| $C=$ , $D=$ | |
| $\omega_{\rm F}/\omega_g$ = |
The rotational speed of the pendulum’s oscillation plane depends on latitude. At a location with northern latitude $\Theta$ ($0 \leqq \Theta \leqq \pi/2$), the time it takes for the oscillation plane to complete one full rotation is $1/\sin\Theta$ days. At the North Pole, it completes one rotation per day, while at the equator, it does not rotate. In my city, Fukuoka (latitude 33.6° N), it completes one rotation in about 1.8 days (1 day and 19 hours).
When the string length $\ell$ is much longer than the oscillation amplitude, the vertical motion (in the $z$-direction) of the mass can be neglected. In this case, the motion of the mass in the horizontal plane is given by 1) \begin{equation}\left\{\begin{array}{rl} x & = C\cos\big((\omega_g+\omega_F)t\big) +D\cos\big((\omega_g-\omega_F)t\big) \\ y & =-C\sin\big((\omega_g+\omega_F)t\big) +D\sin\big((\omega_g-\omega_F)t\big) \end{array}\right. . \tag{1} \end{equation} Here, $g$ is the gravitational acceleration, and $\omega$ is the angular velocity of Earth's rotation: \[ \omega_F := \omega\sin\Theta , \quad \omega_g :=\sqrt{{g\over \ell} + \omega^2\sin^2\Theta}\; . \] $C$ and $D$ are arbitrary constants, and phase indeterminacy is ignored. Usually, the pendulum’s period is much shorter than one day, so $\sqrt{g/\ell} \gg \omega$, thus $\omega_g \approx \sqrt{g/\ell}$.
In particular, when $C = D = C_0/2$, equation (1) becomes \[ x=C_0\cos(\omega_g t)\cos(-\omega_F t), \qquad y=C_0\cos(\omega_g t)\sin(-\omega_F t) \, . \] This corresponds to the pendulum passing through the lowest point ($x = y = 0$), oscillating with $\cos(\omega_g t)$, while the oscillation plane rotates clockwise (with angular velocity $\omega_F$).
In the simulator above, you can use sliders to adjust $C$, $D$, and $\omega_F/\omega_g$ to observe the trajectory of the mass in the horizontal plane.