Pendulum Clock Moved to a Hill Station: Net Rate Error and the Bob Screw
A pendulum clock with a brass rod (α = 1.9 × 10⁻⁵ °C⁻¹) has a period of exactly 2.000 s at sea level, where g = 9.80 m s⁻², and is calibrated at 27 °C. It is carried to a hill station 2240 m above sea level, where g is smaller by the factor (1 − 2h/R) with R = 6.4 × 10⁶ m, and installed in a room held at 17 °C. The watchman predicts the clock will gain time, because 'cold contracts the rod and a shorter pendulum swings faster'. His assistant predicts it will lose about half a minute a day. The officer in charge wants one net figure for the error per day and the exact adjustment to the rating nut, which moves the bob 0.1 mm per turn.
1. Calculate the net error of the clock per day, separating the gravitational and thermal contributions.
Read the worked answer
Gravity. Δg/g = −2h/R = −2(2240)/(6.4 × 10⁶) = −7.0 × 10⁻⁴. From T = 2π√(L/g), ΔT/T = −½(Δg/g) = +3.5 × 10⁻⁴: the period lengthens, so the clock runs slow. Loss = 86 400 × 3.5 × 10⁻⁴ = 30 s per day.
Temperature. ΔL/L = αΔθ = 1.9 × 10⁻⁵ × (17 − 27) = −1.9 × 10⁻⁴. From T ∝ √L, ΔT/T = +½(ΔL/L) = −0.95 × 10⁻⁴: the period shortens, so the clock runs fast: 86 400 × 0.95 × 10⁻⁴ = 8 s per day.
Net. ΔT/T = (3.5 − 0.95) × 10⁻⁴ = 2.55 × 10⁻⁴, giving 86 400 × 2.55 × 10⁻⁴ = 22 s per day.
Verdict on the predictions. The watchman has the sign of the thermal effect right but its magnitude far too small; the assistant's 'half a minute' is the pure gravity answer and overestimates, because cooling cancels roughly a quarter of the gravitational loss. The clock loses about 22 s per day.
2. What adjustment to the bob is needed to restore correct rate, in millimetres and in turns of the nut?
Read the worked answer
Current length. L = gT²/4π² = (9.80 × 4.000)/(4 × 9.8696) = 39.20/39.478 = 0.9930 m.
What must change. The clock is slow, so T is too long by a relative 2.55 × 10⁻⁴. Since T ∝ √L, the length must change by ΔL/L = 2 × (−2.55 × 10⁻⁴) = −5.1 × 10⁻⁴, i.e. the pendulum must be shortened.
Amount. ΔL = 5.1 × 10⁻⁴ × 0.9930 m = 5.1 × 10⁻⁴ m ≈ 0.51 mm.
Turns. At 0.1 mm per turn: 0.51/0.1 ≈ 5 turns, raising the bob (shortening the effective length).
Check. Shortening by 5.1 × 10⁻⁴ in relative length changes the period by −2.55 × 10⁻⁴, i.e. −22 s per day, exactly cancelling the net error. Note the practical points: the adjustment is made with the clock running and re-checked after a day against a reference; and because it must be made at the room temperature of 17 °C, a later return to 27 °C will re-introduce most of the 8 s thermal term.
3. The watchman now suggests letting the pendulum swing through 15° instead of 5° 'to make it hurry up'. Assess this proposal.
Read the worked answer
For small oscillations T = 2π√(L/g) contains no amplitude term, so amplitude is not a rate control at all. The amplitude dependence appears only at second order, as T ≈ T₀(1 + θ₀²/16) with θ₀ in radians - and it makes the period longer, not shorter, so a larger swing slows the clock instead of hurrying it.
Size of the effect. 5° = 0.087 rad → θ₀²/16 ≈ 4.8 × 10⁻⁴ (about 41 s/day); 15° = 0.262 rad → 4.3 × 10⁻³ (about 370 s/day). Changing from 5° to 15° would therefore add roughly 0.38 %, i.e. some 330 s per day of extra loss - fifteen times the 22 s the officer is trying to remove. The proposal is both wrong in direction and disastrous in size.
Additional practical objections. Large amplitudes mean larger air resistance and more energy input from the escapement, so the rate becomes amplitude-dependent and unstable; the arc also stresses the suspension. The correct instrument is the rating nut (length), and the correct procedure is to measure the rate over several days against a reference clock rather than judge it by eye. Amplitude should be kept small and, above all, constant.