The equation of the pendulum is: $$\ddot{\theta}+\frac{g}{l}\sin\theta$$
After some manipulation, we get
$$H=\frac{\dot{\theta}^{2}}{2}-\frac{g}{l}\cos\theta=\mathrm{positive\ constant}$$
Trajectories corresponds to the solution of the equation(above). Evidently, there are three possibilities.
1.$$H<\frac{g}{l}$$
2.$$\frac{-g}{l}\leq H\leq \frac{g}{l}$$
3.$$H>\frac{g}{l}$$
1. For $H<\frac{g}{l}$
(a) When $\theta =\pm \pi$:
$$\dot{\theta}=\pm\sqrt{2(H-\frac{g}{l})} =\pm i\sqrt{2(\frac{g}{l}-H)}$$and we get real solutions.
(b) When $\theta$ = 0:
$$\dot{\theta}=\pm \sqrt{2(H+\frac{g}{l})}=\mathrm{real\ solution}$$
2. Now, for $\frac{-g}{l}<H<\frac{g}{l}$
(a) When $\frac{-g}{l}<H\ \text{and}\ \theta =\pm \pi$:
$$\dot{\theta}=\pm 2\sqrt{2(H-\frac{g}{l})}$$ Since $H>\frac{-g}{l},\ \mathrm{then}\ H-\frac{g}{l}$ is positive. So this case corresponds to real solutions.
(b) When $H>\frac{g}{l}\ \text{and}\ \theta =\pm \pi$:
$$\dot{\theta}=\pm i\sqrt{2(\frac{g}{l}-H)}$$ so the solution is imaginary.
3. There is a final case $H>\frac{g}{l}$ but shall omit it.
How do the different cases of $H$ relate to the trajectories?
The trajectory looks like a figure 8 rotated 90 degrees either direction.
Thanks in advance.