= Solution
For $T=\frac12\dot q^T T\dot q$ and $V=\frac12q^T Vq$, where the <matrices> are symmetric and constant, the <Euler-Lagrange equations> give $T\ddot q+Vq=0$. A <normal mode> $q=v e^{i\omega t}$ therefore solves $(V-\omega^2T)v=0$.
A string segment of horizontal length $b$ has extension $\sqrt{b^2+(\Delta q)^2}-b=(\Delta q)^2/(2b)+O((\Delta q)^4)$. Multiplying by the tension $S$ and using $m=1$ and $S/b=1$, the quadratic <kinetic energy> and <potential energy> are
$$
T=\tfrac12(\dot q_1^2+\dot q_2^2+\dot q_3^2),\qquad V=\tfrac12\{q_1^2+(q_1-q_2)^2+(q_2-q_3)^2+q_3^2\}.
$$
Thus $T=I$ and $V=\begin{pmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{pmatrix}$. Taking $v_i=\sin(ij\pi/4)$, $i=1,2,3$, gives the three <eigenvalues> $2-2\cos(j\pi/4)$, $j=1,2,3$. The <normal mode> frequencies are
$$
\boxed{\omega_1=\sqrt{2-\sqrt2},\qquad\omega_2=\sqrt2,\qquad\omega_3=\sqrt{2+\sqrt2}}.
$$
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