Graphing and linearization for NePhO
A practical walkthrough of linearization, scale choice and graph marking for NePhO experimental questions, with a worked pre-selection example.
By Prabhakar Kafle and Sanam Limbu
Based on a guide by Prabhakar Kafle and Sanam Limbu.
This page is an independent rewrite for this site. The original guide and its figures are not reproduced here.
Why graphing matters
The authors of the source guide observe that graph-based questions in NePhO carry at least 10 marks out of 100. They also notice that, across recent papers, one skill keeps coming back: linearization. This is the authors' observation, not an official mark breakdown.
That makes graph work high-value practice. A clear table, a sensible scale and a well-read slope can earn marks even when the final number is not perfect.
What linearization means
Linearization means rewriting a relationship so that the data can be plotted as a straight line in the form y = mx + c. Once the points lie on a line, the slope and intercept can be matched to the unknown constants in the original equation.
If a relation involves only multiplication, division and powers, taking logarithms can turn it into a straight line. Sometimes taking roots or powers of both sides also helps. For example, if μ = a + b/λ², plotting μ against 1/λ² gives a straight line. The slope is b and the intercept is a.
Worked example: NePhO pre-selection 2017
A capacitor discharges so that V = α e^(−t/T). The measurements are shown below. The task is to find a straight-line form and use it to estimate α and T.
- Take natural logarithms of both sides: ln V = −(1/T)t + ln α.
- Compare with y = mx + c: use y = ln V and x = t. Then the slope is −1/T and the intercept is ln α.
- Process the measured voltages into ln V values.
- Plot ln V against t and draw a best-fit straight line.
| t (s) | V (V) | ln V |
|---|---|---|
| 0 | 4.3 | 1.45 |
| 1 | 2.8 | 1.03 |
| 2 | 1.8 | 0.59 |
| 3 | 1.2 | 0.18 |
| 4 | 0.8 | −0.22 |
| 5 | 0.5 | −0.69 |
| 6 | 0.3 | −1.20 |
With 2 squares = 1 s on the x-axis and 1 square = 0.14 on the y-axis, the points spread across the graph paper. A large triangle on the best-fit line gives a slope of −1.82 / 4.25 ≈ −0.43.
Time constant: since the slope is −1/T, −1/T ≈ −0.43, so T ≈ 2.33 s.
Initial voltage: the intercept is about 1.46, so ln α ≈ 1.46 and α ≈ 4.3 V.
Interactive plot
Switch between the raw measurements and the linearized values. The linearized view shows the best-fit straight line; the original view shows the fitted exponential curve.
Scatter plot of the original measurements: time in seconds on the horizontal axis, and capacitor voltage in volts on the vertical axis. Seven points fall along a decaying fitted curve, V equals 4.3 times e to the power of minus t over 2.33.
The table in the worked example is the text alternative for the plotted values.
Choosing a scale
- Spread the points so the graph covers more than 80% of the paper.
- Choose a scale that goes slightly beyond the data range. For example, allow room for 7 s when the largest reading is 6 s.
- Prefer easy numbers. Use 2 squares = 1 s rather than 2.29 squares = 1 s.
Graph checklist
- Cover more than 80% of the paper.
- Write the scale on the graph.
- Label both axes with the quantity and its unit. For powers of ten, put the multiplier in the axis label, for example “time (×10² s)”.
- Make each point easy to see, for example by circling it.
- Draw a best-fit line with roughly equal numbers of points above and below it.
- Find the slope using a large right-angled triangle and two far-apart points on the line.
Try it yourself
These two prompts were written for this site.
1. Pendulum. The period is T = 2π√(L/g). What should you plot to get a straight line, and what does the slope represent?
Show answer
Plot T² against L. Squaring both sides gives T² = (4π²/g)L, so the slope is 4π²/g and g = 4π² / slope.
2. Power law. For y = a x^n, what should you plot, and how do you find n and a?
Show answer
Plot ln y against ln x. Taking logs gives ln y = n ln x + ln a, so the slope is n and the intercept is ln a. Then a = e^(intercept).