The ODEs
$$
m L \theta^{\prime \prime}+m g \sin \theta=0 \text { and } m L \theta^{\prime \prime}+m g \theta=0
$$
model the motion of an undriven, undamped simple pendulum and of an undriven, undamped linearized pendulum, respectively. Orbital portraits of each ODE show a region of closed orbits encircling the origin. These closed orbits (or cycles) correspond to periodic solutions. Find and compare the periods of the cycles for the simple pendulum and for the linearized pendulum. Follow the outline below in proving the existence of cycles and in studying the periods. [Hint: see also Problem 14.]
- Linearized Pendulum The equation of the linearized pendulum is $m L \theta^{\prime \prime}+m g \theta=0$. Show that all nonconstant solutions are periodic of period $T=2 \pi \sqrt{L / g}$. Show that the corresponding orbits in the $\theta \theta$-state space are elliptical cycles. Choose a value for $L$, plot a portrait of cycles in the state space, plot component graphs, and verify graphically the formula for $T$.
- Closed Orbits of the Simple Pendulum Suppose that the simple pendulum modeled by $m L \theta^{\prime \prime}+m g \sin \theta=0$ is released from rest when $\theta=\theta_0$, where $0<\theta_0<\pi$. Show that the subsequent motion is periodic. [Hint: use Problem 14 to show that orbits are described by the relation $\left(\theta^{\prime}\right)^2=(2 g / L)\left(\cos \theta-\cos \theta_0\right)$, and use symmetries in this relation to show that the orbits are closed, and so represent periodic solutions.]
- Periods of the Simple Pendulum Let $T$ be the period of the orbit with $\theta(0)=\theta_0>0$, $\theta^{\prime}(0)=0$. Show that $T$ is given by
$$
T=4 \sqrt{\frac{L}{2 g}} \int_0^{\theta_0} \frac{d \theta}{\sqrt{\cos \theta-\cos \theta_0}}
$$
[Hint: since $\theta(t)$ initially decreases as $t$ increases, $\theta^{\prime}=-(2 g / L)^{1 / 2}\left(\cos \theta-\cos \theta_0\right)^{1 / 2}$. Show that $\theta$ continues to decrease until the time $t=t_1$ for which $\theta\left(t_1\right)=-\theta_0$.]
- Elliptic Integrals and the Periods of the Simple Pendulum Show that the change of variables $k=\sin \left(\theta_0 / 2\right), \sin \phi=(1 / k) \sin (\theta / 2)$ gives
$$
T=4 \sqrt{\frac{L}{g}} \int_0^{\pi / 2} \frac{d \phi}{\sqrt{1-k^2 \sin ^2 \phi}}
$$
The integral is an elliptic integral of the first kind Its approximate values have been tabulated ${ }^3$ for various values of $k$. For example, if $\theta_0=2 \pi / 3$, then $k=\sqrt{3} / 2$, and the value of the integral is about 2.157 . The corresponding period is about $8.628 \sqrt{L / g}$, quite different from the period of the linearized pendulum, which is $2 \pi \sqrt{L / g} \approx 6.282 \sqrt{L / g}$ Why would you expect the period of the nonlinear pendulum to be greater than the period of the linearized pendulum? Choose various values for $L$ and use an ODE solver to verify the above estimate for the periods.
- Asymptotic Values of the Periods of the Simple Pendulum Explain why $T \rightarrow 2 \pi \sqrt{L / g}$ as $\theta_0 \rightarrow 0$. It is known that $T \rightarrow \infty$ as $\theta_0 \rightarrow \pi$, although a complete mathematical proof of this fact is not given here. Why are these results expected on physical grounds?