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: 9

Answer

$x(t) = (8.0~cm)~cos(\pi~t+\pi)$

Work Step by Step

The general equation for the position of an object in SHM is: $x(t) = A~cos(2\pi~f~t+\phi_0)$ It is known that: A = 8.0 cm f = 0.5 Hz At $t = 0$, the object has its most negative position. This shows that the basic cos-curve is shifted an angle of $\pi$ to the left. Therefore, $\phi_0 = \pi$. We can write the function for this motion as: $x(t) = (8.0~cm)~cos[(2\pi)(0.5)~t+\pi]$ $x(t) = (8.0~cm)~cos(\pi~t+\pi)$
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