College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 10 - Problems - Page 399: 35


The maximum velocity is $3.10~m/s$ The maximum acceleration is $8560~m/s^2$

Work Step by Step

We can find $\omega$: $\omega = 2\pi~f$ $\omega = (2\pi)(440.0~Hz)$ $\omega = 880.0\pi~rad/s$ The amplitude is half the distance between the extreme positions of the motion. We can find the amplitude: $A = \frac{2.24~mm}{2} = 1.12~mm$ We can find the maximum velocity: $v_m = A~\omega$ $v_m = (1.12~\times 10^{-3}~m)~(880\pi~rad/s)$ $v_m = 3.10~m/s$ The maximum velocity is $3.10~m/s$ We can find the maximum acceleration: $a_m = A~\omega^2$ $a_m = (1.12~\times 10^{-3}~m)~(880\pi~rad/s)^2$ $a_m = 8560~m/s^2$ The maximum acceleration is $8560~m/s^2$
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