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A pendulum clock keeps time by counting swings. If the length of the pendulum or the value of g changes even slightly, every swing takes a little longer or shorter, and the error builds up over a day. Exam questions ask how much. This post gives one relation that handles length and g together, works four cases with numbers, and points out where the shortcut fails.
What is a seconds pendulum?
A seconds pendulum takes one second for each swing from one side to the other, so its full period is 2 s. The NCERT Class 11 chapter Oscillations derives T = 2π√(L/g) for small swings in section 13.8. Its Example 13.8 asks for the length of a pendulum "which ticks seconds" and, with g = 9.8 m/s² and T = 2 s, gets L = gT²/4π² ≈ 1 m. Kept to three figures, 9.8/π² = 0.993 m.
The Syllabus for JEE (Main) 2026 lists "simple pendulum: derivation of expression for its time period", and the NEET (UG) 2026 syllabus has the same line. The CBSE Physics curriculum for 2026-27 includes a Class 11 practical to "find the effective length of second's pendulum" from a graph. Our post on measuring g with a simple pendulum covers that graph method; this one is about what changes the period.
How does a small change in L or g change the period?
Take logarithms of T = 2π√(L/g): ln T = ln 2π + ½ ln L − ½ ln g. For small changes this gives
ΔT/T = ½ (ΔL/L) − ½ (Δg/g)
A day has 86,400 s. If each period is longer by a fraction ΔT/T, the clock counts too few swings and loses about 86,400 × ΔT/T seconds per day. A shorter period makes it gain.
Worked example 1: a brass pendulum on a hot day
A clock with a brass pendulum rod keeps correct time at 20 °C. The room warms to 30 °C. NCERT's Thermal Properties of Matter chapter gives α = 1.8 × 10⁻⁵ K⁻¹ for brass in Table 10.1.
ΔL/L = αΔT = 1.8 × 10⁻⁵ × 10 = 1.8 × 10⁻⁴.
ΔT/T = ½ × 1.8 × 10⁻⁴ = 9.0 × 10⁻⁵.
Time lost per day: 86,400 × 9.0 × 10⁻⁵ ≈ 7.8 s.
The rod got longer, the period got longer, so the clock runs slow.
Worked example 2: the same clock carried up a hill
NCERT's Gravitation chapter gives g(h) ≈ g(1 − 2h/R_E) for heights much smaller than Earth's radius, with R_E = 6400 km. Take the clock to a hill station 1.6 km up.
Δg/g = −2 × 1.6 ÷ 6400 = −5.0 × 10⁻⁴.
ΔT/T = −½ × (−5.0 × 10⁻⁴) = +2.5 × 10⁻⁴.
Time lost per day: 86,400 × 2.5 × 10⁻⁴ = 21.6 s.
Working the inverse-square law exactly gives the same 21.6 s to three figures.
Worked example 3: on the Moon, where the shortcut fails
NCERT's Exercise 13.15 gives g on the Moon as 1.7 m/s². That is not a small change: Δg/g = (1.7 − 9.8) ÷ 9.8 ≈ −0.83.
Shortcut: ΔT/T ≈ +0.41, so T ≈ 2.8 s. This is wrong.
Full formula: T = 2 s × √(9.8 ÷ 1.7) ≈ 4.80 s.
A seconds pendulum on the Moon needs L = 1.7 × 2² ÷ 4π² ≈ 0.172 m, about 17 cm.
Use the fractional relation only when the change is a few percent or less.
Worked example 4: fixing a clock that loses 20 s a day
ΔT/T = 20 ÷ 86,400 ≈ 2.31 × 10⁻⁴.
The length must fall by ΔL/L = 2 × 2.31 × 10⁻⁴ ≈ 4.63 × 10⁻⁴.
For a 0.993 m pendulum, ΔL ≈ 0.46 mm. Shorten it by about half a millimetre.
An exact recalculation gives 0.46 mm too, which shows how sensitive a pendulum clock is.
Which mistakes cost marks?
Calling the period 1 s. One second is a single swing; the period is 2 s.
Getting gain and lose backwards. Longer period, fewer swings counted, clock loses.
Using the shortcut for big changes, as on the Moon.
Measuring length to the top of the bob. L runs from the support to the bob's centre.
Expecting mass to matter. T = 2π√(L/g) has no mass in it, and the CBSE practical list includes an experiment with bobs of the same size but different masses to test exactly that.
Large swings. NCERT's derivation uses sin θ ≈ θ, which it shows holds well up to about 20 degrees.
If the logarithm step feels unfamiliar, the Fermi Concept Library covers every concept in physics and mathematics, each with its own lesson, so you can revise small changes and errors before attempting the clock problems.
A practice routine
Day 1: Derive ΔT/T = ½ ΔL/L − ½ Δg/g from memory.
Day 2: Three clock problems: temperature, height and depth. NCERT gives g(d) = g(1 − d/R_E) for a depth d.
Day 3: Two large-change problems (Moon, another planet) using the full formula.
Day 4: One problem in a lift, using our post on a pendulum in an accelerating lift.
Tonight, find how much the brass clock gains on a 10 °C winter morning. You should get about 7.8 s per day.
