 Solution

Put ${\theta }_i={\theta }_i^0+{\eta }_i$. Differentiating the spring length givesThe springs are unextended at equilibrium, so the quadratic potential energy contains only the squares of these first-order extensions:The linearized Euler-Lagrange equations are thereforeThe stiffness matrix is $(k/4)(3I-11^T)$. Its eigenvector $1=(1,1,1{)}^T$ has zero eigenvalue, and every vector whose components sum to zero has eigenvalue $3k/4$. One convenient orthonormal set of normal modes isAny orthonormal basis of the sum-zero plane is equally valid for the two degenerate normal modes. The general small motion isThe linear drift in the zero mode describes a slow uniform rotation. These are angular frequencies; the ordinary frequencies of the two vibrational modes are ${\omega }_1/(2\pi )$.