Pendulum Timing and the Uncertainty Budget
A student uses a simple pendulum to determine g. The length from the pivot to the centre of the bob is measured with a metre rule as 0.842 m, with an estimated reading uncertainty of ±0.002 m. The student times 20 complete oscillations with a stopwatch of resolution 0.01 s and repeats the measurement three times: 36.80 s, 36.92 s and 36.74 s. Reaction time is about 0.2 s. Taking the mean time, the student calculates T and then g = 4π²L/T², obtaining 9.81 m/s². They report the result as (9.81 ± 0.03) m/s², explaining that they added the percentage uncertainty in L (0.24%) to the percentage uncertainty in the stopwatch resolution (0.01/36.82 = 0.027%, doubled because T is squared). The student claims the experiment is precise to about 0.3% and that repeating the timing is unnecessary because the stopwatch is digital.
1. Calculate the mean period T and the value of g from the mean time. Show your working and give g to an appropriate number of significant figures.
Read the worked answer
Mean time: (36.80 + 36.92 + 36.74)/3 = 36.82 s.
Period: T = 36.82/20 = 1.841 s.
T² = 3.389 s².
g = 4π²L/T² = 39.478 × 0.842 / 3.389 = 9.81 m/s².
The data justify 3 significant figures: L is given to 3 s.f. and the period to 4 s.f., so g = 9.81 m/s².
2. Evaluate the student's quoted uncertainty of ±0.03 m/s². Identify the dominant source of uncertainty and decide whether the three timings support a 0.3% precision claim.
Read the worked answer
The quoted ±0.03 m/s² is too small.
The three timings are not identical: their spread is 0.18 s, so a reasonable random uncertainty in the mean 20T is about half the range, 0.09 s. That is 0.09/36.82 = 0.24% in T, and because T is squared it becomes 0.49% in g.
The length contributes 0.002/0.842 = 0.24%. Combining in quadrature: √(0.49² + 0.24²) = 0.55%, giving ±0.05 m/s².
Reaction time is a separate timing uncertainty: 0.2 s on 36.82 s is 0.54%, doubled to about 1.1%, i.e. ±0.11 m/s². The stopwatch resolution (0.01 s) is negligible compared with the observed scatter and reaction time. The dominant uncertainty is therefore the timing, not the metre rule. The 0.3% precision claim is not supported by the three timings.
3. Suggest one improvement that would reduce the dominant uncertainty and explain why it works. Explain also why timing 20 oscillations is better than timing one, but does not remove a systematic error in the length measurement.
Read the worked answer
Improvement: time more oscillations, for example 50 instead of 20, or use a light gate/video timer. Timing more oscillations reduces the fractional effect of reaction time on T by the same factor (20 → 50 reduces it by 2.5), and it also averages random starting/stopping variation over a longer interval.
Timing 20 oscillations is better than timing one because the reaction-time error is divided by 20 when finding T; timing one oscillation would give a timing uncertainty of order 0.2/1.8 ≈ 11%.
However, repeating the timing does not remove a systematic error in L. If the length is measured to the top of the bob rather than its centre, or if there is a parallax or zero error in the rule, every L is shifted by the same amount. Repeating the measurement gives the same shifted value. A different measurement method (fiducial mark at the bob centre, travelling microscope, or plotting T² against L for several lengths) is needed to expose or eliminate that systematic effect.