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For experimentally found pendulum T = 2π sqrt(L/g), Length L = 1.00 m ± 0.01 m, T = 2.00 s ± 0.01 s. Find g and its error.
Ag = 100
Bg = 9.87 m/s²; Δg/g = 0.01 + 2(0.01/2) = 0.02 ≈ 2%
Cg = 1
Dg = 10
Answer & Solution
Correct answer: B. g = 9.87 m/s²; Δg/g = 0.01 + 2(0.01/2) = 0.02 ≈ 2%
g = 4π²L/T². g_calc = 4π² × 1/4 = π² ≈ 9.87 m/s². Δg/g = ΔL/L + 2 ΔT/T = 0.01 + 0.01 = 0.02 = 2%. So g = 9.87 ± 0.2 m/s².
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