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Module: Science Skills and Safety Time allocation: 8 hours

IJSO Training Course Phase III. Module: Science Skills and Safety Time allocation: 8 hours. Objectives:. Describe, distinguish between random uncertainties and systematic errors. Define and apply the concept of significant figures.

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Module: Science Skills and Safety Time allocation: 8 hours

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  1. IJSO Training Course Phase III Module: Science Skills and Safety Time allocation: 8 hours

  2. Objectives: • Describe, distinguish between random uncertainties and systematic errors. • Define and apply the concept of significant figures. • Identify and determine the uncertainties in results calculated from quantities and in a straight-line graph. • Introduce general procedures of writing experimental reports • Laboratory safety and rules.

  3. 1. Significant Digits • Suppose you want to find the volume of a lead cube. You could measure the length l of the side of a lead cube to be 1.76 cm and the volume 13 from your calculator reads 5.451776. The measurement 1.76 cm was to three significant figures so the answer can only be three significant figures. So that the volume = 5.45 cm3. The following rules are applied generally:

  4. All non-zero digits are significant. (e.g., 22.2 has 3 sf) • All zeros between two non-zero digits are significant. (e.g., 1007 has 4 sf) • For numbers less than one, zeros directly after the decimal point are not significant. (e.g., 0.0024 has 2 sf) • A zero to the right of a decimal and following a non-zero digit is significant. (e.g., 0.0500 has 3 sf)

  5. All other zeros are not significant. (e.g., 500 has 1 sf, 17000 has 2 sf) • When multiplying and dividing a series of measurements, the number of significant figures in the answer should be equal to the least number of significant figures in any data of the series.

  6. For example, if you multiply 3.22 cm by 12.34 cm by 1.8 cm to find the volume of a piece of wood, you get an initial answer 71.52264 cm3 from your calculator. However, the least significant measurement is 1.8 cm with 2 sf. Therefore, the correct answer is only 72 cm3. • When adding and subtracting a series of measurements, the answer has decimal places with the least accurate place value in the series of measurements.

  7. For example, what is your answer by adding 24.2 g and 0.51 g and 7.134 g? You get an initial answer 31.844 g from your calculator. However, the least accurate place measurement is 24.2g with only one decimal point. So the answer is 31.8 g.

  8. Exercises: • 1. How many significant figures are indicated by each of the following? (a). 1247 (b). 1007 (c). 0.0345 (d). 2.20 x 107 (e). 62.00 (f). 0.00025 (g). 0.00250(h). sin 45.2o (i). tan-10.24(j). 3.20 x 10-16 (a). 4 (b). 4 (c). 3 (d). 3 (e). 4 (f). 2 (g). 3 (h). 3 (i). 2 (j). 3

  9. 2. (a) Add the following lengths of 3.15 mm and 7.32 cm and 19.2 m. • (b) A rectangular box has lengths of 2.34 cm, 90.66 cm and 3.7 cm. Calculate the volume of the box cm3. (a) 0.00315 + 0.0732 + 19.3 = 19.27635. Thus the answer is 19.3 m. (b) 2.34 x 90.66 x 3.7 = 784.93428. Thus the answer is 780cm3.

  10. 2. Making Measurements • A measurement should always be regarded as an estimate. The precision of the final result of an experiment cannot be better than the precision of the measurements made during the experiment, so the aim of the experimenter should be to make the estimates as good as possible.

  11. There are many factors which contribute to the accuracy of a measurement. Perhaps the most obvious of these is the level of attention paid by the person making the measurements: a careless experimenter gets bad results! However, if the experiment is well designed, one careless measurement will usually be obvious and can therefore be ignored in the final analysis.

  12. Systematic Errors • If a voltmeter is not connected to anything else it should, of course, read zero. If it does not, the "zero error" is said to be a systematic error: all the readings of this meter are too high or too low. The same problem can occur with stop-watches, thermometers etc.

  13. Even if the instrument can not easily be reset to zero, we can usually take the zero error into account by simply adding it to or subtracting it from all the readings. (However, other types of systematic error might be less easy to deal with.) • For this reason, note that a precise reading is not necessarily an accurate reading. A precise reading taken from an instrument with a systematic error will give an inaccurate result.

  14. Random Errors • Try asking 10 people to read the level of liquid in the same measuring cylinder. There will almost certainly be small differences in their estimates of the level. Connect a voltmeter into a circuit, take a reading, disconnect the meter, reconnect it and measure the same voltage again. There might be a slight difference between the readings.

  15. These are random (unpredictable) errors. Random errors can never be eliminated completely but we can usually be sure that the correct reading lies within certain limits. • To indicate this to the reader of the experiment report, the results of measurements should be written as

  16. For example, suppose we measure a length, l to be 25 cm with an uncertainty of 0.1 cm. We write the result as • By this, we mean that all we are sure about is that is somewhere in the range 24.9 cm to 25.1 cm.

  17. A. Quantifying the Uncertainty • The number we write as the uncertainty tells the reader about the instrument used to make the measurement. (As stated above, we assume that the instrument has been used correctly.) Consider the following examples.

  18. Example 1: Using a ruler • The length of the object being measured is obviously somewhere near 4.3 cm (but it is certainly not exactly 4.35 cm). The result could therefore be stated as: • 4.3 cm ± Half the smallest division on the ruler

  19. In choosing an uncertainty equal to half the smallest division on the ruler, we are accepting a range of possible results equal to the size of the smallest division on the ruler. • However, do you notice something which has not been taken into account? A measurement of length is, in fact, a measure of two positions and then a subtraction.

  20. Was the end of the object exactly opposite the zero of the ruler? This becomes more obvious if we consider the measurement again, as shown below.

  21. Notice that the left-hand end of the object is not exactly opposite the 2 cm mark of the ruler. It is nearer to 2 cm than to 2.05 cm, but this measurement is subject to the same level of uncertainty. • Therefore the length of the object is • (6.30 ± 0.05)cm - (2.00 ± 0.05)cm

  22. so, the length can be between • (6.30 + 0.05) - (2.00 - 0.05) and (6.30 - 0.05) - (2.00 + 0.05) • that is, between 4.40 cm and 4.20 cm • We now see that the range of possible results is 0.2 cm, so we write • Length = 4.30 cm ± 0.10 cm • In general, we state a result as

  23. One may record the length of the following red stick to be 5.9 ± 0.1 cm.

  24. Example 2: Using a Stop-Watch • Consider using a stop-watch which measures to 1/100 of a second to find the time for a pendulum to oscillate once. Suppose that this time is about 1s. Then, the smallest division on the watch is only about 1% of the time being measured. We could write the result as • T = 1.00 ± 0.01s • which is equivalent to saying that the time T is between 0.99s and 1.01s.

  25. This sounds quite good until you remember that the reaction-time of the person using the watch might be about 0.1s. Let us be pessimistic and say that the person's reaction-time is 0.15s. Now considering the measurement again, with a possible 0.15s at the starting and stopping time of the watch, we should now state the result as • T = 1.00 s ± (0.01+ 0.3) s

  26. In other words, T is between about 0.7s and 1.3s. We could probably have guessed the answer to this degree of precision even without a stop-watch!

  27. Conclusions from the preceding discussion • If we accept that an uncertainty (sometimes called an indeterminacy) of about 1% of the measurement being made is reasonable, then

  28. B. How many Decimal Places? • Suppose you have a timer which measures to a precision of 0.01s and it gives a reading of 4.58 s. The actual time being measured could have been 4.576 s or 4.585 s etc. However, we have no way of knowing this, so we can only write • t = 4.58s ± 0.01s

  29. We now repeat the experiment using a better timer which measures to a precision of 0.0001 s. The timer might still give us a time of 4.58s but now we would indicate the greater precision of the instrument being used by stating the result as • t = 4.5800 s ± 0.0001 s • So, as a general rule, look at the precision of the instrument being used and state the result to that number of decimal places.

  30. C. How does an uncertainty in a measurement affect the FINAL result? • The measurements we make during an experiment are usually not the final result; they are used to calculate the final result. When considering how an uncertainty in a measurement will affect the final result, it will be helpful to express uncertainties in a slightly different way. Remember, the uncertainty in a given measurement should be much smaller than the measurement itself.

  31. For example, if you write, "I measured the time to a precision of 0.01s", it sounds good: unless you then inform your reader that the time measured was 0.02s! The uncertainty is 50% of the measured time so, in reality, the measurement is useless.

  32. We will define the quantity Relative Uncertainty (sometimes called fractional uncertainty) as follows • (To emphasize the difference, we use the term "absolute uncertainty" where previously we simply said "uncertainty").

  33. Exercises: • 1. If we use the formula x=y/z3 and the percentage uncertainty (relative uncertainty 100%) in y is 3% and in z is 4%, what is it percentage uncertainty in x? • 2. Same as above, but the formula is x=y2/√z ? 1). 15% 2). 8%

  34. We will now see how to answer the question in the title. It is always possible, in simple situations, to find the effect on the final result by straightforward calculations but the following rules can help to reduce the number of calculations needed in more complicated situations.

  35. A few simple examples might help to illustrate the use of these rules. (Rule 2 has, in fact, already been used in the section "Using a Ruler" on page 3.)

  36. Example 1 • Suppose that you want to find the average thickness of a page of a book. We might find that 100 pages of the book have a total thickness of T = 9.0 mm. If this measurement is made using an instrument having a precision of 0.1 mm, then the relative uncertainty is e= 0.1/9.0. Hence, the average thickness of one page, t, is given by t = T/100 = 0.09 mm with an absolute uncertainty 0.09 x e = 0.001mm, or t = 0.090 mm ± 0.001mm. Note: both T and t have only 2 sf.

  37. Example 2 • (a) To find a change in temperature T = T2-T1 , in which the initial temperature T1 is found to be 20°C ± 1°C and the final temperature T2 is found to be 45°C ± 1°C. Then T = 25 ± 2°C. • (b) Now, the initial temperature T1 is found to be 20.2°C ± 0.1°C and the final temperature T2 is found to be 45.23°C ± 0.01°C. Then the calculated value is 25.03 ± 0.11. However, the least decimal place measurement is 20.2 with only one decimal point. So the answer is T = 25.0 ± 0.1°C.

  38. Exercise: • 3. The first part of the trip took 27  3 (s), the second part 14  2 (s). How long time did the whole trip take? How much longer did the first part take compared to the second part? 41  5s , 13  5s

  39. Example 3 • To measure a surface area, S, we measure two dimensions, say, x and y, and then use • S = xy. Using a ruler marked in mm, we measure x = 54 ± 1 mm and y = 83 ± 1 mm. This means the relative uncertainties of x and y are, respectively, 1/54 and 1/83. The relative uncertainty of S is then e = 1/54 + 1/83 = 0.03. The calculated value of the surface area is 4482 with uncertainty 134.46. Thus, the surface area S is 4500 ± 100mm². (2 significant figures)

  40. Exercises: • 4. An object covers 433.07  1.05 (m) in 23.09  1.10 (s). What was the speed? • 5. If using the formula v = u + at we insert u = 6.0  0.4 ms-1, a = 0.200  0.002 ms-2 and time t = 2.00  0.10 s, what will v be?

  41. Example 4 • To find the volume of a sphere, we first find its radius, r (usually by measuring its diameter). We then use the formula: V = 4/3 (π r3) . Suppose that the diameter of a sphere is measured (using an instrument having a precision of ± 0.2mm) and found to be 50.0mm, so r = 25.0mm with relative uncertainty 0.2/50 = 0.004, so r = 25.0 ± 0.1mm. The relative uncertainty of V is 3 x 0.004 = 0.012. The volume of the sphere is V = 65500 ± 800mm3. (3 significant figures)

  42. Exercises: • 6. The dimensions of piece of paper are measured using a ruler marked in mm. The results were x = 60mm and y = 45mm. (a) Rewrite the results of these measurements "correctly". (b) Calculate the maximum and minimum values of the area of the sheet of paper which these measurements give. (c) Express the result of the calculation of area of the sheet of paper in the form: area = A ±A. (a) 60 ± 1mm, 45 ± 1mm.(b) 2806mm2, 2596mm2. (c) 2701 ± 105mm2.

  43. 7. A body is observed to move a distance s = 10m in a time t = 4 s. The distance was measured using a ruler marked in cm and the time was measured using a watch giving readings to 0.1 s a) Express these results "correctly" (that is, giving the right number of significant figures and the appropriate indeterminacy). b) Use the measurements to calculate the speed of the body, including the uncertainty in the value of the speed. Distance: 10 ±0.01 m, time: 4 ± 0.1s. Since, v = 2.5m/s. The uncertainty: 0.065m/s. Therefore, v= 2.5 ±0.1 m/s.

  44. 8. A body which is initially at rest, starts to move with acceleration a. It moves a distance s = 12.00 ±0.12m in a time t = 4.5 ±0.1s. Calculate the acceleration. 1.2 ±0.1 m/s

  45. 9. The diameter of a cylindrical piece of metal is measured to a precision of ±0.02mm. The diameter is measured at five different points along the length of the cylinder. The results are shown below. Units are mm. (i) 7, (ii) 9.4, (iii) 5.6, (iv) 5 and (v) 4.8 (a) Rewrite the list of results "correctly". (b) Calculate the average value of the diameter. (c) State the average value of the diameter in a way which gives an indication of the precision of the manufacturing process used to make the cylinder. (d) Calculate the average value of the area of cross section of the cylinder. State the result as area = A ± A(mm2). (a) (i)7.00 ± 0.02, (ii) 9.40 ±0.02, (iii) 5.60 ±0.02, (iv) 5.00 ± 0.02, and (v) 4.80 ±0.02 (b) 6.36 mm. (c) 0.02mm, 6.36 ± 0.02mm. (d) 31.77 ± 0.20mm2.

  46. 3. Graphs • The results of an experiment are often used to plot a graph. A graph can be used to verify the relation between two variables and, at the same time, give an immediate impression of the precision of the results. When we plot a graph, the independent variable is plotted on the horizontal axis. (The independent variable is the cause and the dependent variable is the effect.)

  47. A. Straight Line Graphs • If one variable is directly proportional to another variable, then a graph of these two variables will be a straight line passing through the origin of the axes. So, for example, Ohm's Law has been verified if a graph of voltage against current (for a metal conductor at constant temperature) is a straight line passing through (0,0). Similarly, when current flows through a given resistor, the power dissipated is directly proportional to the current squared. If we wanted to verify this fact we could plot a graph of power (vertical) against current squared (horizontal). This graph should also be a straight line passing through (0,0).

  48. B. The "Best-Fit" Line • The “best-fit” line is the straight line which passes as near to as many of the points as possible. By drawing such a line, we are attempting to minimize the effects of random errors in the measurements. For example, if our points look like this

  49. The best-fit line should then be • Notice that the best-fit line does not necessarily pass through any of the points plotted.

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