Physics Friday 99

Consider a pendulum clock with a pendulum whose length is made of aluminum. If the average temperature around the clock is
ΔT=5 C° (=5 K) warmer in the summer than in the winter, then how much faster or slower does the clock run in the summer than in the winter?

For small oscillations, the pendulum can be approximated as a harmonic oscillator, giving the formula for the period t (usually uppercase, but here lowercase to avoid confusion with temperature):
,
where l is the length of the pendulum. More specifically, it is the length from the pivot to the pendulum’s center of mass; since most pendula are constructed with a weight on the end much more massive than the length of the arm, this is roughly just the length of the pendulum arm.

The period depends only on the length of the pendulum and the acceleration of gravity; the latter is a constant for this problem. Thus, taking the derivative
,
we see that for a small change of length Δl, we have change in period
,
and dividing by our period, we see our fractional change in period is
,
or half the fractional change of the length.

Now, solid materials undergo thermal expansion, with the fractional change in length proportional to the change in temperature:
,
where α is the coefficient of linear thermal expansion. α is positive, as materials expand when they get warmer. Substituting this into our finding for the fractional change in period

So in the summer, the pendulum will be slightly longer, meaning the period will be slightly longer, and the clock will run slower in the summer than in the winter. Note that the original length l of the pendulum doesn’t affect the result; the proportional expansion is the same.

According to WebElements, the coefficient of linear thermal expansion for aluminum is α=2.31×10-5 (C°)-1.
Thus, for ΔT=5 C°, we have
=5.78×10-5
This is a difference of 1 second per about every 4.81 hours.

[A short discussion of some methods used in real pendulum clocks to prevent this effect can be found here.]