The measured value of the length of a simple pendulum is 20 cm with 2 mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the acc

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The measured value of the length of a simple pendulum is 20 cm with 2 mm accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is N % . The value of N is:

TestHub · Accepted Answer

Time period of a simple pendulum is given by, $T = 2 π (\frac{l}{g})$ .

$\Rightarrow g = \frac{4 π^{2} l}{T^{2}}$

Therefore, fractional error will be

$\frac{Δ g}{g} = \frac{Δ l}{l} + \frac{2 Δ T}{T}$

$= \frac{0.2}{20} + 2 (\frac{1}{40})$

$= \frac{1.2}{20}$

Percentage change $= \frac{1.2}{20} \times 100 = 6 %$

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