The period of oscillation of a simple pendulu | vedclass

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

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$5$

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(C) The formula for the period of a simple pendulum is $T = 2\pi \sqrt{L/g}$.Squaring both sides, we get $T^2 = 4\pi^2 (L/g)$, which implies $g = 4\pi^2 L / T^2$.The relative error in $g$ is given by $\Delta g/g = \Delta L/L + 2(\Delta T/T)$.Given $L = 10 	ext{ cm} = 0.1 	ext{ m}$ and $\Delta L = 1 	ext{ mm} = 0.1 	ext{ cm} = 0.001 	ext{ m}$.So, $\Delta L/L = 0.1 / 10 = 0.01$.The time for $100$ oscillations is $t = 50 	ext{ s}$ with resolution $\Delta t = 1 	ext{ s}$.The period $T = t/100 = 50/100 = 0.5 	ext{ s}$.The error in period is $\Delta T = \Delta t / 100 = 1 / 100 = 0.01 	ext{ s}$.So, $\Delta T/T = 0.01 / 0.5 = 0.02$.Substituting these values into the error formula:$\Delta g/g = 0.01 + 2(0.02) = 0.01 + 0.04 = 0.05$.The percentage error is $0.05 	imes 100\% = 5\%$.

The period of oscillation of a simple pendulum is $T = 2\pi(L/g)^{1/2}$. Measured value of $L$ is $10 \text{ cm}$ known to $1 \text{ mm}$ accuracy and time for $100$ oscillations is $50 \text{ s}$, using a watch of $1 \text{ s}$ resolution. The percentage error in the measurement of $g$ is (in $\%$)

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