Solve the plane pendulum problem using the Hamiltonian approach and show that H is a constant of motion.
Approach : Assuming it means simple pendulum by plane pendulum, I have calculated Hamiltonian as a function of $\theta , l$ (Length of pendulum) ,$ \ p_\theta $
But what does it mean by H is a constant of motion ?
$$H=\frac{p^2}{2m\ell^2}+mg\ell(1-\cos\theta)$$
$H=T+V$ is the total energy of the system. Now Hamilton's equations will be: $$\dot p=-mg\ell\sin\theta\qquad\dot \theta=\frac{p}{m\ell^2}$$ A constant of motion is a quantity that is conserved throughout the motion.
$$ \frac{\mathrm dH(q,p,t)}{\mathrm dt}=\frac{\partial H}{\partial q}\dot q+\frac{\partial H}{\partial p}\dot p+\frac{\partial H}{\partial t}=-\dot p\dot q+\dot q\dot p+\frac{\partial H}{\partial t}=\frac{\partial H}{\partial t} $$ So the Hamiltonian is conserved if it does not depend explicitly on $t$.
If the mass of the pendulum bob is $m$, the length of the "arm" is $l$, and the angle the arm makes with the vertical is $\theta$, the speed $v$ of the bob is
$v = l\dot \theta; \tag 1$
it follows that the kinetic energy is
$T(\dot \theta) = \dfrac{1}{2}mv^2 = \dfrac{1}{2}ml^2 {\dot \theta}^2; \tag 2$
furthermore, if we assume that the "rest" position of the pendulum, where it is hanging "straight down", is at $\theta = 0$, then the potential energy is
$V(\theta) = mg(l - l \cos \theta) = mgl(1 - \cos \theta); \tag 3$
the Lagrangian for this system is then
$L(\theta, \dot \theta) = T(\dot \theta) - V(\theta) = \dfrac{1}{2}ml^2 {\dot \theta}^2 - mgl(1 - \cos \theta); \tag 4$
the cannonical momentum associated with the coordinate $\theta$ is
$p_\theta = \dfrac{\partial L(\theta, \dot \theta)}{\partial {\dot \theta}} = ml^2\dot \theta, \tag 5$
which, according to the usual prescription, leads to the Hamiltonian
$H(p_\theta, \theta) = p_\theta \dot \theta - L(\theta, \dot \theta) = ml^2 {\dot \theta}^2 - \dfrac{1}{2}ml^2 {\dot \theta}^2 + mgl(1 - \cos \theta)$ $=\dfrac{1}{2}ml^2 {\dot \theta}^2 + mgl(1 - \cos \theta) = \dfrac{p_\theta^2}{2ml^2} + mgl(1 - \cos \theta), \tag{6}$
from which we may directly write Hamilton's equations of motion:
${\dot p}_\theta = -\dfrac{\partial H(p_\theta, \theta)}{\partial \theta} = -mgl\sin \theta, \tag 7$
$\dot \theta = \dfrac{\partial H(p_\theta, \theta)}{\partial p_\theta} = \dfrac{p_\theta}{ml^2}; \tag 8$
using (5) in (7) we find
$ml^2 \ddot \theta = -mgl\sin \theta, \tag 9$
or
$\ddot \theta = -\dfrac{g}{l} \sin \theta, \tag{10}$
readily recognizable as the equation of motion of the simple pendulum; on the other hand, (8) yields
$\dot \theta = \dfrac{ml^2 \dot \theta}{ml^2} = \dot \theta, \tag{11}$
an identity.
It is readily seen from (6), again using (5), that
$H(p, q) = \dfrac{p_\theta^2}{2ml^2} + mgl(1 - \cos \theta) = \dfrac{m^2l^4 \dot \theta^2}{2ml^2} + mgl(1 - \cos \theta)$ $= \dfrac{ml^2 \dot \theta^2}{2} + mgl(1 - \cos \theta) = T(\dot \theta) + V(\theta), \tag{12}$
the sum of the kinetic and potential energies of the pendulum,i.e., the net energy of the system. As such, we should expect it to be constant along the trajectories of the equations of motion (7), (8); it remains the same as the system evolves, hence the term, "constant of the motion." Sometimes the fact that $H(p, q)$ is constant along system trajectories is known as the conservation of energy".
We can see this analytically by differentiation $H$ with respect to $t$:
$\dfrac{dH(p, q)}{dt} = \dfrac{\partial H(p_\theta, \theta)}{\partial p_\theta} \dot p_\theta + \dfrac{\partial H(p_\theta, \theta)}{\partial \theta} \dot \theta = \dot \theta \dot p_\theta - \dot p_\theta \dot \theta = 0, \tag{13}$
by vitue of (7) and (8).
"of motion" is a physics idiom, and I'm not sure what it adds to the statement. You want to show $$ \frac{dH}{dt} = \frac{\partial H}{\partial \theta}\dot \theta +\frac{\partial H}{\partial p_{\theta}}\dot p_\theta = 0 $$
