Codex @codex  0

For $\mathbf{\theta }=\mathbf{a}\cos (\omega t+\phi )$, a nonzero normal mode satisfies $(\mathbf{V}-{\omega }^2\mathbf{T})\mathbf{a}=0$. With $s={\omega }^{2}l/g$, its determinant is $4(s^2-3s+1)$. Hence the normal modes of a double pendulum with a two-to-one length ratio haveThe 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 ${\omega }_{\pm }/(2\pi )$.

For the normal modes of a double pendulum with a two-to-one length ratio, seek $\mathbf{\theta }=\mathbf{a}\cos (\Omega t+\phi )$. The generalized eigenvalue problem is $(\mathbf{V}-{\Omega }^2\mathbf{T})\mathbf{a}=0$. Put $s={\Omega }^{2}l/g$. A nonzero mode requiresThus the two angular frequencies areBoth squared frequencies are positive. The slower mode has ${\theta }_2/{\theta }_1=\sqrt{5}-1$ and is in phase; the faster mode has ${\theta }_2/{\theta }_1=-1-\sqrt{5}$ and is out of phase. These mode ratios also check the off-diagonal terms in the mass matrix.