A uniform solid cylinder of mass m rolls without slipping on a rough horizontal table. A light spring of force constant k joins its axle to a fixed…

Let the axle be displaced by $x$ from equilibrium (the constant weight of the block only shifts the equilibrium point). Pure rolling gives $v=\omega R$, so the kinetic energy of the cylinder is $\tfrac12 mv^2+\tfrac12\left(\tfrac12 mR^2\right)\omega^2=\tfrac34 mv^2$. The block moves with the same speed $v$ and carries $\tfrac12 mv^2$; its gravitational potential energy changes linearly in $x$ and is absorbed in the shifted equilibrium. The spring stores $\tfrac12 kx^2$ about equilibrium. Total energy $=\tfrac12\left(\tfrac32 m+m\right)v^2+\tfrac12 kx^2$, so the effective inertia is $M_{eff}=\tfrac52 m$ and $\omega_0=\sqrt{k/M_{eff}}=\sqrt{2k/5m}$. Thus $T=2\pi\sqrt{5m/2k}$. Option (A) omits the block, option (C) treats the cylinder as a sliding mass of inertia $m$ plus the block, and (D) uses a wrong moment of inertia.