# Uncertainty calculation

Here's an exam question that confused me:

Time for bar to swing through 10 oscillations as measured by a stop clock = 15.7 s

Determine the percentage uncertainty in the time t suggested by the precision of the recorded data.

My answer - 0.05/15.7 x 100 = 0.3%

Mark scheme - 0.1/15.7 x 100 = 0.64%

I thought that the absolute uncertainty would be +-0.05 secs, secondly I've read that uncertainties should always be given to 1 significant figure. Could someone clarify this for me?

Your reasoning is good, and I think without any additional information your answer could easily be argued to be correct. It is, however, often the practice that the uncertainty (for 15.7, for example) isn't between [15.65, 15.75] but between [15.6, 15.8]. Which one is correct, I think depends on the context. This page suggests that for a physical measuring device dividing by two is appropriate (i.e. 0.05), but for a digital measuring device you should not (i.e. 0.1). You could easily back this up by suggesting that the stop-watch may not actually round to the nearest 0.1, but instead it just maintains the current tenth until the next tenth is reached, then it increments. For example, if the time was actually 15.78 it might still read 15.7, in which case the uncertainty is indeed 0.1 instead of 0.05.

• Thank you, this makes sense. The time was obtained using a stopwatch, and I'd assumed that it would round to the nearest 0.1, but I can see now that that is not necessarily true. This is a great resource, thank you for your help! Mar 18, 2017 at 20:45

What you have is a digital indicator with a step-width of 0.1 s.

When you look in the JCGM 100 from the BIPM "Evaluation of measurement data — Guide to the expression of uncertainty in measurement", this is called a Type B evaluation of standard uncertainty (see chapter 4.3). If we assume that the value is rounded to the nearest 0.1 s, then we can assume a rectangular probabilty distribution with a half-width of 0.05 s (see chapter 4.3.7 and 4.4.5). This means that the probability of the real value to be in between [15.65, 15.74] is 100 % and 0 % percent outside of that. The best estimate for the standard uncertainty given in chapter for 4.3.7 for such a case is $$u = a/\sqrt{3}$$ with 95% coverage and $$a$$ being the half-width of the step width of the stop watch. This gives $$u=0.05s/\sqrt{3}=0,029s$$ or $$0,37\,\%$$ relative.

As for the significant figures with the uncertainty, you would always include the two first digits other than zero (sorry for this english). Lets say you have $$u=0,0289s$$, then you print it as $$u=0,029s$$ and you include that amount of figures in the result as well, so you would write for your example $$t=15.700s \pm 0.029s$$ or just $$t=15.700(29)s$$.

Since this result differs from the correct answer, I would really be interested in the reasoning behind it, if you have the chance to ask the person responsible.

Measurement uncertainty calculation is a very complex topic btw! In the quoted document are lots of examples for different situations that might help you with future problems!