Let's say you measure the time it takes for 10 oscillations of a mass undergoing simple harmonic motion to within ± 0.01s, what is the uncertainty of the period of one oscillation?
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1$\begingroup$ possible duplicate of Multiple measurements of the same quantity - combining uncertainties $\endgroup$– DavePhDCommented May 15, 2014 at 15:25
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$\begingroup$ I don't think it's a duplicate. The article linked to is about measuring some quantity, say 10 different times and computing the mean. I was referring to taking one measurement of 10 oscillations and dividing by 10. $\endgroup$– 11KilobytesCommented May 15, 2014 at 15:32
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$\begingroup$ When you divide a random variate by a constant, the standard deviation is also divided by the same amount. So the uncertainty in the period of one oscillation becomes 0.001 second, ie, you divide the error by 10. The link DavePhD provided isn't an exact duplicate for the reason you cited, but I'm sure this has been asked before, and in any case it's more of a statistics question than a physics question. $\endgroup$– DumpsterDoofusCommented May 15, 2014 at 15:41
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1$\begingroup$ @DumpsterDoofus that's good as a WAG, but the actual uncertainly is probably more like $\frac{error} {\surd{10}}$ , since you don't know for certain that each oscillation is the same as the others (at least, not in a chaotic situation) $\endgroup$– Carl WitthoftCommented May 15, 2014 at 15:47
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1 Answer
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Given the slew of comments to your question I'll summarise them in an answer to make it simple to read.
If we assume that you know the oscillations are regular and your time for ten of them is $T \pm \Delta T$, then the time for a single oscillation $\tau$ is:
$$ \tau = \frac{T \pm \Delta T}{10} = \frac{T}{10} \pm \frac{\Delta T}{10} $$
So you divide your error by 10 and the error in the time for a single oscillation is $\pm 0.001$ seconds.
Carl's point arises if you measure a single oscillation ten times in separate measurements. In that case if $T$ is now the sum of all ten times, and $\Delta T$ is the error in any single measurement the time for a single oscillation is:
$$ \tau = \frac{T}{10} \pm \frac{\Delta T}{\sqrt{10}} $$
answered May 15, 2014 at 16:04