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Question

(B) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g_{eff}}}$.In a stationary lift,the effective acceleration is $g_{eff}

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

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(B) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g_{eff}}}$.In a stationary lift,the effective acceleration is $g_{eff} = g$,so $T = 2 \pi \sqrt{\frac{l}{g}}$.When the lift accelerates upwards with an acceleration $a = \frac{g}{3}$,the effective acceleration experienced by the pendulum is $g_{eff} = g + a$.Substituting the value of $a$,we get $g_{eff} = g + \frac{g}{3} = \frac{4g}{3}$.The new time period $T^{\prime}$ is given by $T^{\prime} = 2 \pi \sqrt{\frac{l}{g_{eff}}} = 2 \pi \sqrt{\frac{l}{4g/3}}$.Simplifying this,$T^{\prime} = 2 \pi \sqrt{\frac{3l}{4g}} = \frac{\sqrt{3}}{2} \left( 2 \pi \sqrt{\frac{l}{g}} \right)$.Since $T = 2 \pi \sqrt{\frac{l}{g}}$,we have $T^{\prime} = \frac{\sqrt{3}}{2} T$.

The time period of a simple pendulum inside a stationary lift is $T$. When the lift starts accelerating upwards with an acceleration of $\frac{g}{3}$,the time period of the pendulum will be

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