Chapter 7.1, Problem 19P

(a) Consider the equations of motion of an undamped pendulum,

$x\text{'}=y,y\text{'}-{\omega }^2\sin x$;(i)

see Example $1$. Convert this system into a single equation for $dy/dx$, and then show that the trajectories of the undamped pendulum satisfy the equation

$\frac{1}{2}{y}^2+{\omega }^2\left(1-\cos x\right)=c$,(ii)

Where $c$ is a constant of integration.

(b) Multiply Eq. (ii) by $m{L}^2$ and recall that ${\omega }^2=g/L$. Then, by expressing Eq. (ii) in terms of $\theta$, obtain

$\frac{1}{2}m{L}^2{\left(\frac{d\theta }{dt}\right)}^2+mgL\left(1-\cos \theta \right)=E$, (iii)

Where $E=m{L}^{2}c$.

(c) Show that the first term in Eq. (iii) is the kinetic energy of the pendulum and that the second term is the potential energy due to gravity. Thus the total energy $E$ of the pendulum is constant along any trajectory; in other words, the undamped pendulum satisfies the principle of conservation of energy. The value of $E$ is determines by the initial conditions.