wireman Posted September 17, 2012 Report Share Posted September 17, 2012 Please help me out with this one. 1) I measured the time taken for 10 oscillations and got the following resultsTry1 6.84+/-0.02s Try2 6.65+/-0.02s Try3 6.66+/-0.02s Try4 6.56+/-0.02s The average of these results is 6.68 seconds. But how do i calculate the error of the avarage? Is it 0.08s or 0.02s?2) I used a stopwatch for measuring the time taken and its accuracy is 10th of a second. So i basically got times like 20.54 seconds. Now, because of the large error that's possible , can I take the error as 0.1 seconds even though my values are 2 decimals?Thank you. Reply Link to post Share on other sites More sharing options...
Alex . Posted September 17, 2012 Report Share Posted September 17, 2012 Please help me out with this one.1) I measured the time taken for 10 oscillations and got the following resultsTry1 6.84+/-0.02s Try2 6.65+/-0.02s Try3 6.66+/-0.02s Try4 6.56+/-0.02s The average of these results is 6.68 seconds. But how do i calculate the error of the avarage? Is it 0.08s or 0.02s?To calculate the uncertainty for the average you sum the uncertainties and divide by the number of trials:[(±0.02)+(±0.02)+(±0.02)+(±0.02)] / 4 = ±0.022) I used a stopwatch for measuring the time taken and its accuracy is 10th of a second. So i basically got times like 20.54 seconds. Now, because of the large error that's possible , can I take the error as 0.1 seconds even though my values are 2 decimals?If the stopwatch gives values like 20.54s then, I would have thought the uncertainty would be ±0.01s. But I am not entirely sure about that. 1 Reply Link to post Share on other sites More sharing options...
caroline.st18 Posted September 17, 2012 Report Share Posted September 17, 2012 O.o it'S funny coz I just have had this topic on physics lessons, about half an huor ago i think that on this web site everything is well-explained http://www.wikihow.com/Calculate-Mean,-Standard-Deviation,-and-Standard-Erroranyway, good luck Reply Link to post Share on other sites More sharing options...
caroline.st18 Posted September 17, 2012 Report Share Posted September 17, 2012 yep, I totally agree with Alex 1 Reply Link to post Share on other sites More sharing options...
Sword (Hubert Pomorski) Posted September 18, 2012 Report Share Posted September 18, 2012 It would be too easy. You should take under consideration that you are not only limited by equipment, but also by your reflex. So, increasing, as you suggested in point 2), is reasonable if described.My teacher told me to do it like that:Error of the average is equal to error of the measurement divided by square root of repetitions (eg. ) . Although, as I've mentioned, I think more reasonable is +/-0.1 or even +/-0.2. 1 Reply Link to post Share on other sites More sharing options...
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