If the length of a seconds pendulum becomes f | vedclass

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(A) The time period of a seconds pendulum is $T_{1} = 2 \ s$.The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$.Sin

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$\frac{2}{\sqrt{3}} \ s$

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(A) The time period of a seconds pendulum is $T_{1} = 2 \ s$.The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$.Since $2\pi$ and $g$ are constants,we have the relation $T \propto \sqrt{l}$.Given that the new length $l_{2} = \frac{l_{1}}{3}$,we can write the ratio:$\frac{T_{2}}{T_{1}} = \sqrt{\frac{l_{2}}{l_{1}}} = \sqrt{\frac{l_{1}/3}{l_{1}}} = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}}$.Therefore,the new time period is $T_{2} = \frac{T_{1}}{\sqrt{3}} = \frac{2}{\sqrt{3}} \ s$.

If the length of a seconds pendulum becomes $\frac{1}{3}$ of its original length,what will be its new periodic time?

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