= Solution
{c}
The <linearization> at rest at $r=0$ is
$$
\ddot r=\left(\omega^2-\frac ga\right)r.
$$
For $\omega^2<g/a$, small disturbances oscillate. To verify nonlinear <stability>, multiply the full <equation of motion> by $\dot r/a^2$ and integrate to obtain the <first integral>
$$
J=\frac12\left(1+\frac{r^2}{a^2}\right)\dot r^2+\frac12\left(\frac ga-\omega^2\right)r^2.
$$
For $\omega^2<g/a$, $J$ is positive away from the <equilibrium point> in a sufficiently small <neighborhood> of $(r,\dot r)=(0,0)$; its small <level sets> confine the motion. Thus
$$
\boxed{r=0\text{ is stable if }\omega^2<g/a,\text{ and unstable if }\omega^2>g/a.}
$$
At equality, every resting location on the wire is an <equilibrium point>, so a displacement at zero relative <velocity> simply remains there: the equilibrium is \b[neutral with respect to resting displacements]. It is \b[unstable with respect to general perturbations in <phase space>], however. An arbitrarily small outward initial <velocity> $v_0>0$ gives $\dot r=v_0/\sqrt{1+r^2/a^2}>0$ from the <first integral>, eventually taking the bead outside any prescribed small <neighborhood>. Thus strict <phase space> <stability> holds only for $\omega^2<g/a$.
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