= Solution
To quadratic order about $\theta=\phi=0$,
$$
T_2=\frac{ma^2}{2}
\begin{pmatrix}\dot\theta&\dot\phi\end{pmatrix}
\begin{pmatrix}3&1\\1&1\end{pmatrix}
\binom{\dot\theta}{\dot\phi},
$$
and
$$
V_2=\frac{mga}{2}
\begin{pmatrix}\theta&\phi\end{pmatrix}
\begin{pmatrix}2&0\\0&1\end{pmatrix}
\binom{\theta}{\phi}.
$$
For a normal mode, put $c=\omega^2a/g$. The characteristic equation is
$$
\det\begin{pmatrix}2-3c&-c\\-c&1-c\end{pmatrix}
=2c^2-5c+2=0,
$$
so $c=1/2$ or $2$. The <small oscillations of a hoop with a sliding bead> therefore have
$$
\boxed{\omega_1=\frac1{\sqrt2}\sqrt{\frac ga},
\qquad
\omega_2=\sqrt2\sqrt{\frac ga}}.
$$
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