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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?
 
  1. T=2T2T21+T22 
  2. T=2T1T2T21+T23 
  3. T=3T1T2T21+T22 
  4. T=2T1T2T21+T22 


Solution

The correct option is D T=2T1T2T21+T22 
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=2πlgg=4π2lT2
When the lift moves up with uniform acceleration g+a=4π2lT21
When the lift moves down with uniform acceleration ga=4π2lT22
Adding the two 2g=4π2l(T21+T22)T21×T22, substituting g=4π2lT2 and solving T=2T1T2T21+T22 

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