Question

The time period of oscillation of a bar magnet suspended horizontally along the magnetic meridian is ${T_0}$. If this magnet is replaced by another magnet of the same size and pole strength, but with double the mass, the new time period will be :
A. $\dfrac{{{T_0}}}{2}$
B. $\dfrac{{{T_0}}}{{\sqrt 2 }}$
C. $\sqrt 2 {T_0}$
D. $2{T_0}$

Answer 1

Hint: In mechanics and physics, simple harmonic motion is a type of periodic motion in which the restoring force acting on a moving object is inversely proportional to the magnitude of the object's displacement and moves the object toward its equilibrium point. If friction or other energy loss does not stop the oscillation, it can continue for all time.

Formula used:
The time period of bar magnet is given by,
$T = 2\pi \sqrt {\dfrac{I}{{MB}}} $
Here, $T$ is time period of oscillation of bar magnet, $I$ is moment of inertia of bar magnet, $M$ is magnetic moment and $B$ is magnetic field intensity of bar magnet.

Complete step by step solution:
In the question, we have given the time period of an oscillation of a bar magnet that is suspended horizontally along magnetic lines is ${T_0}$. In order for the model to be true, the net force exerted on the object at the pendulum's end must be commensurate to the displacement. As we know, the motion of a simple pendulum may be described by the basic harmonic motion and simple harmonic motion can also be used to describe molecular vibration. The time period of bar magnet is given by,
$T = 2\pi \sqrt {\dfrac{I}{{MB}}} $

Substitute $T = {T_0}$and $I = m{r^2}$in the above formula, then we have:
${T_0} = 2\pi \sqrt {\dfrac{{m{r^2}}}{{MB}}} \,\,\,\,\,....(1)$
Now, let's assume the new time period is $T'$. Consider the given information, the magnet that is replaced by an another magnet of the same size and the pole strength but with the double mass, then:
$T' = 2\pi \sqrt {\dfrac{{2m{r^2}}}{{MB}}} \,\,\,\,\,...(2)$

Compare the equations $(1)$from $(2)$, then we obtain:
$\dfrac{{T'}}{{{T_0}}} = \dfrac{{2\pi \sqrt {\dfrac{{2m{r^2}}}{{MB}}} }}{{2\pi \sqrt {\dfrac{{m{r^2}}}{{MB}}} }} \\$
$\Rightarrow \dfrac{{T'}}{{{T_0}}} = 2\pi \sqrt {\dfrac{{2m{r^2}}}{{MB}}} \times \dfrac{1}{{2\pi }}\sqrt {\dfrac{{MB}}{{m{r^2}}}} \\$
$\Rightarrow \dfrac{{T'}}{{{T_0}}} = \sqrt 2 \\$
$\therefore T' = \sqrt 2 {T_0} $
Therefore, the new time period will be $\sqrt 2 {T_0}$.

Thus, the correct option is C.

Note: Keep in mind that $I = m{r^2}$ is the proper substitute and not $MB = m{r^2}$ while trying to recall the time period formula because $T \propto \sqrt I \propto \sqrt m $. A simple harmonic motion is referred to the movement of a particle along the straight line with an acceleration that is proportional to the particle's distance from a fixed point on the line.