Collection of Solved Problems in Physics

Clock Drift at Higher Temperature

Task number: 1806

• Hint

  Realize what happens with the length of a brass pendulum, if we increase the temperature. What impact does it have on the oscillation period of the pendulum?

• Solution of Hint

  With increasing temperature also increases the length of the pendulum.

  Oscillation period of the pendulum is longer for a longer pendulum, therefore the clock loses time.

• Analysis

  At the temperature of 10 °C, the oscillation period of a brass pendulum is 1 s. At a higher temperature, the length of the pendulum increases due to thermal expansion. The oscillation period of the pendulum increases, because it depends on its length.

• Given values

  t1 = 10 °C	initial temperature
  t2 = 25 °C	increased temperature
  Δτ = ?	change of oscillation period of the pendulum at higher temperature
  τt = ?	total clock drift per one day

  From The Handbook of Chemistry and Physics:

  α = 20·10−6 °C−1

  coefficient of linear thermal expansion of brass

• Solution

  The oscillation period τ₁ of a pendulum at initial temperature t₁ is described as follows:

  \[\tau_1=2\pi\sqrt{\frac{l_1}{g}}\,,\]

  where l₁ is the initial length of the pendulum and g is the acceleration of gravity. The acceleration of gravity does not depend on temperature; however, the length of the pendulum increases with increasing temperature linearly so that at temperature t₂ it can be calculated according to the relation:

  \[l_2=l_1\left(1+\alpha\,\mathrm{\Delta}t\right)=l_1\left[1+\alpha\left(t_2-t_1\right)\right]\,.\]

  This means that the oscillation period increases to τ₂ after we increase the temperature.

  \[\tau_2=2\pi\sqrt{\frac{l_2}{g}}=2\pi\sqrt{\frac{l_1\left[1+\alpha\left(t_2-t_1\right)\right]}{g}}=2\pi\sqrt{\frac{l_1}{g}} \,\sqrt{1+\alpha\left(t_2-t_1\right)}\] \[\tau_2=\tau_1\sqrt{1+\alpha\left(t_2-t_1\right)}\,.\]

  Therefore the clock loses time Δτ every second:

  \[ \mathrm{\Delta}\tau=\tau_2-\tau_1. \]

  We multiply the calculated time lost in one swing Δτ by the number of seconds in one day and thus obtain the time lost per day. For the total time Δτ_t lost we get

  \[ \mathrm{\Delta}\tau_t=24{\cdot}60\cdot60\cdot\mathrm{\Delta}\tau=86{,}400\cdot\mathrm{\Delta}\tau. \]

  Numerical solution

  \[\tau_2=\tau_1\sqrt{1+\alpha\left(t_2-t_1\right)}=1\sqrt{1+19\cdot{10^{-6}}\left(25-10\right)}\dot\,\mathrm{s}=1.00014\,\mathrm{s}\] \[\mathrm{\Delta}\tau=\tau_2-\tau_1=\left(1.00014-1\right)\,\mathrm{s}=0.00014\,\mathrm{s}\] \[\mathrm{\Delta}\tau_t=86{,}400\cdot\mathrm{\Delta}\tau=86{,}400\cdot{0.00014}\,\mathrm{s}\dot=12\,\mathrm{s}\]

• Answer

  After increasing the ambient temperature, the oscillation period increases to a value of about 1.00014 s, which gives a total time lost per day of about 12 s (i.e. one day measured by this clock lasts about 12 seconds longer).

• Note

  Today, we need to measure time very accurately (e.g. because of GPS and other devices). As you have noticed, the precision of pendulum clock depends very strongly on temperature and does not reach required accuracy. For better time measurement, atomic clocks are used today, since they lose only 1 second per 15 million years. Optical clock are even more accurate: they lose one second per a few tens of billions of years!