Q.Obtain the expression for the period of a magnet vibrating in a uniform magnetic field and performing S.H.M.

[!TLDR] $c=\mu B$ in $\tau=-c\theta$, so from the general period template, $T=2\pi\sqrt{I/(\mu B)}$. [!ANSWER] $T=2\pi\sqrt{I/(\mu B)}$

A magnet of magnetic moment $\mu$, freely suspended in a uniform field B, displaced through small angle $\theta$, experiences a restoring torque $\tau=\mu B\sin\theta\approx\mu B\theta$ for small $\theta$ (as derived from the definition of torque on a magnetic dipole in a field). Comparing with the general angular-S.H.M. torque law $\tau=-c\theta$ (Eq. 5.31), the restoring-torque-per-unit-angular-displacement is $c=\mu B$. If I is the magnet's moment of inertia about the suspension axis, the differential equation $I\ddot\theta+c\theta=0$ becomes $I\ddot\theta+\mu B\theta=0$ (Eq. 5.33), confirming angular S.H.M. of angular frequency $\omega=\sqrt{\mu B/I}$. Applying the general period template $T=2\pi\sqrt{\text{displacement}/\text{acceleration per unit displacement}}$ (equivalently $T=2\pi\sqrt{I/c}$ for an angular oscillator), with $c=\mu B$: $T=2\pi\sqrt{\frac{I}{\mu B}}$ This is the standard vibration-magnetometer formula (Eq. 5.34), used to find an unknown B (given $\mu$, I) or an unknown $\mu$ (given B, I). [!ANSWER] $T=2\pi\sqrt{I/(\mu B)}$

Identify $c=\mu B$ from the small-angle torque law, then substitute into the general angular-S.H.M. period formula $T=2\pi\sqrt{I/c}$.

Forgetting the small-angle approximation step, or mixing up I and $\mu B$ inside the square root.