Uncertainties and logarithmic graphs (A2) flashcards

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Absolute uncertainty
The uncertainty in the same unit as the quantity, e.g. 27.5 °C ± 1.0 °C.
Percentage difference
|experimental value − accepted value| ÷ accepted value × 100%. A measure of accuracy.
Error bar
A line through a plotted point showing ± the uncertainty in that value (vertical for y, horizontal for x).
Steepest and shallowest lines
The steepest and least steep straight lines that still pass through every error bar; used to find the uncertainty in a gradient or intercept.
Natural logarithm (ln)
Logarithm to base e ≈ 2.718: ln eˣ = x. Used to linearise exponential relationships.
Base-10 logarithm (log or lg)
Logarithm to base 10: log 10ˣ = x. Either base can be used for power laws, as long as both axes use the same one.
Agreement within uncertainty
A result agrees with an accepted value when the accepted value lies inside value ± uncertainty, i.e. percentage difference < percentage uncertainty.
Uncertainty of a sum or difference
Δy = Δa + Δb for y = a ± b — absolute uncertainties add, even for a subtraction. (No — learn it)
Uncertainty of a product or quotient
%y = %a + %b for y = ab or y = a/b — percentage uncertainties add. (No — learn it)
Uncertainty of a power
%(aⁿ) = n × %a. A square doubles it, a cube triples it, a square root halves it. (No — learn it)
Uncertainty in a gradient
≈ (steepest gradient − shallowest gradient) ÷ 2; percentage = that ÷ best gradient × 100%. (No — learn it)
Log form of y = A e^(kx)
ln y = ln A + kx. Gradient k, intercept ln A. (No — derive it)
Log form of y = k xⁿ
log y = n log x + log k. Gradient n, intercept log k (at log x = 0). (No — derive it)
Linearise V = V₀ e^(−t/RC)
Plot ln V against t: gradient −1/RC, intercept ln V₀.
Linearise A = A₀ e^(−λt)
Plot ln A (background-corrected) against t: gradient −λ, intercept ln A₀; half-life = ln 2 ÷ λ.
Linearise I = k dⁿ to find n
Plot log I against log d: gradient n (no unit); n ≈ −2 for an inverse-square law.
Where to draw steepest and shallowest lines
Usually from the bottom of the first error bar to the top of the last, and from the top of the first to the bottom of the last — checking they pass through every bar.
Axis label for a log
Log of a pure number: ln (V / V), log (T / s). The log itself has no unit.
Effect of a long extrapolation on an intercept
Small differences between the steepest and shallowest gradients become large differences in the intercept, so its uncertainty is large.
Why does a small difference of two readings have a large percentage uncertainty?
The absolute uncertainties add, but the result is small, so the same uncertainty is a large fraction of it.
Which measurement should you improve first?
The one contributing the largest percentage uncertainty — usually the one with the highest power (e.g. a squared diameter).
Percentage difference > percentage uncertainty — what does it suggest?
The result does not agree with the accepted value; a systematic error (or an unconsidered uncertainty) is likely.
Why use a log–log graph rather than y against x⁴?
y against x⁴ assumes the power; log y against log x finds the power as the gradient and tests it.
Why subtract room temperature or background before taking logs?
Only the excess (or corrected count) decays exponentially towards zero; without the correction the log graph curves.
Why use ln rather than log for an exponential?
With ln the gradient is k itself; with base-10 logs the gradient is 0.434k.
a = 2.0 ± 0.1, b = 5.0 ± 0.2. Percentage uncertainty in a/b?
5% + 4% = 9% — quotients add percentages.
Temperatures 18.5 °C and 46.0 °C, each ± 0.5 °C. Uncertainty in the rise?
27.5 °C ± 1.0 °C, i.e. 3.6%.
Diameter 0.28 mm ± 0.01 mm. Percentage uncertainty in the cross-sectional area?
0.01 ÷ 0.28 × 100 = 3.57%; × 2 = 7.1% (area ∝ d²).
g = 9.73 m s⁻² ± 1.4%. Does it agree with 9.81 m s⁻²?
Percentage difference 0.82% < 1.4%: yes, it agrees within the uncertainty.
Steepest gradient 23.0 μF, shallowest 21.4 μF. Uncertainty in C?
(23.0 − 21.4) ÷ 2 = ± 0.8 μF.
ln V against t passes through (0 s, 2.19) and (120 s, −0.45). Time constant?
Gradient = −2.64 ÷ 120 = −0.0220 s⁻¹, so RC = 1 ÷ 0.0220 = 45.5 s.
log T against log r passes through (8.10, 4.579) and (8.85, 5.705). Power n?
n = 1.126 ÷ 0.75 = 1.50.
Write ln 2.85 and ln 76 to the correct precision.
ln 2.85 = 1.047 (3 s.f. → 3 d.p.); ln 76 = 4.33 (2 s.f. → 2 d.p.).
Quoting a final value with its uncertainty
Uncertainty to 1 (or 2) s.f.; value rounded to the same decimal place: (4.6 ± 0.4) × 10⁻⁷ Ω m, not 4.64 ± 0.4.
Intercept of a log–log graph
log k is the value at log x = 0, not where the line meets an axis drawn at log x = 8.0. Use log k = log y − n log x.
Units of a gradient on a ln graph
ln y against t: unit s⁻¹ (1 / unit of x). log y against log x: no unit.
Finishing an agreement question
State both percentages, compare them in one sentence, and conclude — agrees, or does not with a likely systematic error.
Exact numbers in uncertainty calculations
Constants like 2, π, ½ and a count of 20 oscillations carry no uncertainty and add nothing to the percentage.

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