University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 14 - Periodic Motion - Problems - Exercises - Page 460: 14.13


(a) $A = 0.980~m$ (b) $\phi = 1.57~rad$ (c) $x(t) = (0.980)~cos(12.2~t+1.57)$

Work Step by Step

(a) If the spring is neither stretched nor compressed, then all the energy in the system is in the form of kinetic energy. We can find the amplitude. $\frac{1}{2}kA^2 = \frac{1}{2}mv_{max}^2$ $A = \sqrt{\frac{m}{k}}~v_{max}$ $A = \sqrt{\frac{2.00~kg}{300~N/m}}~(12.0~m/s)$ $A = 0.980~m$ (b) If the spring is neither stretched nor compressed at $t=0$, we can say that $x = 0$ and the object is moving in the negative direction. Therefore, the basic cos-curve is shifted to the left an angle of $\frac{\pi}{2}$. Therefore, the phase angle $\phi = \frac{\pi}{2}~rad = 1.57~rad$. (c) We can find the angular frequency. $\omega = \sqrt{\frac{k}{m}}$ $\omega = \sqrt{\frac{300~N/m}{2.00~kg}}$ $\omega = 12.2~rad/s$ We can write the equation for the position. $x(t) = A~cos(\omega~t+\phi_0)$ $x(t) = (0.980)~cos(12.2~t+1.57)$
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