A simple pendulum is executing simple harmoni | vedclass

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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$,which implies $T \propto \sqrt{\ell}$.Let the initial length be

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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$,which implies $T \propto \sqrt{\ell}$.Let the initial length be $\ell_1 = \ell$ and the initial time period be $T_1 = T$.The new length is $\ell_2 = \ell + 0.21\ell = 1.21\ell$.The new time period $T_2$ is given by $\frac{T_2}{T_1} = \sqrt{\frac{\ell_2}{\ell_1}} = \sqrt{\frac{1.21\ell}{\ell}} = \sqrt{1.21} = 1.1$.Thus,$T_2 = 1.1T_1$.The percentage increase in the time period is $\frac{T_2 - T_1}{T_1} 	imes 100 = \frac{1.1T_1 - T_1}{T_1} 	imes 100 = 0.1 	imes 100 = 10\%$.

$A$ simple pendulum is executing simple harmonic motion with a time period $T$. If the length of the pendulum is increased by $21\%$,the percentage increase in the time period of the pendulum of increased length is ..... $\%$

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