University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 14 - Periodic Motion - Problems - Exercises - Page 460: 14.11

Answer

(a) t = 0.150 s (b) t = 0.075 s

Work Step by Step

(a) We can find the angular frequency. $\omega = \frac{2\pi}{T}$ $\omega = \frac{2\pi}{0.900~s}$ $\omega = 6.98~rad/s$ We can write the general equation for the position. $x = A~cos(\omega~t)$ We can find $t$ when $x = \frac{A}{2}$. $x = A~cos(\omega~t)$ $\frac{A}{2} = A~cos(\omega~t)$ $cos(\omega~t) = \frac{1}{2}$ $t = \frac{arccos(\frac{1}{2})}{\omega}$ $t = \frac{\frac{\pi}{3}}{6.98~rad/s}$ $t = 0.150~s$ (b) We can find $t$ when $x = 0$. $x = A~cos(\omega~t)$ $0 = A~cos(\omega~t)$ $cos(\omega~t) = 0$ $t = \frac{arccos(0)}{\omega}$ $t = \frac{\frac{\pi}{2}}{6.98~rad/s}$ $t = 0.225~s$ The time to move from $x = \frac{A}{2}$ to $x = 0$ is $0.225~s-0.150~s$, which is $0.075~s$.
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