Appendix A - Review - A.7 nth Roots; Rational Exponents - A.7 Assess Your Understanding - Page A62: 111

$ 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 $$ T= \frac{ \sqrt{ 3} }{6} \pi\ sec= 0.91\ sec .$$