The Pendulum Practical: Is Reaction Time Really the Culprit?
A secondary physics methods class measures g with a simple pendulum. Each group hangs a 2.54 cm steel bob on thin string, clamps the string to a ring stand, and measures length with a meterstick from the clamp to the top of the bob, obtaining L = 0.828 m. Groups set the release angle at about 25 degrees, time ten complete oscillations with a phone stopwatch, and repeat three times. One group records 18.6 s, 18.7 s, and 18.7 s for ten swings, so T = 1.868 s and g = 4π²L/T² = 9.37 m/s², about 4.5% below the accepted 9.81 m/s². In their conclusion the students write that human reaction time with the stopwatch explains the low value. As they pack up, the instructor asks them to defend that claim with numbers, and to check whether the apparatus contains any error that pushes the result in the same direction every trial.
1. Evaluate the group's reaction-time explanation quantitatively. Does it account for the 4.5% deficit?
Read the worked answer
Reaction time on a stopwatch is roughly 0.2 s and is random in sign, so over a 18.68 s interval it contributes on the order of 0.2/18.68 ≈ 1% to T (about 2% to g, since g ∝ 1/T²), and averaging three trials shrinks the effect further. The three trials spread only 0.1 s, i.e. 0.5% in T, which shows the timing is repeatable, not scattered. A random timing error also cannot bias all trials low. So the explanation fails: something systematic is needed.
Two candidates are visible in the procedure and both bias g downward:
- Length measured to the top of the bob. The pivot-to-center length is 0.828 + 0.0127 = 0.841 m, 1.5% larger than recorded, so g is 1.5% low.
- Amplitude of 25 degrees. The finite-amplitude period is T ≈ T₀(1 + θ₀²/16) with θ₀ = 0.436 rad, giving θ₀²/16 = 0.0119, a 1.2% lengthening of T and a 2.4% reduction in the computed g.
Combined: 9.37 × (0.841/0.828) × (1.0119)² ≈ 9.74 m/s², within about 0.7% of 9.81. The deficit is explained by two systematic effects the students did not control, not by reaction time.
2. The instructor wants a redesign that would let the class state a result with an explicit uncertainty. Describe the changes and the uncertainty budget you would require.
Read the worked answer
Changes, each tied to a specific error identified above:
- Measure L from the clamp to the center of the bob (or measure to the top and add the bob radius, recording which convention was used).
- Keep θ₀ ≤ 5 degrees (θ₀²/16 ≤ 0.0002) so the small-angle period is valid to better than 0.1%.
- Time 20–50 oscillations, or use a photogate at the equilibrium point, so the timing error is divided by the number of periods counted.
- Use a non-stretching string (fishing line) and check that the clamp does not slip during the run.
Uncertainty budget, combined in quadrature and expressed as a percentage of g: length (meterstick ±1 mm on 0.84 m ≈ 0.12%, doubled because g ∝ L), timing (0.2 s on 50 periods of ≈9.3 s ≈ 0.04%, doubled), angle (negligible at 5 degrees), and bob-center location (±1 mm ≈ 0.12%). The dominant terms are the length measurements, giving a combined uncertainty near 0.3–0.4%, i.e. g = 9.80 ± 0.04 m/s² - tight enough to see a real discrepancy with 9.81, which the original design's uncontrolled systematics could not.
3. A colleague claims the amplitude effect is negligible because '25 degrees is a small angle.' Design a short experiment that would settle the question using the class's own apparatus.
Read the worked answer
Hold everything else fixed and vary only the release angle, then look for a trend of the kind the theory predicts.
Procedure: keep L at one measured value, release the bob from 5, 10, 15, 20, and 25 degrees, and time 20 oscillations at each angle (three repeats, averaged). Compute T at each angle and plot T versus θ₀² (θ₀ in radians). The first-order correction T ≈ T₀(1 + θ₀²/16) predicts a straight line whose intercept is T₀ and whose fractional slope is 1/16 = 0.0625 per rad². Over 5 to 25 degrees (θ₀² from 0.0076 to 0.190), the predicted period change is about 1.1%, roughly 0.02 s on a 1.87 s period - resolvable with 20-swing timings.
Interpretation: if the fitted slope is near 0.06, the amplitude correction is real and the 25-degree data must be corrected (or rejected). If the slope is consistent with zero within uncertainty, the correction is below the resolution of this apparatus - but that must be demonstrated, not asserted. Either way the analysis converts a verbal claim into a measured slope with an uncertainty, which is what the standard asks for.