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: 65



Work Step by Step

We can find the phase angle $\phi$ as follows: $\omega_n=\frac{K}{m}$ $\implies \omega_n=\frac{75\times 32.2}{7}$ $\implies \omega_n=18.5742 rad/s$ Now $\phi=tan^{-1}[\frac{2(c/c_{\circ})(\omega_{\circ}/ \omega_n)}{1-(\omega_{\circ}/\omega_n)^2}]$ We plug in the known values to obtain: $\phi=tan^{-1}[\frac{2\times 0.8\times (2/18.5742)}{1-(2/18.5742)^2}]$ $\implies \phi=9.89^{\circ}$
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