[Physics] Hamilton’s equations for a simple pendulum

Question

I don't get how to use Hamilton's equations in mechanics, for example let's take the simple pendulum with
$$H=\frac{p^2}{2mR^2}+mgR(1-\cos\theta)$$
Now Hamilton's equations will be:
$$\dot p=-mgR\sin\theta$$ $$\dot\theta=\frac{p}{mR^2}$$
I know one of the points of Hamiltonian formalism is to get first order diff. equations instead of second order that Lagrangian formalism gives you, but how can I proceed from here without just derivating again wrt. $\dot\theta$ and substituting $\dot p$ to get the same equation that I get with the Lagrangian formulation? Or is that the way to do it? And how could I get the path of the system on the phase space with those equations?

Answer 1

Generally both formulations (Largangian and Hamiltonian) are equivalent, but in your case, if $\theta$ is small, you have a simplified equation for $p$ and you can use a solution ansatz like $e^{i\omega t}$ for both $p$ and $\theta$.

To draw a path in the phase space, you have to solve the equations and/or manage to express $p(\theta)$ or $\theta(p)$.