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A pendulum clock that keeps perfect time in March can be several seconds a day slow in May. Nothing is broken: the rod has grown slightly longer, and a longer pendulum swings more slowly. This is a favourite exam question because it joins two chapters, thermal expansion and simple harmonic motion. This post derives the half-alpha rule, works the gain and loss cases, finds an unknown coefficient from two readings, and shows why the rod material matters.
Which chapters does this use?
The JEE (Main) 2026 syllabus lists thermal expansion and the simple pendulum ("derivation of expression for its time period"), and the NEET (UG) 2026 syllabus lists both too. The JEE (Advanced) 2026 brochure has "thermal expansion of solids, liquids and gases". In NCERT Class 11 (2026-27 reprint), expansion is in Thermal Properties of Matter and the pendulum in Oscillations, Section 13.8. Neither chapter works a pendulum clock example, so this is an exam-style application of the two. Check the current year's syllabus.
Why does a hot clock run slow?
The period of a simple pendulum is T = 2π√(L/g). Heat the rod by Δθ and its length becomes L(1 + αΔθ), where α is the coefficient of linear expansion. NCERT's Table 10.1 gives values of the order of 10⁻⁵ K⁻¹: iron 1.2, brass 1.8 and aluminium 2.5, all × 10⁻⁵ K⁻¹. The new period is
T′ = T√(1 + αΔθ) ≈ T(1 + ½αΔθ)
so the fractional change in period is ½αΔθ. The approximation is safe because αΔθ is tiny.
A clock counts swings. Each swing takes longer, so the clock shows less time than has really passed. Over a day of 86,400 s, the time lost is about
time lost per day ≈ ½αΔθ × 86,400 s
Worked example 1: summer and winter
A clock with an iron pendulum rod (α = 1.2 × 10⁻⁵ K⁻¹) keeps correct time at 25 °C.
At 40 °C (Δθ = +15 K): ½ × 1.2 × 10⁻⁵ × 15 = 9.0 × 10⁻⁵. Time lost per day = 9.0 × 10⁻⁵ × 86,400 ≈ 7.8 s. It runs slow.
At 5 °C (Δθ = −20 K): the rod is shorter, so the clock runs fast. Time gained per day = ½ × 1.2 × 10⁻⁵ × 20 × 86,400 ≈ 10.4 s.
Notice that L and g have cancelled. A long pendulum and a short one with the same rod material lose the same fraction of time.
Worked example 2: finding α and the correct temperature
A clock loses 12 s per day at 40 °C and gains 4 s per day at 20 °C. At what temperature θ₀ does it keep correct time, and what is α for its rod?
Seconds lost per day are proportional to (θ − θ₀), with gains counted as negative losses:
12 = k(40 − θ₀)
−4 = k(20 − θ₀)
Divide: (40 − θ₀)/(20 − θ₀) = −3, so 40 − θ₀ = −60 + 3θ₀, giving θ₀ = 25 °C.
Then k = 12/15 = 0.8 s per day per kelvin. Since k = ½α × 86,400, α = 2 × 0.8/86,400 ≈ 1.85 × 10⁻⁵ K⁻¹. That is close to NCERT's value for brass, so the rod is probably brass. Check: at 20 °C the formula gives 0.8 × (20 − 25) = −4 s, a gain of 4 s. It matches.
Worked example 3: why choose invar?
NCERT notes that invar, an iron-nickel alloy, has a particularly low expansion coefficient; its Table 10.2 gives a volume coefficient of 2 × 10⁻⁶ K⁻¹. For a solid, the linear coefficient is one third of that, about 6.7 × 10⁻⁷ K⁻¹.
For the same 15 K rise as in Example 1: ½ × 6.7 × 10⁻⁷ × 15 × 86,400 ≈ 0.43 s per day. That is about 18 times smaller than the iron rod's 7.8 s. A low-expansion rod keeps the clock far steadier through the seasons.
Which mistakes should you avoid?
Dropping the half. Period goes as √L.
Getting the direction wrong. Hotter means longer, means slower, means the clock loses time.
Using the volume coefficient for the rod's length. Use α, or divide the volume coefficient by 3.
Converting the temperature change to kelvin by adding 273. A change of 15 °C is a change of 15 K.
Using hours instead of seconds for the day. One day is 86,400 s.
For the basic expansion types (holes, flasks and rails), read our guide to thermal expansion problems. For the other way to change a pendulum's period, by changing g, see a pendulum in an accelerating lift.
A practice routine
Derive ½αΔθ from T = 2π√(L/g) without notes, using the binomial approximation.
Solve Example 1 for an aluminium rod and compare it with iron.
Make up two readings (one gain, one loss) and solve for α and θ₀, as in Example 2.
Combine both effects: a clock that is warmer and on a mountain with lower g.
If you can set up the expansion but not the oscillation half, the Fermi Concept Library covers every concept in physics, so you can revise the simple pendulum before returning to the clock.
This week, redo Example 2 with a clock that loses 9 s at 35 °C and gains 3 s at 15 °C.
