The time period of a simple pendulum inside a stationary lift is 5 s. The time period when the lift moves upwards with an acceleration 4 g is ( g = acceleration due to gravity)

Question

The time period of a simple pendulum inside a stationary lift is 5 ​ s. The time period when the lift moves upwards with an acceleration 4 g ​ is ( g = acceleration due to gravity)

TestHub · Accepted Answer

The time period of a simple pendulum in a stationary lift is given by

$T = 2 π \frac{l}{g} = 5$ s

When the lift moves upwards with an acceleration $a = \frac{g}{4}$ , the effective acceleration due to gravity becomes

$g_{e ff} = g + a = g + \frac{g}{4} = \frac{5 g}{4}$

The new time period $T^{'}$ is

$T^{'} = 2 π \frac{l}{g _{e ff}} = 2 π \frac{l}{\frac{5 g}{4}}$

$T^{'} = \frac{2}{5} \times 2 π \frac{l}{g}$

Substituting the value of $T$ , we get

$T^{'} = \frac{2}{5} \times 5 = 2$ s

Answer: $2$ s

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