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: 3

Answer

(a) $A = 13~cm$ (b) $x = 9.0~cm$

Work Step by Step

(a) We can find the amplitude of the oscillation as: $v_{max} = \frac{2\pi~A}{T}$ $A = \frac{v_{max}~T}{2\pi}$ $A = \frac{(40~cm/s)(2.0~s)}{2\pi}$ $A = 12.7~cm = 13~cm$ (b) The equation for the position of the particle is: $x(t) = A~cos(\omega~t) = A~cos(\frac{2\pi~t}{T})$ We can find the glider's position at $t = 0.25~s$ as: $x = (12.7~cm)~cos[\frac{(2\pi)(0.25~s)}{2.0~s}]$ $x = 9.0~cm$
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