A man measures the time period of a pendulum | vedclass

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Question

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.When the lift moves upward with acceleration $a = \frac{g}{4}$,the

Vedclass · Accepted Answer

$\frac{2T}{\sqrt{5}}$

Vedclass · Answer

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.When the lift moves upward with acceleration $a = \frac{g}{4}$,the effective acceleration due to gravity becomes $g' = g + a = g + \frac{g}{4} = \frac{5g}{4}$.The new time period $T'$ is given by $T' = 2\pi \sqrt{\frac{l}{g'}} = 2\pi \sqrt{\frac{l}{5g/4}} = 2\pi \sqrt{\frac{4l}{5g}}$.Comparing $T'$ with $T$,we get $\frac{T'}{T} = \sqrt{\frac{g}{g'}} = \sqrt{\frac{g}{5g/4}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$.Therefore,$T' = \frac{2T}{\sqrt{5}}$.

$A$ man measures the time period of a pendulum $(T)$ in a stationary lift. If the lift moves upward with an acceleration of $\frac{g}{4}$,then the new time period will be:

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