= Solution
In the small-amplitude approximation, $\dot x$ is the speed along the track, and $y=Bx^2$ supplies the <gravitational potential>. The conserved energy is
$$
E=\frac M2\left(1+\frac{r_2^2}{2r_1^2}\right)\dot x^2+MgBx^2.
$$
Differentiating and extending through turning points by continuity gives
$$
\left(1+\frac{r_2^2}{2r_1^2}\right)\ddot x+2gBx=0.
$$
This is <simple harmonic motion> with $\omega_0^2=4gB/\{(r_2/r_1)^2+2\}$. Hence
$$
\boxed{f_0=\frac{\sqrt{gB}}{\pi\sqrt{(r_2/r_1)^2+2}}=0.235\,\mathrm{Hz}.}
$$
The <frequency> is lower than for a sliding point <mass> because energy is also stored in rotation.
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