4.6 Homework In the lab next week you will need to use your knowledge of experim

Question

4.6 Homework In the lab next week you will need to use your knowledge of experimental uncertainties again. However, this time it will be more challenging and there will be harder decisions to make when using the weakest link rule. The following problems will help you be more successful in the lab. Please solve them before coming to lab next week.

1. You are trying to measure the diameter of a circular lab table using a lab meter stick. You measure the diameter in several directions and find the following results: 98.5 cm; 100.0 cm; 97.8 cm. Determine the area of the table. Estimate the uncertainty in the area of the table. What is different about the methods you used to estimate the uncertainty in the diameter and the uncertainty in the area?

2. Eugenia wants to find out how fast she walked along the hiking trail that took her 1.5h to complete (she didn’t have any way of measuring the time that passed during her walk any more precisely than that). She counted the number of steps with a pedometer and got 10000 steps. In order to calculate the distance she walked she estimated the length of her stride. She walked 10 steps three times at her usual rate and measured the distance with the measuring tape (the smallest division is 1 cm) and got 754 cm, 748 cm, 739 cm. Find the length of her stride. What is the length of the hiking trail? What are absolute and relative uncertainties in that length?

3. What is the absolute uncertainty in Eugenia’s time measurement? What is its relative uncertainty? Are these instrumental or random uncertainties?

4. Calculate Eugenia’s average speed. Compare the relative uncertainties in time and distance. Which measurements are more precise? Using the weakest link rule, determine the relative and absolute uncertainties in the speed estimation.

Explanation / Answer

1. given,
   diameter readings = 98.5 cm, 100 cm, 97.8 cm
   so, mean diameter, d = 98.766 cm
   and mean deviation, d(d) = 0.842 cm

   Area, A = pi(d^2)/4 = 7657.457 cm^2
   and dA/A = 2(d(d))/d = 2*0.842/98.766 = 0.01705
   dA = 130.562 cm^2
   hence A = 7657.457 +- 130.562 cm^2

2. time taken, t = 1.5 hr
   number of steps, n = 10,000 steps
   stride length m,eaured in three trials, l = 75.4 cm, 74.8 cm, 73.9 cm
   Average stride length l = 74.7 cm
   mean deviation , dl = 0.38 cm
   also, d(t) = 0.1 hr

   now, speed, v = nl/t = 1000*74.7/1.5 = 49800 cm/hr
   dv/v = sqroot((dl/l)^2 + (dt/t)^2) = sqroot((0.38/1000*74.7)^2 + (0.1/1.5)^2) = 0.0666
   dv = 3320.0001 cm/hr

   hence v = 49800 +- 3320.0001 cm/hr

3. absolute uncertianity in time measurement = dt = 0.1 hr
   relative uncertianity = 0.1/1.5 = 0.06667
   these are instrument errors

4. v = 49800 +- 3320.0001 cm/hr

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