A man measures the period of a simple pendulum inside a stationary lift and finds it be T . If the lift accelerates downwards with an acceleration of g 3 , then the period of the pendulum will be

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When the lift is moving in a downward direction with an acceleration $\frac{g}{3}$ then

$g_{e f f} = g - \frac{g}{3} = \frac{2}{3} g$

Since we know, the time period of a simple pendulum,

$T_{1} = 2 π \sqrt{\frac{l}{g_{e f f}}} = 2 π \cdot \sqrt{\frac{l}{\frac{2}{3} g}}$

$T_{1} = \sqrt{\frac{3}{2}} \cdot 2 π \sqrt{\frac{l}{g}}$

Hence, $T_{1} = \sqrt{\frac{3}{2}} T$ $[∵ T = 2 π \sqrt{\frac{l}{g}}]$

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