Pendulum Clock and Winter Contraction
A physics group at a school in Kathmandu uses a simple pendulum to regulate a laboratory clock. At 20 °C they measure the length from the pivot to the centre of the bob as 99.5 cm and time 40 complete oscillations with a stopwatch, obtaining 80.4 s. The accepted value of g is 9.80 m/s². The pendulum rod is brass, with a linear expansion coefficient of 1.9 × 10⁻⁵ °C⁻¹. In winter the laboratory temperature falls to 5 °C, and the group notices that the clock loses time relative to a radio time signal. A teammate claims the discrepancy at 20 °C already proves the stopwatch is faulty, because the theoretical period for a 99.5 cm pendulum is about 2.00 s. The group must decide whether the 20 °C measurement is acceptable, predict the winter period, and estimate the clock error per day caused by thermal contraction, assuming the pendulum length changes but g remains constant.
1. Calculate the theoretical period at 20 °C and compare it with the measured period. Is the teammate's claim that the stopwatch must be faulty justified? Give at least two other plausible explanations and a practical check.
Read the worked answer
T = 2π√(L/g) = 2π√(0.995/9.80) = 2π√0.10153 = 2π(0.31864) = 2.002 s. Measured T = 80.4/40 = 2.010 s. The difference is 0.008 s, or 0.4%. A hand-operated stopwatch typically has reaction-time uncertainty of about 0.2 s at start and stop; over 40 oscillations that is ±0.4 s total, or ±0.01 s per oscillation, so the discrepancy is within plausible timing error. The claim is too strong. Other explanations: finite amplitude increases the period relative to the small-angle formula; the 99.5 cm length may be measured to the wrong point or the rod/bob may flex; pivot friction and air resistance; stopwatch calibration. Practical check: time 100 oscillations with a photogate or repeat with several students, keep amplitude below about 10°, and measure to the bob centre.
2. Predict the period at 5 °C and the accumulated clock error per day if the clock keeps time with the 20 °C period but the length contracts. State whether the clock runs fast or slow.
Read the worked answer
ΔT = 20 - 5 = 15 °C. ΔL = LαΔT = 0.995 × 1.9 × 10⁻⁵ × 15 = 2.84 × 10⁻⁴ m. L_winter ≈ 0.995284 m. For small changes, ΔT_period/T ≈ ½(ΔL/L). ΔL/L = αΔT = 2.85 × 10⁻⁴; half is 1.43 × 10⁻⁴. Using the 20 °C period 2.002 s, the winter period is 2.002(1 + 1.43 × 10⁻⁴) = 2.00229 s, a lengthening of 2.9 × 10⁻⁴ s per oscillation. Number of oscillations per day = 86400/2.002 ≈ 43,160. Daily error ≈ 43,160 × 2.9 × 10⁻⁴ ≈ 12.5 s. Because the period is longer, the clock completes fewer oscillations per day and runs slow by roughly 12 s per day. If the clock was regulated using the measured 2.010 s rather than the theoretical 2.002 s, the fractional thermal effect is the same, and the error is still about 12–13 s per day.