= Solution
For nonzero $\omega$, the <equilibrium points> satisfy $\sin2\theta=0$. The <equilibrium point> $\theta=0$ is stable: it puts the bead at the point farthest from the rotation axis. Writing $\theta=\eta$ small in the tangential <ordinary differential equation> gives
$$
\ddot\eta=-\omega^2\eta+O(\eta^3),\qquad\boxed{\Omega_{\rm osc}=|\omega|,\quad f_{\rm osc}=\frac{|\omega|}{2\pi}.}
$$
Thus the <small oscillation> is <simple harmonic motion>, with <angular frequency> $|\omega|$ and ordinary <frequency> $|\omega|/(2\pi)$.
The endpoints $\theta=\pm\pi/2$ represent the same rim point $O$. A perturbation there instead satisfies $\ddot\eta=+\omega^2\eta$ and is unstable. Equivalently, the <centrifugal potential> along the hoop is $-\tfrac12\omega^2|\mathbf r|^2=-2a^2\omega^2\cos^2\theta$, whose minimum is at zero and maximum at $O$. If $\omega=0$, every bead position with zero relative <velocity> is an <equilibrium point>, but perturbing that <velocity> produces unrestored drift; there is no nonzero restoring <frequency>.
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