Answer
$$\ddot{\theta}+468 \theta=0
$$
Work Step by Step
Kinematics: Since no slipping occurs, $s_G=0.1 \theta$ hence $s_F=\frac{0.3}{0.1} S_G=0.3 \theta$. Also
$$
\begin{aligned}
& v_G=0.1 \dot{\theta} . \\
& E=T+V \\
& E=\frac{1}{2}\left[(3)(0.125)^2\right] \dot{\theta}^2+\frac{1}{2}(3)(0.1 \theta)^2+\frac{1}{2}(400)(0.3 \theta)^2=\text { const. } \\
& \quad=0.03844 \dot{\theta}^2+18 \theta^2 \\
& 0.076875 \ddot{\theta} \ddot{\theta}+36 \theta \dot{\theta}=0 \\
& 0.076875 \dot{\theta}(\ddot{\theta}+468.29 \theta)=0 \text { Since } 0.076875 \theta \neq 0 \\
& \ddot{\theta}+468 \theta=0
\end{aligned}
$$
