3.2 Measurements, uncertainties and graphs (AS)
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AS practical skills for Unit 3 (WPH13). Combining uncertainties, maximum and minimum gradients and log graphs are A2 skills (Unit 6).
What this page covers
9 points
- Choosing the number and range of readings, and when to repeat
- Significant figures in recorded data and in calculated results
- Uncertainty of a single reading (half the resolution) and of repeated readings (half the range)
- Percentage uncertainty of a single measured quantity
- Random and systematic errors, including zero error
- Spotting, checking and dealing with an anomalous reading
- Plotting graphs with sensible scales, best-fit lines, gradients from a large triangle, and intercepts
- Linearising equations into the form \(y = mx + c\) (for example \(T^2\) against \(l\), \(v^2\) against \(s\), \(R\) against \(l\))
- Evaluating a method and suggesting specific improvements
In one minute
6 points
- Good practical work means taking at least 6 well-spread readings, recording them to the instrument's resolution, estimating how uncertain they are, and turning them into a straight-line graph whose gradient and intercept give a physical quantity.
- Record every reading of one quantity to the same number of decimal places, set by the resolution. Give calculated results to the same number of significant figures as the least precise data (or one more).
- Uncertainty of a single reading = half the resolution; of a set of repeats = half the range. Percentage uncertainty = uncertainty ÷ value × 100%.
- Random errors scatter readings either side of the true value and are reduced by repeating and averaging. Systematic errors shift every reading the same way (for example a zero error) and are not reduced by repeating.
- On a graph, label axes quantity / unit, choose scales so the points cover more than half the grid in each direction, draw one best-fit line, and find the gradient from a large triangle on the line.
- To test an equation, rearrange it into \(y = mx + c\): the gradient and the intercept then give the physical constants.
Contents
15 sections
- How many readings should you take, and over what range?
- How many significant figures should you use?
- How do you estimate the uncertainty in a measurement?
- What are random and systematic errors, and what is a zero error?
- How do you spot and deal with an anomalous reading?
- How do you plot a good graph?
- How do you find a gradient and an intercept?
- How do you turn an equation into a straight-line graph?
- How do you evaluate an experiment and suggest improvements?
- Exam tips for 3.2
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
How many readings should you take, and over what range?
To see how one quantity depends on another, you change the independent variable in steps and measure the dependent variable, keeping the control variables fixed. Four decisions follow.
- Number. For a graph, take at least 6 different values of the independent variable. With fewer points it is hard to judge whether the relationship really is a straight line, and one bad point has too much influence on the line.
- Range. Spread the values over the widest range the apparatus safely allows, for example lengths of wire from 0.100 m to 0.900 m rather than 0.400 m to 0.600 m. A wide range shows the trend clearly and makes the gradient more reliable.
- Spacing. Space the values roughly evenly. If the graph is expected to curve, take extra readings where it changes most quickly.
- Repeats. Repeat each reading (often 3 times, more for hand timings) and calculate a mean. Repeats reduce random error, reveal anomalies and give an uncertainty from the spread. They are pointless when a reading cannot change (a fixed length measured with the same ruler); then just check it once more.
For anything short and repetitive, time many cycles (10 or 20 oscillations) and divide. The reaction-time error is shared over all the cycles, so the percentage uncertainty in the period is much smaller.
Recording a results table
- Every column heading has a quantity and its unit, separated by a solidus: l / m, t / s, T² / s². The cells then hold pure numbers, with no units.
- Independent variable first, then the dependent variable, repeats, mean and processed values (T², 1/l).
- All raw readings of one quantity have the same number of decimal places, matching the resolution of the instrument: 0.100, 0.200, 0.300 m, not 0.1, 0.2, 0.3 m.
How many significant figures should you use?
The figures you write are a claim about how precisely you know a value.
- Recorded data: write down exactly what the instrument shows, to its resolution. A metre rule with a millimetre scale gives 0.452 m, not 0.45 m and not 0.4520 m. A reading of 2.5 cm on calipers that read to 0.01 cm should be written 2.50 cm: the final zero is information.
- Calculated results: give the answer to the same number of significant figures as the least precise piece of data used (one more is usually acceptable). Keep extra figures in the working and round only at the end.
- Means: give a mean to the same number of decimal places as the readings it came from.
- Processed columns (such as T² or 1/l) are given to the same number of significant figures as the raw data, or one more.
A student measures a rectangular block of plastic. Its mass, on a balance reading to 0.1 g, is 86.4 g. Its sides, measured with calipers reading to 0.01 cm, are recorded as 4.02 cm, 2.5 cm and 3.15 cm. (a) Identify the reading that has been recorded incorrectly. (b) The middle side is in fact 2.51 cm. Calculate the density of the plastic to an appropriate number of significant figures.
(a) Every reading from the calipers should have two decimal places. "2.5 cm" hides the resolution of the instrument and should have been written 2.50 cm (or here 2.51 cm). The 2.5 cm reading
(b) Keep extra figures in the working: \(V = 4.02 \times 2.51 \times 3.15 = 31.78\ \text{cm}^3\), so \(\rho = \dfrac{86.4}{31.78} = 2.7183\ldots\ \text{g cm}^{-3}\).
Every measurement has 3 significant figures, so the answer has 3: \(\rho = 2.72\ \text{g cm}^{-3} = 2.72 \times 10^{3}\ \text{kg m}^{-3}\)
Writing all the calculator digits (2.7183377) would claim a precision that the measurements cannot support.
How do you estimate the uncertainty in a measurement?
- Single reading: uncertainty = half the resolution. A reading of 47 mm on a millimetre scale is 47 mm ± 0.5 mm; 1.84 V on a meter with resolution 0.01 V is 1.84 V ± 0.005 V.
- Repeated readings: uncertainty = half the range = (largest − smallest) ÷ 2, after removing any anomaly.
If half the range is smaller than half the resolution (all repeats identical, say), quote half the resolution. For hand timings, reaction time (a few tenths of a second) dwarfs the 0.01 s resolution of a stopwatch, so use the spread of repeats.
Percentage uncertainty
To compare measurements of different sizes, express the uncertainty as a fraction of the value:
\[\text{percentage uncertainty} = \frac{\text{uncertainty}}{\text{measured value}} \times 100\%\]Use the same unit for both; the answer has no unit. It is not in the formulae list, so learn it with "half the resolution" and "half the range". For example, 47 mm ± 0.5 mm is a 1.1% uncertainty, while 1.84 V ± 0.005 V is only 0.27%. The same absolute uncertainty is a larger percentage of a small reading, so make measured quantities as large as practical. At AS you find the percentage uncertainty of one quantity at a time; combining them is a Unit 6 skill.
A student times 10 oscillations of a mass on a spring four times: 8.42 s, 8.61 s, 8.37 s, 8.55 s. (a) Calculate the mean time for 10 oscillations, its uncertainty and its percentage uncertainty. (b) State the period with its uncertainty.
(a) Mean \(= \dfrac{8.42 + 8.61 + 8.37 + 8.55}{4} = 8.4875\ \text{s}\), which is given to the same number of decimal places as the readings: 8.49 s.
Uncertainty = half the range \(= \dfrac{8.61 - 8.37}{2} = 0.12\ \text{s}\). Percentage uncertainty \(= \dfrac{0.12}{8.49} \times 100\% = 1.4\%\).
10T = 8.49 s ± 0.12 s (1.4%)
(b) Divide both the time and its uncertainty by the exact number 10. The percentage uncertainty is unchanged. T = 0.849 s ± 0.012 s (1.4%)
If the student had timed a single oscillation with the same spread of ± 0.12 s, the percentage uncertainty would have been about 14%, ten times larger.
What are random and systematic errors, and what is a zero error?
An error is the difference between a measured value and the true value. The two kinds are fixed in different ways.
- Random error makes repeated readings scatter unpredictably either side of the true value. Sources include reaction time, judging when a swinging pendulum passes a point, small fluctuations in conditions and reading a scale between marks. Reduce it by repeating and averaging, by timing many cycles, or by using a data logger or light gates.
- Systematic error shifts every reading in the same direction by the same amount (or the same fraction). Sources include a zero error, a wrongly calibrated instrument, reading a scale from the same wrong angle every time (parallax) and measuring a pendulum length to the wrong point. Repeating does not reduce it: you must find and remove its cause, or correct the readings.
On a graph, a systematic error that adds the same amount to every value of y moves the whole line up or down. The intercept changes but the gradient does not, so a gradient method cancels this kind of error.
Before use, a newtonmeter reads 0.15 N with nothing attached. A student hangs masses of 0.100 kg, 0.200 kg and 0.300 kg from it and records 1.13 N, 2.11 N and 3.09 N. (a) State the type of error and correct the readings. (b) Show that the gradient of a graph of reading against mass is the same with or without the correction, and state what the gradient represents.
(a) Every reading is too large by the same amount, so this is a systematic (zero) error. Corrected readings: 1.13 − 0.15 = 0.98 N, 2.11 − 0.15 = 1.96 N and 3.09 − 0.15 = 2.94 N.
(b) Uncorrected, using the first and last points: gradient \(= \dfrac{3.09 - 1.13}{0.300 - 0.100} = \dfrac{1.96}{0.200} = 9.8\ \text{N kg}^{-1}\).
Corrected: every value has dropped by 0.15 N, so the differences are unchanged: \(\dfrac{2.94 - 0.98}{0.200} = 9.8\ \text{N kg}^{-1}\). Both gradients are 9.8 N kg⁻¹, which is g; only the intercept (0.15 N) differs.
Dividing a single uncorrected reading by its mass gives 3.09 ÷ 0.300 = 10.3 N kg⁻¹: the zero error has then fed straight into the answer.
How do you spot and deal with an anomalous reading?
An anomalous reading (an anomaly) is one that does not fit the pattern of the others: a repeat that differs from the rest by much more than their spread, or a point well off the best-fit line.
- Check it. If the apparatus is still set up, repeat the measurement for that value. If the new reading fits, it replaces the anomaly.
- If it cannot be repeated, leave it out of the mean or the best-fit line, but keep it in the table, ring it on the graph and say why it was excluded.
- Suggest a cause where you can: misreading a scale, counting one oscillation too many, a loose connection, a kink in a wire.
- Do not discard readings just because they are inconvenient. A point is anomalous only if it is clearly out of line with the others.
A student measures the diameter of a wire at six places with a micrometer of resolution 0.01 mm: 0.37, 0.36, 0.38, 0.46, 0.37, 0.36 (all in mm). Identify the anomaly, then calculate the mean diameter, its uncertainty and its percentage uncertainty.
Anomaly: 0.46 mm differs from the rest by about 0.1 mm, while the others lie within 0.02 mm of each other. Exclude 0.46 mm (and re-measure at that point if possible).
Mean of the remaining five: \(\dfrac{0.37 + 0.36 + 0.38 + 0.37 + 0.36}{5} = 0.368\ \text{mm} \approx\) 0.37 mm
Uncertainty: half the range \(= \dfrac{0.38 - 0.36}{2} = 0.01\ \text{mm}\). This is larger than half the resolution (0.005 mm), so use it: ± 0.01 mm
Percentage uncertainty: \(\dfrac{0.01}{0.368} \times 100\% =\) 2.7%
Including the anomaly would raise the mean to 0.38 mm and make the half-range five times larger (± 0.05 mm instead of ± 0.01 mm). One bad reading would spoil the whole result.
How do you plot a good graph?
Follow the same routine every time.
- Axes: independent variable on the x-axis, dependent variable on the y-axis, unless the question tells you which graph to plot. Then plot exactly that.
- Labels: label each axis quantity / unit, the same as the table headings.
- Scales: choose scales so the plotted points occupy more than half the grid in each direction. Use convenient steps of 1, 2 or 5 units (or 10, 20, 50 …) per large square, never 3 or 7, which make plotting and reading off slow and error-prone. The origin need not be included unless you need the intercept.
- Plotting: mark each point with a small, neat cross, accurate to within half a small square.
- Best-fit line: one thin straight line (or smooth curve) with the points scattered evenly either side along its length. Do not join the dots, and do not force the line through the origin or through the first point unless the physics demands it.
Choosing 3 N per large square, or letting the points fill only a corner of the grid, makes plotting slow and the gradient imprecise. Check: do the points fill more than half the grid in both directions, and can every value be read without a calculator?
How do you find a gradient and an intercept?
For a straight best-fit line, the gradient is found from two points on the line:
\[\text{gradient} = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}\]where \((x_1, y_1)\) and \((x_2, y_2)\) are read from the best-fit line. This is not in the formulae list, so learn it, and remember that the gradient has units: (unit of y) ÷ (unit of x).
- Draw a large triangle: its hypotenuse should cover more than half of the drawn line. A small triangle magnifies the effect of reading errors.
- Use points on the line, ideally where it crosses grid lines, not data points (unless a data point lies exactly on the line).
- Show the triangle on the graph and write the two coordinate pairs in your working.
- Give the gradient with a unit and to a sensible number of significant figures (2 or 3).
The intercept is the value of y where the line crosses the y-axis at x = 0. If the x-axis does not start at zero, you cannot read it directly; calculate it from \(c = y - mx\) using any point on the line and the gradient you found.
A student measures the resistance R of different lengths l of a wire of diameter 0.315 mm, using an ohmmeter connected with crocodile clips.
| l / m | 0.100 | 0.200 | 0.300 | 0.400 | 0.500 | 0.600 | 0.700 | 0.800 | 0.900 |
|---|---|---|---|---|---|---|---|---|---|
| R / Ω | 1.06 | 1.62 | 2.34 | 2.90 | 3.58 | 4.79 | 4.75 | 5.45 | 6.03 |
(a) Identify the anomaly. (b) Use the graph to find the gradient. (c) Explain what the intercept shows. (d) Use \(R = \rho l / A\) to determine the resistivity of the metal.
(a) Compare each point with the line through the others. At 0.600 m the line gives about 4.2 Ω, but 4.79 Ω was recorded; it is even larger than the 0.700 m reading. The 0.600 m reading is anomalous.
(b) Large triangle from 0.100 m to 0.900 m, values read from the line: gradient \(= \dfrac{6.04 - 1.05}{0.900 - 0.100} = \dfrac{4.99}{0.800} =\) 6.24 Ω m⁻¹
(c) The line does not pass through the origin: the intercept is about 0.4 Ω. This is a systematic error: the resistance of the clips, leads and contacts (or a zero error of the ohmmeter) is added to every reading.
(d) \(R = \left(\dfrac{\rho}{A}\right) l\), so gradient \(= \rho / A\) and \(\rho = \text{gradient} \times A\). Area \(A = \dfrac{\pi d^2}{4} = \dfrac{\pi (0.315 \times 10^{-3})^2}{4} = 7.79 \times 10^{-8}\ \text{m}^2\).
\(\rho = 6.24 \times 7.79 \times 10^{-8} =\) \(4.86 \times 10^{-7}\ \Omega\ \text{m}\)
Using the gradient rather than a single R/l value removes the systematic contact resistance from the result.
How do you turn an equation into a straight-line graph?
Your eye can judge a straight line but not whether a curve is a parabola or a square root, and a line gives a gradient and an intercept to turn into constants. So rearrange the equation into the form
\[y = mx + c\]Whatever multiplies the variable on the x-axis is the gradient; the constant term is the intercept. The axes can carry processed values (T², v², 1/l). If the points lie on a straight line, the data support the relationship; if the line should pass through the origin, check that it does.
| Equation | Plot y | against x | Gradient | Intercept |
|---|---|---|---|---|
| \(R = \rho l / A\) (+ contact resistance) | R | l | \(\rho / A\) | contact resistance (ideally 0) |
| \(T = 2\pi\sqrt{l/g}\) | T² | l | \(4\pi^2 / g\) | 0 |
| \(v^2 = u^2 + 2as\) | v² | s | 2a | u² |
| \(s = \tfrac{1}{2}at^2\) (from rest) | s | t² | \(\tfrac{1}{2}a\) | 0 |
| \(V = \varepsilon - Ir\) | V | I | −r | ε |
| \(f = v / 2l\) (string fundamental) | f | 1/l | v/2 | 0 |
For a pendulum, T against l is a curve; squaring gives \(T^2 = (4\pi^2/g)\,l\), a straight line through the origin.
A student times 20 oscillations of a simple pendulum for seven lengths l, measured from the point of suspension to the centre of the bob. The period is given by \(T = 2\pi\sqrt{l/g}\).
| l / m | 20T / s | T / s | T² / s² |
|---|---|---|---|
| 0.200 | 17.88 | 0.894 | 0.799 |
| 0.300 | 22.02 | 1.101 | 1.21 |
| 0.400 | 25.30 | 1.265 | 1.60 |
| 0.500 | 28.40 | 1.420 | 2.02 |
| 0.600 | 31.02 | 1.551 | 2.41 |
| 0.700 | 33.62 | 1.681 | 2.83 |
| 0.800 | 35.86 | 1.793 | 3.21 |
(a) Explain why a graph of T² against l is plotted. (b) Use the graph to determine g. (c) Explain the effect on the graph and on g if every length had been measured to the top of the bob instead.
(a) Square both sides: \(T^2 = \left(\dfrac{4\pi^2}{g}\right) l\). Compared with \(y = mx + c\), y is T², x is l, the gradient is \(4\pi^2/g\) and the intercept is 0. A straight line through the origin tests the equation, and g comes from the gradient.
(b) Large triangle from l = 0.200 m to 0.800 m, read from the line: gradient \(= \dfrac{3.22 - 0.80}{0.600} = 4.03\ \text{s}^2\ \text{m}^{-1}\).
Rearrange gradient \(= 4\pi^2/g\): \(g = \dfrac{4\pi^2}{4.033} =\) 9.79 m s⁻²
(c) Every l would be too small by the bob's radius, a systematic error. The line shifts sideways and no longer passes through the origin, but the gradient, and so g, is unchanged.
Write the rearranged equation, then state which quantity is y, which is x, and what the gradient (and intercept) represent. "It gives a straight line" alone does not explain how you get the constant.
How do you evaluate an experiment and suggest improvements?
Improvements must target the real weakness and be specific enough to carry out.
- Find the biggest source of uncertainty, usually the measurement with the largest percentage uncertainty, and aim your improvement at it.
- Be specific. Not "be more careful" or "use better equipment", but "use light gates connected to a data logger to remove reaction time", "time 20 oscillations instead of 5", or "measure the diameter with a micrometer instead of a ruler".
- Random errors: more repeats, more cycles, larger measured quantities (smaller percentage uncertainty), more readings over a wider range.
- Systematic errors: check and correct zero errors, view scales at eye level to avoid parallax, use a fiducial mark, measure to the correct point (for example the centre of the bob).
- Comparing with an accepted value: state whether your result is close to it. A result that is consistently too high or too low points to a systematic error rather than a random one.
Exam tips for 3.2
Check every table you write: quantity / unit in each heading, no units in the cells, and every reading of one quantity to the same number of decimal places as the instrument's resolution.
For a gradient, draw a triangle covering more than half the line, read both corners from the line itself, and show the values you read. Then give the gradient with its unit.
Carry at least one extra figure through a calculation, then give the final answer to the number of significant figures of the least precise data (2 or 3 in most questions). A long string of calculator digits is not "more accurate".
One reading: half the resolution. Several repeats: half the range, with any anomaly removed. If both are available, quote the larger, and show the calculation that produced it.
When explaining an error, name its type and its effect: a zero error is systematic, makes every reading too large (or too small) by the same amount, and changes the intercept of a graph but not its gradient.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| More decimal places in an answer make it more accurate. | A result cannot be more precise than the data it came from. | Give results to the s.f. of the least precise data (or one more). |
| Repeating readings removes all errors. | A systematic error shifts every repeat by the same amount, so the mean is still wrong. | Repeats reduce random error; systematic error needs its cause removed or a correction. |
| A stopwatch reading to 0.01 s gives timings with an uncertainty of 0.005 s. | Human reaction time varies by far more than 0.01 s. | Use half the range of repeated timings. |
| The best-fit line must pass through the origin. | Systematic errors and the physics (for example \(v^2 = u^2 + 2as\)) can give a non-zero intercept. | Draw the line that best fits the points, then interpret the intercept. |
| The gradient can be found from any two data points. | Individual points carry random error; two close points magnify it. | Use a large triangle on the best-fit line. |
| An anomaly should be included because it is "real data". | A reading far outside the pattern usually comes from a mistake, and it distorts the mean or the line. | Repeat it if possible; otherwise exclude it, ring it and say why. |
How this topic connects
3.1 Planning experiments and measuring instruments
Resolution, zero error and parallax start with choosing and reading the instrument. This page takes those readings forward into uncertainties and graphs.
3.3 The AS core practicals
Every core practical (free fall, Young modulus, resistivity, internal resistance and the rest) ends with a table, a linearised graph and an evaluation using exactly these skills.
Units 1 and 2, and Unit 6
The equations being linearised come from mechanics (\(v^2 = u^2 + 2as\)), materials and electricity (\(R = \rho l / A\), \(V = \varepsilon - Ir\)) and waves (\(f = v/2l\)). Practical skills at A2 (Unit 6) build on this page with combined percentage uncertainties, maximum and minimum gradient lines, and logarithmic graphs.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| uncertainty = ½ × resolution | single reading | No — learn it |
| uncertainty = ½ × (largest − smallest) | set of repeated readings | No — learn it |
| percentage uncertainty = (uncertainty ÷ value) × 100% | comparing uncertainties of different measurements | No — learn it |
| \(\text{gradient} = \Delta y / \Delta x\); \(c = y - mx\) | straight-line graphs | No — learn it |
| \(v^2 = u^2 + 2as\) | linearised: v² against s | Yes |
| \(R = \rho l / A\) | linearised: R against l | Yes |
| \(T = 2\pi\sqrt{l/g}\) | linearised: T² against l | No at AS — given in the question when needed |
Check your understanding
1. A digital stopwatch reads to 0.01 s. Why is the uncertainty in a hand timing much larger than 0.005 s, and how should you estimate it?
Human reaction time in starting and stopping the watch varies by tenths of a second, far more than the resolution. Repeat the timing and use half the range of the repeats.
2. A thermometer with 1 °C divisions reads 23 °C. State the uncertainty and the percentage uncertainty.
± 0.5 °C (half the resolution). Percentage uncertainty \(= \dfrac{0.5}{23} \times 100\% = 2.2\%\).
3. Repeat readings of a length are 12.4, 12.6, 12.5, 13.9 and 12.5 cm. Give the mean and its uncertainty.
13.9 cm is anomalous and is excluded. Mean \(= \dfrac{12.4 + 12.6 + 12.5 + 12.5}{4} = 12.5\ \text{cm}\); uncertainty \(= \dfrac{12.6 - 12.4}{2} = \pm 0.1\ \text{cm}\).
4. Which graph would you plot to find a and u from \(v^2 = u^2 + 2as\), and how?
v² on the y-axis against s on the x-axis. Gradient = 2a, so a = gradient ÷ 2; intercept = u², so u = √intercept.
5. Classify each as random or systematic: (a) a balance that reads 0.3 g with nothing on it; (b) variation in judging when a pendulum passes its lowest point; (c) always reading a meniscus from slightly above.
(a) Systematic (zero error). (b) Random. (c) Systematic (parallax in the same direction every time).
Specification checklist
- (implementation) I can comment on the number of readings, their range and spacing, and whether repeats are needed.
- (implementation) I can record readings in a table with quantity / unit headings and a consistent number of decimal places set by the resolution.
- (implementation) I can spot an anomalous reading, explain how to check it, and deal with it correctly.
- (analysis) I can give calculated results to an appropriate number of significant figures, with the correct unit.
- (analysis) I can estimate the uncertainty of a single reading as half the resolution and of repeated readings as half the range, and calculate a percentage uncertainty.
- (analysis) I can distinguish random from systematic errors and recognise and correct a zero error.
- (analysis) I can plot a graph with sensible scales and a best-fit line, and find a gradient from a large triangle and an intercept.
- (analysis) I can rearrange an equation into the form \(y = mx + c\) to choose a straight-line graph, and get physical quantities from its gradient and intercept.
- (evaluation) I can identify the main source of uncertainty and suggest specific improvements to a method.