Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 14 - Exercises and Problems - Page 270: 17

Answer

The answer is below.

Work Step by Step

a) We know that the amplitude is the number in front of the cosine, so it is $1.3 \ cm$. b) $ \lambda= \frac{2\pi}{.69}=9.11\ cm$ c) Using the equation, we find $\omega = 31s$. Period equals 2 times pi divided by omega. This gives: $ T = \frac{2\pi}{31}=.2s$ d) $v=\frac{\omega}{k}=\frac{31}{.69}=45\ cm$ e) Negative x direction
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.