JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Elasticity & SHM

Picture the dipole as two masses $m$ at the ends of a rod of length $d$, free to turn about its centre; its moment of inertia about that centre is $I=2m{\left(\frac{d}{2}\right)}^2=\frac{m{d}^2}{2}$. When it is displaced by a small angle $\theta$ from the field direction, the restoring torque is $\tau =pE\sin \theta =qEd\sin \theta$, and for small $\theta$ we use $\sin \theta \approx \theta$. Newton's rotational law $\tau =I\alpha$ becomes $qEd\theta =\frac{m{d}^2}{2}\alpha$, so $\alpha =\frac{2qE}{md}\theta$. This is the signature of simple harmonic motion, with ${\omega }^2$ equal to the coefficient of $\theta$, so $\omega =\sqrt{\frac{2qE}{md}}$.