Simple pendulum: is the experiment wrong, or just uncorrected?
During a Class 12 practical examination, a student is asked to determine g with a simple pendulum. He clamps a fine thread to a rigid retort stand and ties a brass bob of diameter 2.0 cm to the free end. Using a metre scale, he measures the distance from the clamp to the top of the bob as 84.0 cm and records this as the length of the pendulum. He pulls the bob aside so that the thread makes about 15° with the vertical, releases it, and times 20 complete oscillations with a stopwatch. The stopwatch reads 37.2 s. Taking π² = 9.87, he calculates g and writes 9.59 m/s² on his answer sheet, adding the remark that the experiment is "wrong because the standard value is 9.8". His examiner tells him that at least two corrections must be applied before the result can be judged, and that one of the two corrections is worth far more than the other. She also points out that his stopwatch has a human reaction-time limitation of roughly 0.2 s per press.
1. Reproduce the student's numerical result from the recorded data, and state whether that raw number can be used to judge whether the experiment succeeded.
Read the worked answer
The arithmetic is correct even though the physics is not yet settled.
T = 37.2/20 = 1.860 s, so T² = 3.4596 s². With L as recorded = 0.840 m:
g = 4π²L/T² = (4 × 9.87 × 0.840)/3.4596 = 33.163/3.4596 = 9.59 m/s².
That number alone cannot decide whether the experiment failed. Both errors present are systematic, not random:
- measuring to the top of the bob makes L too small (lowers g);
- releasing from 15° makes T too long (lowers g again).
Both biases push in the same direction, so 9.59 is a biased estimate, not evidence about the accepted value. The correct order is: remove known biases → then compare with 9.81 → then quote an uncertainty. Declaring the experiment "wrong because the standard value is 9.8" inverts the logic of measurement.
2. Apply both corrections and decide whether the corrected result is consistent with g = 9.81 m/s², stating the uncertainty.
Read the worked answer
Correction 1 (length). The pendulum length runs to the centre of the bob: L = 0.840 + (2.0/2)/100 = 0.850 m.
Correction 2 (amplitude). For a finite swing, T ≈ T₀(1 + θ²/16) with θ in radians. 15° = 0.2618 rad, θ² = 0.0685, θ²/16 = 0.00428, so the measured period is about 0.43% too long.
T₀ = 1.860/1.00428 = 1.8521 s, T₀² = 3.4303 s² g = (4 × 9.87 × 0.850)/3.4303 = 33.556/3.4303 = 9.78 m/s².
Uncertainty. Reaction time ≈ 0.2 s out of 37.2 s is 0.54% in T, i.e. about 1.1% in g = ±0.10 m/s². So g = 9.78 ± 0.10 m/s².
The interval contains 9.81, so the data are consistent with the standard value. The student's remark should be withdrawn: nothing in the data indicates a failed experiment, only an uncorrected one.
3. The student proposes two improvements: (a) release from 5° instead of 15°; (b) time 50 oscillations instead of 20. Which one actually reduces the uncertainty in g more, and by roughly what factor?
Read the worked answer
The two proposals attack different things.
(a) Smaller amplitude. At 5° = 0.0873 rad, θ²/16 ≈ 0.00048, i.e. a 0.05% bias instead of 0.43%. But he can already remove a 0.43% bias by calculation, so the residual gain is negligible.
(b) More oscillations. 50 oscillations take about 93 s, so a 0.2 s reaction error is 0.215% of the interval instead of 0.54%. The uncertainty in g falls from about 1.1% to about 0.43%, i.e. ±0.04 m/s² - a factor of roughly 2.5 better.
Since the random timing error dominates the error budget, (b) is the change that matters; (a) is good practice but is not the limiting factor. The best answer is to do both, and in addition measure the length to the centre of the bob with a vernier calliper rather than a metre scale.