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 impli

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(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 $g = 4\pi^2 \frac{l}{T^2}$.The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2 \frac{\Delta T}{T}$.Given values are $l = 20.0 	ext{ cm}$,$\Delta l = 1 	ext{ mm} = 0.1 	ext{ cm}$.Total time for $100$ oscillations is $t = 90 	ext{ s}$ with a resolution $\Delta t = 1 	ext{ s}$.The period $T = \frac{t}{100} = 0.9 	ext{ s}$,and the error in period $\Delta T = \frac{\Delta t}{100} = \frac{1}{100} = 0.01 	ext{ s}$.Now,the percentage error is $\frac{\Delta g}{g} 	imes 100 = \left( \frac{\Delta l}{l} 	imes 100 ight) + 2 \left( \frac{\Delta T}{T} 	imes 100 ight)$.Substituting the values: $\frac{\Delta g}{g} 	imes 100 = \left( \frac{0.1}{20.0} 	imes 100 ight) + 2 \left( \frac{0.01}{0.9} 	imes 100 ight)$.$= 0.5\% + 2 	imes 1.11\% = 0.5\% + 2.22\% = 2.72\%$.Rounding to the nearest integer,we get $3\%$.

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