Question

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 ?

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

Answer 1

The correct option is A Motion remains $S.H.M$ with time period =$2T$

Time period of oscillation is,

$T=2\pi \sqrt{\frac{I}{M{B}_H}}$

$T=2\pi \sqrt{\frac{m{K}^2}{M{B}_H}}......\left(1\right)$

After quadrupling the mass, the new time period is,

$T^{{}^{\prime }}=2\pi \sqrt{\frac{4m{K}^2}{M{B}_H}}......\left(2\right)$

On dividing (2) $÷$ (1) we get,

$T^{{}^{\prime }}=2T$

It continues to perform $S.H.M$ in the same field with time period twice the initial time period.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.