= Solution
For the <normal modes of a double pendulum with a two-to-one length ratio>, seek $\boldsymbol\theta=\mathbf a\cos(\Omega t+\varphi)$. The <generalized eigenvalue problem> is $(\mathbf V-\Omega^2\mathbf T)\mathbf a=0$. Put $s=\Omega^2l/g$. A nonzero mode requires
$$
0=\det\begin{pmatrix}4-8s&-2s\\-2s&1-s\end{pmatrix}=4(s^2-3s+1).
$$
Thus the two angular frequencies are
$$
\boxed{\Omega_\pm=\sqrt{\frac{g}{2l}(3\pm\sqrt5)},\qquad f_\pm=\frac{\Omega_\pm}{2\pi}.}
$$
Both squared frequencies are positive. The slower mode has $\theta_2/\theta_1=\sqrt5-1$ and is in phase; the faster mode has $\theta_2/\theta_1=-1-\sqrt5$ and is out of phase. These mode ratios also check the off-diagonal terms in the mass matrix.
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