Q.Prove that under certain conditions a magnet vibrating in uniform magnetic field performs angular S.H.M.

[!TLDR] Restoring torque on a displaced magnet is $\tau=\mu B\sin\theta\approx\mu B\theta$ for small $\theta$, i.e. $\tau\propto\theta$ and opposite in direction -- the defining conditio …

A bar magnet of magnetic moment $\mu$, freely suspended in a uniform field B, aligns with the field at equilibrium. Displaced through a small angle $\theta$, the field exerts a restoring torque of magnitude $\tau=\mu B\sin\theta$ on it, always directed so as to reduce $\theta$ back towards zero. For SMALL $\theta$ (the 'certain conditions' of the question), $\sin\theta\approx\theta$, so $\tau\approx\mu B\theta$, and including the direction, $\tau=-\mu B\theta$. If I is the magnet's moment of inertia about the suspension axis, Newton's second law for rotation gives $I\frac{d^2\theta}{dt^2}=-\mu B\theta$ (Eq. 5.33), i.e. $\frac{d^2\theta}{dt^2}+\frac{\mu B}{I}\theta=0$ -- structurally identical to the S.H.M. differential equation $\ddot x+\omega^2x=0$, with $\theta$ in place of x …

Apply the small-angle approximation to $\tau=\mu B\sin\theta$ to get $\tau\propto-\theta$, then substitute into $\tau=I\ddot\theta …

Omitting the 'small angle' qualifier -- for large $\theta$, $\sin\theta$ is not proportional to $\theta$, and the m …