A student times 20 complete oscillations of a pendulum three times, obtaining 27.8 s, 28 s and 28.2 s. (a) Calculate the mean period. (b) Explain why timing 20 oscillations is better than timing one. (c) State how to define the start and finish consistently and one variable to keep fixed when repeating.

Question

Accepted Answer

1.4 s; time many oscillations; use the same reference crossing and direction; keep pendulum length fixed. The period is the duration of one complete oscillation. First average the repeated total times: (27.8 + 28 + 28.2)/3 = 28 s. This reduces the influence of random variation between trials. Divide the mean total time by the number of complete oscillations: T = 28/20 = 1.4 s. A complete oscillation returns to the same position moving in the same direction. Start and stop as the bob crosses a fixed marker in the same direction. Timing many oscillations makes a roughly fixed start–stop reaction-time error a smaller fraction of the measured interval. Keep the length from the support to the bob's centre unchanged, and use the same small release angle without pushing. Repeats alone do not remove a consistently miscounted oscillation or a systematic timing error.

Timing oscillations: trial 3

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