## Engineering Mechanics: Statics & Dynamics (14th Edition)

$x^{..}+333x=0$
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: $x^{..}+333x=0$