While performing an experiment involving a bar pendulum, we had to measure the time period of one oscillation by measuring the time taken for 30 oscillations. There is some confusion regarding calculation of error. If the least count of the stopwatch is 0.01s, will the error in time period be 0.01s or 0.01/30s? Can the error in time period be less than the least count of the stopwatch?
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$\begingroup$ Given your set-up, the error in the time period cannot be less than the resolution of the instrument you're measuring time. Why are supposing otherwise? $\endgroup$– Mozibur UllahCommented Nov 11, 2018 at 8:48
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$\begingroup$ Because for 30 oscillations,the error is 0.01s.But for 1 oscillation shouldn't it be less? $\endgroup$– Potato GirlCommented Nov 11, 2018 at 10:01
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$\begingroup$ Why should it be less? Taking the average will give improved accuracy but you're not going to get below the inherent accuracy of your stopwatch. $\endgroup$– Mozibur UllahCommented Nov 11, 2018 at 10:29
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1 Answer
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In the simple model of uncertainty that I'm guessing you're using, if the stopwatch reads (for example) $13.71\ \mathrm{s}$, that means the actual time it measured is something between $13.7100000\ldots$ seconds and $13.7200000\ldots$ seconds. The size of that range is $0.01\ \mathrm{s}$, so the uncertainty of the time measured by the stopwatch is $0.01\ \mathrm{s}$. And since you're assuming the time measured by the stopwatch is the time taken for 30 oscillations of the pendulum, that means the uncertainty in the time taken for 30 oscillations is $0.01\ \mathrm{s}$.
However, when you start talking about the uncertainty in the time taken for one oscillation of the pendulum, that's a different quantity than the time taken for 30 oscillations, and accordingly it will have a different uncertainty - in this case, less. In a rough sense, you can think of it like this: the uncertainty that comes from the precision of the stopwatch means that your time measurement might vary by $0.01\ \mathrm{s}$, but if you time 30 cycles of the pendulum, that $0.01\ \mathrm{s}$ variation has to be "spread out" over all 30 cycles. The details get a little complicated but the net effect is that each cycle gets a fraction of the whole uncertainty.
