A pendulum performs S.H.M. with period 3 second in a stationery lift. If lift moves up with acceleration 3 g , the period of pendulum is [ g = acceleration due to gravity ]

T = 3 a = 3 g T ′ = 2 π g + 3 g ℓ ∴ T T ′ = 4 g /3 g = 2 3 T ′ = 3 × 2 3 = 2 3 = 1.5 sec

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Physics - SHM Question with Solution | TestHub

PhysicsSHMCalculation of time periodMedium2 minPYQ_2020

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A pendulum performs S.H.M. with period $3$ second in a stationery lift. If lift moves up with acceleration $\frac{g}{3}$ , the period of pendulum is $[g =$ acceleration due to gravity $]$

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Solution:

$T = 3 a = \frac{g}{3}$ $T^{'} = 2 π \frac{ℓ}{g + \frac{g}{3}}$ $∴ \frac{T ^{'}}{T} = \frac{g}{4 g /3} = \frac{3}{2}$ $T^{'} = 3 \times \frac{3}{2} = \frac{3}{2} = 1.5 sec$

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

⏱ 2mℹ️ Source: PYQ_2020

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