= Solution
\b[Constant azimuth.] If $\phi$ is constant, $\dot\phi=0$ and the motion is a planar <simple pendulum>:
$$
\boxed{\ddot\theta+\frac gl\sin\theta=0.}
$$
For the <small-angle approximation>, $\sin\theta\simeq\theta$, giving <simple harmonic motion> with
$$
\boxed{\omega_{\mathrm{plane}}=\sqrt{\frac gl},\qquad T_{\mathrm{plane}}=2\pi\sqrt{\frac lg}.}
$$
The harmonic <angular frequency> requires small <amplitude>. At finite libration amplitude $\theta_{\max}<\pi$, the exact <oscillation period> is $4\sqrt{l/g}\,K(\sin(\theta_{\max}/2))$, where $K$ is the <complete elliptic integral of the first kind>; the nonlinear pendulum does not have the same amplitude-independent period.
\b[Constant polar angle.] If $\theta=\theta_0$ is nonzero and constant, the bob traces a horizontal <circle>: this is a <conical pendulum>. Setting $\ddot\theta=0$ and dividing by $\sin\theta_0$ gives
$$
\Omega^2\cos\theta_0=\frac gl,\qquad
\boxed{\dot\phi=\Omega=\pm\sqrt{\frac{g}{l\cos\theta_0}},\qquad 0<\theta_0<\frac\pi2.}
$$
The sign distinguishes the two senses of rotation. The exactly downward configuration $\theta_0=0$ is stationary; its azimuth is undefined and it must not be obtained by dividing by $\sin\theta_0$. A nontrivial constant-angle orbit of this gravity-only system needs $\cos\theta_0>0$ even though the supporting rod can also push.
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