 and 2 have opposite equal displacements and bead 3 remains at its equilibrium position. In the second term beads 1 and 3 do this while bead 2 stays fixed. The last term moves all three identically, leaving spring extensions unchanged: $D$ is a rigid phase shift and $C$ a uniform drift around the wire. The sketches show positive modal amplitudes; all arrows reverse half a vibration period later.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2017/paper-1-c12-ring-mode-shapes.png]
{title=Original displacement sketches for two independent equal-frequency bead modes and the rigid-rotation zero mode; arrows indicate displacement along the ring rather than its radius}
The two nonzero-<frequency> <eigenvectors> are not unique, because any independent linear combinations in their degenerate two-dimensional <eigenspace> form an equally valid basis. In particular the displayed pair is independent but not orthogonal; one could use $(1,-1,0)$ and $(1,1,-2)$ instead. The zero <eigenspace> is one-dimensional, so its vector is fixed up to scale and sign.
\b[Specified release.] Zero initial velocities permit $\phi=\psi=0$ and require $C=0$. At $t=0$, the three displacement conditions become $A+B+D=a$, $-A+D=0$, $-B+D=0$. Therefore one convenient set of constants is
$$
\boxed{A=B=D=\frac a3,\quad C=0,\quad\phi=\psi=0,
\quad\omega=\sqrt{3k/m}.}
$$
Equivalently,
$$
\boxed{x_1(t)=\frac a3[1+2\cos(\omega t)],\qquad
x_2(t)=x_3(t)=\frac a3[1-\cos(\omega t)].}
$$
Their sum remains $a$, as required by zero initial total tangential momentum. Alternative amplitude signs with compensating phase shifts represent the same motion.
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