A simple pendulum has a time period t. Its ti | vedclass

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Question

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$.Initially,the effective acceleration due to gravity is $g = 9

Vedclass · Accepted Answer

$t \sqrt{\frac{9.8}{12.8}}$

Vedclass · Answer

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$.Initially,the effective acceleration due to gravity is $g = 9.8 \ m/s^2$,so $t = 2\pi \sqrt{\frac{l}{9.8}}$.When the lift moves upwards with an acceleration $a = 3 \ m/s^2$,the effective acceleration becomes $g' = g + a = 9.8 + 3 = 12.8 \ m/s^2$.The new time period $t'$ is $t' = 2\pi \sqrt{\frac{l}{12.8}}$.Dividing the two equations: $\frac{t'}{t} = \frac{2\pi \sqrt{l/12.8}}{2\pi \sqrt{l/9.8}} = \sqrt{\frac{9.8}{12.8}}$.Therefore,$t' = t \sqrt{\frac{9.8}{12.8}}$.

$A$ simple pendulum has a time period $t$. Its time period in a lift which is moving upwards with an acceleration of $3 \ m/s^2$ is:

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