Time period and frequency of angular SHM

Since angular acceleration is $\alpha=d^2\theta/dt^2$, the torque equation $I\alpha=-\kappa\theta$ becomes $\dfrac{d^2\theta}{dt^2}=-\dfrac{\kappa}{I}\theta$, which has exactly the form of the SHM differential equation $d^2y/dt^2=-\omega^2y$. Comparing the two directly identifies the angular frequency of this torsional/angular oscillator as $\omega=\sqrt{\kappa/I}\ \text{rad s}^{-1}$. Using $\omega=2\pi f$, the frequency of oscillation is $f=\dfrac{1}{2\pi}\sqrt{\dfrac{\kappa}{I}}$ Hz, and using $T=1/f$, the time period is $T=2\pi\sqrt{\dfrac{I}{\kappa}}$ seconds. This is the pattern used throughout the unit: whenever a system's equation of motion reduces to (second derivative) $=-\omega^2\times$(di …