PHYSICS LABORATORY: Measuring Instrument Circus

Transcription

1 PHYSICS LABORATORY: Measuring Instrument Circus Mr Smith s data/answers to questions Remember that ultimately it is the experimenter s discretion about how many decimal places to report in the instrument readings. Generally we report according to the smallest graduation. However, if the smallest graduation on the instrument is wide enough to estimate beyond the smallest graduation with confidence, then you can do so. Anyway, the value reported will be within the stated uncertainty if the uncertainty is determined correctly. 1. Diameter of a marble The three measurements were taken along three different axes of the marble. d 1 d 2 d Uncertainty in stated result: d max d min = = 0.06 mm So the reported answer is: d = ± 0.06 mm Other causes of error (other than instrument uncertainty): It is difficult to get the calipers precisely along the axis (from one side to the other); therefore the measurement may have been less than the true diameter of the marble. Assumptions made in results: That the marble is a perfect sphere (i.e, the marble has a consistent diameter all the way around it.) The calipers are sensitive enough to detect this imperfection. 2. Surface area of a marble Instrument used: Calipers;, smallest graduation = 0.01 mm. Using the value of the diameter from #1, d = ± 0.06 mm, then r = 8.06 ± 0.03 mm. Surface area of a sphere is given by : S = 4πr 2 % unc in r 2 = (0.03/8.06) x 2 = and S = mm 2 Since x = 5.715, the reported answer, with proper significant figures, is: S = 816 ± 6 mm 2 Other causes of error, assumptions made in results: Same as in #1.

2 3. Volume of a marble Instrument used: Calipers, smallest graduation = 0.01 mm Using the value of the diameter from #1, d = ±0.06 mm, then r = 8.06 ±0.03 mm. Surface area of a sphere is given by : V = (4/3)πr 3 % unc in r 3 = (0.03/8.06) x 3 = and S = mm 3 Since x = , the reported answer, with proper significant figures, is: S = 2190 ± 30 mm 3 Other causes of error, assumptions made in results: Same as in #1. 4. Mass of a grain of rice Since it is impossible to measure with accuracy the mass of one grain, you must measure many, and divide by that number. I measured the mass of 20 grains, and tried to get ones more or less the same size, since they varied greatly in size. 20m 1 20m 0m 3 Ave 20m Ave m ± 0.01 g ± 0.01 g ± 0.01 g ± 0.01 g Instrument used: Digital scale; smallest graduation = 0.01 g. Uncertainty in stated result: 20m max 20m min = = 0.015/20 = g So the reported answer is: m = ± g I have been liberal with significant figures here, in order for the answers to make sense. Other causes of error (other than instrument uncertainty): The scale is quite sensitive to air currents in the room. Assumptions made in results: That the grains of rice are all the same size and mass. Clearly they are not, so it is possible that the stated value is not true of any of them, since it is an average.

3 5. Diameter of a copper wire The three measurements were taken along different lengths of the wire. d 1 d 2 d Uncertainty in stated result: d max d min = = 0.03 mm So the reported answer is: d = 0.75 ± 0.03 mm Other causes of error (other than instrument uncertainty): Slight bends in the wire may have affected the results, since there is a limit to how tightly the calipers can pinch the wire and straighten it out. Assumptions made in results: That the wire is exactly the same thickness along its entire length. 6. Time for ball to fall 2 meters t 1 t 2 t 3 Ave t ± 0.3 s ± 0.3 s ± 0.3 s Instrument used: Digital stopwatch; smallest graduation is 0.01 s, but as always, we report uncertainties in times as ±0.3 s. Note how significant the instrument error is for these short times, nearly 50%! Uncertainty in stated result: t max t min = = 0.05 mm So the reported answer is: d = 0.62 ± 0.05 mm Other causes of error (other than instrument uncertainty): Reaction time is the major factor, as well as estimating the height from which the ball is dropped (parallax is an issue here). Air currents in the room could have played a role, since ping-pong balls are light. Assumptions made in results: That our reaction time is consistent for each trial. No way!

4 7. Thickness of a sheet of paper It is impossible to measure the thickness of one sheet using the laboratory calipers, so you must measure many, and divide by that number. I measured the thickness of 20 sheets, at different spots around the edges of the papers. 20x 1 20x 2 20x 3 Ave 20x Ave x Uncertainty in stated result: 20x max 20x min = = 0.02/20 = mm So the reported answer is: d = ± mm I have been liberal with significant figures here, in order for the answers to make sense. Other causes of error (other than instrument uncertainty): None. Assumptions made in results: That sheets of paper are of uniform thickness over their entire area. 8. Mass of a 1 kg mass For this exercise, I used a triple beam balance. Although we use digital scales more often, the advantage of the triple beam balance is that you can use it to measure masses up to 2.6 kg. Most digital scales do not go this high. Therefore, it is worthwhile to practice using this instrument. I measured the mass of 3 different 1 kg masses. m 1 m 2 m 3 ± 0.1 g ± 0.1 g ± 0.1 g Instrument used: Triple beam balance; smallest graduation = 0.1 g. With this instrument, I felt it was possible to report the readings with confidence beyond the smallest graduation (so to the nearest 0.01 g). Uncertainty in stated result: m max m min = = g So the reported answer is: m = ± mm Other causes of error (other than instrument uncertainty): The masses used to balance the beam themselves have an error, for sure. Also, parallax and trying to estimate when the lines line up is an issue. There is significant error here. Assumptions made in results: That the 1 kg masses really are all 1 kg, and that they are all the same amounts. Are we measuring the accuracy of the 1 kg mass, or the triple beam balance here?

5 9. The area of the rectangle It does not make sense to take 3 different readings of the sides (you ll interpret the reading the same every time). Therefore, I measured the top and bottom (2 length measurements) and the left and right sides (2 width measurements). L 1 L 2 Ave L W 1 ± 0.10cm W 2 Ave W Instrument used: Ruler; smallest graduation = 0.1 cm. With this ruler, I felt it was possible to report the readings with confidence beyond the smallest graduation (so to the nearest 0.01 cm). Uncertainties in measured values: Lmax Lmin = = cm Area = L x W = cm 2 Wmax Wmin = = cm % unc in L: /5.84 = % unc in W: /2.59 = % unc in A: = Abs unc in A: (15.126)( ) = So the reported answer is: A = 15.1 ± 0.2 cm 2 Note that we really have to round the final uncertainty up, to match the number of decimal places in the answer, which is determined by the number of significant figures. Other causes of error (other than instrument uncertainty): It is hard to tell where to measure the line from the lines on the paper have a thickness that introduce a potential error. Measure from the middle, the left edge, or the right edge? Assumptions made in results: That the rectangle really is a rectangle (remember this is photocopy).