= Normal modes of an equilateral spring triangle
{title2=$\omega^2=0,0,0,3k/(2m),3k/(2m),3k/m$}
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 $u_i=Jr_i$, 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 $3\times6$ <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 $BB^T=(3/2)I+(1/2)\mathbf1\mathbf1^T$, with <eigenvalues> $3,3/2,3/2$. The nonzero <eigenvalues> of $B^TB$ are the same, so the stiffness matrix $kB^TB$ 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.
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