How can you read from a phase space diagram how quickly a path gets traversed? I'm learning about Hamiltonian mechanics and it is quite interesting. However I'm trying to understand how to see how quickly a path in phase space gets traversed. How do you read from a phase space diagram the period or the speed of which a particle travels this phase space path? For example a pendulum:

 A: You need to find a pair of generalized variable $q$ and its time derivative $\dot{q}$ (or, if you are using 100% Hamiltonian formalism, the generalized momentum associated $p$ to retrieve $\dot{q}$).
Then take a line in the phase-space passing through the point you wish as the initial point of the trajectory, namely the initial conditions of the system $q_k^0$, $\dot{q}_k^0$. This can be written in a parametric form as a function of a parameter $s$
$q_i = Q_i(s;q_k^0, \dot{q}_k^0)$
$\dot{q_i} = \dot{Q_i}(s;q_k^0, \dot{q}_k^0)$
Remembering that $\dot{q} = \dfrac{d q}{dt}$, with a change of variable $s(t)$, it's possible to write
$\dot{q}_i(s) = \dfrac{d q_i}{dt}(s) = \dfrac{d q_i}{d s} \dfrac{d s}{d t} = q_i'(s) \dfrac{ds}{dt}$,
and rearranging separating the variables
$dt = \dfrac{q'_i(s)}{\dot{q}_i(s)}ds$,
and eventually integrating along the path from the initial point identified by
the value of the parameter $s = s_0$ and time $t = t_0$ (related by the change of variable $s_0 = s(t_0)$), to the final point identified by the value $s = s_1$ and time $t = t_1$,
$\Delta t = t_1 - t_0 = \displaystyle \int_{t_0}^{t_1} dt = \int_{s_0}^{s_1} \dfrac{q'_i(s)}{\dot{q}_i(s)}ds$.
Example - small amplitude oscillation of a pendulum.
Let's do an example on the small amplitude oscillations of a pendulum, to find a solution analytically. We're taking a system whose phase-diagram is the one of your question, and the evolution is described by the small ellipses (in the small oscillation limit), around the origin.
The non-linear equation governing the free evolution of the system reads
$m L^2 \ddot{\theta} + m L g \sin \theta = 0$,
and the linearized equation around the stable equilibrium $\overline{\theta} = 0$ reads
$m L^2 \ddot{\theta} + m L g \theta = 0$$\qquad \rightarrow \qquad$
$\ddot{\theta} + \dfrac{g}{L} \theta = 0$$\qquad \rightarrow \qquad$
$\ddot{\theta} + \Omega \theta = 0$,
being $T = \dfrac{2 \pi}{\Omega}$ the period of the oscillation. Now we can choose $q = \theta$ and define $\dot{q} = \omega$ as the generalized coordinate and its time derivative, and write the second-order ODE as a system of 2 first-order ODEs, namely
$\left\{ \begin{array} \\ \dot{\theta} = \omega  \\  \dot{\omega} =  -\Omega^2 \theta \end{array}\right.$
Now, we have to eliminate time, as an example multiplying the two equations together, to get
$0 = \omega \dot \omega + \Omega^2 \theta \dot{\theta} = \dfrac{d}{dt} \left(\dfrac{1}{2}\omega^2 + \Omega^2 \dfrac{1}{2} \theta^2 \right)$,
that tell us that the content of the parentheses is a constant of the motion (its the mechanical energy here), and thus we can write
$2E = \omega^2 + \Omega^2 \theta^2 = \dot{q}^2 + \Omega^2 q^2$.
This is the equation of an ellipse in the phase-space, that can be written in parametric form (put the parametric form into the implicit one as a proof), as
$q(s) = \dfrac{\sqrt{2E}}{\Omega} \cos s$
$\dot{q}(s) = \sqrt{2E} \sin s$.
Now, we are ready to evaluate time between two points on the same line in the phase-space,
$\Delta t = t_1 - t_0 = \displaystyle \int_{t_0}^{t_1} dt = \int_{s_0}^{s_1} \dfrac{q'(s)}{\dot{q}(s)} ds = - \int_{s_0}^{s_1} \dfrac{\frac{\sqrt{2E}}{\Omega} \sin s}{\sqrt{2E} \sin s} ds  = - \int_{s_0}^{s_1} \dfrac{1}{\Omega}  ds = -\dfrac{\Delta s}{\Omega}$.
Few examples:

*

*full oscillation: $s_1 = s_0 + 2 \pi$, $\Delta t = \frac{2\pi}{\Omega}$,

*half oscillation: $s_1 = s_0 + \pi$, $\Delta t = \frac{\pi}{\Omega}$
A: For a particle in a 1D potential (i.e. in a 2D phase space), the period $T(E)=A^{\prime}(E)$ is the derivative of the phase space area $A(E)$ of the orbit wrt. the orbit's energy $E$, cf. above comment by knzhou, see e.g. my Phys.SE answer here.
If e.g the potential vanishes at $x=0$, then the energy $E=\frac{p(x=0)^2}{2m}$ can be read off from the orbit's intercept with the $p$-axis.
Hence OP's requested information can in principle be geometrically deduced from the phase space diagram.
