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: 2

Answer

(a) A = 13 cm (b) x = 9.2 cm

Work Step by Step

(a) We can find the angular frequency as: $T = \frac{2\pi}{\omega}$ $\omega = \frac{2\pi}{T}$ $\omega = \frac{2\pi}{2.0~s}$ $\omega = \pi~rad/s$ We can find the amplitude as: $v_{max} = A~\omega$ $A = \frac{v_{max}}{\omega}$ $A = \frac{0.40~m/s}{\pi~rad/s}$ $A = 0.13~m = 13~cm$ (b) We can find $x$ when $t = 0.25~s$; $x(t) = A~cos(\omega~t)$ $x = (0.13~m)~cos[(\pi)(0.25~s)]$ $x = 0.092~m = 9.2~cm$
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