The time period of a simple pendulum in a sta | vedclass

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(C) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{\ell}{g_{	ext{eff}}}}$.$(a)$ When the lift is stationary $(a = 0)$,the ef

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$\sqrt{\frac{6}{7}} T$

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(C) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{\ell}{g_{	ext{eff}}}}$.$(a)$ When the lift is stationary $(a = 0)$,the effective acceleration is $g_{	ext{eff}} = g$. Thus,$T = 2 \pi \sqrt{\frac{\ell}{g}}$.$(b)$ When the lift accelerates upwards with $a = \frac{g}{6}$,the effective acceleration becomes $g_{	ext{eff}} = g + a = g + \frac{g}{6} = \frac{7g}{6}$.The new time period $T^{\prime}$ is given by:$T^{\prime} = 2 \pi \sqrt{\frac{\ell}{g_{	ext{eff}}}} = 2 \pi \sqrt{\frac{\ell}{7g/6}} = 2 \pi \sqrt{\frac{6\ell}{7g}}$.Comparing this with the original time period $T$,we get:$T^{\prime} = \sqrt{\frac{6}{7}} \left( 2 \pi \sqrt{\frac{\ell}{g}} ight) = \sqrt{\frac{6}{7}} T$.

The time period of a simple pendulum in a stationary lift is $T$. If the lift accelerates with $\frac{g}{6}$ vertically upwards,then the new time period will be:
(where $g =$ acceleration due to gravity)

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The time period of a simple pendulum in a stationary lift is $T$. If the lift accelerates with $\frac{g}{6}$ vertically upwards,then the new time period will be:(where $g =$ acceleration due to gravity)