Question

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

• A. Motion remains S.H.M. with a time period $=2T$
• B. Motion remains S.H.M. with a time period $=4T$
• C. Motion remains S.H.M. and the period remains nearly constant
• D. Motion remains S.H.M. with a time period $=\frac{T}{2}$

Answer 1

The correct option is A

Motion remains S.H.M. with a time period $=2T$

Step 1: Given Data

Time period $=T$

Let $I$ be the moment of inertia.

Let the initial mass of the magnet be $m$.

The final mass $m\text{'}=4m$

Let the length of the magnet be $L$.

Let $M$ be the magnetic moment.

Let $H$ be the magnetic intensity.

Step 2: Time Period for initial mass

We know that the moment of inertia of a cuboid is given as

$I=\frac{m{L}^2}{12}$

We know that the time period is given as

$T=2\mathrm{\pi }\sqrt{\frac{\mathrm{I}}{\mathrm{MH}}}$

$\therefore T\propto \sqrt{I}$

$\Rightarrow T\propto \sqrt{\frac{m{L}^2}{12}}$

$\Rightarrow T\propto \sqrt{m}$ $\left(1\right)$

Step 3: Time Period for final mass

For the second mass, we can write the time period as

$T\text{'}\propto \sqrt{m\text{'}}$

$\Rightarrow T\text{'}\propto \sqrt{4m}$

$\Rightarrow T\text{'}\propto 2\sqrt{m}$

From equation $\left(1\right)$ we get,

$T\text{'}=2T$

Therefore, the motion remains S.H.M. with a time period $=2T$.

Hence, the correct answer is option (A).