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: 8

Answer

$x(t) = (4.0~cm)~cos(8\pi~t+\frac{3\pi}{2})$

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: $A = 4.0 cm$ $f = 4.0 Hz$ At $t = 0$, the object passes through the equilibrium point moving to the right. This shows that the basic cos-curve is shifted an angle of $\frac{3\pi}{2}$ to the left. Therefore, $\phi_0 = \frac{3\pi}{2}$ We can write the function for this motion as: $x(t) = (4.0~cm)~cos[(2\pi)(4.0)~t+\frac{3\pi}{2}]$ $x(t) = (4.0~cm)~cos(8\pi~t+\frac{3\pi}{2})$
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