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Question:

The period of oscillation of a simple pendulum is T=2π√(L/g). Measure of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. The accuracy in the determination of g?

2.7

3

2.5

2

Solution:

Δg/g = (ΔL/L) + 2(ΔT/T)
ΔL = 1 mm = 0.1 cm
ΔT = 1 s
T = 90 s
L = 20.0 cm
Δg/g = (0.1/20) + 2(1/90) = 0.005 + 0.0222 = 0.0272
Percentage error in g = 0.0272 × 100 = 2.72%
Therefore, the accuracy in the determination of g is approximately 2.7%