Suppose I wish to obtain the period of an oscillation, $T$, so I measure the time for $n$ oscillations to obtain $nT$. Suppose I use a stopwatch, so the quantity $nT$ has an uncertainty of around 0.1 seconds. How do I estimate the uncertainty in $T$?

Using the standard equation for propagating the error in $z = ax\pm b$, I'd get $\Delta T = (1/n)(\Delta (nT))$. However, I believe this is only valid when the measurements are independent, which here they're clearly not. This also implies the uncertainty would fall to zero if I took enough measurements: this doesn't seem intuitive, since there will always be some sampling uncertainty.

Is the best way to approach this to learn about a Bayesian approach, or can I still do something sensible with the traditional statistical techniques that I am familiar with?


1 Answer 1


$\delta (nT ) = \delta n *T + n * \delta T $

$\delta n = 0$


$\delta (nT) = n* \delta T$

And therefore

$\delta T = \delta (nT) /n$

In other words, your error in $T$ is just your measured error divided by $n$. So your calculations were right. This makes an intuitive sense because if we imagine that the error in the measurement is constant, then for every oscillation we measure at once, the errors are also measured all together, and we assume that they add cumulatively.

  • $\begingroup$ I would argue that $\delta n$ would not be equal to zero, since there is only a finite precision at which you can observe and react. $\endgroup$
    – fibonatic
    Dec 4, 2014 at 10:51
  • $\begingroup$ Which is encompassed in $\delta T$. $n$ is simply a count of how many oscillations are being measured, and if you do the experiment properly, you're never unsure of how many oscillations you're measuring. You could make the opposite assumption, that $\delta T$ is zero and $\delta n$ is nonzero, although that's usually not helpful. But assuming them both to be non-zero just overcomplicates things and doesn't wind up being helpful at all. $\endgroup$
    – MattS
    Dec 4, 2014 at 16:47
  • $\begingroup$ but it is always good to keep an openmind when it comes to error analysis and additional sources for errors from assumptions you have made. For instance when your stopwatch has an display accuracy of 1/100th of a second, it would be unrealistic to take that as to total contribution to the error. $\endgroup$
    – fibonatic
    Dec 4, 2014 at 16:54

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