Timing g with a simple pendulum: which uncertainty actually matters?
A student determines g using a simple pendulum. The distance from the pivot to the centre of the bob is measured with a metre rule as 0.842 m; the rule has a 1 mm resolution. The bob diameter is 22 mm, measured with vernier calipers. The student times 20 complete oscillations with a stopwatch of resolution 0.01 s and obtains 36.72 s. The reaction time is estimated as 0.20 s. Five repeat timings differ by 0.18 s in total. A graph of T² against L for several lengths is a straight line through the origin. The final value of g differs from 9.81 m/s² by 1.5%. The student must decide whether to quote the uncertainty in the period using the stopwatch resolution, the reaction time, or the repeat spread, and whether the metre rule or the stopwatch dominates the uncertainty in g.
1. Calculate the period and estimate the percentage uncertainty in T from the stopwatch resolution, the reaction time and the repeat spread. Which estimate should be quoted?
Read the worked answer
The period is
T = 36.72 / 20 = 1.836 s.
- Stopwatch resolution: ±0.01 s over 20 oscillations gives ±0.0005 s per oscillation, i.e. 0.0005/1.836 ≈ 0.03%. This is negligible.
- Reaction time: if the total interval is uncertain by about ±0.20 s, the period uncertainty is ±0.010 s, i.e. 0.54%. If the start and stop delays were independent and both ±0.20 s, the worst case would be ±0.40 s, giving ±1.1%; in practice the same observer tends to cancel part of this.
- Repeat spread: 0.18 s over 20 oscillations is 0.009 s per oscillation, i.e. 0.49%. The standard uncertainty of the mean of five repeats would be smaller, about 0.08 s in the total, or 0.004 s per period.
The resolution is not the limiting factor. The honest quoted random uncertainty is about 0.5%, based on the repeat spread, with reaction time treated as a possible systematic effect of similar size. Quoting the stopwatch resolution alone would badly understate the uncertainty.
2. Combine this with the length uncertainty and decide whether the 1.5% discrepancy from 9.81 m/s² is significant. What checks should be made?
Read the worked answer
Since g = L(2π/T)², the fractional uncertainty is
Δg/g = ΔL/L + 2ΔT/T.
The metre rule gives ΔL/L = 0.001/0.842 ≈ 0.12%, provided the length is genuinely measured to the centre of the bob. Using ΔT/T ≈ 0.49% gives Δg/g ≈ 0.12% + 2(0.49%) ≈ 1.1%. Using the reaction-time estimate of 0.54% gives ≈1.2%. The observed 1.5% discrepancy is therefore only slightly larger than the estimated uncertainty; it is not strong evidence of a genuine error.
A more important check is the effective length. If the student measured to the top of the bob instead of its centre, the length is too large by the bob radius, 11 mm. That is 11/842 ≈ 1.3% in L and hence about 1.3% in g, enough to explain the discrepancy. Other checks: finite-amplitude correction (about θ²/16, only 0.2% at 10°), pivot position, and whether the pendulum swings in a plane. The through-origin graph supports the T² ∝ L model but does not remove a constant length offset by itself; a genuine offset would normally show as an intercept.
3. The T² against L graph is a straight line through the origin. What does this confirm and what does it not confirm?
Read the worked answer
For a simple pendulum, T² = (4π²/g)L. A straight line through the origin therefore confirms the expected proportionality and allows g to be found from the gradient: g = 4π² / gradient. It also argues against a constant additive error in L, because such an error would produce a non-zero intercept.
It does not confirm that the absolute value of g is correct. A systematic error in timing that scales all periods equally, or a wrong effective length that scales all lengths, could still leave a straight line through the origin with the wrong gradient. Nor does it test small-angle behaviour or the independence of mass. To validate the value, the student should compare the gradient result with an independent method and check the effective length and timing procedure.