A bar magnet is oscillating in the earth's magnetic field with a period $T$. What happens to its period and motion, if its mass is quadrupled?

The time period $T$ of a bar magnet oscillating in a uniform magnetic field $B$ is given by the formula $T = 2\pi \sqrt{\dfrac{I}{MB}}$, where $I$ is the moment of inertia of the magnet and $M$ is its magnetic dipole moment.

For a bar magnet of mass $m$, length $l$, and width $w$, the moment of inertia about the axis of oscillation is $I = \dfrac{m(l^2 + w^2)}{12}$.

When the mass is quadrupled, the new mass $m' = 4m$. Assuming the dimensions of the magnet remain unchanged, the new moment of inertia becomes $I' = \dfrac{(4m)(l^2 + w^2)}{12} = 4I$.

The new time period $T'$ is given by $T' = 2\pi \sqrt{\dfrac{I'}{MB}} = 2\pi \sqrt{\dfrac{4I}{MB}} = 2 \times 2\pi \sqrt{\dfrac{I}{MB}} = 2T$.

Since the restoring torque $\tau = -MB \sin \theta \approx -MB \theta$ for small oscillations, the motion remains simple harmonic.

Answer: Motion remains simple harmonic with time period $2 T$.