A student measures (length in m, period in s) (0.4, 1.27), (0.6, 1.55), (0.8, 1.8), (1, 2.01), (1.2, 2.2). Model: T = 2π√(L/g). (a) Explain a T²-against-L plot. (b) Use L = 1.00 m to estimate g. (c) Give a reliable complete method and controls. (d) Explain graph extraction of g and a non-zero intercept. (e) Explain why large amplitude invalidates the model.

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Accepted Answer

At 1.00 m, g ≈9.77 m/s²; T²–L gradient is 4π²/g, so g = 4π²/gradient. Squaring gives T² = (4π²/g)L, a straight line through the origin. This linearisation makes model testing and extracting g clearer than a curved T–L plot. At L = 1.00 m and T = 2.01 s, g = 4π²L/T² = 9.77 m/s². Measure pivot to bob centre, use a small fixed release angle without push, time 10–20 complete oscillations from the same directional crossing, divide by count and repeat each length. Use several well-spaced lengths, a rigid stand, fiducial marker and no draught. Plot T² against L; gradient m gives g = 4π²/m, with steep/shallow acceptable lines estimating uncertainty. A non-zero intercept suggests timing offset, effective-length zero error or miscounting. At large angles, period depends on amplitude because the small-angle approximation fails.

Pendulum model investigation

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Pendulum period

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Pendulum period — supplied values
Length / m	Period / s
0.4	1.27
0.6	1.55
0.8	1.8
1	2.01
1.2	2.2