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 - Exercises and Problems: 4

Answer

(a) A = 20 cm (b) f = 0.25 Hz

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 two cycles are completed in 8 seconds. Therefore, the period $T = 4~s$. This means that the frequency is: $f = \frac{1}{T} = \frac{1}{4~s} = 0.25~Hz$
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