Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 14 - Oscillations - Conceptual Questions: 8

Answer

(a) $\phi_0 = \frac{7\pi}{6}$ (b) At point 1: $phase = \frac{5\pi}{6}$ At point 2: $phase = \frac{3\pi}{2}$ At point 3: $phase = \frac{\pi}{6}$

Work Step by Step

(a) The general equation for the velocity of the particle is: $v(t) = -A~\omega~sin(\omega~t+\phi_0)$ We can find $\phi_0$ when $v(0) = \frac{v_{max}}{2}$ $\frac{v_{max}}{2} = -A~\omega~sin(\omega~t+\phi_0)$ $sin(\phi_0) = \frac{-1}{2}$ $\phi_0 = arcsin(\frac{-1}{2})$ $\phi_0 = \frac{7\pi}{6}, \frac{11\pi}{6}$ In the graph, we can see that the velocity is increasing just after $t=0$. Therefore $\phi_0 = \frac{7\pi}{6}$. (b) We can find the phase at point 1: $phase = \frac{7\pi}{6}-\frac{2\pi}{6} = \frac{5\pi}{6}$ We can find the phase at point 2: $phase = \frac{7\pi}{6}+\frac{\pi}{3} = \frac{3\pi}{2}$ We can find the phase at point 3: $phase = \frac{7\pi}{6}+\pi = \frac{13\pi}{6}= \frac{\pi}{6}$
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