1.3 Measurements and uncertainties — Learn

Prerequisites: SI units and prefixes from Section 1.1; resolution, accuracy and error from Section 1.2; rearranging a formula; calculating a percentage.

By the end of this section you can: state the uncertainty in a reading and in a mean; convert between absolute and percentage uncertainty; combine uncertainties through a sum, a product, a quotient and a power; quote a result with its uncertainty; and decide whether a result agrees with an accepted value.

1. The uncertainty in a reading

A measurement is not a single number. It is a value with an uncertainty: a statement of the range in which the true value is expected to lie. A length written as $0.800 \pm 0.001\,\text{m}$ means that the true length is believed to lie between 0.799 m and 0.801 m.

The uncertainty in a single reading is half the smallest division of the instrument. A metre rule marked in millimetres gives a reading uncertainty of 0.5 mm.

When a quantity is found from two readings, the two reading uncertainties add. The reading uncertainty is an absolute uncertainty, because it carries the same unit as the quantity.

2. Repeated readings: the mean and the half-range

A measurement is usually repeated. The mean of the readings is the best estimate of the true value.

The uncertainty is found from the spread of the readings. The range is the largest reading minus the smallest reading, and the uncertainty is half the range. This is called the half-range convention.

Worked example 1. The time for 20 oscillations is measured three times: $35.90\,\text{s}$ , $35.85\,\text{s}$ and $35.95\,\text{s}$ . Find the mean and the uncertainty.

\begin{aligned} \text{mean} &= \frac{35.90 + 35.85 + 35.95}{3} = 35.90\,\text{s} \\ \text{range} &= 35.95 - 35.85 = 0.10\,\text{s} \\ \text{uncertainty} &= \frac{0.10}{2} = 0.05\,\text{s} \end{aligned}

The time for 20 oscillations is $35.90 \pm 0.05\,\text{s}$ .

Worked example 2. The pendulum length is found from two metre-rule readings: $0.000\,\text{m}$ at the pivot and $0.800\,\text{m}$ at the centre of the bob. Each reading has an uncertainty of 0.0005 m. Find the length and its uncertainty.

\begin{aligned} L &= 0.800 - 0.000 = 0.800\,\text{m} \\ \Delta L &= 0.0005 + 0.0005 = 0.001\,\text{m} \end{aligned}

The length is $0.800 \pm 0.001\,\text{m}$ . A difference adds the absolute uncertainties, exactly as a sum does.

3. Percentage uncertainty

An absolute uncertainty is hard to judge on its own. An uncertainty of 1 mm is tiny on a pendulum and enormous on a wire. The percentage uncertainty puts the uncertainty in proportion:

\text{percentage uncertainty} = \frac{\text{absolute uncertainty}}{\text{value}} \times 100\%

Worked example 3. Find the percentage uncertainty in the time for 20 oscillations, $35.90 \pm 0.05\,\text{s}$ .

\text{percentage uncertainty} = \frac{0.05}{35.90} \times 100\% = 0.14\%

The pendulum period is $T = \dfrac{35.90}{20} = 1.795\,\text{s}$ . Dividing by 20 divides the uncertainty by 20 as well, so $T = 1.795 \pm 0.0025\,\text{s}$ . The percentage uncertainty is unchanged, 0.14%. The percentage uncertainty in the length is

\frac{0.001}{0.800} \times 100\% = 0.125\%

4. Combining uncertainties

The rule depends on the arithmetic.

Adding or subtracting quantities: add the absolute uncertainties. For masses of $52.4 \pm 0.1\,\text{g}$ and $128.7 \pm 0.1\,\text{g}$ , a mass difference carries $0.1 + 0.1 = 0.2\,\text{g}$ .
Multiplying or dividing quantities: add the percentage uncertainties.
Raising a quantity to a power: multiply the percentage uncertainty by the power.

Worked example 4. The acceleration due to gravity is found from

g = \frac{4\pi^2 L}{T^2}

The length has a percentage uncertainty of 0.125% and the period has a percentage uncertainty of 0.139%. Find the percentage uncertainty in $g$ .

\begin{aligned} \text{percentage uncertainty in } T^2 &= 2 \times 0.139\% = 0.278\% \\ \text{percentage uncertainty in } g &= 0.125\% + 0.278\% \\ &= 0.40\% \quad (2\ \text{s.f.}) \end{aligned}

The constant $4\pi^2$ carries no uncertainty, so it adds nothing.

The value of $g$ is

g = \frac{4\pi^2 \times 0.800}{1.795^2} = 9.802\,\text{m}\,\text{s}^{-2}

so the absolute uncertainty is $9.802 \times \dfrac{0.40}{100} = 0.04\,\text{m}\,\text{s}^{-2}$ .

5. Quoting a result

The uncertainty is quoted to one significant figure. The value is quoted to the same decimal place as the uncertainty. For the pendulum,

g = 9.80 \pm 0.04\,\text{m}\,\text{s}^{-2}

The three figures in 9.80 matter, because the uncertainty is in the second decimal place. Writing $g = 9.8021 \pm 0.04\,\text{m}\,\text{s}^{-2}$ would overstate the precision of the measurement.

To decide whether a result agrees with an accepted value, compare the range with the accepted value. Here the measured range is 9.76 to 9.84, and the accepted value 9.81 lies inside it. The measurement agrees within its uncertainty.

Watch out. A small percentage uncertainty does not prove an accurate result. A systematic error shifts every reading the same way, so the readings can be tightly grouped and still be wrong. The pendulum length, for example, is measured to the centre of the bob. A consistent mistake there would not show up in the uncertainty at all, but it would move $g$ away from 9.81 m s⁻². Re-check by asking what could be wrong in the same way every time.

The conventions this book uses

A single reading: uncertainty $=$ half the smallest division.
A quantity from two readings: the absolute uncertainties add.
Repeated readings: mean, and uncertainty $=$ half the range.
Multiply or divide: add percentage uncertainties. A power multiplies the percentage uncertainty.
Quote the uncertainty to one significant figure, and the value to the same decimal place.

The awarding bodies explain these rules in slightly different words, so always state the rule you have used.

6. Try it: how many oscillations?

The uncertainty in a hand-timed measurement comes mostly from reaction time, about 0.2 s. Timing many oscillations and dividing by the number shares that uncertainty over more oscillations.

Predict first. Do not move the slider yet. Write down your answers.

If you time 50 oscillations instead of 1, what happens to the percentage uncertainty in the period?
If you time 50 oscillations instead of 20, does the period itself change?
The period is squared in the equation for $g$ . What does that do to the percentage uncertainty in $g$ ?

Test. Set the slider to 1, then to 20, then to 50. Watch the percentage uncertainty in the period and in $g$ .

Explain. Why does timing more oscillations reduce the percentage uncertainty in the period, but not change the period? Write two or three sentences. Use the idea that the uncertainty of 0.2 s is in the total time, and that the period is the total time divided by the number of oscillations.

Progress check

Return to the chapter problem.

The time for 20 oscillations is $35.90 \pm 0.05\,\text{s}$ . Find the period $T$ and its absolute uncertainty.
Use $g = \dfrac{4\pi^2 L}{T^2}$ with $L = 0.800 \pm 0.001\,\text{m}$ to find $g$ and its percentage uncertainty.
Quote $g$ with its absolute uncertainty, and decide whether it agrees with $9.81\,\text{m}\,\text{s}^{-2}$ .

If question 1 goes wrong, remember that dividing the time by 20 divides its uncertainty by 20 as well. If question 2 goes wrong, find the percentage uncertainty in the length and in the period first, then double the second one.

You should now be able to

state the reading uncertainty from the smallest division of an instrument;
find the mean and the half-range uncertainty of repeated readings;
convert an absolute uncertainty into a percentage, and back again;
combine uncertainties through a sum, a difference, a product, a quotient and a power;
quote a result with its uncertainty to the right number of significant figures;
decide whether a measurement agrees with an accepted value, and explain why a small uncertainty does not prove accuracy.