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

$T=2 \sqrt{ 2} \pi\ sec= 8.89\ sec .$
The period $T$ of a pendulum is given by $$T=2\pi\sqrt{\frac{l}{32}}=2\pi\sqrt{\frac{64}{32}}=2 \sqrt{ 2} \pi\ sec .$$ Thus, we have: $$T=2 \sqrt{ 2} \pi\ sec= 8.89\ sec .$$