Pendulum Precision Under a Time Limit: Which Change Actually Buys Accuracy?
A physics practical group must report a value of g from a simple pendulum. The string runs from the clamp to the top of the bob and measures 0.988 m on a metre scale whose smallest division is 1 mm. The bob is a steel sphere of diameter 24.0 mm, measured with a vernier calliper of least count 0.02 mm. Twenty oscillations, timed from the equilibrium position with a stopwatch whose stated least count is 0.01 s, take 40.2 s; the student's reaction time is about 0.2 s. The release amplitude is roughly 15 degrees.
Three groups share the apparatus and the slot ends in 20 minutes. The group can afford exactly one of: (i) moving the clamp so the string-to-bob-top length becomes 1.488 m; (ii) timing 50 oscillations at the present length; (iii) adding the bob radius to the length - a change a teammate dismisses as 'too small to matter'. The demonstrator will only accept a value carrying a defensible uncertainty.
1. Build the uncertainty budget for g = 4π²L/T² from these data. Compute g, the percentage uncertainty, and rank options (i) and (ii) by precision gained per minute of lab time.
Read the worked answer
Value first. With the radius added, L = 0.988 + 0.012 = 1.000 m, and T = 40.2/20 = 2.010 s:
g = 4π²(1.000)/(2.010)² = 39.478/4.040 = 9.77 m s⁻².
Budget.
- Length: ΔL ≈ 1 mm scale division → 0.001/1.000 = 0.10%.
- Time: the reaction time, not the 0.01 s least count, is the real limit → 0.2/40.2 = 0.50% in the 20-oscillation reading, and this is the same fractional error in T.
- Since g ∝ T⁻², the time term doubles: Δg/g = 0.10% + 2(0.50%) = 1.1%, i.e. g = 9.77 ± 0.11 m s⁻².
Option (i), longer length. L becomes 1.500 m, so T scales as √1.5 → T ≈ 2.46 s and 20 oscillations take 49.2 s. Time term 0.2/49.2 = 0.41%; length term 0.067%. Δg/g = 0.067 + 0.82 = 0.89%. Gain: a factor 1.25, paid for with roughly 10 minutes of resetting, plumbing and re-levelling.
Option (ii), more oscillations. 50 oscillations take 100.5 s, so the reaction error is diluted: 0.2/100.5 = 0.20%, giving Δg/g = 0.10 + 0.40 = 0.50%. Gain: a factor 2.2, and it costs only about 100 s per run.
Judgement. Option (ii) dominates: the dominant error is in timing, and lengthening L attacks it only through √L, while counting more swings attacks it linearly. The honest caveat is that a single 50-swing run cannot reveal scatter, so the group should use the remaining time for two repeats rather than one very long run, and quote the spread.
2. The stopwatch says its least count is 0.01 s. Explain why the group must not use 0.01/40.2 as the timing uncertainty, and state what they should use instead. Should the reaction-time error be treated as random or systematic?
Read the worked answer
A least count is a resolution figure: it says the display can be read to 0.01 s, not that a human can start and stop the watch to 0.01 s. Taking 0.01 s as the uncertainty would give 0.025% on the timing and imply Δg/g ≈ 0.15%, which is roughly seven times more optimistic than reality - an error budget that would later make the result look 'wrong' when compared with 9.81 m s⁻².
The binding limit is the operator. Reaction time to a visual cue is typically 0.15–0.25 s, so 0.2 s is the defensible figure. Note that in Δg/g = ΔL/L + 2Δt/t, Δt is propagated once (the total timing error for the 20-swing interval), not once per oscillation - 0.2/40.2 = 0.50%, not 20 × 0.50%.
Random or systematic? Mostly systematic. The same student starting the watch at the bob's passage and stopping it at the 20th passage carries the same personal lag at both ends, and these largely cancel; what remains is the run-to-run fluctuation of that lag, which is smaller. The group can demonstrate this cheaply: time ten separate 20-swing runs and take the standard deviation of the ten values. If that spread is much smaller than 0.2 s, the systematic part is indeed cancelling and the budget should use the measured standard deviation, not the crude 0.2 s. Either way the number must come from repeated measurement, never from the manufacturer's least count.
3. A teammate argues the 15-degree release angle 'does not matter'. Quantify the small-angle error and decide whether the run must be repeated or can be corrected, given the group's uncertainty.
Read the worked answer
For a simple pendulum the period at finite amplitude is
T = T₀(1 + θ₀²/16 + …),
with θ₀ in radians. Here θ₀ = 15° = 0.262 rad, so θ₀²/16 = 0.0685/16 = 0.0043, i.e. the measured period is about 0.43% longer than the small-angle period.
Because g ∝ T⁻², the extracted g is low by 2 × 0.43% = 0.86%, about 0.084 m s⁻² - smaller than the quoted ± 0.11 m s⁻², but of the same order, so it is not honest to call it negligible. Uncorrected, the run gives 9.77 m s⁻²; multiplying by the correction factor (1 + θ₀²/16)² = 1.0086 gives 9.85 m s⁻².
Decision. The correction is analytic, its size is known, and it sits inside the uncertainty, so repeating the run is a poor use of the remaining 20 minutes. The right move is to apply the factor, state it explicitly as a systematic correction, and then attack the dominant term - timing - by counting more oscillations. What is not acceptable is either ignoring the amplitude or, worse, quoting the uncorrected number as if it were the small-angle result.
For future runs, keep θ₀ ≤ 5°: the error then falls to (0.087)²/16 = 0.05%, which is genuinely ignorable. Note also that the teammate's instinct ('it is only a few degrees') is exactly the reasoning that produces unexplained 1% discrepancies between groups using the same apparatus.