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

A freely suspended bar magnet in a uniform field performs angular SHM about the field direction, giving a period analogous to a torsional pendulum.

Consider a bar magnet of magnetic moment $M$ and moment of inertia $I$ (about the vertical suspension axis), free to rotate in a uniform magnetic field $B$. When displaced by a small angle $\theta$ from equilibrium (aligned with $B$), the restoring torque is

$\tau = -MB\sin\theta \approx -MB\theta \quad (\text{small } \theta)$

By Newton's law for rotation, $\tau = I\ddot\theta$, so

$I\ddot\theta = -MB\theta \;\Rightarrow\; \ddot\theta = -\dfrac{MB}{I}\theta$

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