The period of an oscillating simple pendulum | vedclass

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Question

(A) Given the formula for the time period: $T = 2 \pi \sqrt{\frac{\ell}{g}}$.Squaring both sides: $T^2 = 4 \pi^2 \frac{\ell}{g}$,which implies $g = 4

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$0.2$

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(A) Given the formula for the time period: $T = 2 \pi \sqrt{\frac{\ell}{g}}$.Squaring both sides: $T^2 = 4 \pi^2 \frac{\ell}{g}$,which implies $g = 4 \pi^2 \frac{\ell}{T^2}$.The relative error in $g$ is given by: $\frac{\Delta g}{g} = \frac{\Delta \ell}{\ell} + 2 \frac{\Delta T}{T}$.Given: $\ell = 100 	ext{ cm} = 1 	ext{ m}$,$\Delta \ell = 1 	ext{ mm} = 0.001 	ext{ m}$.So,$\frac{\Delta \ell}{\ell} = \frac{0.001}{1} = 0.001$.The time for $100$ oscillations is $t = 100 	imes T = 100 	imes 2 = 200 	ext{ s}$.The error in measuring $100$ oscillations is $\Delta t = 0.1 	ext{ s}$.The error in the period $T$ is $\Delta T = \frac{\Delta t}{100} = \frac{0.1}{100} = 0.001 	ext{ s}$.So,$\frac{\Delta T}{T} = \frac{0.001}{2} = 0.0005$.Substituting these into the relative error formula: $\frac{\Delta g}{g} = 0.001 + 2(0.0005) = 0.001 + 0.001 = 0.002$.The percentage error is $\frac{\Delta g}{g} 	imes 100 = 0.002 	imes 100 = 0.2 \%$.

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