Q.What is meant by angular harmonic oscillation? Compute the time period of angular harmonic oscillation.

Step 1. Definition. When a rigid body free to rotate about a fixed axis is displaced from its equilibrium (mean) orientation -- the orientation where the net restoring torque is zero -- and released, it undergoes angular harmonic oscillation if the resulting restoring torque is proportional to the angular displacement and directed to bring it back.

Step 2. The restoring torque. If $\vec\theta$ is the angular displacement from equilibrium, the restoring torque is $\vec\tau=-\kappa\vec\theta$, where $\kappa$ (kappa) is the restoring torsion constant of the suspension (e.g. a torsion fibre), the torque needed to produce unit angular displacement.

Step 3. Equation of motion. With $I$ the moment of inertia of the body about the axis and $\vec\alpha=d^2\vec\theta/dt^2$ the angular acceleration, Newton's second law for rotation gives $\vec\tau=I\vec\alpha=-\kappa\vec\theta$, i.e. $\dfrac{d^2\theta}{dt^2}=-\dfrac{\kappa}{I}\theta$.

Step 4. Comparing with SHM. This has exactly the form $d^2y/dt^2=-\omega^2y$, so $\omega^2=\kappa/I$, giving $\omega=\sqrt{\kappa/I}$.

Step 5. Time period. Using $T=2\pi/\omega$, the time period of angular harmonic oscillation is $T=2\pi\sqrt{I/\kappa}$ seconds (and correspondingly frequency $f=\dfrac{1}{2\pi}\sqrt{\kappa/I}$).

Define angular SHM via the restoring torque law, apply the rotational form of Newton's second law, compare to the SHM equation, and read off omega and hence T.

• Using linear mass m instead of moment of inertia I in the final formula.
• Forgetting the negative sign in the torque law, which would give an unstable (non-oscillatory) equation instead.