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: 34

Answer

(a) $T = 1.4~s$ (b) $A = 0.30~m$

Work Step by Step

(a) We can find the period as: $T = 2\pi~\sqrt{\frac{m}{k}}$ $T = 2\pi~\sqrt{\frac{1.0~kg}{20~N/m}}$ $T = 1.4~s$ (b) We can find the amplitude as: $A = \sqrt{x^2+\frac{v^2}{\omega^2}}$ $A = \sqrt{x^2+\frac{v^2}{(k/m)}}$ $A = \sqrt{(0.20~m)^2+\frac{(-1.0~m/s)^2}{(20~N/m)/(1.0~kg)}}$ $A = 0.30~m$
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