THANK YOU in advance! 1. Method question: If a stopwatch were used to measure ju
ID: 1793693 • Letter: T
Question
THANK YOU in advance!
Explanation / Answer
1.
Assuming the stopwatch has a least count of 0.1 sec.
That error appears once per measurement, so if we try to measure one cycle the error will me max which is 0.1 sec per cycle. But if we measure 20 cycles and then divide the total time by 20 the error is also getting divided by 20 for each cycle. So now the error per cycle is much less which is 0.1/20=0.005sec.
I would suggest that you take the time for at least 20 or more oscillations and then divide the total time by the number of oscillations. More the number of oscillations better will be the accuracy of your measurement. This is because of the error 0.1sec appears only once in one single measurement so if we take 50 oscillations, the error for the time period of 50 oscillations will be 0.1 sec and for one oscillation will be 0.1/50 sec = 0.002sec.
Theoretically, this has no limit but, i.e. we can keep on increasing the number of oscillations per measurement to decrease the error per oscillation. But practically we've got better things to do than taking time period measurements all day.
A sweet spot will be 50 oscillations but if you want more accuracy you can do a 100 or 200, it's totally up to you.
2.
As I mentioned earlier taking 50 oscillations will definitely improve the accuracy than taking 10 or 20 oscillations.
suppose the error is - 'd'sec.
for 20 oscillations the error per oscillation gets reduced to d/20 sec.
and for 50 oscillations the error per oscillation is d/50 sec.
so the error reduces by 0.4 times.
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