Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 13 - Exercises and Problems: 24

Answer

$v_{max}=0.021\frac{m}{s}$ $a_{max}=4239\frac{m}{s^2}$

Work Step by Step

We can find the angular frequency as follows: $\omega=2\pi f$ We plug in the known values to obtain: $\omega=2(3.1416)(32768)$ $\omega=205887\frac{rad}{s}$ Now, we can determine the maximum velocity and maximum acceleration $v_{max}=A\omega$ $\implies v_{max}=(100\times 10^{-9})(205887)$ $v_{max}=0.021\frac{m}{s}$ $a_{max}=A\omega^2$ $\implies a_{max}=(100\times 10^{-9})(205887)^2$ $a_{max}=4239\frac{m}{s^2}$
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