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(B) The time period of a simple pendulum at rest is given by $T = 2\pi \sqrt{\frac{l}{g}}$.When the lift accelerates upwards with an acceleration $a$,

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$3$

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(B) The time period of a simple pendulum at rest is given by $T = 2\pi \sqrt{\frac{l}{g}}$.When the lift accelerates upwards with an acceleration $a$,the effective acceleration due to gravity becomes $g_{eff} = g + a$.The new time period is $T' = 2\pi \sqrt{\frac{l}{g+a}}$.Given that $T' = T/2$,we have $\frac{T}{2} = 2\pi \sqrt{\frac{l}{g+a}}$.Dividing the expression for $T'$ by $T$,we get $\frac{1}{2} = \sqrt{\frac{g}{g+a}}$.Squaring both sides,$\frac{1}{4} = \frac{g}{g+a}$.This implies $g + a = 4g$,so $a = 3g$.

$A$ simple pendulum is suspended from the ceiling of a lift. When the lift is at rest,its time period is $T$. With what acceleration should the lift be accelerated upwards in order to reduce its period to $T/2$ (in $g$)? ($g$ is acceleration due to gravity).

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