Measuring and the reliability of a measurement

One formula or one step of reasoning. Core exercises first, the rest is extra practice.

1. Exercise 3-1Core

   A small ball's falling time is measured four times and the results are $0.52\,\text{s}$, $0.48\,\text{s}$, $0.55\,\text{s}$ and $0.49\,\text{s}$. What is the mean of the series?

   sCheck
   Answer

   $0.51\,\text{s}$

   Show the solution

   Given

   $t_1 = 0.52\,\text{s}$, $t_2 = 0.48\,\text{s}$, $t_3 = 0.55\,\text{s}$, $t_4 = 0.49\,\text{s}$; the number of measurements is $n = 4$

   Find

   the mean of the series $\bar{t}$ in the unit $\text{s}$

   Principle

   Every measurement has a random error, which throws the result in one direction or another. When the results are added up and the sum is divided by the number of measurements, the deviations partly cancel each other. So the mean is a better estimate than a single result.

   Formula

   $\bar{t}$$= \dfrac{t_1 + t_2 + t_3 + t_4}{n}$

   Conversions

   All times are in seconds, so there is no need to convert units.

   Substitution

   The sum is

   $0.52\,\text{s} + 0.48\,\text{s} + 0.55\,\text{s} + 0.49\,\text{s}$

   $= 2.04\,\text{s}$

   so

   $\bar{t}$$= \dfrac{2.04\,\text{s}}{4}$

   $= 0.51\,\text{s}$

   Answer

   $0.51\,\text{s}$

   Two significant figures like the measurement results (the divisor $4$ is an exact count).

   Reasonableness

   The mean must be between the smallest result $0.48\,\text{s}$ and the largest result $0.55\,\text{s}$, and $0.51\,\text{s}$ is. Check backwards: $4 \cdot 0.51\,\text{s} = 2.04\,\text{s}$ is the sum of the results.

2. Exercise 3-2Core

   The length of a reference piece is known exactly: it is $20.0\,\text{cm}$. A tape measure gives the length of the piece as $20.3\,\text{cm}$.

   What is the measurement error with its sign?

   cmCheck
   Answer

   $0.3\,\text{cm}$

   Show the solution

   Given

   true value (reference length) $x_\text{true} = 20.0\,\text{cm}$, measurement result $x = 20.3\,\text{cm}$

   Find

   the measurement error $x - x_\text{true}$ with its sign in the unit $\text{cm}$

   Principle

   The measurement error is the difference between the measurement result and the true value. The sign tells the direction: plus means that the result is too large, and minus that it is too small.

   Formula

   The measurement error is the difference between the result and the true value: $\text{measurement error} = x - x_\text{true}$

   Conversions

   Both lengths are in centimetres, so no conversions are needed.

   Substitution

   $x - x_\text{true}$$= 20.3\,\text{cm} - 20.0\,\text{cm}$

   $= +0.3\,\text{cm}$

   Answer

   $0.3\,\text{cm}$

   That is, the result is too large. The difference is known only to one decimal place, because both lengths are given with one decimal place.

   Reasonableness

   The result $20.3\,\text{cm}$ is larger than the true value $20.0\,\text{cm}$, so the error must be positive. Relative to the length the error is

   $\dfrac{0.3\,\text{cm}}{20.0\,\text{cm}}$$\approx 1.5\,\%$

   which is a large error for a tape measure (a tape is usually accurate to about a millimetre), so the cause is probably systematic, for example a worn start of the tape or a wrong zero.

3. Exercise 3-3

   The time of ten swings of a pendulum is measured five times with a hand stopwatch. Each result is marked as a dot on the time axis in the picture. Read the smallest and the largest result from the picture.

   What is half of the range of the results?

   sCheck
   Answer

   $0.35\,\text{s}$

   Show the solution

   Given

   the five results of the figure on the time axis; the smallest result $t_\text{min} = 12.2\,\text{s}$ and the largest $t_\text{max} = 12.9\,\text{s}$

   Find

   half of the range $\Delta t$ in the unit $\text{s}$

   Principle

   The scatter of the results tells the size of the random error. The simplest estimate is half of the range, that is, half of the difference between the largest and the smallest result, because a single result can deviate from the middle in either direction. Only the two extremes determine the range, so the other points are not needed.

   Reading the graph

   The vertical lines of the axis are $0.1\,\text{s}$ apart and the numbers $0.2\,\text{s}$ apart. The leftmost point is on the line $12.2\,\text{s}$, so $t_\text{min} = 12.2\,\text{s}$. The rightmost point is on the line that is one interval to the right of the number $12.8$, so $t_\text{max} = 12.9\,\text{s}$. The other points are at $12.4\,\text{s}$, $12.6\,\text{s}$ and $12.7\,\text{s}$.

   Formula

   $\Delta t$$= \dfrac{t_\text{max} - t_\text{min}}{2}$

   Conversions

   All times are in seconds, so no conversions are needed.

   Substitution

   $\Delta t$$= \dfrac{12.9\,\text{s} - 12.2\,\text{s}}{2}$

   $= \dfrac{0.7\,\text{s}}{2}$

   $= 0.35\,\text{s}$

   Answer

   $0.35\,\text{s}$

   Is exactly half of the range. The uncertainty is stated with one significant figure, so in a report it is $0.4\,\text{s}$. Both answers are accepted.

   Reasonableness

   In the figure the extreme points are seven squares apart, that is, $0.7\,\text{s}$, and half of it is smaller than the range, as it should be. The middle of the range is $12.55\,\text{s}$, and $12.55\,\text{s} \pm 0.35\,\text{s}$ gives exactly the interval $12.2\,\text{s}$–$12.9\,\text{s}$, onto which all the points fall.

4. Exercise 3-4

   A kitchen scale shows a reading of $1.2\,\text{g}$ when empty, although there is nothing on it. An object is placed on the scale, and the scale shows $48.7\,\text{g}$.

   What is the mass of the object?

   gCheck
   Answer

   $47.5\,\text{g}$

   Show the solution

   Given

   the reading of the scale when empty (zero reading) $m_0 = 1.2\,\text{g}$, the reading with the object $m_\text{reading} = 48.7\,\text{g}$

   Find

   the mass of the object $m$ in the unit $\text{g}$

   Principle

   The scale adds the same amount $1.2\,\text{g}$ to every reading, because it already shows it when empty. This is a systematic error, and it is corrected by subtracting the zero reading from the reading.

   Formula

   $m = m_\text{reading} - m_0$

   Conversions

   Both readings are in grams, so no conversions are needed.

   Substitution

   $m$$= 48.7\,\text{g} - 1.2\,\text{g}$

   $= 47.5\,\text{g}$

   Answer

   $47.5\,\text{g}$

   Three significant figures; both readings are known to one decimal place.

   Reasonableness

   The scale shows too much, so the true mass must be smaller than the reading, and $47.5\,\text{g} < 48.7\,\text{g}$. Check: $47.5\,\text{g} + 1.2\,\text{g} = 48.7\,\text{g}$.

5. Exercise 3-5

   The smallest scale division of a ruler is $1.0\,\text{mm}$, and the uncertainty of a reading is estimated to be half of a scale division.

   How large is the uncertainty of the reading?

   mmCheck
   Answer

   $0.5\,\text{mm}$

   Show the solution

   Given

   the smallest scale division $d = 1.0\,\text{mm}$

   Find

   the uncertainty of the reading $\Delta x$ in the unit $\text{mm}$

   Principle

   A reading is taken from the scale according to the nearest scale mark. In a simple situation the uncertainty of the reading is estimated as half of a scale division; the total uncertainty can be larger if, for example, the edge of the object is unclear or the scale is viewed from the side.

   Formula

   $\Delta x = \dfrac{d}{2}$

   Conversions

   The scale division is already in millimetres, so no conversions are needed.

   Substitution

   $\Delta x$$= \dfrac{1.0\,\text{mm}}{2}$

   $= 0.5\,\text{mm}$

   Answer

   $0.5\,\text{mm}$

   ; the error estimate is a rough estimate, so it is stated with one significant figure.

   Reasonableness

   If $23.0\,\text{mm}$ is read from a ruler, on the basis of the reading the length is estimated to be between $22.5\,\text{mm}$ and $23.5\,\text{mm}$, and this interval is one scale division, that is, $1.0\,\text{mm}$ wide. So the uncertainty of the reading is smaller than a scale division, but the total uncertainty of the measurement can be larger than this.

6. Exercise 3-6

   Which of the following situations is an example of a systematic error?

   Choose an option.

   • A student looks at the scale sometimes from the left and sometimes from the right, and the readings vary accordingly.
   • A student records the result $3.42\,\text{s}$ as $34.2\,\text{s}$.
   • A hand-held stopwatch is stopped sometimes too early and sometimes too late, so the times vary from one measurement to the next.
   • The zero point of a thermometer is wrong, so it reads about $1.5\,{}^\circ\text{C}$ in ice water every time.

   Answer
   I have done my solution.
   Answer

   The zero point of a thermometer is wrong, so it reads about $1.5\,{}^\circ\text{C}$ in ice water every time.

   Show the solution

   Find

   which of the situations is a systematic error.

   Principle

   A systematic error is almost the same in direction and size in every measurement, for example a wrong zero point of an instrument. A random error varies at random in either direction, and a gross error is a single mistake that shows up as a result differing from the others.

   Model answer

   The systematic error is the wrong zero point of the thermometer: in ice water the thermometer reads about $1.5\,{}^\circ\text{C}$ too high every time, so all readings shift in the same direction. Repeating and averaging do not remove it, but comparison with a known value (ice water is $0\,{}^\circ\text{C}$) reveals it, and the readings are corrected by subtracting the error. The varying moment of stopping the clock and the varying viewing direction are random errors, which the mean reduces. The wrongly recorded number $34.2\,\text{s}$ is a gross error, which is checked and corrected.

7. Exercise 3-7

   The width of a wooden block is measured five times: $5.41\,\text{cm}$, $5.47\,\text{cm}$, $5.39\,\text{cm}$, $5.44\,\text{cm}$ and $5.43\,\text{cm}$. The error estimate is taken to be half of the range and it is rounded to one significant figure. Give the mean rounded so that its precision matches the error estimate.

   cmCheck
   Answer

   $5.43\,\text{cm}$

   Show the solution

   Given

   $x_1 = 5.41\,\text{cm}$, $x_2 = 5.47\,\text{cm}$, $x_3 = 5.39\,\text{cm}$, $x_4 = 5.44\,\text{cm}$, $x_5 = 5.43\,\text{cm}$; $n = 5$

   Find

   the mean $\bar{x}$ in $\text{cm}$, rounded to the precision of the error estimate $\Delta x$

   Principle

   The result is given in the form $\bar{x} \pm \Delta x$. The error estimate is rounded to one significant figure (two if the first figure is 1), and the mean is rounded to the same decimal place, because the figures after the decimal place of the error estimate are already uncertain.

   Formula

   $\bar{x}$$= \dfrac{x_1 + x_2 + x_3 + x_4 + x_5}{n}$

   $\Delta x$$= \dfrac{x_\text{max} - x_\text{min}}{2}$

   Conversions

   All lengths are in centimetres, so no conversions are needed.

   Substitution

   The sum of the results is

   $5.41\,\text{cm} + 5.47\,\text{cm} + 5.39\,\text{cm} + 5.44\,\text{cm} + 5.43\,\text{cm}$

   $= 27.14\,\text{cm}$

   so

   $\bar{x}$$= \dfrac{27.14\,\text{cm}}{5}$

   $= 5.428\,\text{cm}$

   $\Delta x$$= \dfrac{5.47\,\text{cm} - 5.39\,\text{cm}}{2}$

   $= 0.04\,\text{cm}$

   Answer

   $5.43\,\text{cm}$

   Because the error estimate $0.04\,\text{cm}$ is a one-figure number at the hundredths place (the first figure is not 1). The result is $5.43\,\text{cm} \pm 0.04\,\text{cm}$.

   Reasonableness

   The interval $5.43\,\text{cm} \pm 0.04\,\text{cm}$, that is $5.39\,\text{cm}$–$5.47\,\text{cm}$, covers all five results. The result $5.428\,\text{cm}$ would be too precise, because it would imply knowing the width to the thousandths, although the error estimate is in hundredths.

8. Exercise 3-8

   The thickness of a plate is measured with two instruments: a ruler whose smallest scale division is $1.0\,\text{mm}$, and a digital calliper whose display changes in steps of $0.01\,\text{mm}$. The uncertainty of the reading is estimated for both as half of the smallest reading interval.

   How many times smaller is the uncertainty of the calliper reading than that of the ruler?

   timesCheck
   Answer

   $100$

   Show the solution

   Given

   the smallest scale division of the ruler $d_\text{r} = 1.0\,\text{mm}$, the smallest step of the calliper display $d_\text{c} = 0.01\,\text{mm}$

   Find

   how many times smaller the uncertainty of the calliper reading $\Delta x_\text{c}$ is than that of the ruler $\Delta x_\text{r}$

   Principle

   The uncertainty of each reading is estimated as half of the smallest reading interval: when reading a scale, the reading is placed between the marks, and a digital display rounds the reading to the nearest step. A smaller reading interval means a smaller reading uncertainty, so the calliper is the more precise instrument. This is an estimate of the reading; the total uncertainty of a measurement can also be affected by the instrument and the method.

   Formula

   $\Delta x = \dfrac{d}{2}$ for both instruments, and the comparison is the ratio $\Delta x_\text{r} : \Delta x_\text{c}$

   Conversions

   Both intervals are in millimetres, so no conversions are needed.

   Substitution

   $\Delta x_\text{r}$$= \dfrac{1.0\,\text{mm}}{2}$

   $= 0.5\,\text{mm}$

   $\Delta x_\text{c}$$= \dfrac{0.01\,\text{mm}}{2}$

   $= 0.005\,\text{mm}$

   so $0.5\,\text{mm} : 0.005\,\text{mm} = 100$

   Answer

   $100$

   Times smaller, one significant figure like the step $0.01\,\text{mm}$. The reading uncertainty of the calliper is a hundredth of that of the ruler.

   Reasonableness

   The ratio is the same as the ratio of the reading intervals $1.0\,\text{mm} : 0.01\,\text{mm} = 100$, because halving cancels out. With a ruler the width is read to half a millimetre and with a calliper to a hundredth of a millimetre, which matches everyday experience that a calliper is more precise.

Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.

1. Exercise 3-9Core

   The rolling time of a ball is measured six times: $1.84\,\text{s}$, $1.91\,\text{s}$, $1.78\,\text{s}$, $1.88\,\text{s}$, $1.96\,\text{s}$ and $1.82\,\text{s}$. The estimate of random error is taken to be half of the range.

   What percentage of the mean is it?

   %Check
   Answer

   $5\,\%$

   Show the solution

   Given

   $t = 1.84\,\text{s}$, $1.91\,\text{s}$, $1.78\,\text{s}$, $1.88\,\text{s}$, $1.96\,\text{s}$ and $1.82\,\text{s}$; $n = 6$, $t_\text{min} = 1.78\,\text{s}$, $t_\text{max} = 1.96\,\text{s}$

   Find

   the ratio of the estimate of random error $\Delta t$ to the mean $\bar{t}$ as a percentage

   Principle

   The estimate of random error is half of the range, and the precision of the result is best described by its ratio to the result, the relative uncertainty. The mean is taken as the result, because it reduces the effect of random error.

   Formula

   $\bar{t}$$= \dfrac{t_1 + \dots + t_6}{n}$

   $\Delta t$$= \dfrac{t_\text{max} - t_\text{min}}{2}$

   $\varepsilon$$= \dfrac{\Delta t}{\bar{t}} \cdot 100\,\%$

   Conversions

   The times are in seconds and the ratio has no unit. It is multiplied by a hundred to get percent.

   Substitution

   $\bar{t}$$= \dfrac{11.19\,\text{s}}{6}$

   $= 1.865\,\text{s}$

   $\Delta t$$= \dfrac{1.96\,\text{s} - 1.78\,\text{s}}{2}$

   $= 0.09\,\text{s}$

   so

   $\varepsilon$$= \dfrac{0.09\,\text{s}}{1.865\,\text{s}} \cdot 100\,\%$

   $= 4.83\,\%$

   Answer

   $5\,\%$

   One significant figure like the error estimate $0.09\,\text{s}$ (calculator value $4.83\,\%$).

   Reasonableness

   Rough estimate: $0.09\,\text{s}$ is about $\tfrac{1}{20}$ of the time $1.87\,\text{s}$, that is, about $5\,\%$. For a hand-timed measurement a random error of a few percent is usual.

2. Exercise 3-10Core

   The zero end of a tape measure is worn. A rod exactly $1.000\,\text{m}$ long is measured with the tape three times, and the readings are $1.012\,\text{m}$, $1.009\,\text{m}$ and $1.011\,\text{m}$.

   By how many millimetres too much does the tape show on average?

   mmCheck
   Answer

   $11\,\text{mm}$

   Show the solution

   Given

   the length of the rod $L_0 = 1.000\,\text{m}$ and the readings of the tape $L_1 = 1.012\,\text{m}$, $L_2 = 1.009\,\text{m}$, $L_3 = 1.011\,\text{m}$

   Find

   the average error of the tape $\Delta L_\text{sys}$ in the unit $\text{mm}$

   Principle

   The rod is the same length all the time, and the worn start of the tape adds the same amount to every reading: this is a systematic error. The readings vary slightly because of random error, so the error is estimated as the difference between the mean of the readings and the true value.

   Formula

   $\bar{L} = \dfrac{L_1 + L_2 + L_3}{3}$ and $\Delta L_\text{sys} = \bar{L} - L_0$

   Conversions

   The result comes out in metres, and $1\,\text{m} = 1000\,\text{mm}$, so it is multiplied by a thousand.

   Substitution

   The sum is $1.012\,\text{m} + 1.009\,\text{m} + 1.011\,\text{m} = 3.032\,\text{m}$, so

   $\bar{L}$$= \dfrac{3.032\,\text{m}}{3}$

   $= 1.01067\,\text{m}$

   $\Delta L_\text{sys}$$= 1.01067\,\text{m} - 1.000\,\text{m}$

   $= 0.01067\,\text{m}$

   $= 10.67\,\text{mm}$

   Answer

   about $11\,\text{mm}$ (calculator value $10.67\,\text{mm}$): the readings vary by about $\pm 1.5\,\text{mm}$, so the mean should not be stated more precisely than a millimetre.

   Reasonableness

   Each reading is $9\,\text{mm}$–$12\,\text{mm}$ too large, so the mean $11\,\text{mm}$ is between these. The tape shows about $1\,\%$ too much, and the sign of the error is positive, because the readings exceed the true value.

3. Exercise 3-11

   The length of a glass tube is measured three times, and the results have been recorded in different units: $12.4\,\text{cm}$, $0.125\,\text{m}$ and $123\,\text{mm}$.

   What is the mean in centimetres?

   cmCheck
   Answer

   $12.4\,\text{cm}$

   Show the solution

   Given

   $L_1 = 12.4\,\text{cm}$, $L_2 = 0.125\,\text{m}$, $L_3 = 123\,\text{mm}$

   Find

   the mean $\bar{L}$ in the unit $\text{cm}$

   Principle

   A mean can be calculated only from numbers in the same unit, because numbers in different units cannot be added. So convert all results to centimetres first.

   Formula

   $\bar{L} = \dfrac{L_1 + L_2 + L_3}{3}$

   Conversions

   $0.125\,\text{m} = 12.5\,\text{cm}$, because $1\,\text{m} = 100\,\text{cm}$, and $123\,\text{mm} = 12.3\,\text{cm}$, because $10\,\text{mm} = 1\,\text{cm}$.

   Substitution

   The sum is $12.4\,\text{cm} + 12.5\,\text{cm} + 12.3\,\text{cm} = 37.2\,\text{cm}$, so

   $\bar{L}$$= \dfrac{37.2\,\text{cm}}{3}$

   $= 12.4\,\text{cm}$

   Answer

   $12.4\,\text{cm}$

   Three significant figures like the measurement results.

   Reasonableness

   The mean is between the smallest result $12.3\,\text{cm}$ and the largest result $12.5\,\text{cm}$. If the numbers had not been converted, the sum $12.4 + 0.125 + 123$ would give the absurd mean $45.2$, so the conversion is necessary.

4. Exercise 3-12

   The same length is measured five times with two instruments. Instrument A gives the results $5.2\,\text{cm}$, $5.6\,\text{cm}$, $5.4\,\text{cm}$, $5.3\,\text{cm}$ and $5.5\,\text{cm}$. Instrument B gives the results $5.40\,\text{cm}$, $5.43\,\text{cm}$, $5.41\,\text{cm}$, $5.42\,\text{cm}$ and $5.44\,\text{cm}$.

   How many times as large is half of the range of the results of instrument A compared with that of instrument B?

   timesCheck
   Answer

   $10$

   Show the solution

   Given

   instrument A: $5.2$, $5.6$, $5.4$, $5.3$ and $5.5\,\text{cm}$; instrument B: $5.40$, $5.43$, $5.41$, $5.42$ and $5.44\,\text{cm}$

   Find

   the ratio $k = \dfrac{\Delta x_\text{A}}{\Delta x_\text{B}}$, where $\Delta x$ is half of the range

   Principle

   Half of the range describes the size of the random error, that is, how well the measurement repeats. A smaller half means better repeatability, so the ratio tells how much better one instrument repeats.

   Formula

   $\Delta x$$= \dfrac{x_\text{max} - x_\text{min}}{2}$

   for both instruments, and

   $k$$= \dfrac{\Delta x_\text{A}}{\Delta x_\text{B}}$

   Conversions

   All lengths are in centimetres. In the ratio the units cancel, so no conversions are needed.

   Substitution

   $\Delta x_\text{A}$$= \dfrac{5.6\,\text{cm} - 5.2\,\text{cm}}{2}$

   $= 0.2\,\text{cm}$

   $\Delta x_\text{B}$$= \dfrac{5.44\,\text{cm} - 5.40\,\text{cm}}{2}$

   $= 0.02\,\text{cm}$

   so

   $k$$= \dfrac{0.2\,\text{cm}}{0.02\,\text{cm}}$

   $= 10$

   Answer

   $10$

   Times; the differences $0.4\,\text{cm}$ and $0.04\,\text{cm}$ are one-figure numbers, so the ratio is given with one significant figure.

   Reasonableness

   The results of instrument B fit in an interval $0.04\,\text{cm}$ wide and the results of instrument A in one $0.4\,\text{cm}$ wide, that is, ten times wider. Good repeatability still does not tell whether the mean of instrument B is accurate.

5. Exercise 3-13

   The time of ten revolutions of a record on a turntable was measured eight times with a hand-held stopwatch, and the results are in the table. Take half of the range as the error estimate of the time of 10 revolutions and round the mean to the precision of the error estimate. Give the time of one revolution in the form $\bar{T} \pm \Delta T$.

   Time of 10 revolutions of the record

   $n$	$t_{10}\,(\text{s})$
   1	18.12
   2	18.31
   3	18.05
   4	18.27
   5	18.19
   6	18.35
   7	18.10
   8	18.22

   sCheck
   Answer

   $\bar{T} = 1.820\,\text{s}$

   Show the solution

   Given

   the eight times $t_{10}$ (10 revolutions) of the table: $18.12\,\text{s}$, $18.31\,\text{s}$, $18.05\,\text{s}$, $18.27\,\text{s}$, $18.19\,\text{s}$, $18.35\,\text{s}$, $18.10\,\text{s}$ and $18.22\,\text{s}$; $n = 8$

   Find

   the mean time of one revolution $\bar{T}$ in the unit $\text{s}$

   Principle

   A time taken by hand has a random error, so the result is the mean and the error estimate is half of the range. The time of one revolution is obtained from the time of ten revolutions by dividing both the mean and the error estimate by the exact number $10$.

   Formula

   $\bar{t}_{10}$$= \dfrac{t_1 + \dots + t_8}{n}$

   $\;\Delta t_{10}$$= \dfrac{t_\text{max} - t_\text{min}}{2}$

   $\;\bar{T} = \dfrac{\bar{t}_{10}}{10}$ and $\Delta T = \dfrac{\Delta t_{10}}{10}$

   Conversions

   Not needed: all times are in seconds.

   Substitution

   The sum is $145.61\,\text{s}$, so

   $\bar{t}_{10}$$= \dfrac{145.61\,\text{s}}{8}$

   $= 18.20125\,\text{s}$

   $\Delta t_{10}$$= \dfrac{18.35\,\text{s} - 18.05\,\text{s}}{2}$

   $= 0.15\,\text{s}$

   Thus

   $\bar{T}$$= \dfrac{18.20125\,\text{s}}{10}$

   $= 1.820125\,\text{s}$

   $\Delta T$$= \dfrac{0.15\,\text{s}}{10}$

   $= 0.015\,\text{s}$

   Answer

   $\bar{T} = 1.820\,\text{s}$

   That is $1.82\,\text{s}$. The error estimate $0.015\,\text{s}$ is a two-figure number because the first figure is 1, so the result is rounded to thousandths: $T = 1.820\,\text{s} \pm 0.015\,\text{s}$.

   Reasonableness

   The usual speed of a turntable is $33.33$ revolutions per minute, so one revolution takes $60\,\text{s} : 33.33 = 1.800\,\text{s}$. The measured $1.820\,\text{s} \pm 0.015\,\text{s}$ has the right order of magnitude, but $1.800\,\text{s}$ falls just outside that interval; the turntable may therefore be turning slightly slowly, or the timing may have a small systematic error. Timing ten revolutions reduces the relative share of the reaction time: the error estimate $0.15\,\text{s}$ is only about $0.8\,\%$ of the time $18.2\,\text{s}$.

6. Exercise 3-14

   Two groups measured the same amount of water, whose volume is known accurately: $50.0\,\text{ml}$. The results are in the table. Calculate the deviation of the mean of group A's results from the known value, with its sign.

   Which series of measurements is more repeatable and which is closer to the true value? Justify with the half-ranges and the means.

   Measured volumes

   $n$	$V_\text{A}\,(\text{ml})$	$V_\text{B}\,(\text{ml})$
   1	50.9	49.4
   2	51.1	50.5
   3	50.8	49.8
   4	51.0	50.3
   5	50.9	50.1

   mlCheck
   Answer

   $+0.94\,\text{ml}$

   Show the solution

   Given

   known volume $V_\text{true} = 50.0\,\text{ml}$; the results of group A $50.9\,\text{ml}$, $51.1\,\text{ml}$, $50.8\,\text{ml}$, $51.0\,\text{ml}$, $50.9\,\text{ml}$ and the results of group B $49.4\,\text{ml}$, $50.5\,\text{ml}$, $49.8\,\text{ml}$, $50.3\,\text{ml}$, $50.1\,\text{ml}$

   Find

   the deviation of the mean of series A $\bar{V}_\text{A} - V_\text{true}$ with its sign in $\text{ml}$; which series is more repeatable and which is closer to the true value

   Principle

   Repeatability tells about random error, and it is described by the half-range $\Delta V$. The difference between the mean and the known value tells about systematic error, which repeating does not reduce.

   Formula

   $\bar{V}$$= \dfrac{V_1 + \dots + V_5}{n}$

   $\;\Delta V$$= \dfrac{V_\text{max} - V_\text{min}}{2}$

   and deviation $\bar{V}_\text{A} - V_\text{true}$

   Conversions

   Not needed: all volumes are in millilitres.

   Substitution

   $\bar{V}_\text{A}$$= \dfrac{254.7\,\text{ml}}{5}$

   $= 50.94\,\text{ml}$

   so the deviation is $50.94\,\text{ml} - 50.0\,\text{ml} = 0.94\,\text{ml}$. In addition

   $\Delta V_\text{A}$$= \dfrac{51.1\,\text{ml} - 50.8\,\text{ml}}{2}$

   $= 0.15\,\text{ml}$

   $\Delta V_\text{B}$$= \dfrac{50.5\,\text{ml} - 49.4\,\text{ml}}{2}$

   $= 0.55\,\text{ml}$

   $\bar{V}_\text{B} = 50.02\,\text{ml}$.

   Answer

   $+0.94\,\text{ml}$

   The readings are three-figure numbers, and since the error estimate $\Delta V_\text{A} = 0.15\,\text{ml}$ is to hundredths, the deviation is also given in hundredths. Series A is more repeatable ($0.15\,\text{ml} < 0.55\,\text{ml}$), but all its results are over $50\,\text{ml}$, so it has a systematic error. The mean of series B is closer to the true value, although its scatter is larger.

   Reasonableness

   All of A's results are on the same side of the true value, and the deviation $0.94\,\text{ml}$ is a multiple of the scatter $\Delta V_\text{A} = 0.15\,\text{ml}$, so the deviation is probably not random. Good repeatability therefore does not guarantee a correct result.

7. Exercise 3-15

   The period of a pendulum is about $1.5\,\text{s}$. In hand timing, starting and stopping the clock cause, because of reaction time, an uncertainty of about $0.2\,\text{s}$ in the whole measured time (an estimate that does not depend on the length of the measured time).

   You plan a measurement in which $N$ consecutive swings are timed at once and the time is divided by $N$.

   At least how many swings must be timed so that the relative uncertainty of the period is at most $1\,\%$? Also explain why timing a longer time reduces the effect of reaction time.

   swingsCheck
   Answer

   $14$

   Show the solution

   Given

   period $T \approx 1.5\,\text{s}$; uncertainty due to reaction time for the whole measured time $\Delta t = 0.2\,\text{s}$ (an estimate); required relative uncertainty at most $1\,\%$

   Find

   the minimum number of swings $N$

   Principle

   Reaction time acts once at the start and once at the end, whether the measured time is short or long. When $N$ swings are timed and the time is divided by $N$, the uncertainty is divided by $N$ as well.

   Formula

   $\Delta T = \dfrac{\Delta t}{N}$ and the requirement $\dfrac{\Delta T}{T} \le 0.01$, so $\dfrac{\Delta t}{N \cdot T} \le 0.01$, from which $N \ge \dfrac{\Delta t}{0.01 \cdot T}$

   Conversions

   Not needed: all times are in seconds, and $1\,\% = 0.01$.

   Substitution

   $N$$\ge \dfrac{0.2\,\text{s}}{0.01 \cdot 1.5\,\text{s}}$

   $= 13.33\ldots$

   Answer

   At least $14$ swings must be timed, because $N$ is an integer and $13$ is not enough ($0.2 : (13 \cdot 1.5) = 1.03\,\%$).

   Reasonableness

   Check: $N = 14$ gives

   $\Delta T$$= 0.2\,\text{s} : 14$

   $= 0.014\,\text{s}$

   $0.014\,\text{s} : 1.5\,\text{s} = 0.95\,\%$, so the requirement is met. Timing a single swing would give the uncertainty $0.2 : 1.5 = 13\,\%$.

   Model answer

   The error from reaction time does not grow with the length of the time; it is about the same at the start and at the end. A long measurement is divided among many swings, so the share per swing becomes smaller. Repeating also reduces random error through the mean.

Optional: application, two ideas together or a new situation where you reason for yourself.

1. Exercise 3-16

   In hand timing the stopwatch is always started $0.18\,\text{s}$ too late and always stopped $0.12\,\text{s}$ too late (the delays are the same every time).

   What is the systematic error of the timing, that is, the reading minus the true time? Give the answer with its sign.

   sCheck
   Answer

   $-0.06\,\text{s}$

   Show the solution

   Given

   starting delay $\tau_1 = 0.18\,\text{s}$ (the stopwatch starts too late), stopping delay $\tau_2 = 0.12\,\text{s}$ (the stopwatch stops too late)

   Find

   the systematic error $\Delta t_\text{sys}$ = reading − true time, with its sign, in the unit $\text{s}$

   Principle

   The delays are the same in every measurement, so they are systematic. A stopwatch that starts late lags behind and the reading is too short, and a stopwatch that stops late shows too much. So the delays act in opposite directions.

   Formula

   reading $= t_\text{true} - \tau_1 + \tau_2$, so the error is $\Delta t_\text{sys} = \tau_2 - \tau_1$

   Conversions

   Both delays are in seconds, so no conversions are needed.

   Substitution

   $\Delta t_\text{sys}$$= 0.12\,\text{s} - 0.18\,\text{s}$

   $= -0.06\,\text{s}$

   Answer

   $-0.06\,\text{s}$

   Two decimal places like the delays. The minus sign tells that the reading is too short.

   Reasonableness

   The starting delay is larger than the stopping delay, so the reading comes out short, and the sign is negative. The delays cancel each other almost completely: the error is only a third of the starting delay. If the reading were $3.44\,\text{s}$, the true time would be $3.50\,\text{s}$.

2. Exercise 3-17

   There are 250 similar sheets in a stack of paper. The thickness of the stack is measured with a ruler whose smallest scale division is $1.0\,\text{mm}$, and the uncertainty of the reading is estimated to be half of a scale division.

   What is the error estimate of the thickness of one sheet, when all the sheets are equally thick? Give the answer in micrometres, where one micrometre is a thousandth of a millimetre.

   µmCheck
   Answer

   $2\,\mu\text{m}$

   Show the solution

   Given

   the number of sheets $n = 250$, the smallest scale division $d = 1.0\,\text{mm}$

   Find

   the error estimate of the thickness of one sheet $\Delta h_1$ in the unit $\mu\text{m}$

   Principle

   The thickness of the stack is the sum of $n$ equally thick sheets, $h = n h_1$. The thickness of the stack has a reading error, and because the number of sheets is exact, the error of the thickness of one sheet is the error of the stack divided by $n$.

   Formula

   $\Delta h = \dfrac{d}{2}$ and $\Delta h_1 = \dfrac{\Delta h}{n}$, so $\Delta h_1 = \dfrac{d}{2n}$

   Conversions

   The result comes out in millimetres, and $1\,\text{mm} = 1000\,\mu\text{m}$, so it is multiplied by a thousand.

   Substitution

   $\Delta h_1$$= \dfrac{1.0\,\text{mm}}{2 \cdot 250}$

   $= \dfrac{0.5\,\text{mm}}{250}$

   $= 0.0020\,\text{mm}$

   $= 2.0\,\mu\text{m}$

   Answer

   $2\,\mu\text{m}$

   One significant figure, because the error estimate is rounded to one significant figure; the number of sheets is an exact number.

   Reasonableness

   The thickness of ordinary copy paper is about $0.1\,\text{mm} = 100\,\mu\text{m}$, so the relative uncertainty is about $2\,\%$. Measuring one sheet directly with a ruler would give an uncertainty of $0.5\,\text{mm}$, that is, about $500\,\%$ of its thickness, so measuring the stack reduces the relative uncertainty to a 250th.

3. Exercise 3-18

   The length of a steel rod is known exactly from a reference measurement: it is $8.00\,\text{cm}$. The rod is measured five times with vernier callipers, and the readings are $8.24\,\text{cm}$, $8.27\,\text{cm}$, $8.25\,\text{cm}$, $8.29\,\text{cm}$ and $8.25\,\text{cm}$. Calculate the estimate of the systematic error, that is, the difference between the mean and the true value.

   cmCheck
   Answer

   $0.26\,\text{cm}$

   Show the solution

   Given

   true value $x_\text{true} = 8.00\,\text{cm}$; readings $8.24$, $8.27$, $8.25$, $8.29$ and $8.25\,\text{cm}$; $n = 5$

   Find

   the estimate of the systematic error $\Delta x_\text{sys}$ in the unit $\text{cm}$

   Principle

   The readings are close together but all clearly above the true value. The mean reduces random error, so its difference from the true value describes the systematic error that remains.

   Formula

   $\bar{x}$$= \dfrac{x_1 + \dots + x_5}{n}$

   $\Delta x_\text{sys} = \bar{x} - x_\text{true}$

   Conversions

   All lengths are in centimetres, so no conversions are needed.

   Substitution

   The sum is

   $8.24\,\text{cm} + 8.27\,\text{cm} + 8.25\,\text{cm} + 8.29\,\text{cm} + 8.25\,\text{cm}$

   $= 41.30\,\text{cm}$

   so

   $\bar{x}$$= \dfrac{41.30\,\text{cm}}{5}$

   $= 8.26\,\text{cm}$

   $\Delta x_\text{sys}$$= 8.26\,\text{cm} - 8.00\,\text{cm}$

   $= 0.26\,\text{cm}$

   Answer

   $0.26\,\text{cm}$

   Two significant figures; the difference is to two decimal places.

   Reasonableness

   The estimate of random error

   $\dfrac{8.29\,\text{cm} - 8.24\,\text{cm}}{2}$

   $= 0.025\,\text{cm}$

   $\approx 0.03\,\text{cm}$

   is about a tenth of the systematic error, so the error is dominated by the systematic part, for example a wrong zero. If $0.26\,\text{cm}$ is subtracted from all the readings, they are $7.98\,\text{cm}$–$8.03\,\text{cm}$, that is, close to the true value.

4. Exercise 3-19

   The travel time of a cart from one end of a track to the other was measured eight times with a hand-held stopwatch, and the results are in the table. The laboratory notebook has a note at the sixth measurement: “I looked away for a moment and stopped the clock late.”

   May the result $2.97\,\text{s}$ be left out? Justify with the notebook note and calculate the mean $\bar{t}$ that you can justifiably use.

   What if there were no note and no reason could be found for the deviation? Then calculate the mean of all eight results and compare the half-range with and without the result.

   Travel time of the cart

   $n$	$t\,(\text{s})$
   1	2.41
   2	2.38
   3	2.44
   4	2.40
   5	2.43
   6	2.97
   7	2.39
   8	2.42

   sCheck
   Answer

   $2.41\,\text{s}$

   Show the solution

   Given

   the eight times of the table; the sixth result $2.97\,\text{s}$ differs from the others, and the laboratory notebook has a note at it that the clock was stopped late

   Find

   may the result be left out, the mean $\bar{t}$ in $\text{s}$, and what would be done if no reason had been recorded

   Principle

   A clearly deviating result is checked, but it is not removed merely because it deviates. It may be left out only if the reason for the gross error is found and the omission is justified and recorded. Here the notebook note gives the reason: the clock was stopped late. If no reason is found, the result stays in the data and the deviation is described in the uncertainty.

   Formula

   $\bar{t}$$= \dfrac{t_1 + \dots + t_n}{n}$

   where $n = 7$ remaining results

   Conversions

   Not needed: all times are in seconds.

   Substitution

   The sum of the results is

   $2.41\,\text{s} + 2.38\,\text{s} + 2.44\,\text{s} + 2.40\,\text{s} + 2.43\,\text{s} + 2.39\,\text{s} + 2.42\,\text{s}$

   $= 16.87\,\text{s}$

   so

   $\bar{t}$$= \dfrac{16.87\,\text{s}}{7}$

   $= 2.41\,\text{s}$

   Answer

   The result may be left out, because the notebook note gives the reason and the omission is recorded; the mean is $2.41\,\text{s}$, three significant figures like the measurements. The half-range is $(2.44\,\text{s} - 2.38\,\text{s}) : 2 = 0.03\,\text{s}$. If no reason had been found, the result would stay in the data: the mean would be $19.84\,\text{s} : 8 = 2.48\,\text{s}$ and the half-range $(2.97\,\text{s} - 2.38\,\text{s}) : 2 = 0.30\,\text{s}$, ten times as large. The result would be given with this uncertainty, the deviation would be mentioned, and it would be worth repeating the measurement.

   Reasonableness

   With the deviating result the mean shifts by $0.07\,\text{s}$, which is over twice the half-range $0.03\,\text{s}$. The omission is based on the recorded reason and not on the size of the deviation $2.97\,\text{s} - 2.41\,\text{s} = 0.56\,\text{s}$: on the basis of the notebook note the result is a gross error.

5. Exercise 3-20

   Two groups measured the density of the same aluminium object. Group A got the result $\rho_\text{A} = 2.74 \pm 0.05\,\text{g/cm}^3$ and group B the result $\rho_\text{B} = 2.60 \pm 0.05\,\text{g/cm}^3$ (the uncertainties are the groups' estimates).

   A student reasons as follows. “The results differ by only $0.14\,\text{g/cm}^3$, that is about $5\,\%$, so they are consistent.”

   Find the error in the reasoning and use the uncertainty intervals to judge whether the results are consistent. How large is the gap between the intervals?

   g/cm³Check
   Answer

   $0.04\,\text{g/cm}^3$

   Show the solution

   Given

   $\rho_\text{A} = 2.74\,\text{g/cm}^3$, $\Delta\rho_\text{A} = 0.05\,\text{g/cm}^3$; $\rho_\text{B} = 2.60\,\text{g/cm}^3$, $\Delta\rho_\text{B} = 0.05\,\text{g/cm}^3$; the student's conclusion: the difference is about $5\,\%$, so the results are consistent

   Find

   the error in the student's reasoning, whether the results are consistent, and the gap $E$ between the uncertainty intervals in $\text{g/cm}^3$

   Principle

   The consistency of two results is judged with uncertainty intervals, not with the percentage by which the results differ. If the intervals $\rho \pm \Delta\rho$ overlap, a difference cannot be shown; if they do not, the difference is greater than the stated uncertainties explain. The uncertainties are estimates, not certain limits.

   Formula

   the gap is the difference between the difference of the results and the sum of the uncertainties:

   $E$$= |\rho_\text{A} - \rho_\text{B}| - (\Delta\rho_\text{A} + \Delta\rho_\text{B})$

   Conversions

   Not needed: all values are in $\text{g/cm}^3$.

   Substitution

   $E$$= |2.74\,\text{g/cm}^3 - 2.60\,\text{g/cm}^3| - (0.05\,\text{g/cm}^3 + 0.05\,\text{g/cm}^3)$

   $= 0.14\,\text{g/cm}^3 - 0.10\,\text{g/cm}^3$

   $= 0.04\,\text{g/cm}^3$

   Answer

   $0.04\,\text{g/cm}^3$

   Two decimal places like the results. The student's error is to judge consistency by the percentage difference: the intervals $2.69$–$2.79\,\text{g/cm}^3$ and $2.55$–$2.65\,\text{g/cm}^3$ do not overlap, and a gap of $0.04\,\text{g/cm}^3$ is left between them. According to these estimates the difference is greater than the stated uncertainties explain, so the results differ from each other; the cause is worth looking for, for example in a systematic error. The difference still does not prove either result wrong.

   Reasonableness

   The uncertainties are about $2\,\%$ of the results, so the $5\,\%$ difference is over twice as large as they are. If B were $2.66\,\text{g/cm}^3$, the intervals would overlap although the results would still differ by $0.08\,\text{g/cm}^3$.