= Solution
Use the ordered displacement <vector> $q=(x_1,y_1,x_2,y_2,x_3,y_3,x_4,y_4)^T$, and define $K$ by $V=\tfrac12q^TKq$. For an unstretched <spring>, the extension to first order is the <dot product> of the relative <displacement> with its equilibrium <unit vector>. Summing its squared extensions gives the <central spring stiffness matrix>. The four side <springs> and two diagonal <springs> give
$$
V=\frac k2\left[(x_1-x_2)^2+(y_1-y_3)^2+(y_2-y_4)^2+(x_3-x_4)^2+\frac{(x_4-x_1-y_4+y_1)^2}2+\frac{(x_3-x_2+y_3-y_2)^2}2\right].
$$
Therefore
$$
K=\frac k2\begin{pmatrix}
3&-1&-2&0&0&0&-1&1\\
-1&3&0&0&0&-2&1&-1\\
-2&0&3&1&-1&-1&0&0\\
0&0&1&3&-1&-1&0&-2\\
0&0&-1&-1&3&1&-2&0\\
0&-2&-1&-1&1&3&0&0\\
-1&1&0&0&-2&0&3&-1\\
1&-1&0&-2&0&0&-1&3
\end{pmatrix}.
$$
The <kinetic energy> is $\tfrac m2\dot q^T\dot q$, so a <normal mode> satisfies $Kq=m\omega^2q$. Keeping the factor $1/2$ in the definition of $V$ fixes the normalization of $K$ and the resulting frequencies.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2003/paper-2-square-spring-modes.png]
{title=Rigid rotation, breathing and modes with reflection symmetry}
{description=Original displacement diagrams. Grey lines show all six equilibrium springs; arrows show representative mode vectors. The rotation has zero frequency, the breathing mode has squared frequency 4k/m, and the two other modes have squared frequency 2k/m.}
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