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

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

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 .$

Work Step by Step

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 $$ T= \frac{ \sqrt{ 3} }{6} \pi\ sec= 0.91\ sec .$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.