A set of resistance values follows R = R_0(1 + αθ) with temperature θ. State the axes for a straight-line plot and identify the gradient and intercept. [2 marks]

SingaporePhysicsSyllabus dot point

How do we linearise a physical relationship so that a straight-line graph yields the quantities we want from its gradient and intercept?

Rearrange a physical relationship into straight-line form y = mx + c, plot the appropriate variables, and extract physical quantities from the gradient and intercept

A focused answer to the H2 Physics Measurement learning outcome on graphical analysis. Linearising relationships into y = mx + c form, choosing axes, and reading physical quantities from gradient and intercept in Paper 3 and Paper 4.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to take a physical relationship that may be non-linear, rearrange it into the straight-line form $y = mx + c$ , decide what to plot on each axis, and then extract the physical quantity you want from the gradient and the intercept. This skill is central to the Paper 3 data-based question and to the analysis section of the Paper 4 practical.

The answer

Why a straight line is the goal

A straight line is the easiest graph to interpret: its gradient and intercept are read directly and a best-fit line averages out random scatter. So the strategy for any relationship is to manipulate it into the form:

y = mx + c

where $m$ is the gradient and $c$ is the $y$ -intercept, then plot the combinations that play the roles of $y$ and $x$ .

Linearising common relationships

The key is to identify which grouping of variables behaves like $y$ , which like $x$ , and what the gradient and intercept then represent.

$v = u + at$ is already linear: plot $v$ against $t$ ; gradient $= a$ , intercept $= u$ .
$T^2 = \dfrac{4\pi^2}{g} L$ for a pendulum: plot $T^2$ against $L$ ; gradient $= \dfrac{4\pi^2}{g}$ .
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ for a lens: plot $\dfrac{1}{v}$ against $\dfrac{1}{u}$ ; intercept $= \dfrac{1}{f}$ .

Power laws and the log-log plot

If $y = a x^n$ , taking logarithms gives:

\lg y = n \lg x + \lg a

Plot $\lg y$ against $\lg x$ . The gradient is the power $n$ and the intercept is $\lg a$ , so $a = 10^{\text{intercept}}$ . A straight log-log line confirms the power-law form.

Exponentials and the log-linear plot

If $y = y_0 e^{-kx}$ (radioactive decay, capacitor discharge), take natural logs:

\ln y = \ln y_0 - kx

Plot $\ln y$ against $x$ . The gradient is $-k$ and the intercept is $\ln y_0$ . This is the standard way to find a decay or time constant from data.

Reading the gradient correctly

The gradient must be calculated from two well-separated points on the best-fit line, not from a single data point:

\text{gradient} = \frac{y_2 - y_1}{x_2 - x_1}

Always carry units through the gradient, because the physical quantity often comes from the gradient's units as much as its value.

Worked example

A capacitor discharges through a resistor so that the voltage follows $V = V_0 e^{-t/RC}$ . A student records $V$ against $t$ and wants the time constant $RC$ . Show how to find it graphically, given two points on the linearised line: $(t = 0,\ \ln V = 2.30)$ and $(t = 8.0\ \text{s},\ \ln V = 0.70)$ .

Step 1: Linearise the relationship

Take natural logs: $\ln V = \ln V_0 - \dfrac{1}{RC} t$ . This is $y = mx + c$ with $y = \ln V$ , $x = t$ , gradient $= -\dfrac{1}{RC}$ .

Step 2: Calculate the gradient from the two points

m = \frac{0.70 - 2.30}{8.0 - 0} = \frac{-1.60}{8.0} = -0.20\ \text{s}^{-1}

Step 3: Relate the gradient to the time constant

-\frac{1}{RC} = -0.20 \implies RC = \frac{1}{0.20} = 5.0\ \text{s}

Step 4: State the result

The time constant is $RC = 5.0\ \text{s}$ , and the intercept ( $\ln V_0 = 2.30$ ) gives $V_0 = e^{2.30} = 10\ \text{V}$ .

Examples in context

Example 1. Finding $g$ from a pendulum. Rearranging $T = 2\pi\sqrt{L/g}$ to $T^2 = \dfrac{4\pi^2}{g}L$ and plotting $T^2$ against $L$ gives a straight line through the origin. Measuring its gradient $G$ yields $g = \dfrac{4\pi^2}{G}$ . The straight-line method averages many trials and exposes any systematic offset as a non-zero intercept.

Example 2. Confirming an inverse-square field. Suspecting $I \propto \dfrac{1}{d^2}$ for a point source, a student plots $I$ against $\dfrac{1}{d^2}$ . A straight line through the origin confirms the inverse-square law, while a log-log plot of $\lg I$ against $\lg d$ with gradient $-2$ provides independent confirmation of the power.

Try this

Q1. The relationship $E_k = \tfrac{1}{2}mv^2$ is to be tested by varying $v$ . State what you would plot to obtain a straight line through the origin, and what the gradient represents. [2 marks]

Cue. Plot $E_k$ against $v^2$ ; gradient $= \tfrac{1}{2}m$ .

Q2. A set of resistance values follows $R = R_0(1 + \alpha\theta)$ with temperature $\theta$ . State the axes for a straight-line plot and identify the gradient and intercept. [2 marks]

Cue. Plot $R$ against $\theta$ ; intercept $= R_0$ , gradient $= R_0\alpha$ , so $\alpha = \dfrac{\text{gradient}}{\text{intercept}}$ .

Q3. Data are suspected to follow $y = ax^3$ . Describe a logarithmic plot to confirm this and find $a$ . [3 marks]

Cue. Plot $\lg y$ against $\lg x$ ; a straight line of gradient $3$ confirms the cube law, and $a = 10^{\text{intercept}}$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksA spring obeys

T = 2\pi\sqrt{\dfrac{m}{k}}

, where

T

is the period of oscillation of a mass

m

and

k

is the spring constant. (a) Show how to plot a straight-line graph to determine

k

. (b) State what you would plot on each axis and how

k

is found from the gradient.

Show worked answer →

(a) Square both sides to remove the root: $T^2 = \dfrac{4\pi^2}{k} m$ .

This is in the form $y = mx + c$ with $y = T^2$ , $x = m$ , gradient $= \dfrac{4\pi^2}{k}$ and intercept $= 0$ .

(b) Plot $T^2$ (vertical axis) against $m$ (horizontal axis). The graph is a straight line through the origin with gradient $\dfrac{4\pi^2}{k}$ .

Rearranging: $k = \dfrac{4\pi^2}{\text{gradient}}$ .

Markers reward the squaring step, the clear identification of which variable is $y$ and which is $x$ , the correct gradient expression, and the final rearrangement for $k$ .

Original4 marksThe current-voltage relationship for a component is suspected to be

I = aV^n

. Explain how a logarithmic plot can test this relationship and find both

a

and

n

Show worked answer →

Take logarithms of both sides: $\lg I = \lg a + n \lg V$ .

This is in the form $y = mx + c$ with $y = \lg I$ , $x = \lg V$ , gradient $= n$ and intercept $= \lg a$ .

Plot $\lg I$ against $\lg V$ . If the data fall on a straight line, the power-law relationship holds. The gradient gives $n$ directly, and the constant is found from $a = 10^{\text{intercept}}$ .

Markers reward taking logs of both sides, identifying the gradient as $n$ and the intercept as $\lg a$ , and recovering $a$ by taking the antilog of the intercept.