In a stationary lift,the time period of a sim | vedclass

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

Vedclass · Accepted Answer

$\frac{2}{\sqrt{3}} T$

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(B) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g}}$.When the lift accelerates downwards with an acceleration $a = \frac{g}{4}$,the effective acceleration due to gravity $g_{	ext{eff}}$ becomes $g - a$.$g_{	ext{eff}} = g - \frac{g}{4} = \frac{3g}{4}$.The new time period $T_1$ is given by $T_1 = 2 \pi \sqrt{\frac{L}{g_{	ext{eff}}}} = 2 \pi \sqrt{\frac{L}{\frac{3g}{4}}}$.$T_1 = 2 \pi \sqrt{\frac{4L}{3g}} = 2 \pi \sqrt{\frac{L}{g}} 	imes \sqrt{\frac{4}{3}}$.Since $T = 2 \pi \sqrt{\frac{L}{g}}$,we have $T_1 = T 	imes \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} T$.

In a stationary lift,the time period of a simple pendulum is $T$. If the lift starts accelerating downwards with an acceleration of $\frac{g}{4}$,then the new time period of the pendulum will be:

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