The period of oscillation of a simple pendulu | vedclass

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(A) The formula for the period of a simple pendulum is $T = 2 \pi \sqrt{\frac{L}{g}}$.Squaring both sides,we get $T^2 = 4 \pi^2 \frac{L}{g}$,which imp

Vedclass · Accepted Answer

$1.13$

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(A) The formula for the period of a simple pendulum is $T = 2 \pi \sqrt{\frac{L}{g}}$.Squaring both sides,we get $T^2 = 4 \pi^2 \frac{L}{g}$,which implies $g = \frac{4 \pi^2 L}{T^2}$.The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$.Given values are $L = 1.0 	ext{ m}$,$\Delta L = 1 	ext{ mm} = 0.001 	ext{ m}$,$T = 1.95 	ext{ s}$,and $\Delta T = 0.01 	ext{ s}$.Substituting these values: $\frac{\Delta g}{g} = \frac{0.001}{1.0} + 2 	imes \frac{0.01}{1.95}$.$\frac{\Delta g}{g} = 0.001 + 0.010256 = 0.011256$.To find the percentage error,multiply by $100$: $0.011256 	imes 100 \approx 1.13 \%$.

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