Past exam of the mathematical methods II course of the University of Oxford / 2013 / 8 / e / Solution

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 ${\omega }_{\pm }/(2\pi )$.