The time period of a simple pendulum measured in a stationary lift is T. What will happen to its time period when the lift falls freely?

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Solution

$The relation for time period is, T = 2 π \sqrt{\frac{l}{g}} According to question, when the lift falls freely, downward acceleration is g . So, New time period, T' = 2 π \sqrt{\frac{l}{g - g}} = \infty Since time period is infinity, so it will not oscillate at all .$

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The time period of a simple pendulum measured in a stationary lift is T. What will happen to its time period when the lift falls freely?

The relation for time period is,T=2πlgAccording to question, when the lift falls freely, downward acceleration is g.So,New time period, T'=2πlg-g=∞Since time period is infinity, so it will not oscillate at all.

$The relation for time period is, T = 2 π \sqrt{\frac{l}{g}} According to question, when the lift falls freely, downward acceleration is g . So, New time period, T' = 2 π \sqrt{\frac{l}{g - g}} = \infty Since time period is infinity, so it will not oscillate at all .$