Combining time-period measurements

Suppose you want to measure the time-period of a certain physical process, for example the period of a pendulum. You start by measuring the time of one period. You want a more precise estimate of the period so you measure a few more individual periods. After a while you find it tedious to start and stop your stopwatch and you record the time of a larger number of periods. You may change this number of measured periods yet again. 

Is it possible to use all these data in an optimal way to estimate the period of the pendulum? Yes!

Pendulum in the lobby of the Physics Department at the University of Oslo

Bernt Rostad, CC BY 2.0, via Wikimedia Commons

We want to estimate the mean period \(\bar{T}\) and its uncertainty \(\sigma_T\) with \(N\) measurements of the time \(t_i\) for a number of periods per measurement \(n_i\), where each time measurement has the same uncertainty \(\sigma_t\).


The mean period can be estimated by minimizing the \(\chi^2\) of the measurements with respect to the mean period \(\bar{T}\)\(\chi^2 = \sum_{i=1}^N {(t_i-n_i\bar{T})^2\over \sigma_t^2}\), by setting the first derivative to zero: \({\partial \chi^2\over\partial\bar{T}} = 0 = \sum_{i=1}^N{-2 n_i(t_i-n_i \bar{T})\over \sigma_t^2}\). This can be solved for the mean period, \(\bar{T} ={\sum_{i=1}^Nn_i t_i\over \sum_{i=1}^N n_i^2}\).

To find the uncertainty on the mean period we use the second derivative of the \(\chi^2\)\({1\over\sigma_T^2} = {1\over 2}{\partial^2\chi^2\over \partial \bar{T}^2} = \sum_{i=1}^N{n_i^2\over \sigma_t^2}\), which results in \({\sigma_T^2} = {\sigma_t^2\over \sum_{i=1}^N n_i^2}\).

The time resolution per measurement might be known for the instrument being used, or it can be estimated from the standard deviation with respect to the product of  \(\bar{T}\) and the number of periods per measurement: \(\sigma_t^2\simeq {1\over N-1}\sum_{i=1}^N (t_i-n_i\bar{T})^2.\)

These formulae for variable \(n_i\) simplify to well-known results for two typical cases of constant \(n_i\):

  1. Perform \(N\) measurements of 1 period at a time (\(n_i=1\)): \(\bar{T}\pm\sigma_T = {\sum_{i=1}^N t_i\over N} \pm {\sigma_t\over\sqrt{N}}\)
  2. Perform \(N\) measurements of a constant number of periods at a time (\(n_i = m\)): \(\bar{T}\pm\sigma_T = {1\over m}{\sum_{i=1}^N t_i\over N} \pm {1\over m}{\sigma_t\over\sqrt{N}}\)

The two cases above make it clear that the best strategy is to measure the time for a large number of periods, rather than making many measurements of one period at a time.

By Alex Read
Published Dec. 9, 2020 12:39 PM - Last modified Dec. 9, 2020 12:39 PM