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 calculated as:

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

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 find 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 B: $\sqrt{2}$