The period of a simple pendulum gets doubled | vedclass

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(B) We know that the time period $T$ of a simple pendulum is given by the formula:$T = 2 \pi \sqrt{\frac{l}{g}}$From this relation,we can see that $T

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its length is made four times.

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(B) We know that the time period $T$ of a simple pendulum is given by the formula:$T = 2 \pi \sqrt{\frac{l}{g}}$From this relation,we can see that $T \propto \sqrt{l}$.Let the initial time period be $T_1$ with length $l_1$,and the final time period be $T_2$ with length $l_2$.We want the period to be doubled,so $T_2 = 2 T_1$.Using the ratio:$\frac{T_2}{T_1} = \sqrt{\frac{l_2}{l_1}}$Substituting $T_2 = 2 T_1$:$2 = \sqrt{\frac{l_2}{l_1}}$Squaring both sides:$4 = \frac{l_2}{l_1} \Rightarrow l_2 = 4 l_1$Therefore,the period of a simple pendulum is doubled when its length is made four times the original length.

The period of a simple pendulum gets doubled when

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