The period of oscillation of a simple pendulu | vedclass

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(B) 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 impl

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$0.4$

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(B) 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 $l = 80	ext{ cm}$ and $\Delta l = 0.1	ext{ cm}$, so $\frac{\Delta l}{l} = \frac{0.1}{80} = 0.00125$.The total time for $50$ oscillations is $t = 50T = 50 	imes 1.5 = 75	ext{ s}$.The least count of the stopwatch is $\Delta t = 0.1	ext{ s}$.The error in the period $T$ is $\Delta T = \frac{\Delta t}{50} = \frac{0.1}{50} = 0.002	ext{ s}$.Thus, $\frac{\Delta T}{T} = \frac{0.002}{1.5} = \frac{2}{1500} = 0.00133$.Substituting these values into the error formula: $\frac{\Delta g}{g} = 0.00125 + 2(0.00133) = 0.00125 + 0.00266 = 0.00391$.The percentage error is $\frac{\Delta g}{g} 	imes 100 \approx 0.391\% \approx 0.4\%$.

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