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

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$\sqrt{3/2} T$

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(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$.When the lift is stationary,$g_{eff} = g$.When the lift is descending with an acceleration $a = g/3$,the effective acceleration $g_{eff}$ is given by $g_{eff} = g - a = g - g/3 = 2g/3$.The new time period $T'$ is given by $T' = 2\pi \sqrt{\frac{\ell}{g_{eff}}} = 2\pi \sqrt{\frac{\ell}{2g/3}} = 2\pi \sqrt{\frac{3\ell}{2g}}$.Since $T = 2\pi \sqrt{\frac{\ell}{g}}$,we can write $T' = \sqrt{\frac{3}{2}} 	imes (2\pi \sqrt{\frac{\ell}{g}}) = \sqrt{\frac{3}{2}} T$.

$A$ lift is descending with an acceleration of $g/3$. What will be the time period of a simple pendulum suspended from its ceiling if its time period in a stationary lift is $T$?

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