Lesson 5 of 5
Paper 4: planning and data handling
I can distinguish random and systematic errors, including zero error, and relate them to precision and accuracy.Syllabus
Syllabus 9478, 1(f). Distinguish random and systematic errors, including zero error, and relate them to precision and accuracy.
I can assess derived uncertainty by adding absolute or relative uncertainties, or by numerical substitution.Syllabus
Syllabus 9478, 1(g). Assess derived uncertainty by adding absolute or relative uncertainties, or by numerical substitution.
Make a guess. You are not marked.
A student plots T2 against l for a simple pendulum and gets a straight line through the origin. What does the gradient give?
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4 pi2 / g, so g is 4 pi2 divided by the gradient.
Squaring T = 2 pi sqrt(l / g) gives T2 = (4 pi2 / g) l. Comparing with y = mx, the gradient is 4 pi2 / g.
Paper 4, the practical paper, is 20% of your H2 grade. Planning is worth 4% and the rest rewards measuring, presenting data and analysing it, partly in spreadsheet software. The same few methods appear every year.
1. Turn the relationship into a straight line
A straight line is easy to test and its gradient and intercept give you constants. Rearrange the equation into the form y = mx + c. Then plot y against x.
| Equation | Plot | Gradient | Intercept |
|---|---|---|---|
| T = 2π√(l/g) | T2 against l | 4π2/g | 0 |
| V = E - Ir | V against I | -r | E |
| y = kxn | lg y against lg x | n | lg k |
| I = I0e-μx | ln I against x | -μ | ln I0 |
Use lg against lg for a power law with an unknown power n. Use ln against x for an exponential. A logarithm has no unit, so label the axis with the quantity divided by its unit, such as lg (T / s).
2. Read the gradient and intercept properly
- Use a large triangle that covers more than half the line. Read both corners from the line and write their coordinates in your working.
- Check where your x-axis starts. If it does not start at zero, the line does not cross the y-axis on your grid. Find c by putting one point on the line into c = y - mx.
- Give the gradient a unit where it has one: the y-unit divided by the x-unit.
Worked analysis
Find the power n in T = kmn
The period T of a mass m oscillating on a spring was measured for six masses. Each T is the mean time for 20 oscillations divided by 20.
| m / kg | T / s | lg (m / kg) | lg (T / s) |
|---|---|---|---|
| 0.100 | 0.398 | -1.000 | -0.400 |
| 0.200 | 0.561 | -0.699 | -0.251 |
| 0.300 | 0.690 | -0.523 | -0.161 |
| 0.400 | 0.793 | -0.398 | -0.101 |
| 0.500 | 0.890 | -0.301 | -0.051 |
| 0.600 | 0.972 | -0.222 | -0.012 |
Taking lg of both sides gives lg T = n lg m + lg k. So plot lg T against lg m. The points lie on a straight line, which confirms a power law.
Best-fit line: lg T = 0.499 lg m + 0.099
n = gradient = 0.499 ≈ 0.50
lg k = 0.099, so k = 100.099 = 1.26 s kg-0.5
n = 0.50 means T is proportional to √m. For a spring, T = 2π√(m/ks), so 2π/√ks = 1.26 and the spring constant ks = (2π/1.26)2 = 25 N m-1.
3. Find and quote the uncertainty
For repeated readings, take the uncertainty as half the range. Three timings of 20 oscillations, 15.84 s, 15.92 s and 15.88 s, have a mean of 15.88 s and an uncertainty of (15.92 - 15.84)/2 = 0.04 s. So T = 15.88/20 = (0.794 ± 0.002) s.
On a graph with error bars, draw the best-fit line and a worst acceptable line: the steepest or shallowest line that still passes through all the error bars. The uncertainty in the gradient is the difference between the two gradients. Quote the uncertainty to 1 significant figure, and round the value to the same decimal place.
Then use the uncertainty rules to carry the uncertainty into any quantity you calculate from it.
4. Expect to use a spreadsheet
In Paper 4 you must process data in spreadsheet software. Practise each of these until it is quick:
- Import an xlsx data file, or type in your own readings.
- Write a formula once and copy it down a column, such as =LOG10(A2) or =LN(B2).
- Convert degrees to radians before using SIN, COS or TAN.
- Plot a labelled graph, set both axis scales, add a linear trendline and display its equation.
- Find an area under a curve by adding trapezium strips, and a gradient at a point from Δy/Δx over a small interval.
Spreadsheet tasks in other chapters are labelled Paper 4 practice. Try the motion data workbook first, because it explains importing and copying formulas step by step.
5. Write a plan that earns every mark
A planning question asks you to describe an experiment you do not carry out. Use these headings:
- Variables: the independent variable you change, the dependent variable you measure, and the controlled variables with how you keep each constant.
- Diagram: a labelled diagram of a workable set-up.
- Procedure: numbered steps that say how and with what you measure each quantity.
- Analysis: the graph you plot, why it should be straight, and how you get the answer from its gradient or intercept.
- Risks: a real hazard of this set-up and the precaution that reduces it.
- Reliability: detail that improves accuracy, such as a fiducial marker, repeats or a check for zero error.
Worked plan
Test whether T = kmn for a mass on a spring
Variables. Independent: the mass m on the spring. Dependent: the period T. Controlled: use the same spring throughout, and release each mass from the same small displacement.
Set-up. Clamp the spring to a stand, and clamp the base of the stand to the bench. Hang a mass holder from the spring. Fix a pointer (fiducial marker) level with the equilibrium position of the mass.
Procedure. Measure m on a balance. Pull the mass down by about 2 cm and release it. Start the stopwatch as the mass passes the marker, then time 20 oscillations. Repeat the timing and find the mean, then T = mean time / 20. Repeat for at least six masses from 0.100 kg to 0.600 kg.
Analysis. Plot lg (T / s) against lg (m / kg). If the line is straight, the relationship is a power law. Its gradient is n and its y-intercept is lg k.
Risks. A falling mass or a toppling stand could injure your feet. Clamp the stand to the bench and do not stretch the spring beyond its elastic limit.
Reliability. Timing 20 oscillations makes your reaction time a small fraction of the total. Timing from the marker, where the mass moves fastest, makes the start and stop clearer.
Try it step by step
Work it out in steps: Choose the graph that gives n as its gradient, then Gradient n, then Intercept lg k.
Show the answer
Problem 1.
Choose the graph that gives n as its gradient: Taking logs: lg T = n lg m + lg k, so plot lg T against lg m. The gradient is n and the intercept is lg k.
Gradient n: n = (0.10 - (-0.20)) / (-0.40 - (-1.00)) = 0.30 / 0.60 = 0.50
Intercept lg k: At lg m = 0: lg k = -0.20 + 0.50 x 1.00 = 0.30, so k = 100.30 = 2.0
Problem 2.
Choose the graph that gives n as its gradient: Plot lg T against lg m.
Gradient n: n = (0.40 - 0.10) / (0.00 - (-1.00)) = 0.30
Intercept lg k: The second point has lg m = 0, so lg k = 0.40
Problem 3.
Choose the graph that gives n as its gradient: lg T against lg m, because lg T = n lg m + lg k is a straight line with gradient n.
Gradient n: n = (0.75 - 0.25) / (0.50 - (-0.50)) = 0.50
Intercept lg k: At lg m = 0, halfway between the points: lg k = 0.50
Check your understanding
A student suspects that T = km^n, where k and n are unknown constants. Which graph lets her find n from its gradient?
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lg T against lg m
Taking lg of both sides gives lg T = n lg m + lg k. This has the form y = mx + c, so the gradient is n and the y-intercept is lg k.