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 461: 14.26


$v = 0.377~m/s$ $a = -0.615~m/s^2$

Work Step by Step

We can find the angular frequency. $T = \frac{2\pi}{\omega}$ $\omega = \frac{2\pi}{T}$ $\omega = \frac{2\pi}{3.20~s}$ $\omega = 1.96~rad/s$ We can find the speed when x = 0.160 m. $v = \omega~\sqrt{A^2-x^2}$ $v = (1.96~rad/s)~\sqrt{(0.250~m)^2-(0.160~m)^2}$ $v = 0.377~m/s$ We can find the acceleration when x = 0.160 m. $a = -\omega^2~x$ $a = -(1.96~rad/s)^2(0.160~m)$ $a = -0.615~m/s^2$
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