Oscillations of Three Masses on a Circle

Question

I am working on problem 10 (a) in the 3rd edition of Goldstein's Classical Mechanics:

Three equal mass points have equilibrium positions at the vertices of an equilateral triangle. They are connected by equal springs that lie along the arcs of the circle circumscribing the triangle. Mass points and springs are constrained to move only on the circle, so that, for example, the potential energy of a spring is determined by the arc length covered. Determine the eigenfrequencies and normal modes of small oscillations in the plane.

I write the potential energy as

\begin{equation*} V = \frac{k}{2}\left[R\theta_{2}-R\theta_{1}\right]^{2} + \frac{k}{2}\left[R\theta_{3}-R\theta_{2}\right]^{2} + \frac{k}{2}\left[R\theta_{3}-R\theta_{1}\right]^{2} \end{equation*}

where $R$ is the radius of the circumscribing circle and $\theta_{i}$ is the angular displacement from equilibrium of the $i^{\mathrm{th}}$ mass, the equilibrium angles being $0$, $2\pi/3$, and $4\pi/3$. The kinetic energy is

\begin{equation*} T = \frac{mR^{2}}{2}\left(\dot{\theta}_{1}^{2}+\dot{\theta}_{2}^{2}+\dot{\theta}_{3}^{2}\right). \end{equation*}

This gives rise to the secular equation

\begin{equation*} \left|\mathbf{V}-\omega^{2}\mathbf{T}\right| = 0 \implies \omega^{2}\in \left\{0,\frac{3k}{m}\right\} \end{equation*}

where

\begin{equation*} \mathbf{V} = \left(\begin{array}{ccc} 2kR^{2} & -kR^{2} & -kR^{2}\\ -kR^{2} & 2kR^{2} & -kR^{2}\\ -kR^{2} & -kR^{2} & 2kR^{2} \end{array}\right)\hspace{1pc}\mbox{ and }\hspace{1pc}\mathbf{T} = \left(\begin{array}{ccc} mR^{2} & 0 & 0\\ 0 & mR^{2} & 0\\ 0 & 0 & mR^{2} \end{array}\right) \end{equation*}

Physically, though, it seems like there should be three eigenfrequencies. The zero eigenfrequency corresponds to uniform rotation of all the masses together. But shouldn't there be two non-zero frequencies, corresponding to (1) one mass being stationary while the other two oscillate out of phase and (2) two masses oscillating in phase and the third oscillating out of phase? Incidentally, I'm not quite sure about the third term in the potential energy, given that the angles are $2\pi$-periodic, but the way $V$ is written, it vanishes at the equilibrium and so does the derivative of $V$ with respect to each $\theta_{i}$ at equilibrium, so it seems alright.

Answer 1

I didn't try to replicate your approach but instead took my own using $x$ instead of $\theta$ (which gives the same result). I am pretty sure this problem simplifies to a linear three-mass-three-spring system with the following structure

$$ \bf \ddot X = [M] X $$ where $\mathbf {X} = [x_1 \ x_2 \ x_3]^T$ and

$$ \mathbf{M} = \left[ \begin{array}{ccc} (-k_1-k_3)/m_1 & k_1/m_1 & k_3/m_1 \\ k_1/m_2 & (-k_1-k_2)/m_2 & k_2/m_2 \\ k_3/m_3 & k_2/m_3 & (-k_2-k_3)/m_3 \end{array} \right] $$

Setting $m_1 = m_2 = m_3$ and $k_1 = k_2 = k_3$ we have

$$ \mathbf{M} = \left[ \begin{array}{ccc} (-2k)/m & k/m & k/m \\ k/m & (-2k)/m & k/m \\ k/m & k/m & (-2k)/m \end{array} \right] $$

then obtain by eigen-decomposition, as you did (give or take a '-' sign),

$\omega^2 = [0 \ -3k/m \ -3k/m]$

with corresponding eigenvectors

$[1 \ 1 \ 1]^T \ \ [-1 \ 1 \ 0]^T$ and $[-1 \ 0 \ 1]^T$.

So I guess our first instinct is not correct and there is only one nonzero frequency in this case. However, based on some inspection, if we relax the condition that $m_1 = m_2 = m_3$ and/or $k_1 = k_2 = k_3$, with some help from symbolic software like MAPLE or MATLAB, we will get two different eigenfrequencies as we break the three-fold symmetry.

For example, if I numerically set $m_1 = m_3 = 1$ but $m_2 = 2$ and $k_1 = k_2 = k_3 = 1$ then I obtain

$\omega^2 = [0 \ -2 \ -3]$

with corresponding eigenvectors

$[1 \ 1 \ 1]^T \ \ [-1 \ -1 \ \ 1]^T$ and $[-1 \ 0 \ 1]^T$.

Which is more along the lines of what you were expecting to see.