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NEET 2024 · Simple Pendulum

Simple Pendulum question from NEET 2024

Q32NEET 2024Easy

Simple Pendulum

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:

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Answer: (b)

$T = 2\pi\sqrt{\frac{l}{g}}$ is independent of mass.
$l' = \frac{l}{2} \Rightarrow T' = \frac{T}{\sqrt{2}} = \frac{\sqrt{2}}{2}T$
Comparing with $\frac{x}{2}T$ : $x = \sqrt{2}$ .
So the answer is (b) $\sqrt{2}$ .

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