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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.When the lift is stationary,$g_{eff} = g$,so $T = 2\pi \sqrt{

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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.When the lift is stationary,$g_{eff} = g$,so $T = 2\pi \sqrt{\frac{L}{g}}$.When the lift falls freely,it experiences weightlessness,meaning the effective acceleration due to gravity $g_{eff} = 0$.Substituting $g_{eff} = 0$ into the formula,the time period $T' = 2\pi \sqrt{\frac{L}{0}} = \infty$.Since frequency $f = \frac{1}{T'}$,we get $f = \frac{1}{\infty} = 0$.Therefore,the frequency of oscillation is zero.

$A$ simple pendulum is attached to the roof of a lift. If the time period of oscillation when the lift is stationary is $T$,then the frequency of oscillation when the lift falls freely will be

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