 Solution

The initial displacement lies in an invariant subspace of axial motions. Let the three opposite vertex pairs have positionsFor a vertex on the positive $x$ axis, the two springs towards the $y$ vertices contribute $-k(q_x+q_y)$ in the $x$ direction; the two towards the $z$ vertices contribute $-k(q_x+q_z)$. Transverse components cancel in pairs. Cyclically,The matrix is $I+11^{\mathsf{T}}$. It has eigenvalue $4$ on the breathing vector $(1,1,1)$, and eigenvalue $1$ on the orthogonal plane $q_x+q_y+q_z=0$. This identifies the axial normal modes of an octahedral spring framework without a large characteristic determinant.

Put ${\omega }_0=\sqrt{k/m}$ and decompose the initial vector asSince all initial velocities are zero, only cosine time factors occur. The complete motion to first order in the displacements iswith the vertex positions given by the first displayed formula. No translation or rotation is excited: the initial displacements are orthogonal to those zero modes and their initial velocities vanish. The two checks ${ϵ}_1={ϵ}_2$ and ${ϵ}_1+2{ϵ}_2=0$ respectively isolate pure breathing and pure traceless axial oscillation. Nonlinear corrections beyond the small-displacement approximation have been omitted.