= Solution
For $\boldsymbol\theta=\mathbf a\cos(\omega t+\varphi)$, a nonzero <normal mode> satisfies $(\mathbf V-\omega^2\mathbf T)\mathbf a=0$. With $s=\omega^2l/g$, its determinant is $4(s^2-3s+1)$. Hence the <normal modes of a double pendulum with a two-to-one length ratio> have
$$
\boxed{\omega_-^2=\frac{g}{2l}(3-\sqrt5),\qquad\frac{\theta_2}{\theta_1}=\sqrt5-1,}
$$
$$
\boxed{\omega_+^2=\frac{g}{2l}(3+\sqrt5),\qquad\frac{\theta_2}{\theta_1}=-1-\sqrt5.}
$$
The slow mode is in phase and the fast mode out of phase. The general small motion is their superposition with independent amplitudes and phases; ordinary frequencies are $\omega_\pm/(2\pi)$.
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