A man measures the period of simple pendulum inside a stationary lift and finds it to be T second. If the lift accelerates downwards with acceleration of g4, the period of oscillation will be
A
T×√32s
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B
T×2√3s
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C
T2s
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D
√Ts
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Solution
The correct option is BT×2√3s Effective acceleration due to gravity when the lift accelerates downwards, g′=g−g4=3g4 Time period T=2π√(lg) ...... (i) T′=2π√(lg′)=2π√(4l3g)=2π×2√3√(lg) ....(ii) Now dividing Eq. (ii) by Eq. (i), we get T′T=2√3