Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 15 - Oscillations - Problems - Page 438: 42a

Answer

L = 0.500 meters

Work Step by Step

Given: $\theta$ = 0.0800cos(4.43t + $\phi$} $m$ = 60g = 0.06 Kg Use the equation L = $\frac{g}{\omega^2}$ and recognize that $\theta$ = 0.0800cos(4.43t + $\phi$}, = 0.0800cos($\omega$t + $\phi$}: Thus $\omega$ = 4.43 and: L = $\frac{g}{\omega^2}$ = $\frac{9.81}{4.43^2}$ = 0.500 meters
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.