1.2.3Foundations in Physics
Determining Uncertainties from Graphs
5 min read 9 parts Lesson 7 of 19 1 quiz
1.2.3 Determining Uncertainties from Graphs — AQA A Level Physics 7408 notes: Use error bars and best and worst fit lines to find the gradient and intercept of a straight-line graph with uncertainties.A straight-line graph uses many measurements together, so the gradient and inte…
- line of best fit
- worst acceptable line
- Worked example 1
- g = 9.8 ± 0.5 m s−2
- Worked example 2
- Worked example 3
Section 1 of 9Why we use graphs
A straight-line graph uses many measurements together, so the gradient and intercept are determined more reliably than from one calculation.
- where y and x are the plotted quantities, m is the gradient and c is the intercept.
Plot with the independent variable on the horizontal axis, label each axis with quantity and unit, choose a scale that makes the data fill more than half of the graph, and mark points with small, clear crosses.
Section 2 of 9Error bars
An error bar shows the uncertainty in a plotted value.
- and you draw bars only for that one;
- if a question says that uncertainties in x are negligible, only vertical bars are used.
If the error bars are given as percentages, compute the absolute uncertainty for each point separately, since it varies with the value. When you plot a derived quantity such as T2, the percentage uncertainty in T2 is double that in T.
Section 3 of 9Best-fit and worst-fit lines
The line of best fit passes as near as possible to all points, with points scattered evenly on both sides, and must pass through the error bars (and through the origin only if theory says so, and the data supports it). A good best-fit line should pass through most error bars.
The worst acceptable line is the line with the greatest (or least) gradient that still passes through all the error bars. Draw two such lines: one steeper and one shallower than the best-fit line, usually pivoting about the points at the ends of the data set so that they pass through the top of the error bar at one end and the bottom at the other.
Then:
Section 4 of 9Calculating the gradient
Calculate the gradient using a large triangle: choose two points on the line (not data points, unless they lie on it) as far apart as possible, and use their coordinates read from the line.
- and the percentage uncertainty in g equals that in the gradient.
The same method applies to V against I (gradient gives resistance, or in a cell internal resistance), force against extension (spring constant) and many more experiments. The y-intercept may have a physical meaning, such as the emf of a cell.
Section 5 of 9Checking against theory
If the accepted value of a constant lies within the range given by your gradient and its uncertainty, the experiment agrees with theory within the stated uncertainty.
- where you must present data with uncertainties and justify conclusions.
Section 6 of 9Practical tips for drawing and reading graphs
A few habits make graph work reliable.
Choose the scales so that the plotted points fill at least half of the grid in both directions, and use simple scale steps such as 1, 2 or 5 units per large square, never 3 or 7. Read the coordinates of your gradient triangle to half a small square. When you state a conclusion, compare the experimental value with the accepted value and say whether it lies within the range implied by the uncertainty, which is the only valid test of agreement.
Section 7 of 9Worked examples
A graph of T2 against L for a pendulum has a best-fit gradient of 4.05 s2 m−1. The worst-fit lines have gradients of 4.25 and 3.85 s2 m−1. Find g and its uncertainty.
Gradient = 4.05 ± 0.20 (half of 4.25 − 3.85), a percentage uncertainty of 4.9%.
g = 4π2/gradient = 39.48/4.05 = 9.75 m s−2.
Uncertainty 4.9% of 9.75 = 0.48, so g = 9.8 ± 0.5 m s−2, which agrees with 9.81 m s−2.
The length of a pendulum is measured as 0.800 m with a 0.2% uncertainty, and the period is 1.80 s ± 0.02 s. Find the percentage uncertainty in T2 for a graph's error bar.
Percentage uncertainty in T = 0.02/1.80 = 1.1%.
For T2: 2 × 1.1% = 2.2%. T2 = 3.24 s2, so the error bar is ±0.07 s2.
A V–I graph for a cell has a best-fit y-intercept of 1.52 V, and the worst-fit line meets the axis at 1.47 V. State the emf and its uncertainty.
Uncertainty = 1.52 − 1.47 = 0.05 V.
emf = 1.52 ± 0.05 V (3.3%).
Section 8 of 9Key terms
| Term | Meaning |
|---|---|
| Error bar | A bar through a data point showing the uncertainty in that value. |
| Best-fit line | The line that passes close to all points with scatter on both sides. |
| Worst-fit line | The steepest or shallowest line that still passes through all error bars. |
| Gradient | Δy/Δx of a line; units are the unit of y divided by the unit of x. |
| Intercept | The y value where the line crosses the y-axis. |