• anonymous
In the first video lecture, I am confused about how to calculate the uncertainty measurement in the second and third examples for the apple-dropping experiment. In the first example, you add .003 m to 3.0 m and subtract .003 m from 1.5 m. When you take the ratio of the results, you get 2.006. So, the uncertainty is .006. I get that. However, I'm confused about the second and third calculations. Using the same process, how do you get .002 and .008?
MIT 8.01 Physics I Classical Mechanics, Fall 1999
  • katieb
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  • anonymous
You do a similar approach. You check the difference between the most 'inaccurate' and most 'accurate' measurements. Thus, to calculate the ratio of t1/t2, that is the sqrt(h1/h2). We already know that the ratio h1/h2 is 2.000 with an uncertainty of 0.006. Thus, the most 'inaccurate' ratio of h1/h2 possible is 2.006. You take the sqrt of that and get 1.416. The most 'accurate' ratio of h1/h2 possible is 2.000. You take the sqrt of that and get 1.414. The difference is 0.002, which is the 0.002 uncertainty you are asking about. As for the 0.008 uncertainty. That comes from the uncertainties for the times the professor recorded. t1 was recorded as 0.781 seconds with an uncertainty of 0.002. The most 'inaccurate' time possible is 0.783 s, or 0.779 s. t2 was recorded as 0.551 seconds with an uncertainty of 0.002 as well. The most 'inaccurate' time possible is 0.553 or 0.549 s. To calculate the uncertainty in t1/t2, we look at the maximum 'error' possible' That means we select 0.783 for t1 (because it is the upper bound) and 0.549 for t2 (because it is the lower bound; this choice will encompass the largest possible difference in our measurements due to uncertainty. t1/t2 using 0.783/0.549 gives 1.426. t1/t2 using the most 'accurate' times possible was 0.781/0.551, giving 1.417. The difference between this is 0.008 (roughly, if you account for the rounding of numbers and such). That's where the errors come from. I hope this helps!
  • anonymous
Thanks, you where help me too!

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