A pendulum performs S.H.M. with period sqrt{3 | vedclass

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(B) The time period of a simple pendulum in a stationary lift is given by $T = 2\pi \sqrt{\frac{\ell}{g}} = \sqrt{3} \ s$. When the lift moves upward

Vedclass · Accepted Answer

$1.5$

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(B) The time period of a simple pendulum in a stationary lift is given by $T = 2\pi \sqrt{\frac{\ell}{g}} = \sqrt{3} \ s$. When the lift moves upward with an acceleration $a = \frac{g}{3}$,the effective acceleration due to gravity becomes $g_{eff} = g + a = g + \frac{g}{3} = \frac{4g}{3}$. The new time period $T^{\prime}$ is given by $T^{\prime} = 2\pi \sqrt{\frac{\ell}{g_{eff}}} = 2\pi \sqrt{\frac{\ell}{4g/3}}$. Dividing $T^{\prime}$ by $T$,we get $\frac{T^{\prime}}{T} = \sqrt{\frac{g}{4g/3}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$. Substituting the value of $T = \sqrt{3} \ s$,we get $T^{\prime} = \sqrt{3} 	imes \frac{\sqrt{3}}{2} = \frac{3}{2} = 1.5 \ s$.

$A$ pendulum performs $S.H.M.$ with period $\sqrt{3} \ s$ in a stationary lift. If the lift moves up with acceleration $\frac{g}{3}$,the period of the pendulum is $[g=$ acceleration due to gravity $]$. (in $s$)

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