Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 15 - Oscillations - Exercises and Problems: 7

Answer

(a) $\phi = \frac{5\pi}{3}$ (b) $v = 0.136~m/s$ (c) $v_{max} = 0.157~m/s$

Work Step by Step

(a) The general equation for the motion is: $x(t) = A~cos(\omega~t+\phi)$ At $t = 0$: $x = A~cos(\phi) = \frac{A}{2}$ $cos(\phi) = \frac{1}{2}$ $\phi = arccos(\frac{1}{2})$ $\phi = \frac{\pi}{3}, \frac{5\pi}{3}$ Since the basic cos-curve is shifted to the left $\frac{5\pi}{3}$, the phase constant is $\phi = \frac{5\pi}{3}$ (b) From the graph, we can see that one cycle is completed in 4 seconds. Therefore, the period $T = 4~s$. We can find the angular frequency. $T = \frac{2\pi}{\omega}$ $\omega = \frac{2\pi}{T}$ $\omega = \frac{2\pi}{4~s}$ $\omega = 1.57~rad/s$ From the graph, we can see that the amplitude is 0.10 meters. We can find the velocity at t = 0: $v(t) = -A~\omega~sin(\omega~t+\phi)$ $v = -A~\omega~sin(\phi)$ $v = -(0.10~m)(1.57~rad/s)~sin(\frac{5\pi}{3})$ $v = 0.136~m/s$ (c) We can find the maximum speed as: $v_{max} = A~\omega$ $v_{max} = (0.10~m)(1.57~rad/s)$ $v_{max} = 0.157~m/s$
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.