PX285 - D4 - roller-coaster revisited

revisiting the roller-coaster problem
the lagrangian:

L = \frac{1}{2} m {\dot{s}}^{2} - m g h (s)

the hamiltonian:

H = \dot{s} \frac{\partial L}{\partial \dot{s}} - L = \frac{1}{2} m {\dot{s}}^{2} + m g h (s) = E

seeking an equation for the trajectory, ie, an ODE for $\dot{s} :$

\begin{matrix} {\dot{s}}^{2} = \frac{2}{m} (E - m g h (s)) \\ \frac{d s}{d t} = \pm \sqrt{\frac{2}{m}} \sqrt{E - m g h (s)} \\ \pm \sqrt{\frac{m}{2}} \int_{s_{0}}^{s_{1}} \frac{d s^{'}}{\sqrt{E - m g h (s)}} = \int_{0}^{t} d t^{'} \\ ⟹ f (s) = t \end{matrix}

therefore, $t$ as a function of $s$ is obtained, which can then be used to obtain $s$ as a function of $t :$

s = f^{- 1} (t)

consider a promising particular case, a cycloid:

h (s) = \frac{1}{2} c s^{2}

t = \pm \sqrt{\frac{m}{2}} \int_{s_{0}}^{s_{1}} \frac{d s^{'}}{\sqrt{E - m g (\frac{1}{2} c s^{2})}}

- try: $s = \sqrt{\frac{2 E}{m g c}} \sin θ$
$⟹ d s^{'} = \sqrt{\frac{2 E}{m g c}} \cos θ d θ$

\frac{m g c}{2} s^{' 2} = \frac{m g c}{2} \frac{2 E}{m g c} \sin^{2} θ = E \sin^{2} θ

\begin{aligned} t & = \pm \sqrt{\frac{m}{2}} \int_{θ (s_{0})}^{θ (s)} \frac{\sqrt{\frac{2 E}{m g c}} \cos θ d θ}{\sqrt{E (1 - \sin^{2} θ)}} \\ = \pm \sqrt{\frac{m}{2}} \sqrt{\frac{2}{m g c}} \int_{θ (s_{0})}^{θ (s)} d θ \\ = \pm \sqrt{\frac{1}{g c}} [θ (s_{0}) - θ (s)] \end{aligned}

define $ω^{2} = g c :$

\begin{matrix} ω t = \pm (θ (s) - θ (s_{0})) \\ ω t \pm θ (s_{0}) = \pm θ (s) \\ \sin (ω t + ϕ) = \sin θ = s \sqrt{\frac{m g c}{2 E}} \\ ∴ s = \sqrt{\frac{2 E}{m g c}} \sin (ω t + ϕ) \end{matrix}

this is an equation of simple harmonic motion
at maximum height:

s (0) = \sqrt{\frac{2 E}{m g c}} \sin ϕ = s_{0}