Question: 37. 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 :
(1) \sqrt{3}
(2) \sqrt{2}
(3) 2 \sqrt{3}
(4) 4
Answer: Option (2)
Explanation:
The time period of a simple pendulum is given by
T = 2\pi \sqrt{\frac{l}{g}}.
The time period depends only on the length l of the pendulum and acceleration due to gravity g,
and is independent of the mass of the bob.
Let the original length be l.
Then the original time period is T = 2\pi \sqrt{\frac{l}{g}}.
The new length is made half of the original length,
so l' = \frac{l}{2}.
The new time period is T' = 2\pi \sqrt{\frac{l'}{g}} = 2\pi \sqrt{\frac{l}{2g}}.
Taking the ratio of new time period to original time period,
\frac{T'}{T} = \sqrt{\frac{l/(2g)}{l/g}} = \sqrt{\frac{1}{2}}.
Thus, T' = \frac{1}{\sqrt{2}} T.
Given that the new time period is \frac{x}{2} times the original time period,
\frac{x}{2} = \frac{1}{\sqrt{2}}.
Solving, x = \sqrt{2}.