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: 61



Work Step by Step

The amplitude of vibration can be determined as follows: $\omega_n=\sqrt{\frac{K}{m}}$ $\implies \omega_n=\sqrt{\frac{2\times 800}{450}}$ $\implies \omega_n=1.8856rad/s$ We know that $T=\frac{1}{f}$ $\implies T=\frac{2\pi}{\omega_{\circ}}$ $\implies 15Km/h\times \frac{1000m/Km}{3600s/h}=\frac{4\omega_{\circ}}{2\pi}$ This simplifies to: $\omega_{\circ}=6.5rad/s$ Now $x=|\frac{\delta_{\circ}}{1-(\frac{\omega_{\circ}}{\omega_n})^2}|$ We plug in the known values to obtain: $x=4.53\times 10^{-6}m=4.53mm$
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