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We're testing the period of a pendulum in physics class by measuring the time it takes to complete 10 periods then dividing that by 10. Our timing equipment measures to the nearest 100th of a second.

There's a big debate at school over whether or not we can go into precision of 1000ths if, for example, we time 10.00 seconds for 10 periods then divide by 10 to get 1.000s per period (preserving 4 sig digs). Yes, the timer cannot measure beyond 100ths, but the rules of sig digs dictate that division does not change the number of sig digs.

It also makes sense anyway because a difference of .001 seconds per period should be detectable if aggregated across 10 periods and measured with precision to 100ths of a second. But a lot of students and teachers do not accept that we can have more precision using "math tricks". Who is right?

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  • $\begingroup$ Of course you can get more precision. It's called statistics. The more trial measurements you take the better your certainty is. $\endgroup$ – Brandon Enright Apr 9 '14 at 2:04
  • $\begingroup$ @BrandonEnright They are not exactly trials. If you take 5 separate trials then average them, you should still get hundredths of a second, but a trial where you go 10 seconds or beyond then divide by 10 is a bit different. $\endgroup$ – sudo Apr 9 '14 at 2:21
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Yes, if you measure 10 periods at four digits precision, then after dividing by the 10 (an exact integer) you're still good to four digits.

Imagine if instead of a pendulum, you were measuring radio waves at several MHz. You measure let's say exactly one billion cycles using your timer good to 100ths of a second. Divide by one billion. How good should the result be?

As for the naysayers, just what is their argument, if they have one? Saying anything about "math tricks" is just name-calling. What is their logic? They have none, I suspect.

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  • $\begingroup$ The "logic" is that, as they see it, there should be no way to get more than 100ths of a second precision using 100ths of a second equipment. Plus it seems weird that doing this is different from taking 5 trials that would all be under 10 seconds and averaging them. So yeah, it's not really logical, but I'm facing opposition on this. (Luckily, not a very important issue). $\endgroup$ – sudo Apr 9 '14 at 2:19
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    $\begingroup$ I would want to know the precision with which the clock measures 10 seconds. Then decide if it is a sum of 10 trails in which case your "noise" will decrease as the square root of the number of trials. What to do if it is one trial? It is a great subject and I have an entire book here somewhere that covers it. But before that, if it is a simple pendulum then this effort might be a waste. If the pendulum is not isochronous, the period DOES change with amplitude and every swing has a different period. $\endgroup$ – C. Towne Springer Apr 9 '14 at 4:27
  • $\begingroup$ @C.TowneSpringer Surprisingly, it hasn't changed over time by even a 100th of a second in my tests, but I only have a small amount of uncertain and imprecise (not to 1000ths) data on that in particular, so it might be a waste anyway. That's to say that the question was more theoretical, and I can't forget about it ever since I posed it in class ;) $\endgroup$ – sudo Apr 9 '14 at 4:53
  • $\begingroup$ You have some error in the measurement of a deci-period of 10 swings. Can you spread this error over the 10 swings. Guess what? "Measurements and their Uncertainties" by Hughes and Hase does not give a nice answer. It is either deci-period/10 + error/10 or deci-period/10 + error/sqrt(10). This is a common enough problem that Dr. Google should be able to find it. Searching and trying to derive...... $\endgroup$ – C. Towne Springer Apr 9 '14 at 5:25

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