A uniform disc of mass m and radius R is pivoted smoothly at its centre of mass. A light spring of stiffness k is attached with the disc tangentially as shown in the following figure. Find the angular frequency in rad/s of torsional oscillations of the disc.

$(\mathrm{T}\text{ake }\mathrm{m}=5\mathrm{kg}\text{ and }\mathrm{K}=10\mathrm{N}/\mathrm{m})$

If we twist (rotate) the disc through a small clockwise angle $\theta ,$ the spring will be deformed (compressed) by a distance $\mathrm{x}=\mathrm{R}\theta .$ Hence, the spring force ${\mathrm{F}}_{\mathrm{s}}=\mathrm{kx}=\mathrm{k}\left(\mathrm{R}\theta \right)$ will produce a restoring torque.

Restoring torque: $\mathrm{\tau }=-{\mathrm{F}}_{\mathrm{s}}\mathrm{R}\text{ where }{\mathrm{F}}_{\mathrm{s}}=\mathrm{kR\theta }$ 

This gives $\tau =-{\mathrm{kR}}^2\theta$ It means after removing the external (applied) torque, the
restoring torque rotates the disc with an angular acceleration $\alpha$ which will bring the spring-disc system back to its original state. Newton’s law of rotation (or torque equation): Applying Newton’s second law of rotation, we have 

$\tau =I_{\mathrm{C}}\alpha$ Where $\tau =-k{R}^2\theta$ This gives $\alpha =-\frac{{\mathrm{kR}}^2\theta }{{\mathrm{I}}_{\mathrm{c}}}\text{ where }{\mathrm{I}}_{\mathrm{c}}=\frac{{\mathrm{mR}}^2}{2}$

Then $\alpha =-\frac{2\mathrm{k}}{\mathrm{m}}\theta$ Comparing the above equation with $\alpha =-{\omega }^2\theta ,$ we have $\mathrm{\omega }=\sqrt{\frac{2\mathrm{k}}{\mathrm{m}}}$ 

After substituting the values we get $\omega =2\mathrm{rad}/\mathrm{s}.$