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 675: 49



Work Step by Step

We can determine the required amplitude as follows: $K=\frac{F}{\delta}$ $\implies K=\frac{18}{0.014}=17.928rad/s$ Now $x=\frac{F_{\circ/K}}{1-(\frac{\omega_{\circ}}{\omega_n})^2}$ We plug in the known values to obtain: $x=\frac{0.015}{1-(\frac{2\times 2\pi}{17.98})^2}$ $\implies x=0.0295m=29.5mm$
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