1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – johnsmith4725 Mar 18 '17 at 20:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.