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

Published by Pearson

# Chapter 14 - Oscillations - Exercises and Problems - Page 402: 5

#### Answer

(a) A = 10 cm (b) f = 0.5 Hz

#### Work Step by Step

(a) The motion moves between -10 cm and 10 cm. Therefore, the amplitude is 10 cm. (b) From the graph, we can see that two cycles are completed in 4 seconds. Therefore, the period $T = 2~s$. This means that the frequency is: $f = \frac{1}{T} = \frac{1}{2~s} = 0.5~Hz$

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