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 - Page 404: 42

Answer

(a) $10.1\mu m$ (b) $64m/s$

Work Step by Step

(a) We can determine the maximum amplitude as follows: $F_{max}=m\omega^2 A$ $\implies F_{max}=m(2\pi f)^2A$ $\implies F_{max}=4\pi ^2mf^2 A$ This can be rearranged as: $A=\frac{F_{max}}{4\pi^2mf^2}$ We plug in the known values to obtian: $A=\frac{40000N}{4\pi^2(0.10\times 10^{-3}Kg)(1.0\times 10^6Hz)^2}$ This simplifies to: $A=10.1\mu m$ (b) The maximum speed of the disk can be determined as $v_{max}=2\pi fA$ We plug in the known values to obtain: $v_{max}=2\pi (1.0\times 10^6Hz)(10.1\times 10^{-6}m)$ $\implies v_{max}=64m/s$
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