A man measures the period of a simple pendulu | vedclass

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(C) In a stationary lift,the time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.When the lift accelerates upwards with an acce

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$\frac{2T}{\sqrt{5}}$

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(C) In a stationary lift,the time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.When the lift accelerates upwards with an acceleration $a = \frac{g}{4}$,the effective acceleration due to gravity becomes $g_{eff} = g + a = g + \frac{g}{4} = \frac{5g}{4}$.The new time period $T'$ is given by $T' = 2\pi \sqrt{\frac{l}{g_{eff}}} = 2\pi \sqrt{\frac{l}{5g/4}} = 2\pi \sqrt{\frac{4l}{5g}}$.We can rewrite this as $T' = \sqrt{\frac{4}{5}} 	imes (2\pi \sqrt{\frac{l}{g}}) = \frac{2}{\sqrt{5}} T$.

$A$ man measures the period of a simple pendulum inside a stationary lift and finds it to be $T$ seconds. If the lift accelerates upwards with an acceleration of $\frac{g}{4}$,then the period of the pendulum will be

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