Pendulum Timing with an Outlier and a Large Release Angle
A student determines g with a simple pendulum. The length from the pivot to the centre of the bob is 0.992 m, measured with a metre rule. She times 20 complete oscillations four times and records 39.80 s, 39.95 s, 40.05 s and 41.60 s, then averages all four results. The stopwatch reads to 0.01 s and her reaction time is about 0.20 s. She releases the bob from about 25° and notices that the period is slightly longer than when she releases it from 5°. She reports g = (9.6 ± 0.1) m/s², states that the uncertainty is set by the stopwatch resolution, and attributes the remaining difference from 9.81 m/s² to air resistance. The bob is spherical, 40 mm in diameter, and the string is tight and straight during the timing. g is taken as 9.81 m/s² in the accepted value.
1. Identify the outlier and calculate g from the remaining data. Compare with 9.81 m/s².
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The 41.60 s reading is 1.6 s above the others (about 4%), far outside the 0.25 s spread of the first three. It almost certainly comes from miscounting an oscillation or a delayed stop, so it should be rejected, not averaged in.
Mean of the three consistent readings: (39.80 + 39.95 + 40.05)/3 = 39.933 s.
Period T = 39.933/20 = 1.9967 s.
g = 4π²L/T² = (4 × 9.870 × 0.992)/(1.9967²) = 39.16/3.987 = 9.82 m s⁻².
This agrees with 9.81 m s⁻² to about 0.1%, so the accepted value is well supported once the outlier is removed.
2. Critically assess the student's uncertainty statement. Which source dominates, and how could it be reduced?
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The stopwatch resolution of 0.01 s is negligible: over 40 s it is 0.025%. Reaction time (~0.20 s) enters at both start and stop, so it largely cancels; the random part is about 0.2/40 = 0.5% in T, i.e. about 1% in g.
The length measurement (±1 mm on 992 mm) is 0.1% in L, so about 0.1% in g. The dominant systematic is the release angle: at 25° the period is about 1.2% too long, and because g ∝ 1/T² that makes g about 2.4% too low (≈0.24 m s⁻²).
To reduce it, release from less than 5° (θ²/16 < 0.05%) and measure the length to the centre of the bob, for example with a vernier caliper and a fiducial mark. Repeating and rejecting outliers also matters.
3. Explain why a 25° release makes the period longer, estimate the size of the effect, and decide whether it explains the student's 9.6 m/s² result.
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For large amplitudes the restoring force is not proportional to displacement, so the motion is not perfectly simple harmonic. The standard correction is T ≈ T0(1 + θ0²/16), with θ0 in radians.
θ0 = 25° = 0.436 rad, so θ0²/16 = 0.190/16 = 0.0119. The period is about 1.2% longer than the small-angle value.
Since g = 4π²L/T², a 1.2% increase in T lowers g by about 2.4%: 9.82 × 0.976 ≈ 9.59 m s⁻². That is very close to the student's 9.6 m s⁻², so the large release angle, together with the outlier, explains her low result.
Air resistance would also lower the measured g slightly, but the damping effect is far smaller (typically well under 0.1% for a metal bob over 20 swings), so it is not the main cause.