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PhysicsSHMCalculation of time periodMedium2 minPYQ_2024

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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:NEET 2024

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Solution:

$T^{'} = 2 π \frac{ℓ ^{'}}{g} where ℓ^{'} = \frac{ℓ}{2} T = 2 π \frac{ℓ}{g} T^{'} = \frac{x}{2} T 2 π \frac{ℓ}{2 g} = \frac{x}{2} 2 π \frac{ℓ}{g} \frac{1}{2} = \frac{x}{2} \Rightarrow x = 2$

Stream:NEETSubject:PhysicsTopic:SHMSubtopic:Calculation of time period

⏱ 2mℹ️ Source: PYQ_2024

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