Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 14 - Oscillations - Exercises and Problems: 54

Answer

For a simple pendulum of a mass on a string: $\omega = \sqrt{\frac{mgd}{I}} = \sqrt{\frac{g}{L}}$

Work Step by Step

For a mass on a string of length $L$, the moment of inertia is $I = mL^2$. The distance $d$ from the center of mass to the pivot is $L$. We can find an expression for the angular frequency as: $\omega = \sqrt{\frac{mgd}{I}}$ $\omega = \sqrt{\frac{mgL}{mL^2}}$ $\omega = \sqrt{\frac{g}{L}}$
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.