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Planning from a given relationship

Edexcel IAL Physics · WPH16 Practical Skills in Physics II

Unit 6 planning questions usually give you an equation and ask how to test it or find a constant. Start from the formula and work outwards.

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Start from the formula

1. Identify the quantity you can change (independent) and the one you measure (dependent).

2. Link them in one clear sentence using the formula.

3. Decide what to plot so the graph is a straight line, and what the gradient or intercept gives.

4. State exactly how many measurements to take and over what range, and whether readings are repeats for a mean or extra values for the graph.

5. Add techniques for each measurement and the control variables.

Exponential relationships

If a quantity decays exponentially, plot its natural log against the other variable. For damped oscillations, the amplitude after n oscillations of period T is

A = A_0 e^-λnT, so ln A = -λT n + ln A_0

Plot ln A against n: gradient = -λT. Examiners did not accept log A (base 10) when the question used e.

Reading a graph for a 'half' value

To find when the amplitude has halved, use ln(A_0/2) = ln A_0 - ln 2 on the ln A axis, and read off n. Do not halve the value of ln A_0. Then time = nT. Do not try to time 'until it halves' directly: n need not be a whole number.

Worked example

For damped oscillations of period 0.80 s, a graph of ln A against n has gradient -0.0462. Find λ and the time for the amplitude to halve.

Gradient = -λT, so λ = 0.0462 / 0.80 = 0.0578 s^-1
Halving: ln 2 = 0.0462 n, so n = 15.0 oscillations
t = nT = 15.0 x 0.80 = 12.0 s

Answer: λ = 0.0578 s^-1; the amplitude halves after about 12 s

Key idea: Start from the formula: what you change, what you measure, the straight-line graph and what its gradient gives, how many readings over what range. For exponentials, plot ln against the variable.

Check yourself

A graph of ln A against n for damped oscillations of period 1.20 s has gradient -0.0350. Find λ in s^-1, to 3 s.f.
Show answer
0.0292 s^-1. λ = 0.0350 / 1.20 = 0.0292 s^-1.
For that graph (gradient -0.0350 per oscillation), after how many oscillations has the amplitude halved? Give 3 s.f.
Show answer
19.8 oscillations. n = ln 2 / 0.0350 = 19.8.
Using n = 19.8 oscillations and T = 1.20 s, find the time for the amplitude to halve, in s, to 3 s.f.
Show answer
23.8 s. t = nT = 19.8 x 1.20 = 23.8 s.
The initial amplitude is 12.0 cm. What value of ln(A / cm) should you look for on the graph when the amplitude has halved? Give 2 d.p.
Show answer
1.79. ln 6.0 = 1.792, so 1.79. Not half of ln 12.0 (which would be 1.24).
A quantity is predicted to follow y = y_0 e^-kx. Which graph tests this?
- y against x, a straight line
- log y against log x
- ln y against x, a straight line of gradient -k
- y against e^x
Show answer
ln y against x, a straight line of gradient -k. Taking natural logs gives ln y = -kx + ln y_0.
Which statement about readings would gain credit in a plan?
- Record the amplitude after every 5 oscillations until n = 40, giving 8 readings
- Take many readings
- Take readings until the oscillation stops
- Take a few readings for a mean
Show answer
Record the amplitude after every 5 oscillations until n = 40, giving 8 readings. It gives a specific number and range for the graph.
A plan for damped oscillations spends most of its words on measuring the spring's extension. Why is this weak?
- Extension is too hard to measure
- It is a safety hazard
- Extension is not a variable in the formula being tested
- Extension is always constant
Show answer
Extension is not a variable in the formula being tested. Examiners penalised plans that focused on quantities the formula does not use.

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Last updated 8 November 2026. Spotted a mistake? Tell us.