= Normal modes of three beads connected by springs on a circular hoop
{title2=$\omega^2=0,\;3k/m,\;k/m+2k/M$}
Consider three beads of <masses> $m,m,M$ on a horizontal hoop, connected in cyclic order by identical <springs> constrained to follow the arcs. Their common natural length affects only a constant in the <potential energy>, since the sum of the three arc lengths is fixed. About the equally spaced configuration, angular displacements $q_i$ have
$$
T=\frac{R^2}{2}(m\dot q_1^2+m\dot q_2^2+M\dot q_3^2),\qquad
V-V_0=\frac{kR^2}{2}\big[(q_2-q_1)^2+(q_3-q_2)^2+(q_1-q_3)^2\big].
$$
The <generalized eigenvalue problem> has the rigid <rotation> $(1,1,1)$ as a <zero mode>. Exchanging the equal <masses> separates an antisymmetric <normal mode> $(1,-1,0)$, with squared <angular frequency> $3k/m$, from a symmetric <normal mode> $(1,1,-2m/M)$, with squared <angular frequency> $k/m+2k/M$. The last <mode shape> has zero total <angular momentum>. The hoop radius cancels from both the <kinetic energy> and the restoring quadratic form. These formulas use arc-constrained <springs>; straight chord <springs> have a different <potential energy>.
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