## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$T= \frac{ \sqrt{ 3} }{6} \pi\ sec= 0.91\ sec .$
Since $1 ft=12 in$, then the period $T$ of a pendulum is given by $$T=2\pi\sqrt{\frac{l}{32}}=2\pi\sqrt{\frac{8/12}{32}}=\sqrt{\frac{1}{3}} \frac{\pi}{2} \ sec\\ = \frac{ \sqrt{ 3} }{6} \pi\ sec .$$ Thus, we have $$T= \frac{ \sqrt{ 3} }{6} \pi\ sec= 0.91\ sec .$$