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: 51



Work Step by Step

We can determine the maximum speed as follows: $\omega_n=\sqrt{\frac{K}{m}}$ $\omega_n=\sqrt{\frac{800}{40}}=2\sqrt{5}rad/s$ We know that $y=\frac{F_{\circ}/K}{1-(\frac{\omega_{\circ}}{\omega_n})^2}cos 2t$ We plug in the known values to obtain: $y=\frac{100/800}{1-(\frac{2}{2\sqrt{5}})^2}cos 2t$ This simplifies to: $y=0.15625\space cos2t$ $\implies y^{.}=0.15625(-2sin2t)$ $\implies y^{.}=-0.3125\space sin2t$ Thus, the maximum speed is $v_{max}=0.3125m/s$
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