Answer
$ T=2 \sqrt{ 2} \pi\ sec= 8.89\ sec
.$
Work Step by Step
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:
$$ T=2 \sqrt{ 2} \pi\ sec= 8.89\ sec
.$$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
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: 109
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
