The period of a simple pendulum measured inside a stationary lift is found to be T. If the lift starts accelerating upwards with acceleration of g/3, then the time period of the pendulum is
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a
T3
b
T3
c
32T
d
3 T
answer is C.
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Detailed Solution
For stationary lift T1=2πlgFor ascending lift with acceleration a, T2=2πlg+a⇒T1T2=g+ag⇒T1T2=g+g3g=43 ⇒T2=32T