Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 22 - Vibrations - Section 22.2 - Energy Methods - Problems - Page 661: 30



Work Step by Step

We can determine the required differential equation of motion as follows: $V=V_e+V_g$ $\implies V=2(\frac{1}{2}Kx^2)$ $\implies V=500x^2$ We know that $\frac{d}{dt}(T+V)=0$ $\implies \frac{d}{dt}(\frac{3}{2}x^2+500x^2)=0$ $3x^{.}x^{..}+500\times 2xx^{.}=0$ This simplifies to: $x^{..}+333x=0$
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