The two 'outliers' that weren't: timing a pendulum
Pendulum depth study. A Year 11 Physics student determines g using a 50 g brass bob on inextensible string, timing ten complete oscillations with a stopwatch and dividing by ten. Length is measured with a metre rule from the clamp to the top of the bob, because the bob's centre is hard to mark. Her results:
| L (m) | 0.200 | 0.400 | 0.600 | 0.800 | 0.900 | 1.000 |
|---|---|---|---|---|---|---|
| g (m s⁻²) | 9.46 | 9.45 | 9.67 | 9.69 | 9.66 | 9.68 |
For the first two lengths the swing amplitude was "about 30°"; after a teacher's reminder the remaining four were run at about 5°. The local accepted value is 9.794 m s⁻². In her report she writes: "The 0.200 m and 0.400 m results are outliers, so I deleted them. Mean g = 9.68 ± 0.03 m s⁻². This is 1.2% below the accepted value, which suggests either local gravity differs or the accepted value needs revision."
1. Evaluate the decision to delete the first two results and the way the uncertainty is reported.
Read the worked answer
The deletion reaches the right conclusion for the wrong reason, and the uncertainty statement is not defensible.
For θ = 30° = 0.524 rad, the large-amplitude correction gives T ≈ T₀(1 + θ²/16) = 1.0172 T₀, so g measured = g/(1.0172)² = 9.794/1.0347 = 9.47 m s⁻². The two "outliers" sit at 9.46 and 9.45 - they are not random scatter, they are exactly what the small-angle approximation predicts when it is violated. Treating them as outliers hides a known systematic effect; the correct move is to exclude them from the mean and report them with the mechanism, or correct them to 9.47 and show they then agree with the rest.
Second, ±0.03 is the scatter of the four retained readings (range 0.03), so it describes random error only. The dominant residual is a systematic 1.2% shortfall, which is many times larger. Quoting ±0.03 implies the accepted value lies outside the student's interval, but the real problem is her apparatus, not local gravity. She has no basis for questioning the accepted value until the systematic is identified and removed.
2. Recommend one change to her method that would most reduce the remaining 1.2% shortfall, and justify it numerically.
Read the worked answer
The shortfall is consistent with reaction-time lag. At L = 0.600 m, T ≈ 1.555 s, so ten swings take about 15.55 s. A combined lag of roughly 0.10 s (about 0.05 s at each press) is 0.64% of that time; because g ∝ 1/T², a 0.64% overstatement of T lowers g by about 1.3% - the whole discrepancy. The lag is a fixed time, so its fractional effect shrinks as the timed interval grows.
Best single change: time 20 (or 30) complete oscillations instead of ten. At 20 swings the same 0.10 s lag is 0.32% of the interval, halving the bias in g to roughly 0.6%; at 30 swings it falls to about 0.4%. This costs nothing and needs no new equipment, whereas a light gate would remove the error entirely but may not be available.
Two supporting points: the amplitude problem is already solved at 5° (θ²/16 = 0.0005, negligible), so amplitude is no longer the limiting error; and simply repeating the ten-swing timing improves reliability but not accuracy, because the lag recurs identically each time.
3. A classmate suggests re-plotting the four retained points as T² against L and reading g from the gradient, because "a graph fixes measurement errors in L". Assess how the clamp-to-top-of-bob measurement would show up on that graph.
Read the worked answer
Measuring to the top of the bob makes every recorded L too short by the same amount, say 0.015 m: L_true = L_meas + 0.015. Substituting into T² = (4π²/g)L_true gives
T² = (4π²/g)L_meas + (4π²/g)(0.015).
With 4π²/g ≈ 4.08 s² m⁻¹, the predicted intercept is about +0.06 s², and the gradient is unchanged, so g read from the gradient would be correct despite the error. A constant offset in L moves the line sideways relative to the true origin, which is exactly what an intercept tests.
But the classmate's confidence is misplaced in practice. The retained points span only 0.60–1.00 m, and 0.06 s² is under 2% of the smallest T² (2.4 s²) - comparable to the timing scatter - so a four-point fit would only hint at the offset. More importantly, the timing lag is multiplicative, not additive: it scales with T² and therefore tilts the gradient rather than shifting the intercept. The graph would leave the 1.2% bias intact.