If the mass of the bob in a simple pendulum is increased to thrice it…

The correct answer is option (2) : $\sqrt{2}$

The period of oscillation, $T$, of a simple pendulum is determined by the formula :

$T=2\pi \sqrt{\frac{L}{g}}$

where :

$L$ is the length of the pendulum

$g$ is the acceleration due to gravity

The mass of the bob does not factor into the equation for the period.

Let's first denote the original length of the pendulum as $L$ and the original period of oscillation as $T_1$. Hence,

$T_1=2\pi \sqrt{\frac{L}{g}}$

When the length of the pendulum is halved, the new length $L'$ would be $\frac{L}{2}$. Thus, the new period $T_2$ can be calculates as :

$T_2=2\pi \sqrt{\frac{\frac{L}{2}}{g}}=2\pi \sqrt{\frac{L}{2g}}=2\pi \left(\frac{1}{\sqrt{2}}\right)\sqrt{\frac{L}{g}}=\frac{1}{\sqrt{2}}.2\pi \sqrt{\frac{L}{g}}=\frac{T_1}{\sqrt{2}}$

We are given that the new period $T_2$ is $\frac{x}{2}T_1$. Therefore, we can set up the equation :

$\frac{T_1}{\sqrt{2}}=\frac{x}{2}T_1$

To din the value of $x$, we solve for $x$ :

$\frac{1}{\sqrt{2}}=\frac{x}{2}$

Multiplying both sides by 2 :

$\frac{2}{\sqrt{2}}=x$

Simplify to :

$x= \sqrt{2}$

Hence, the correct answer is :

Option 2: $\sqrt{2}$