Pendulum Practical: When Is a Result Wrong?
During a first-year mechanics practical, a team determines g with a simple pendulum. The bob is a 2.0 cm steel sphere; the thread is clamped, and the length from the clamp to the bob's centre is recorded as 0.800 m using a metre scale with a 1 mm least count. A digital stopwatch with 0.01 s resolution is used to time 20 complete oscillations; three readings are 35.80 s, 35.84 s and 35.78 s. One member times a single oscillation as 1.79 s, computes g = 9.86 m/s², and concludes that the apparatus is faulty because the accepted value is 9.81 m/s². Another proposes doubling the pendulum to 1.600 m so that the timing error 'halves'. The manual requires a value with an uncertainty statement, not a single number. The team has 40 minutes left; repeating the whole experiment is possible but would use all of it.
1. From the 20-oscillation data, calculate T and g, and state whether the result is consistent with 9.81 m/s². Show the uncertainty reasoning rather than just substituting numbers.
Read the worked answer
Mean time for 20 oscillations = (35.80 + 35.84 + 35.78)/3 = 35.807 s, so T = 1.7903 s. Then g = 4π²L/T² = 39.478 × 0.800 / (1.7903)² = 31.583 / 3.2053 = 9.85 m/s². The stopwatch resolution is 0.01 s, but the realistic operator reaction is about 0.2 s; on a 35.8 s interval that is 0.56% in T and about 1.1% in g (because g depends on T²). The length contributes 1 mm/800 mm = 0.13%, negligible in comparison. So g = 9.85 ± 0.11 m/s². The gap from 9.81 is 0.04 m/s², under half a standard uncertainty. The result is consistent; the apparatus is not shown to be faulty. The correct conclusion is a measured value with uncertainty, not 'wrong'.
2. Evaluate both proposals: double the length to 1.600 m, or count more oscillations at the original length. Which reduces the uncertainty more, and what is the dominant error?
Read the worked answer
Doubling L increases T by √2, so the fractional timing error falls by about √2: from 0.56% to 0.40%, and the error in g from about 1.1% to 0.8%. The length error does halve from 0.13% to 0.06%, but it was never dominant. Counting 50 oscillations at the original length divides the timing error by 2.5: 0.56% becomes 0.22% in T, and about 0.45% in g. Counting more oscillations is therefore clearly better and avoids disturbing the setup. The dominant error is the start/stop reaction time, not the metre-scale least count and not the formula. The small-angle condition should also be respected (amplitude below about 10°), otherwise the simple pendulum formula itself becomes a systematic error.
3. Why is the single-oscillation value of 1.79 s unacceptable, and what does it show about instrument resolution versus operator error?
Read the worked answer
If T is taken from one oscillation, a 0.2 s reaction error is 0.2/1.79 = 11% in T, which becomes about 22% in g: g ≈ 9.9 ± 2.2 m/s². That cannot distinguish 9.81 from almost anything. The stopwatch's 0.01 s resolution is not the limiting factor; human reaction is roughly twenty times larger. A high-resolution display can therefore give a false sense of accuracy. The remedy is to time a long interval containing many oscillations and divide, which reduces the operator contribution without changing the instrument.