The time period of a simple pendulum inside a | vedclass

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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.For a stationary lift,$g_{eff} = g$,so $T_1 = 2\pi \sqrt{\fra

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$1.5 \ s$

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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.For a stationary lift,$g_{eff} = g$,so $T_1 = 2\pi \sqrt{\frac{L}{g}} = \sqrt{3} \ s$.When the lift moves upwards with an acceleration $a = g/3$,the effective acceleration is $g_{eff} = g + a = g + g/3 = 4g/3$.The new time period $T_2$ is given by $T_2 = 2\pi \sqrt{\frac{L}{4g/3}} = 2\pi \sqrt{\frac{3L}{4g}} = \sqrt{\frac{3}{4}} 	imes (2\pi \sqrt{\frac{L}{g}})$.Substituting $T_1 = \sqrt{3} \ s$,we get $T_2 = \frac{\sqrt{3}}{2} 	imes \sqrt{3} = \frac{3}{2} = 1.5 \ s$.

The time period of a simple pendulum inside a stationary lift is $\sqrt{3} \ s$. When the lift moves upwards with an acceleration $g/3$,the new time period will be ($g=$ acceleration due to gravity).

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