Normal modes of an equilateral spring triangle (${\omega }^2=0,0,0,3k/(2m),3k/(2m),3k/m$)

Three equal point masses $m>0$ move in a plane and are connected pairwise by ideal springs of equal positive stiffness $k$. At an unstretched equilateral triangle equilibrium, let $r_i$ be the vertex positions relative to the centre, and let $u_i$ be small displacements. If $e_{ij}=(r_i-r_j)/|r_i-r_j|$, the quadratic potential energy is $(k/2){\sum }_{i<j}[e_{ij}\cdot (u_i-u_j){]}^2$, and the kinetic energy is $(m/2){\sum }_i|{\dot{u}}_i{|}^2$.

There are three independent zero-frequency motions: two common translations and the infinitesimal rigid rotation $u_i=J{r}_i$, where $J(x,y)=(-y,x)$. The uniform radial motion $u_i=r_i$ is the breathing normal mode, with ${\omega }^2=3k/m$. The two shear modes are $u_i=(r_{ix},-r_{iy})$ and $u_i=(r_{iy},r_{ix})$, each with ${\omega }^2=3k/(2m)$.

For a compact proof of the spectrum, form the $3\times 6$ matrix $B$ taking the six displacement components to the three first-order spring extensions, orienting the three edges cyclically. Its row norms squared are two, and every pair of rows has inner product $1/2$. Thus $B{B}^T=(3/2)I+(1/2)11^T$, with eigenvalues $3,3/2,3/2$. The nonzero eigenvalues of $B^{T}B$ are the same, so the stiffness matrix $k{B}^{T}B$ has the three positive values in the title and a three-dimensional kernel of a linear map. The three rigid motions span that kernel. A zero mode has time factor $A+Bt$ rather than a nonconstant periodic oscillation.