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 676: 59



Work Step by Step

The required angular velocity can be calculated as follows: $K=\frac{W}{\delta}$ $K=\frac{150}{1in\times \frac{1}{12}ft/in}=1800lb/ft$ As $\omega_n=\sqrt{\frac{K}{m}}$ $\implies \omega_n=\sqrt{\frac{Kg}{W}}$ $\implies \omega_n=\sqrt{\frac{1800\times 32.2}{150}}$ $\implies \omega_n=19.7rad/s$ $F_{\circ}=mr\omega_{\circ}^2=\frac{W}{g}r\omega_{\circ}^2$ $F_{\circ}=\frac{0.25}{32.2}\times \frac{10in}{12in/ft}$ $\implies F_{\circ}=2.588lb$ The amplitude is given as $x=|\frac{F_{\circ}/K}{1-(\frac{\omega_{\circ}}{\omega_{n}})^2}|$ We plug in the known values to obtain: $\frac{0.25in}{12in/ft}=|\frac{2.588/1800}{1-(\frac{\omega_{\circ}}{19.7})^2}|$ This simplifies to: $\omega_{\circ}=18.9rad/s\approx 19.0rad/s$
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