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A normal mode has all coordinates following one fixed displacement pattern with a common time factor. Its normal frequency is that factor's oscillation frequency; normal mode coordinates are amplitudes in a basis of such patterns, making the linear equations independent. A zero-frequency mode can instead have constant velocity along a symmetry direction.

For the unit circle, each angular displacement equals its corresponding arc displacement. The kinetic energy and change in potential energy areAny common spring prestress changes only the constant $V_0$, because the sum of the three arc lengths is fixed. The Euler-Lagrange equations giveThese are the normal modes of three beads connected by springs on a circular hoop, with third mass $M=\mu m$. For $\mu >0$, their three mode vectors and squared angular frequencies areThe zero mode rotates the entire configuration without stretching a spring. The antisymmetric mode moves the equal masses oppositely while leaving the third fixed. In the symmetric mode the equal masses move together against the third, with zero total angular momentum. These vectors are orthogonal in the mass inner product with matrix $m\mathrm{diag}(1,1,\mu )$.

Writing $\mathbf{\theta }=Q_0{v}_0+Q_A{v}_A+Q_S{v}_S$ gives independent equations ${\ddot{Q}}_0=0$, ${\ddot{Q}}_A+{\omega }_A^2{Q}_A=0$, ${\ddot{Q}}_S+{\omega }_S^2{Q}_S=0$. In particular $Q_0=A+Bt$, retaining possible slow common rotation. For $\mu =1$, the two positive frequencies coincide at $\sqrt{3k/m}$; their two-dimensional space is all patterns with ${\theta }_1+{\theta }_2+{\theta }_3=0$.

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 areThe springs themselves are constrained to the hoop, so these are arc lengths. If their common natural length is ${\ell }_0$, Hooke's law gives $V=\frac{k}{2}{\sum }_i({\ell }_i-{\ell }_0{)}^2$. Since the arc lengths have fixed sum $2\pi$, all terms linear in the $q_i$ cancel. Dropping the constant,Thus the unspecified common natural length does not affect the motion. The mass and restoring matrices in the generalized eigenvalue problem areExchange of the two equal masses separates symmetric and antisymmetric normal modes. Direct multiplication gives the complete setThe 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 $M=2m$.

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, $\mathbf{x}=\mathbf{q}\cos (\omega t+\phi )$, 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.

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 isFor example, differentiating the potential with respect to $x_1$ gives $k(2x_1-x_2-x_3)$. The three Euler-Lagrange equations therefore areThe 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.

Frequencies. The stiffness matrix divided by $k$ isIt annihilates $1$. 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 areThe repeated frequency is a degeneracy of normal modes, while the zero frequency expresses free rigid rotation of all beads along the wire.

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 $\ddot{Q}+{\omega }^{2}Q=0$, with $\omega =\sqrt{3k/m}$. The uniform coordinate satisfies ${\ddot{Q}}_0=0$. HenceThe six freely specifiable real constants $A,\varphi ,B,\psi ,C,D$ 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.

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.

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.

Specified release. Zero initial velocities permit $\varphi =\psi =0$ 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 isEquivalently,Their sum remains $a$, as required by zero initial total tangential momentum. Alternative amplitude signs with compensating phase shifts represent the same motion.