= Solution
The <Euler-Lagrange equations> from the quadratic <Lagrangian> are
$$
12\ddot\theta+2\ddot\phi+\frac{4g}{a}\theta=0,
\qquad2\ddot\theta+\ddot\phi+\frac ga\phi=0.
$$
For a <normal mode> proportional to $e^{i\omega t}$, let $\lambda=\omega^2a/g$. Nonzero amplitudes require
$$
\det\begin{pmatrix}4-12\lambda&-2\lambda\\-2\lambda&1-\lambda\end{pmatrix}
=4-16\lambda+8\lambda^2=0.
$$
Hence
$$
\boxed{\omega_\pm^2=\frac ga\left(1\pm\frac1{\sqrt2}\right).}
$$
The corresponding amplitude ratios are $\Phi/\Theta=2\lambda/(1-\lambda)$. The lower-frequency mode has \b[$\Phi/\Theta=2(\sqrt2-1)$], with the angles in phase; the higher-frequency mode has \b[$\Phi/\Theta=-2(1+\sqrt2)$], with opposite signs. The small hoop's spin remains zero in both modes.
Back to article page
