Pendulum Measurement of g: Uncertainty and Design
A student determines g with a simple pendulum. The string is 1.000 m long, measured with a metre rule of 1 mm resolution from the pivot to the centre of a 50 g bob; the uncertainty in locating the bob’s centre is about ±2 mm. She times 20 complete oscillations with a stopwatch of 0.01 s resolution and repeats five times: 40.12, 40.08, 40.15, 40.10, 40.13 s. She uses g = 4π²L/T², with T = (total time)/20. A second student times a single oscillation and gets 2.01 s. The first student reports g = 9.81 m/s² with no uncertainty. The teacher asks for a defensible uncertainty statement and for an experimental-design critique. The bob is released from about 5° from vertical, and the student assumes the small-angle approximation is exact.
1. From the five times, calculate the mean period and g. Show your values.
Read the worked answer
Mean total time = (40.12 + 40.08 + 40.15 + 40.10 + 40.13)/5 = 40.116 s. Period T = 40.116/20 = 2.0058 s. Then g = 4π²L/T² = 4π² × 1.000/(2.0058)² = 39.478/4.023 = 9.81 m/s².
2. Estimate the standard uncertainty in g from the timing and length measurements. Which source dominates? State the result as g ± u(g).
Read the worked answer
Deviations from the mean time are +0.004, −0.036, +0.034, −0.016, +0.014 s. The sample standard deviation is s = √(Σd²/(5−1)) = 0.027 s. The standard error of the mean is u(t) = s/√5 = 0.012 s. Relative timing uncertainty is u(T)/T = 0.012/40.116 = 3.0 × 10⁻⁴. Since g ∝ T⁻², the timing contribution to u(g)/g is 2 × 3.0 × 10⁻⁴ = 6.0 × 10⁻⁴.
Length uncertainty is about 2 mm = 0.002 m, so u(L)/L = 2.0 × 10⁻³. Combined relative uncertainty: u(g)/g = √[(6.0 × 10⁻⁴)² + (2.0 × 10⁻³)²] = 2.1 × 10⁻³. Thus u(g) = 9.81 × 2.1 × 10⁻³ = 0.021 m/s², so report g = (9.81 ± 0.02) m/s². The length measurement dominates.
3. The second student argues that timing one oscillation is equivalent because the stopwatch resolution is only 0.01 s. Critique this and suggest two improvements that would reduce the dominant uncertainty.
Read the worked answer
Timing one oscillation is not equivalent. With 0.01 s resolution, one period has relative resolution about 0.01/2.0 = 0.005, or 0.5%, which alone gives about 1% in g-much worse than the 0.2% length uncertainty. Timing 20 oscillations reduces the resolution contribution by a factor of 20.
Improvements: (1) measure the length more precisely and define the endpoint-use a vernier caliper or a jig to locate the bob’s centre, since length dominates; (2) use a fiducial mark at the equilibrium position and time many oscillations, or use a photogate; (3) repeat for several lengths and plot T² against L, obtaining g from the slope 4π²/g, which also tests for a systematic offset in L. Keep the amplitude small because the small-angle formula is an approximation.