= Solution
Put $\theta=\theta_e+\delta$ at the stable <equilibrium>. Since $M\sin\theta_e=m$, linearization of <9/d> gives
$$
(M+m)R\ddot\delta+Mg\cos\theta_e\,\delta=0.
$$
Thus
$$
\boxed{\omega_0=\sqrt{\frac{Mg\cos\theta_e}{(M+m)R}}=\sqrt{\frac{g\sqrt{M^2-m^2}}{(M+m)R}},\qquad f_0=\frac{\omega_0}{2\pi}.}
$$
As $m\to0$, the result approaches $\sqrt{g/R}$, the <simple pendulum> value. As $m\uparrow M$, the restoring curvature tends to zero and this small-amplitude <frequency> tends to zero; the endpoint itself is not a stable quadratic minimum.
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