When we find the mean of a quantity, we are finding a value that is representative of a number of values of that quantity. When we find the
mean age of a class of students, we are calculating an age that can be used to represent all the ages in that particular class. Here we are finding a mean of a set of separate values which are not related to one another. We might, on the other hand, find a mean value for a single quantity like the diameter of a piece of glass tubing, which could have slightly different values when measured at different places along the tube. In both cases the mean value would be representative of one ‘thing’ – the class in one case and the diameter of the tubing in the other.
There might, for instance, be slight variation in the external diameter of a piece of glass tubing from place to place along its length. In this case and in the one before, the mean is a representative value of a quantity, the ages in the class or the diameter of the glass tubing.
With the class of students, we have a population of fixed size and so we have a fixed number of ages to add together and divide by the number of students in the class. Our result will therefore be a mean of one fixed value. In the case of the glass tubing, however, we could add together any number of diameters and then divide the sum by that number to find the mean. In the first case we would obtain a mean of definite value. In the second, the value obtained would depend on how many values of the diameter we use and where along the tube the diameter is measured. We could, therefore, in this latter case obtain a number of different values for the mean.
mean ❯
It will be clear that the larger the number of diameter measurements taken, the better and the more representative will be the mean value that we find.
Worked example 2.6
A student used a digital stopwatch, reading to the nearest 0.1 s, to measure the time for ten oscillations of a simple pendulum. The student, however, has a reaction time of 0.2 s. She measures the time for ten swings as 20.4 s. (i) What is the maximum percentage uncertainty in this value?
(ii) Calculate the maximum percentage error in the time for one swing. (iii) Calculate the mean time for one swing to an appropriate number of
sig. figs.
What would have been the maximum percentage uncertainty if the student had taken the time for 20 swings?
Solution
Since the student was just as likely to start timing too early as she was to start too late, we must add the two uncertainties of 0.2 s.
(i) Time for 10 swings = 20.4 s ± 0.4 s
= 20.4 s ± 2% (since 0.4 / 20.4 = 0.02 = 2%) (ii) Time for 1 swing = (20.4 s ± 0.4 s) / 10 = 2.04 s ± 0.04 s
= 2.04 ± 2%
Note that although the maximum percentage uncertainty in the time for one swing is the same, i.e. 2%, the actual error is only (0.04 s. Had the student measured the time for only one swing, the error would have been as much as 0.4 s, ten times as large (spreading the error!).
(iii) The maximum error in one swing = 2 × 0.04 s = 0.08 s, which we will call 0.1 s.
Note carefully
(i) The starting time is 0 s ± 0.2 s, and the stopping time is 20.4 s ± 0.2 s. So the time interval = (20.4 – 0) ± 0.4 s = (20.4 ± 0.04) s.
(ii) As the number of swings timed increases by a given factor, the uncertainty in the time for one swing reduces by the same factor. By using a large number of swings, we are ‘spreading’ the uncertainty, and the effect of the uncertainty on each swing is correspondingly reduced.
(iii) If the student had used 20 swings, she would most probably have obtained a time of about 2 × 20.4 s. This time would, however, have had the same uncertainty of ± 0.4 s. The maximum percentage uncertainty in the time for one swing would therefore have been
40.8 × ± 0.4 s
20 = 2.04 ± 0.02 s
This gives an uncertainty of ±1%, half as much as before. By increasing the number of swings by a factor of 2, we have reduced the uncertainty in the value of one swing by ½. It therefore is the best to time as many swings as is practicable.
The oscillations of a simple pendulum will be studied in some depth in the next chapter.
The degree of significance must therefore reflect this fact and so we must express the time for one swing to the nearest tenth (0.1) of a second and not to the nearest 0.01 s.
So the time for one swing should be stated as 2.0 s (nearest tenth of a second) since 2.04 corrected to the nearest tenth is 2.0.
Chapter summary
• We judge the accuracy of the result obtained for the value of a quantity by the closeness or otherwise of the result to the ‘true’ value.
• We judge the precision of the result obtained for the value of a quantity by the width of the range of values within which the true value lies.
• The precision of an instrument is the maximum error that is likely to be made when taking a reading with that instrument.
• The degree of precision of an analogue instrument is generally taken as the value of the smallest subdivision on the scale of the instrument.
• The precision of a digital instrument is the value to which the instrument ‘reads correct’. It is the maximum error possible in a reading given by that instrument. • The overall actual error or uncertainty in a sum or a difference of a set of quantities
is the sum of the actual errors in the individual quantities. Errors are always added, never subtracted.
• Generally, in a number expressed to a certain degree of significance, the last digit gives the maximum uncertainty of the result.
• A useful guiding principle (or a ‘rule of thumb’) in deciding how many significant figures to use is that the degree of significance of the result should be no greater than that of the least precise of the values used in the calculation. Therefore you should use the same number of significant figures as that in the least precise of the values used in the calculation.
• The ‘rule of thumb’ is applied as follows:
1 Examine the values in the numerical expression to see which of them is the least precise. The least precise value is the one with the least number of significant figures. (It is showing the largest fractional or percentage error.)
2 Express the final result to the same number of significant figures as that contained by the least precise of the measurements used.
• In recording a reading given by an instrument, we must use a degree of significance that reflects the degree of precision of that instrument. This degree of significance is shown by the number of significant figures used to represent the value.
• The mean value of a property possessed by an object is the value that may be taken to represent that property for the whole object.
• If we are to obtain a mean value of a quantity that is near to the ‘true’ value, we must calculate it from a number of readings of that quantity that is as large as is practicable.
• When carrying out practical work, we should take care when measuring those quantities in which the percentage uncertainty is greatest.
Answers to ITQs
ITQ1 The limits of uncertainty are ±0.05 g; range 6.65–6.75 g; yes.
ITQ2 Because any mass it measures must be given as a multiple of 5. So the reading will be 265 g.
ITQ3 To be sure that there is only one line of sight which is vertical and passes through the pointer and on to the correct reading.
ITQ4 The uncertainty would have been half as large, i.e., about 12%.
ITQ5 (a) Case (iv); (b) (i) 10; (ii) 0.01; (iii) 0.0001; (v) 10; (c) greatest in (i); (v); least in (iii).
ITQ6 384 400 km; 384 400 km. First value is more precise; limits of error are ±50 km.
ITQ7 0.5 cm
ITQ8 (i) 0.142 mm; (ii) 0.142 mm ± 0.0005 mm
ITQ9 (i) All resistors marked in this way will have a resistance within the range 9.8 ohms to 10.2 ohms.
(ii) Yes, since the degree of tolerance is an indication of the spread of the values or how different the smallest is from the largest.
(iii) Yes, since it means that the values are all very close to one another.
Examination-style questions
1 (i) A digital clock gives one ‘tick’ every second. What is the degree of precision of the clock?
(ii) A stop-clock gives 50 ‘ticks’ in 25 seconds. What is the maximum uncertainty in the reading of a time interval measured with this clock, disregarding reaction error? What are its limits of uncertainty?
(iii) A student measured the time for 10 oscillations of a simple pendulum as 14.0 s, using the stop-clock mentioned in (ii) above. The student has a reaction time of 0.2 s and she is known to have started and stopped the watch too late.
(a) What is her maximum reaction error in taking the measurement for 10 oscillations and for 1 oscillation?
(b) What is the maximum precision error in the time for 10 oscillations and for 1 oscillation?
(c) Calculate the total error in the time for 10 oscillations and for 1 oscillation. (d) Hence write down the time for 1 oscillation, to an appropriate number of
significant figures.
2 A sample of a crystalline chemical compound is found to have a mass of 4.2 g when weighed on a certain balance. After being heated for a very long time to remove the water of crystallisation, the compound is found to have a mass of 3.9 g.
(i) What is the precision of the balance used? (ii) Write down the limits of error of the readings.
(iii) Calculate the mass of water driven off during heating with its limits of error.
(iv) Comment on the suitability, or otherwise, of this balance for an experiment such as this. 3 I measured the inside diameter of a cylinder as 3.42 cm and the outside diameter as
4.60 cm.
(i) Calculate the maximum thickness and the minimum thickness of the cylinder wall. (ii) State the value of the thickness with its limits of error.
4 The width of a roughly rectangular field is measured at five different points along its length, and the length is measured at five different points along its width. The values obtained were as follows:
Width/m 24.2 24.2 24.0 24.4 24.0
Length/m 39.8 40.0 40.0 39.9 40.1
Calculate a value, to an appropriate number of significant figures, for: (i) the mean width;
(ii) the mean length; (iii) the perimeter; (iv) the area of the field.
5 (i) Write down the range within which each of the values represented below might be considered to lie: (a) 642 m (b) 6000 nm (c) 3000.0 K (d) 0.003 J (e) 1.23 × 10–3 mg
(ii) What is the maximum error (or uncertainty) implied in each of the values given above, as a percentage of the value?
(iii) Which of the values is: (a) the least precise? (b) the most precise?
6 John was asked to find the diameter of a very fine wire using a 30 cm rule graduated in millimetres and a piece of glass tubing. To do this, he wound a length of the wire over the glass tubing closely and tightly so that the turns touched one another. He then measured the distance occupied by 42 turns of the wire along the tube as 14 mm.
(i) What was the value of the smallest subdivision of the rule?
(ii) Express the diameter of the wire to an appropriate number of significant figures. 7 (i) Express:
(a) 0.0002 g to the nearest mg (b) 0.00742 g to the nearest 10–2 g
(c) 0.39 mg to the nearest microgram (d) 4.6 cm3 to the nearest 0.5 cm3
(e) 1623 kg to the nearest 5 kg (f) 16 249 m to the nearest km.
(ii) Suggest a situation to which each of the answers might apply.
(iii) Each of the values above was the reading given by an instrument that measured to the degree of precision indicated. Calculate the maximum percentage error in each of the measurements and hence arrange the measurements in order of decreasing precision.
understand the nature of parallax errors and systematic errors, and how toreduce their effects
understand the nature of the skills you will develop by doing practical work plan, design, carry out, record readings and compile reports on practicalactivities
understand the skills associated with:– observation, recording and reporting (O/R/R)
– measurement and manipulation (M/M)
– planning and designing (P/D)
– analysis and interpretation (A/I)
manipulative skills (psychomotor skills) knowledge skills (cognitive skills) planning designing observing handling information selecting manipulating data discrimination reporting concluding handling equipment manipulating equipment measuring
experiments and investigations experimental procedure