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: 55



Work Step by Step

We can find the required amplitude as follows: $K=\frac{mg}{\delta}$ $K=\frac{25\times 9.81}{0.05}=4905N/m$ $\omega_n=\sqrt{\frac{K}{m}}$ $\implies \omega_n=\sqrt{\frac{4905}{25}}=14 rad/s$ and $F_{\circ}=mr\omega_{\circ}^2$ $\implies F_{\circ}=3.5\times 0.1\times (18)^2=113.4N$ Now $x=\frac{F_{\circ}/K}{1-(\frac{\omega_{\circ}}{\omega_n})^2}$ We plug in the known values to obtain: $x=\frac{35/4905}{1-(\frac{18}{14})^2}$ $\implies x=0.0355m=35.5mm$
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