Chapter 7.1, Problem 22P

(a)

To determine

To prove: The sequence of equations $\frac{1}{2}m{L}^2\frac{d}{d\theta }\left[{\left(\frac{d\theta }{dt}\right)}^2\right]=-mgL\sin \theta$, $\frac{1}{2}m{\left(L\frac{d\theta }{dt}\right)}^2=mgL\left(\cos \theta -\cos \alpha \right)$, $dt=-\sqrt{\frac{L}{2g}}\frac{d\theta }{\sqrt{\cos \theta -\cos \alpha }}$ where $m{L}^2\frac{d^2\theta }{d{t}^2}+mgL\sin \theta =0$ is the formula for natural period of an undamped nonlinear pendulum. That is obtained by setting $c=0$ in equation of motion $\frac{d^2\theta }{d{t}^2}+\frac{c}{mL}\frac{d\theta }{dt}+\frac{g}{L}\sin \theta =0$.

Also, give the reason behind the negative square root chosen in the last equation.

(b)

To determine

To prove: The formula $\frac{T}{4}=-\sqrt{\frac{L}{2g}}{\displaystyle \underset{\alpha }{\overset{0}{\int }}\frac{d\theta }{\sqrt{\cos \theta -\cos \alpha }}}$, where $T$ is the natural period of oscillation.

(c)

To determine

To prove: The elliptical integral $T=4\sqrt{\frac{L}{g}}{\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{d\varphi }{\sqrt{1-k^2{\sin }^2\varphi }}}$ by using the identities $\cos \theta =1-2{\sin }^2\left(\frac{\theta }{2}\right)\text{and}\cos \alpha =1-2{\sin }^2\left(\frac{\alpha }{2}\right)$ followed by a change of variable $\sin \left(\frac{\theta }{2}\right)=k\sin \varphi \text{with}k=\sin \left(\frac{\alpha }{2}\right)$.

(d)

To determine

The values of $T$, by evaluating the integral in expression for $T$ and compare with graphical estimate obtain in problem 20 as given below: