The time period of oscillation of a simple pendulum is $T = 2 π \sqrt{(l / g)}$ .measured value of length is 20cm known to have 1mm accuracy and time for 100 oscillation of the pendulum is found to be 90s using wrist watch of 1s resolution.what is the accuracy in determining the value of acceleration due to gravity?

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Solution

$T =$ Time period of one oscillation
$t =$ Time period of 100 oscillation
For 100 oscillation, $t = 100 T$
Take ln and differentiate.
$\frac{Δ t}{t} = \frac{Δ T}{T} = \frac{2}{90} . . . . . (1)$
Time period of one oscillation
$T = 2 π \sqrt{\frac{L}{g}}$
$g = 4 π^{2} \frac{L}{T^{2}}$
Take ln on both side and differentiate
$\frac{Δ g}{g} = \pm (\frac{Δ L}{L} + 2 \frac{Δ T}{T})$
$\frac{Δ g}{g} = \pm (\frac{0.1}{20} + \frac{2}{90}) == \pm 2.7$
Hence, error in gravity is $\pm 2.7$ %

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