 Solution

The first two vectors displace every particle identically in the horizontal or vertical direction. Every relative displacement is zero, so every spring has zero extension to first order. The third vector is $(y_i,-x_i)$ evaluated at the equilibrium points, which is an infinitesimal clockwise rotation. For any edge the relative displacement is perpendicular to that edge, again giving zero extension.

These are the rigid-motion zero modes of a spring framework: two translations and one infinitesimal rotation. Each has $Kq=0$, so $\boxed{\omega =0}$. In the linearized dynamics its amplitude has the form $A+Bt$, expressing free rigid motion rather than a restoring oscillation. Exact finite rigid rotations also preserve every spring length; the mode vector is their first-order tangent.