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Uncertainty in derived quantities

Key idea: For separately varying inputs, add absolute uncertainties for sums and differences. For products and quotients, add fractional uncertainties; a power multiplies the fractional uncertainty.

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  • GCE A-Level H2 Physics 2027
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Most results in physics are not measured directly but calculated: a speed from a distance and a time, a density from a mass and a volume. Each measured input has an uncertainty, so the calculated result has one too. This lesson shows how to find it, either with simple rules that add uncertainties or by working out the largest and smallest possible values.

Sums and differences: add absolute uncertainties

If q = a + b or q = a-b, then

Δ q = Δ a + Δ b

These are worst-case allowances for separately measured inputs that can vary within their stated ranges. Take the worst case. For a sum, the largest value is (a + Δ a) + (b + Δ b), which is Δ a + Δ b above the best value. For a difference, the largest value is (a + Δ a)-(b-Δ b), which is again Δ a + Δ b above. So the absolute uncertainties add, even when the quantities are subtracted.

This makes a small difference between two similar measurements very uncertain. For a = (0.250±0.001) m and b = (0.235±0.001) m, the difference a-b = (0.015±0.002) m has a 13% uncertainty, although each length is known to better than 0.5%.

Products, quotients and powers: add fractional uncertainties

If q = ab or q = a/b, then, provided the uncertainties are small,

(Δ q)/q = (Δ a)/a + (Δ b)/b

and if q = aⁿ,

(Δ q)/q = |n|(Δ a)/a

Here is why. If a is 1% too large and b is 2% too large, then ab is about 3% too large: (1.01)(1.02) = 1.0302. The percentages add because the errors multiply. A square, a² = a × a, doubles the fractional uncertainty, and a square root halves it. An exact number, such as π or the 1/2 in (1/2)mv², adds nothing.

Absolute uncertainties have units; fractional and percentage uncertainties do not. To use the product rule, find the fractional uncertainty first, then multiply by q to get Δ q.

Worked example 1

Uncertainty in a speed

Problem

A trolley travels d = (5.00±0.05) m in t = (2.00±0.02) s. Find its average speed and the uncertainty, first with the rule and then by working out the largest and smallest possible speeds.

Show full solution

  1. Apply the quotient rule

    Method

    Add the fractional uncertainties of d and t.

    Reason

    Speed is a quotient, v = d/t.

    Working

    Answer

    v = 5.00/2.00 = 2.50 m s⁻¹ and (Δ v)/v = 0.05/5.00 + 0.02/2.00 = 0.010 + 0.010 = 0.020

    So Δ v = 0.020 × 2.50 = 0.05 m s⁻¹, and v = (2.50±0.05) m s⁻¹.

  2. Work out the limits

    Working

    Answer

    The largest speed uses the largest distance and the smallest time: 5.05/1.98 = 2.55 m s⁻¹. The smallest uses the smallest distance and the largest time: 4.95/2.02 = 2.45 m s⁻¹. To this precision, both are about 0.05 m s⁻¹ from 2.50 m s⁻¹, agreeing with the small-uncertainty approximation.

Input limits for a calculated speed

For time from 1.98 to 2.02 seconds, both possible-speed curves decrease. Distance 4.95 metres gives the lower curve and 5.05 metres the upper. The central inputs give 2.50 metres per second. The overall extremes are about 2.4505 and 2.5505 metres per second.

Scroll across the figure to read all labels.

For time from 1.98 to 2.02 seconds, both possible-speed curves decrease. Distance 4.95 metres gives the lower curve and 5.05 metres the upper. The central inputs give 2.50 metres per second. The overall extremes are about 2.4505 and 2.5505 metres per second.For time from 1.98 to 2.02 seconds, both possible-speed curves decrease. Distance 4.95 metres gives the lower curve and 5.05 metres the upper. The central inputs give 2.50 metres per second. The overall extremes are about 2.4505 and 2.5505 metres per second.
Calculated v = d/t for separately varying distance and time intervals. The largest speed uses the largest distance and smallest time. These curves are not a motion history.
Open full-size graph
What to noticeThe axes zoom in on the uncertainty intervals and do not start at zero. Compare opposite input limits for a quotient.
View figure data
Figure data: Input limits for a calculated speed
SeriesAllowed measured time, t (s)Possible calculated speed, v (m s⁻¹)
d = 4.95 m1.9802.500
d = 4.95 m1.9842.495
d = 4.95 m1.9882.490
d = 4.95 m1.9922.485
d = 4.95 m1.9962.480
d = 4.95 m2.0002.475
d = 4.95 m2.0042.470
d = 4.95 m2.0082.465
d = 4.95 m2.0122.460
d = 4.95 m2.0162.455
d = 4.95 m2.0202.450
d = 5.05 m1.9802.551
d = 5.05 m1.9842.545
d = 5.05 m1.9882.540
d = 5.05 m1.9922.535
d = 5.05 m1.9962.530
d = 5.05 m2.0002.525
d = 5.05 m2.0042.520
d = 5.05 m2.0082.515
d = 5.05 m2.0122.510
d = 5.05 m2.0162.505
d = 5.05 m2.0202.500
Central inputs2.0002.500

The graph shows every speed the inputs allow. At a fixed distance, a larger time gives a smaller speed, so the highest speed is at the top left and the lowest at the bottom right. Working out the limits like this is called numerical substitution. It is the safest method when the uncertainties are large or the formula is awkward, because the rules above are only approximations for small uncertainties.

Guided practice 2

A difference times a force

About 4 min

Problem

The moment M = (a-b)F is calculated from a = (0.120±0.001) m, b = (0.100±0.001) m and F = (3.00±0.03) N. Find the central difference and its absolute uncertainty, the fractional uncertainty in M, then M and its absolute uncertainty. Enter the absolute uncertainty before rounding the reported result.

Work through one operation at a time

Show the steps Optional working
Unit: m
Unit: N m
Show the steps Optional working
Unit: m
Unit: m
Unit: N m

Hints

Hint 1

the difference first

Which operation must you complete before multiplying by F?

Hint 2

then the product

M is a product of (a-b) and F, so add their fractional uncertainties.

Show solution step by step

  1. The difference

    Reason

    Absolute uncertainties add for a difference.

    Working

    Answer

    a-b = (0.020±0.002) m, a fractional uncertainty of 0.10.

  2. The product

    Working

    Answer

    M = (0.020)(3.00) = 0.060 N m and (Δ M)/M = 0.10 + 0.03/3.00 = 0.11, so Δ M = 0.11 × 0.060 = 0.0066 N m. The result is (0.060±0.007) N m.

Check your understanding 1

The area of a circle is found from its radius, r = (2.00±0.05) cm, using A = π r². Find the area with its percentage and absolute uncertainties. Enter the absolute uncertainty to two significant figures before rounding the final reported result.

Unit: cm²
Unit: %
Show the steps Optional working
Unit: cm²
Unit: %
Unit: cm²
Show answer

A = π(2.00)² = 12.6 cm². The radius has a 2.5% uncertainty, and squaring doubles it, so the area has a 5.0% uncertainty: Δ A = 0.050 × 12.57 = 0.63 cm². The area is (12.6±0.6) cm². π is exact, so it adds no uncertainty.

Spot the mistake 3

A square root halves the percentage uncertainty

About 4 min

Learner claim

A speed is calculated from v = square root of 2gh, with h = (2.00±0.01) m and g taken as exact. A learner says v has the same 0.5% uncertainty as h. Find the percentage uncertainty in v before rounding.

Calculate the percentage uncertainty

Unit: %
Show solution step by step

  1. Find the percentage uncertainty in h

    Reason

    The power rule works with fractional or percentage uncertainties.

    Working

    Answer

    0.01/2.00 × 100% = 0.5%

  2. Apply the power

    Working

    Answer

    v ∝ h^(1/2), so the percentage uncertainty in v is 1/2 × 0.5% = 0.25%.

Check your understanding 2

A quantity x = 2.0±1.0 is used to calculate y = 1/x. Find the largest and smallest values of y. Compare the asymmetry with the public answer and explain why the fractional rule is unreliable in the next activity.

Show answer

x lies between 1.0 and 3.0, so y lies between 1/3.0 = 0.33 and 1/1.0 = 1.0, with a best value of 0.50. The range is lopsided: + 0.50 but only -0.17. The rule assumes a small uncertainty, but 50% is not small, so it cannot describe this range. Numerical substitution can.

Check your understanding 3

Why does the small-uncertainty rule fail to describe y = 1/x for x = 2.0±1.0? Use the central value and limits in your explanation.

Show answer

The best value is 0.50, with limits 0.33 and 1.0: the deviations are about −0.17 and +0.50. The input uncertainty is 50%, so a linear small-change approximation is poor. Numerical substitution exposes the asymmetric range rather than imposing a symmetric allowance.

Check your understanding 4

A learner calculates q = x-x, where both terms are the same measurement x, and gives Δ q = 2Δ x. What is wrong?

Show answer

x-x is exactly zero whatever the value of x, so q has no uncertainty. The rule assumes the two quantities can be wrong in different directions, but one measurement cannot be too large and too small at once. The rules apply to separately measured quantities.

Check your understanding 5

The lab calculates cylinder density from ρ = 4m/(π d²h). Use one reading, a metre rule for the diameter and vernier callipers for the length. Before opening the model, predict how changing only the diameter tool to a micrometer changes the diameter’s contribution to percentage uncertainty.

Diameter contribution after changing to a micrometer
Show answer

The diameter allowance drops from ±1 mm to ±0.01 mm, so its fractional contribution falls. The power remains two, and division still adds a positive worst-case percentage allowance.

Test the input contributions

Select Cylinder density, set Repeat readings to 1, choose the metre rule for diameter and vernier callipers for length. Record the input contributions, then change only the diameter tool to the micrometer. The lab uses the larger of its stated instrument allowance and half the reading range; it does not calculate statistical uncertainty in a mean.

ρ = 4m / (πd²h), so %ρ ≈ %m + 2%d + %h. d readings: 12.7 mm; h readings: 30 mm; m readings: 31.9 g. Here % denotes percentage uncertainty. Each input uses the larger of its stated instrument allowance and half the reading range, not statistical uncertainty in a mean. Uniform-cylinder model with separately varying input allowances; product and power estimates are first-order. The approximate result is ρ = 8.4 ± 0.4 g cm⁻³.

Result
ρ = 8.4 ± 0.4 g cm⁻³
Estimated percentage uncertainty
5.2 %
Input contributions
m: 0.31%; d²: 1.6%; h: 3.3%
Rounded input estimates
d: 12.7 ± 0.1 mm; h: 30 ± 1 mm; m: 31.9 ± 0.1 g
Input allowance model
Larger of stated instrument allowance and half the reading range
Model assumptions
Uniform cylinder; separately varying allowances; first-order estimate

Try this

0 of 4 done

  1. Measure the diameter with a metre rule, then with a micrometer. Watch the d² bar. (not done yet)

  2. Get the percentage uncertainty in the density below 1%. (not done yet)

  3. Take five or more repeat readings of the diameter with the micrometer. (not done yet)

  4. Make the spring's extension large enough that its uncertainty is under 5%. (not done yet)

Check your understanding 6

Record the two sets of contributions. Did the largest contribution change? Explain why a better diameter instrument eventually gives little benefit unless another measurement improves. You may keep your table on paper.

Show answer

Initially, diameter contributes about 15%, far above mass and length (each about 0.3%). With the micrometer, diameter contributes about 0.16%, while length and mass remain about 0.3% each. The total falls from about 16% to about 0.8%; length becomes the largest contribution. Further improvements to diameter leave the mass and length allowances, so those inputs now limit improvement. Compare your table, changed contribution, largest contribution and explanation.

Exam-style question 1

The acceleration of free fall from a pendulum

4 marks

Examination question

A pendulum of length l = (0.800±0.002) m takes (36.0±0.2) s for 20 oscillations. The acceleration of free fall is found from g = 4π²l/T².

(a) Calculate g. [1 mark]

(b) Find the absolute uncertainty in g, and state g with its uncertainty. [3 marks]

Write your working before viewing the mark scheme

Show the steps Optional working
Unit: s
Unit: m s⁻²
Show the steps Optional working
Unit: s
Show the steps Optional working
Unit: s
Unit: m s⁻²
Unit: m s⁻²
Show solution step by step

  1. Calculate g

    1 mark

    Method

    Find the period, then substitute.

    Reason

    The period is the time for one oscillation.

    Working

    Answer

    T = 36.0/20 = 1.80 s, so g = 4π²(0.800)/(1.80)² = 9.75 m s⁻².

  2. Combine the fractional uncertainties

    2 marks

    Method

    Add the fractional uncertainty in l to twice that in T.

    Reason

    g is proportional to l and to T⁻²; 4π² is exact.

    Working

    Answer

    (Δ T)/T = 0.2/36.0 = 0.0056, the same as for the total time. So (Δ g)/g = 0.002/0.800 + 2(0.2/36.0) = 0.0025 + 0.0111 = 0.0136.

  3. State the result

    1 mark

    Working

    Answer

    Δ g = 0.0136 × 9.75 = 0.13 m s⁻², so g = (9.75±0.13) m s⁻². Keep two significant figures in the uncertainty because its first digit is 1; match the value to its decimal place.

Check your understanding 7

Show the complete pendulum calculation on paper or below. Explain why timing 20 oscillations changes the absolute period uncertainty but not its fractional uncertainty, and report g with matching decimal places.

Show answer

T = 36.0/20 = 1.80 s and Δ T = 0.2/20 = 0.01 s, so Δ T/T = 0.2/36.0. The count is exact. g = 4π²(0.800)/(1.80)² = 9.75 m s⁻². Add 0.002/0.800 + 2(0.2/36.0) = 0.0136; multiplying by the unrounded g gives 0.13 m s⁻². Report (9.75±0.13) m s⁻². Compare the period, power factor, fraction-to-absolute step and final decimal places.

Timing 20 oscillations instead of one keeps the same 0.2 s timing uncertainty but spreads it over a time twenty times longer, so the fractional uncertainty in T is twenty times smaller.

Common mistakes

  • Subtracting uncertainties when quantities are subtracted. For separately varying inputs, absolute uncertainties add.
  • Adding absolute uncertainties for a product or quotient. Use fractional uncertainties.
  • Forgetting to multiply by the power: T² doubles the fractional uncertainty in T.
  • Giving an uncertainty to an exact number such as π or 1/2.
  • For the largest value of a quotient, using the largest denominator instead of the smallest.

Before you move on

Check your understanding 8

Two separately measured lengths are a = (4.00±0.10) m and b = (2.00±0.20) m. Find the uncertainty in a-b and estimate the percentage uncertainty in a/b. Then find the largest possible value of a/b by substitution.

Unit: m
Show the steps Optional working
Unit: %
Show answer

For the difference, add absolute uncertainties: a-b = (2.00±0.30) m. For the quotient, add fractional uncertainties: 0.10/4.00 + 0.20/2.00 = 0.125, or 12.5%. Its largest possible value is 4.10/1.80 = 2.28, using the largest numerator and smallest denominator.

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