A student plots a graph of extension x against force F and draws a best-fit line
Measurements & Uncertainties - OCR A-Level Physics
Key Definition
Worst acceptable lines
The two lines on a graph that show the steepest and shallowest gradients still consistent with the error barsLines added to plotted points to show the absolute uncertainty in each reading.. The spread between them sets the uncertainty in the gradient and in the intercept of the line of best fit.
The two lines on a graph that show the steepest and shallowest gradients still consistent with the error barsLines added to plotted points to show the absolute uncertainty in each reading.. The spread between them sets the uncertainty in the gradient and in the intercept of the line of best fit.
Diagram pending
Graph of extension x against force F. Five plotted points with vertical error bars. Three straight lines through the data: solid line of best fit (gradient 0.0125 m N⁻¹), dashed steepest acceptable line (0.0138 m N⁻¹), dashed shallowest acceptable line (0.0112 m N⁻¹). The worst lines just touch the corners of the error-bar boxes.
Will be replaced with a GeoGebra SVG in stream 2.
Worked example. A student plots extensionThe increase in length of a spring or wire when a force is applied. Measured in metres (m). $x$ against force $F$ for a spring and draws a line of best fit. The gradients of the best-fit and worst acceptable lines are below. Find the percentage uncertainty in the gradient, then in the spring constant $k$.
- Best-fit gradient $= 0.0125 \text{ m N}^{-1}$.
- Steepest acceptable line gradient $= 0.0138 \text{ m N}^{-1}$.
- Shallowest acceptable line gradient $= 0.0112 \text{ m N}^{-1}$.
$$\Delta(\text{gradient}) = \frac{0.0138 - 0.0112}{2} = 0.0013 \text{ m N}^{-1}$$
$$\% \text{ uncertainty in gradient} = \frac{0.0013}{0.0125} \times 100\% = 10.4\%$$
- The gradient is $\frac{1}{k}$, the reciprocal of the spring constantThe force per unit extension of a spring. A measure of its stiffness. Measured in N m⁻¹..
- So $k = \frac{1}{0.0125} = 80 \text{ N m}^{-1}$, with the same percentage uncertainty of 10.4%.
- Absolute uncertaintyThe uncertainty in the same units as the measurement, written as ± a value. in $k = \frac{10.4}{100} \times 80 = 8.3 \text{ N m}^{-1}$.
- Final answer: $k = 80 \pm 8 \text{ N m}^{-1}$. The uncertainty is rounded to 1 s.f. and the result to match.
Common Mistake
MEDIUM
Wrong: Subtracting the two worst gradients without dividing by 2, giving an uncertainty twice the correct size. Or quoting the spring constant as $80.0 \pm 8.3 \text{ N m}^{-1}$ with mismatched significant figures.
Right: $\Delta(\text{gradient}) = \frac{\text{steepest} - \text{shallowest}}{2}$. The uncertainty rounds to 1 s.f., then the value rounds to match the last meaningful digit of the uncertainty.
Right: $\Delta(\text{gradient}) = \frac{\text{steepest} - \text{shallowest}}{2}$. The uncertainty rounds to 1 s.f., then the value rounds to match the last meaningful digit of the uncertainty.
Examiner Tips and Tricks
- If the gradient is the reciprocal of the quantity you want (here $\frac{1}{k}$), the percentage uncertainty carries over unchanged. Multiply by the value, not the percentage, to get the absolute uncertainty.
- OCR practical papers expect the working line by line: gradient, $\Delta$gradient, percentage uncertainty, value, absolute uncertainty, final answer with units.