The time period T of a simple pendulum of len | vedclass

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(B) Given,$T = 2 \pi \sqrt{\frac{l}{g}}$.Taking the natural logarithm on both sides,we get $\ln T = \ln(2 \pi) + \frac{1}{2} \ln l - \frac{1}{2} \ln g

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$1 \%$

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(B) Given,$T = 2 \pi \sqrt{\frac{l}{g}}$.Taking the natural logarithm on both sides,we get $\ln T = \ln(2 \pi) + \frac{1}{2} \ln l - \frac{1}{2} \ln g$.Differentiating both sides,we get $\frac{dT}{T} = \frac{1}{2} \frac{dl}{l}$.Given that the length is increased by $2 \%$,so $\frac{dl}{l} = 0.02$.Substituting this value,we get $\frac{dT}{T} = \frac{1}{2} 	imes 0.02 = 0.01$.Therefore,the percentage change in the time period is $\frac{dT}{T} 	imes 100 = 0.01 	imes 100 = 1 \%$.Hence,the approximate change in the time period is $1 \%$.

The time period $T$ of a simple pendulum of length $l$ is given by $T=2 \pi \sqrt{\frac{l}{g}}$. If the length is increased by $2 \%$,then the approximate change in the time period is:

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