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