Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 13 - Exercises and Problems - Page 238: 34


.20 meters

Work Step by Step

We simplify the equation for energy to solve for the Amplitude and find: $A = \sqrt{\frac{2E}{4\pi^2f^2m}}$ $A = \sqrt{\frac{2(.51)}{4\pi^2(1.2)^2(.45)}}=\fbox{.20 meters}$
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.