Chapter 11.11, Problem

The period of a pendulum with length L that makes a maximum angle ${\theta }_0$ with the vertical is

$T=4\sqrt{\frac{L}{g}}{\displaystyle {\int }_0^{\pi /2}\frac{dx}{\sqrt{1-k^2{\sin }^{2}x}}}$

where $k=\sin \left(\frac{1}{2}{\theta }_0\right)$ and g is the acceleration due to gravity. (In Exercise 7.7.42 we approximated this integral using Simpson’s Rule.)

(a) Expand the integrand as a binomial series and use the result of Exercise 7.1.50 to show that

$T=2\pi \sqrt{\frac{L}{g{2}^2}}\left[1+\frac{1^2}{2^2}{k}^2+\frac{1^2{3}^2}{2^2{4}^2}{k}^4+\frac{1^2{3}^2{5}^2}{2^2{4}^2{6}^2}{k}^6+\cdot \cdot \cdot \right]$

If ${\theta }_0$ is not too large, the approximation $T=2\pi \sqrt{L/g}$, obtained by using only the first term in the series, is often used. A better approximation is obtained by using two terms:

$T=2\pi \sqrt{\frac{L}{g}}\left(1+\frac{1}{4}{k}^2\right)$

(b) Notice that all the terms in the series after the first one have coefficients that are at most $\frac{1}{4}$ Use this fact to compare this series with a geometric series and show that

$2\pi \sqrt{\frac{L}{g}}\left(1+\frac{1}{4}{k}^2\right)\le T\le 2\pi \sqrt{\frac{L}{g}}\frac{4-3k^2}{4-4k^2}$

(c) Use the inequalities in part (b) to estimate the period of a pendulum with $L=1$ meter and ${\theta }_0={10}^{\circ }$. How does it compare with the estimate $T\approx 2\pi \sqrt{L/g}$ What if ${\theta }_0={42}^{\circ }$?