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Having distinguished random from systematic errors in Lesson 5, we now put numbers on the idea of uncertainty. When a physicist writes:
l = 1.250 ± 0.005 m
they are making a precise, mathematically meaningful statement: "I believe the true length of this rod lies somewhere in the interval 1.245 m to 1.255 m, and my best single estimate of it is 1.250 m". This lesson teaches you how to derive, write and interpret such statements.
We cover four things: the distinction between precision and accuracy (already met qualitatively in Lesson 5, now made numerical); resolution and how it bounds the uncertainty of a single reading; absolute vs percentage uncertainty; and the number of significant figures appropriate for a quoted uncertainty.
As we saw with the archery target analogy, precision is about the spread of repeated measurements, and accuracy is about how close they are to the true value. Numerically:
These are independent. A set of measurements can be precise but inaccurate (tight cluster offset from the true value — large systematic error), accurate but imprecise (scattered around the true value — small systematic error but large random error), or both, or neither.
The resolution of an instrument is the smallest change in the quantity being measured that the instrument can distinguish. For a digital instrument, the resolution is the last displayed digit. For an analogue instrument, it is typically the smallest scale division.
State the measurement and its absolute uncertainty for each of the following.
(a) A 30 cm ruler reading 148 mm. (b) A digital voltmeter displaying 3.27 V. (c) A human-operated stopwatch showing 14.58 s. (d) A vernier caliper reading 12.35 mm (resolution 0.05 mm).
Solution:
(a) 148 ± 0.5 mm (half the smallest division) (b) 3.27 ± 0.01 V (last digit) (c) 14.58 ± 0.2 s — or sometimes ±0.1 s; the resolution of the display is misleading because human reaction time dominates. (d) 12.35 ± 0.025 mm or equivalently ±0.03 mm (half the smallest division = 0.025 mm)
When you take N repeated measurements, the best estimate of the true value is the mean:
x̄ = (x₁ + x₂ + ... + x_N) / N
The uncertainty in the mean can be estimated in a simple way sufficient for A-Level as:
uncertainty in mean ≈ (x_max − x_min) / 2
That is, half the range. This is conservative and easy to remember. (A more rigorous statistical treatment uses the standard error, but OCR accepts the half-range method.)
A student times 10 oscillations of a pendulum six times and records:
14.5, 14.8, 14.3, 14.7, 14.6, 14.4 seconds
Calculate the mean time for 10 oscillations and its absolute uncertainty.
Solution:
Mean = (14.5 + 14.8 + 14.3 + 14.7 + 14.6 + 14.4) / 6 = 87.3 / 6 = 14.55 s
Range = 14.8 − 14.3 = 0.5 s
Uncertainty = 0.5 / 2 = ±0.25 s
Quoted result: (14.55 ± 0.25) s or, to appropriate significant figures, (14.5 ± 0.3) s.
The period of a single oscillation is T = 14.55 / 10 = 1.455 s, with uncertainty 0.25/10 = 0.025 s, so T = (1.46 ± 0.03) s.
This simple trick — dividing by N to reduce the uncertainty — is why OCR practical methods always recommend measuring a "bunch" of oscillations, fringes or periods, rather than one at a time.
Uncertainty can be quoted in two equivalent ways:
Both forms carry the same information. Which you use depends on the context.
| Context | Preferred form |
|---|---|
| Stating a single measurement with a single uncertainty | Absolute |
| Comparing uncertainties across different quantities | Percentage |
| Propagating uncertainties through multiplication or division (Lesson 7) | Percentage |
| Propagating uncertainties through addition or subtraction | Absolute |
| Reporting a final answer | Absolute, with appropriate sig figs |
percentage uncertainty = (absolute uncertainty / measured value) × 100%
and
absolute uncertainty = (percentage uncertainty / 100) × measured value
(a) A length is measured as 45.0 ± 0.5 cm. What is the percentage uncertainty?
(0.5 / 45.0) × 100 = 1.11% ≈ 1.1%
(b) A time is measured with 2% uncertainty, and the best value is 3.2 s. What is the absolute uncertainty?
(2/100) × 3.2 = 0.064 s ≈ 0.06 s
Quoted result: (3.20 ± 0.06) s.
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