Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 22 - Vibrations - Section 22.6 - Electrical Circuit Analogs - Problems - Page 677: 63


$12.2rad/s$ $7.07rad/s$

Work Step by Step

We can determine the two possible values of $\omega$ as follows: $\omega_n=\frac{K}{m}$ $\omega_n=\frac{2\times 2500}{50}=10rad/s$ We know that $y_{max}=\frac{\delta_{\circ}}{1-(\frac{\omega_{\circ}}{\omega_n})^2}$ We plug in the known values to obtain: $\pm 0.4=\frac{0.2}{1-(\frac{\omega_{\circ}}{10})^2}$ $\implies \pm 0.4\times (1-(\frac{\omega_{\circ}}{10})^2)=0.2$ This simplifies to: $\omega_{\circ}=10\sqrt{1\pm 0.5}$ This simplifies for two values: $\implies \omega_{\circ}=5\sqrt{6}=12.2rad/s$ and $\omega_{\circ}=5\sqrt{2}=7.07rad/s$
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