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: 5

Answer

(a) The amplitude is 20 cm. (b) $f = 0.125~Hz$ (c) $\phi = \frac{\pi}{3}$

Work Step by Step

(a) The motion moves between -20 cm and 20 cm. Therefore, the amplitude is 20 cm. (b) From the graph, we can see that one cycle is completed in 8 seconds. Therefore, the period $T = 8~s$. We can find the frequency. $f = \frac{1}{T} = \frac{1}{8~s} = 0.125~Hz$ (c) 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 $\frac{\pi}{3}$, the phase constant is $\phi = \frac{\pi}{3}$.
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