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A student determines g using the formula g = 4*pi²*L/T², where L is approximately 1 m. The error in L is deltaL (in cm), and for period T, the student times n oscillations with a stopwatch of least count deltaT (in seconds) and also makes a human error of 0.1 s. For which of the following experimental setups will the measurement of g be most accurate?

1. deltaL = 0.5 cm, deltaT = 0.1 s, n = 20
2. deltaL = 0.5 cm, deltaT = 0.1 s, n = 50
3. deltaL = 0.5 cm, deltaT = 0.01 s, n = 20
4. deltaL = 0.1 cm, deltaT = 0.05 s, n = 50

Correct answer: deltaL = 0.1 cm, deltaT = 0.05 s, n = 50

Solution

Relative error: dg/g = dL/L + 2*(delta_T_total)/T. Here delta_T_total (error in T per oscillation) = (deltaT + 0.1)/n. T ~ 2 s for L=1m. Compute for each option and select minimum.

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