If we consider the Hamiltonian for the simple harmonic oscillator given by,
$$H(p,x) = \frac{p^2}{2m}+\frac{kx^2}{2}$$
where $m$ is the mass, $k$ is the stiffness and $p$ is the momentum, then the equations of motion for the oscillator can be written as,
$$\frac{dp}{dt}= -\frac{\partial H}{\partial x},\,\,\,\,\,\,\,\,\,\frac{dx}{dt}=\frac{\partial H}{\partial p}$$
The equations of motion in the Lagrange subsidiary form can be written as,
$$\frac{dp}{-kx} =\frac{dx}{(p/m)}=\frac{dt}{1}$$
How can one solve the subsidiary equations to show that the trajectory of the oscillator is,
$$x(t) = A\,\text{sin}(\omega t+\phi)$$
where $\omega = \sqrt \frac{k}{m}$ and $A$ and $\phi$ are constants.
A more useful rearrangement of the respective Hamilton's equations is $\dot{p}=-kx,\,\dot{x}=\frac{p}{m}$ (but you can also obtain these from the subsidiary form) so $\ddot{x}=-\omega^2 x$.