 Solution

For $T=\frac{1}{2}{\dot{q}}^{T}T\dot{q}$ and $V=\frac{1}{2}{q}^{T}Vq$, 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 $(V-{\omega }^{2}T)v=0$.

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 areThus $T=I$ and $V=\left(\begin{array}{ccc}2& -1& 0\\ -1& 2& -1\\ 0& -1& 2\end{array}\right)$. 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