8867 / 2027
Quantities and Measurement overview

Lesson 5 of 5

Data-based questions: graphs and gradients

  • I can distinguish random and systematic errors, including zero error, and relate them to precision and accuracy.Syllabus

    Syllabus 8867, 1(f). Distinguish random and systematic errors, including zero error, and relate them to precision and accuracy.

  • I can assess derived uncertainty by adding absolute or relative uncertainties, or by numerical substitution.Syllabus

    Syllabus 8867, 1(g). Assess derived uncertainty by adding absolute or relative uncertainties, or by numerical substitution.

Make a guess. You are not marked.

A student plots T2 against l for a simple pendulum and gets a straight line through the origin. What does the gradient give?

  1. 4 pi2 / g, so g is 4 pi2 divided by the gradient.
  2. The period of the pendulum for a length of 1 m.
  3. g itself, because the graph is a straight line.
Show the answer

4 pi2 / g, so g is 4 pi2 divided by the gradient.

Squaring T = 2 pi sqrt(l / g) gives T2 = (4 pi2 / g) l. Comparing with y = mx, the gradient is 4 pi2 / g.

H1 has no practical paper, but Paper 2 has one or two data-based questions worth 15 to 20 marks. They give you a table of results and ask you to choose a graph, find a gradient and judge the uncertainty.

1. Turn the relationship into a straight line

A straight line is easy to test and its gradient and intercept give you constants. Rearrange the equation into the form y = mx + c. Then plot y against x.

Common ways to get a straight line
EquationPlotGradientIntercept
T = 2π√(l/g)T2 against l4π2/g0
V = E - IrV against I-rE
y = kxnlg y against lg xnlg k
I = I0e-μxln I against x-μln I0

Use lg against lg for a power law with an unknown power n. Use ln against x for an exponential. A logarithm has no unit, so label the axis with the quantity divided by its unit, such as lg (T / s).

2. Read the gradient and intercept properly

  • Use a large triangle that covers more than half the line. Read both corners from the line and write their coordinates in your working.
  • Check where your x-axis starts. If it does not start at zero, the line does not cross the y-axis on your grid. Find c by putting one point on the line into c = y - mx.
  • Give the gradient a unit where it has one: the y-unit divided by the x-unit.

Worked analysis

Find the power n in T = kmn

The period T of a mass m oscillating on a spring was measured for six masses. Each T is the mean time for 20 oscillations divided by 20.

Mass, period and their logarithms
m / kgT / slg (m / kg)lg (T / s)
0.1000.398-1.000-0.400
0.2000.561-0.699-0.251
0.3000.690-0.523-0.161
0.4000.793-0.398-0.101
0.5000.890-0.301-0.051
0.6000.972-0.222-0.012

Taking lg of both sides gives lg T = n lg m + lg k. So plot lg T against lg m. The points lie on a straight line, which confirms a power law.

Best-fit line: lg T = 0.499 lg m + 0.099
n = gradient = 0.499 ≈ 0.50
lg k = 0.099, so k = 100.099 = 1.26 s kg-0.5

n = 0.50 means T is proportional to √m. For a spring, T = 2π√(m/ks), so 2π/√ks = 1.26 and the spring constant ks = (2π/1.26)2 = 25 N m-1.

3. Find and quote the uncertainty

For repeated readings, take the uncertainty as half the range. Three timings of 20 oscillations, 15.84 s, 15.92 s and 15.88 s, have a mean of 15.88 s and an uncertainty of (15.92 - 15.84)/2 = 0.04 s. So T = 15.88/20 = (0.794 ± 0.002) s.

On a graph with error bars, draw the best-fit line and a worst acceptable line: the steepest or shallowest line that still passes through all the error bars. The uncertainty in the gradient is the difference between the two gradients. Quote the uncertainty to 1 significant figure, and round the value to the same decimal place.

Then use the uncertainty rules to carry the uncertainty into any quantity you calculate from it.

Try it step by step

Work it out in steps: Choose the graph that gives n as its gradient, then Gradient n, then Intercept lg k.

  1. Problem 1. A student suspects T = k m^n. Two points on the best-fit line of lg (T / s) against lg (m / kg) are (-1.00, -0.20) and (-0.40, 0.10). Find n and the intercept lg k.
  2. Problem 2. For the same model, two points on the best-fit line of lg T against lg m are (-1.00, 0.10) and (0.00, 0.40). Find the intercept lg k.
  3. Problem 3. Two points on the best-fit line of lg T against lg m are (-0.50, 0.25) and (0.50, 0.75). Choose the graph, then find n and lg k.
Show the answer

Problem 1.

Choose the graph that gives n as its gradient: Taking logs: lg T = n lg m + lg k, so plot lg T against lg m. The gradient is n and the intercept is lg k.

Gradient n: n = (0.10 - (-0.20)) / (-0.40 - (-1.00)) = 0.30 / 0.60 = 0.50

Intercept lg k: At lg m = 0: lg k = -0.20 + 0.50 x 1.00 = 0.30, so k = 100.30 = 2.0

Problem 2.

Choose the graph that gives n as its gradient: Plot lg T against lg m.

Gradient n: n = (0.40 - 0.10) / (0.00 - (-1.00)) = 0.30

Intercept lg k: The second point has lg m = 0, so lg k = 0.40

Problem 3.

Choose the graph that gives n as its gradient: lg T against lg m, because lg T = n lg m + lg k is a straight line with gradient n.

Gradient n: n = (0.75 - 0.25) / (0.50 - (-0.50)) = 0.50

Intercept lg k: At lg m = 0, halfway between the points: lg k = 0.50

Check your understanding

A student suspects that T = km^n, where k and n are unknown constants. Which graph lets her find n from its gradient?

  1. T squared against m
  2. T against m
  3. lg T against lg m
Show the answer

lg T against lg m

Taking lg of both sides gives lg T = n lg m + lg k. This has the form y = mx + c, so the gradient is n and the y-intercept is lg k.