Question

# A simple pendulum in a lift moving up with a uniform acceleration ‘a’ has a time period T1 and when the lift is moving down with the same acceleration has a time period T2. If the lift were stationary, find the time period of that pendulum?

A
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B
T=2T2T21+T22
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C
T=2T1T2T21+T23
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D
T=3T1T2T21+T22
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Solution

## The correct option is A The effective acceleration due to gravity when the lift moves up is (g+a) and when the lift moves down it is (g-a) For a simple pendulum in a stationary lift T=2π√lg⇒g=4π2lT2 When the lift moves up with uniform acceleration g+a=4π2lT21 When the lift moves down with uniform acceleration g−a=4π2lT22 Adding the two 2g=4π2l(T21+T22)T21×T22, substituting g=4π2lT2 and solving T=√2T1T2√T21+T22

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