Physics - SHM Question with Solution | TestHub

The moment of inertia of the system about the pivot (center of the rod) is due to the two point masses. Since the rod is massless, $I=m(L/2{)}^2+m(L/2{)}^2=2m(L/2{)}^2$. Substituting the given values, $I=2(3)(2/2{)}^2=6(1{)}^2=6$ kg m${}^2$.

When the rod is displaced by a small angle $\theta$, the spring is stretched (or compressed) by a distance $x=d\theta$. The restoring force exerted by the spring is $F=kx=kd\theta$. The restoring torque about the pivot is $\tau =Fd=(kd\theta )d=k{d}^2\theta$. Thus, the angular spring constant is $K_{angular}=k{d}^2=6(1{)}^2=6$ N m/rad. The time period of angular simple harmonic motion is $T=2\pi \sqrt{\frac{I}{K_{angular}}}=2\pi \sqrt{\frac{6}{6}}=2\pi \sqrt{1}=2\pi$ seconds. The value of $\frac{T^2}{{\pi }^2}=\frac{(2\pi {)}^2}{{\pi }^2}=\frac{4{\pi }^2}{{\pi }^2}=4$.