Question

If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is $\frac{x}{2}$ times its original time period. Then the value of x is :

• A. $\sqrt3$
• B. $\sqrt2$
• C. $2\sqrt3$
• D. 4

Answer 1

Answer: (B)

$\sqrt2$

Answer 2

The time period of a pendulum is:
$T = 2\pi\sqrt{\frac{L}{g}}$.
If the length $L$ is halved, the new time period becomes:
$T_new = 2\pi\sqrt{\frac{L/2}{g}} = 2\pi \cdot \frac{1}{\sqrt{2}}\sqrt{\frac{L}{g}} = \frac{T}{\sqrt{2}}$.
Comparing with $\frac{x}{2}$:
$\frac{T}{\sqrt{2}} = \frac{x}{2}T \implies x = \sqrt{2}$.