= Solution
Along the <holonomic constraint> $z=x^2/(4a)$, we have $\dot z=x\dot x/(2a)$ and hence speed squared $(1+x^2/(4a^2))\dot x^2$. The smooth wire's <normal reaction> is perpendicular to the permitted <velocity> and does no work. <Conservation of mechanical energy>, with release from rest at height $h$, gives
$$
\frac12m\left(1+\frac{x^2}{4a^2}\right)\dot x^2+\frac{mgx^2}{4a}=mgh.
$$
The <turning points> are $x=\pm x_0$ with $x_0=2\sqrt{ah}$. One full oscillation has twice the travel time between them:
$$
T=2\int_{-x_0}^{x_0}
\sqrt{\frac{1+x^2/(4a^2)}{2g[h-x^2/(4a)]}}dx.
$$
Set $x=x_0u$ and $\beta=h/a$. After cancelling the scale factors,
$$
T=2\sqrt{\frac{2a}{g}}\int_{-1}^1
\sqrt{\frac{1+\beta u^2}{1-u^2}}du,
$$
so the <bead on a parabolic wire> has
$$
\boxed{G(\beta)=2\sqrt2\int_{-1}^1\sqrt{\frac{1+\beta u^2}{1-u^2}}du.}
$$
For $\beta\ll1$, expand the numerator uniformly on $[-1,1]$:
$$
\sqrt{1+\beta u^2}=1+\frac12\beta u^2+O(\beta^2).
$$
The integrable endpoint weight gives $\int_{-1}^1(1-u^2)^{-1/2}du=\pi$ and $\int_{-1}^1u^2(1-u^2)^{-1/2}du=\pi/2$, for example by $u=\sin\theta$. Therefore the requested small-oscillation period is
$$
\boxed{T=2\pi\sqrt{\frac{2a}{g}}
\left[1+\frac{h}{4a}+O\left((h/a)^2\right)\right].}
$$
The leading frequency is $\sqrt{g/(2a)}$, and the first finite-amplitude correction increases the period. At $h=0$ the ring is stationary; the expression is the limiting period of nonzero small oscillations.
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