A man measures the period of a simple pendulu | vedclass

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(C) The time period of a simple pendulum in a stationary lift is given by $T = 2\pi \sqrt{\frac{l}{g}}$,where $l$ is the length of the pendulum and $g

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The time period of a simple pendulum in a stationary lift is given by $T = 2\pi \sqrt{\frac{l}{g}}$,where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.When the lift accelerates upwards with an acceleration $a = g/4$,the effective acceleration due to gravity becomes $g_{eff} = g + a = g + g/4 = 5g/4$.The new time period $T'$ is given by $T' = 2\pi \sqrt{\frac{l}{g_{eff}}}$.Substituting the value of $g_{eff}$,we get $T' = 2\pi \sqrt{\frac{l}{5g/4}} = 2\pi \sqrt{\frac{4l}{5g}} = \sqrt{\frac{4}{5}} 	imes (2\pi \sqrt{\frac{l}{g}})$.Since $T = 2\pi \sqrt{\frac{l}{g}}$,we have $T' = \sqrt{\frac{4}{5}} T = \frac{2}{\sqrt{5}} T$.

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

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