Free Pearson Edexcel International A Level Physics lesson
Graphs, logarithms and uncertainty
Extract constants without concealing the limits of the data.
What you’ll learn
- Linearise power and exponential relations.
- Use best-fit gradients and intercepts.
- Combine measurement uncertainties and evaluate agreement.
A useful graph separates gradient from intercept
Choose transformed axes from the model before plotting. If T² = (4π²/k)m, plotting T² vertically against mass horizontally should give a straight line with gradient 4π²/k. A nonzero intercept can reveal an effective oscillating mass or an offset; forcing every line through the origin discards this information.
Use clearly labelled quantities and units with a scale that fills much of the available graph and is easy to read. Draw a best-fit line consistent with the data rather than joining every point. For its gradient use two widely separated points on the fitted line, not necessarily two measured points. A large triangle reduces the fractional influence of reading its coordinates.
Inspect residuals as well as the appearance of a line. A systematic curve can reveal model failure or a missing correction even when correlation is high. Give constants to precision supported by the measurements and state whether an intercept is compatible with zero within uncertainty. A visually straight trend alone cannot establish exact proportionality.
y = mx + c; gradient m = Δy/Δx; for T² versus mass, k = 4π²/mgradient
Symbols and units
- mgradient is a fitted slope, not object mass
- T is oscillation period in s
- M is oscillating mass in kg
- k is spring constant in N m⁻¹.
How the equation explains the idea
- For a mass–spring system, T = 2π√(M/k). Squaring gives T² = (4π²/k)M, so the slope of T² against M is 4π²/k.
- Invert the slope relation to obtain k = 4π²/mgradient. An additional effective spring mass can create an intercept while preserving the ideal slope; do not confuse intercept with spring constant.
Conditions. Ideal spring with stiffness k = 40 N m⁻¹ and effective extra mass 0.050 kg. T² = (4π²/k)(m + 0.050 kg); the plotted zero attached mass is a model extrapolation.
Attached mass / kg; Period squared / s².
The slope is π²/10 ≈ 0.987 s² kg⁻¹. The intercept is about 0.0493 s²; dividing intercept by slope gives 0.050 kg. Ignoring the intercept biases estimates based on a single T²/m ratio.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
A spring from its graph
- A fitted T²–mass line passes through (0.10 kg, 0.20 s²) and (0.50 kg, 1.00 s²).
- Gradient = (1.00 − 0.20)/(0.50 − 0.10) = 2.0 s² kg⁻¹.
- k = 4π²/2.0 = 19.7 N m⁻¹.
The graph gradient is not the spring constant; the model relates them reciprocally.
Keep this idea. Derive the axis choice and gradient meaning from the model before calculating a constant.
Distinguish a power law from an exponential
For a power law y = Axⁿ, taking logs gives ln(y/yref) = n ln(x/xref) plus a constant. A log–log graph therefore has gradient n. Pressure against volume at fixed gas amount and temperature should give exponent −1. Logarithm arguments must be dimensionless; writing reference units explicitly keeps the transformation meaningful.
An exponential y = y₀e^(−kt) instead gives ln(y/yref) = ln(y₀/yref) − kt. Plot log y against untransformed time to obtain slope −k. A log–log plot and a semi-log plot answer different model questions. Changing consistently from natural logs to base-10 leaves a power exponent unchanged, but an exponential slope needs the corresponding conversion factor.
Do not take a logarithm of zero or a negative corrected reading. An additive background must be estimated and subtracted before testing an exponential signal, and uncertainty grows relatively large when the corrected value approaches zero. Inspect the transformed residuals and retain the original data so that a visually neat straight line cannot hide an inappropriate correction.
y = Axⁿ ⇒ log y = n log x + constant; y = y₀e^(−kt) ⇒ ln(y/yref) = constant − kt
Symbols and units
- n is dimensionless power-law exponent
- k is exponential decay constant in s⁻¹ when time t is in s
- xref and yref are positive fixed reference quantities in the same units as x and y.
How the equation explains the idea
- For y = Axⁿ, logarithms of dimensionless ratios give log(y/yref) = n log(x/xref) + constant. A log–log slope therefore gives n.
- For y = y₀ exp(−kt), ln(y/yref) = ln(y₀/yref) − kt. A natural-log-versus-time slope gives −k. A base-10 logarithm would instead give −k/ln 10.
Conditions. Two dimensionless-ratio power laws, y/yref = (x/xref)² and y/yref = (x/xref)⁻¹. Only positive original x and y values are represented; logarithmic coordinates can be negative.
ln(x / xref) / dimensionless; ln(y / yref) / dimensionless.
The slopes are +2 and −1. A negative log coordinate means the original value is below its reference, not negative. An exponential decay instead becomes linear on log-y versus ordinary time axes.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
Two straight-line interpretations
- On a log–log pressure–volume plot, Δln p = −0.693 and Δln V = +0.693 give slope −1.
- On a log-voltage–time plot, a slope of −0.20 s⁻¹ gives RC = 5.0 s.
- The first gradient is dimensionless; the second has units s⁻¹.
Read the axes before naming the physical meaning of a logarithmic slope.
Keep this idea. Use log–log for a power exponent and log-y against time for an exponential rate.
Carry uncertainty through a result
State how each measurement uncertainty is estimated. For the specification’s single-reading convention, use half the instrument resolution; for repeated readings a half-range can describe the observed spread. Other known effects, such as reaction time or calibration, may dominate those simple estimates. A resolution contribution is not a guarantee of total accuracy.
For a conservative small-uncertainty estimate, add absolute uncertainties for sums and differences, and add percentage uncertainties for products and quotients. Powers multiply the percentage contribution by the magnitude of the power. Thus for g = 4π²l/T², the approximate percentage uncertainty is percentage uncertainty in l plus twice that in T. Correlated offsets may cancel in a difference and require physical thought.
State the result and absolute uncertainty to compatible decimal places, then compare overlapping plausible ranges or a difference with its uncertainty. A small discrepancy inside uncertainty does not prove exact agreement; a large one prompts inspection of method and assumptions. To improve the result, identify the largest contribution and target it instead of simply adding decimal places to every measurement.
For z = xᵃyᵇ: δz/|z| ≈ |a|δx/|x| + |b|δy/|y|; half-range = (maximum − minimum)/2
Symbols and units
- δx, δy and δz are positive absolute uncertainties
- a and b are dimensionless powers in z = xᵃyᵇ
- fractions use magnitudes in denominators.
How the equation explains the idea
- For small uncertainties in a worst-case linear approximation, fractional contributions add with absolute exponents: δz/|z| ≈ |a|δx/|x| + |b|δy/|y|.
- For z = x²/y with 2% in x and 3% in y, the estimate is 2 × 2% + 3% = 7%. This conservative rule differs from root-sum-square propagation of independent statistical standard uncertainties.
Worked example
A pendulum estimate
- Length uncertainty is 0.5% and period uncertainty is 0.8%.
- For g ∝ l/T², percentage uncertainty is 0.5 + 2 × 0.8 = 2.1%.
- At g = 9.7 m s⁻², absolute uncertainty is about 0.20 m s⁻².
Report approximately 9.7 ± 0.2 m s⁻² under this conservative model.
Keep this idea. Propagate the measurement uncertainties using the powers in the physical equation.
Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.
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