AQA A-Level Physics Paper 1, June 2025: Question 3

5 marks · Hard difficulty · Practical Techniques & Data Analysis

Draw an acceleration-displacement graph for a mass-spring system undergoing SHM and identify the point where kinetic energy is halved.

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Question

Question 03 shows text stating that a mass–spring system is undergoing simple harmonic motion with maximum velocity 0.39 m s⁻¹ and period 0.81 s. Part 03.1 asks to draw a graph of acceleration against displacement on Figure 4, including suitable units and scales on both axes for 4 marks. Figure 4 is an empty grid of square graph paper with major and minor gridlines. Part 03.2 asks to label with a 'P' a point on the graph where the mass has 50% of its maximum kinetic energy for 1 mark.
Question text

03 A mass–spring system is undergoing simple harmonic motion (SHM).

The maximum velocity of the mass is 0.39 m s–1 and the period of oscillation is 0.81 s.

03.1 Draw, on Figure 4, a graph of acceleration against displacement for the system.

Label each axis with a suitable unit and scale.

[4 marks]

Figure 4

03.2 Label, with a P, a point on your graph where the mass has 50% of its maximum

kinetic energy.

[1 mark]

Mark scheme

Show the mark scheme Mark scheme for 03.1: 1st mark for finding omega = 2pi/T or amplitude = 0.05 m; 2nd mark for using a = -omega^2 x or maximum acceleration = 3.0 m s⁻²; 3rd mark for a straight line through the origin with negative gradient through all four quadrants; 4th mark for line ending at (0.05, -3.0) and (-0.05, 3.0). Diagram shows axes labeled acceleration / m s⁻² and displacement / m with a line passing from top-left to bottom-right. Mark scheme for 03.2: 1 mark for placing P at (maximum displacement)/sqrt(2), approximately 3.54 x 10⁻² m and 2.1 m s⁻².

Question Answers Additional comments/Guidance Mark AO

03.1 2π Expect to see: 4 2 × AO1

Evidence of ω = OR 0.05 m seen

T 2 × AO2

OR

V 1

A =

their ω

Use of a = −ω2x OR 3.0 m s-2 seen

Straight line through origin with negative gradient featuring

all four quadrants 3

Condone axes either way round

Single straight line ends at (0.05,-3.0)

OR Award 4 for correct graph with correct values

Single straight line ends at (0.05, 3.0) 4

Evidence for 1 and 2 can be seen on the

graph.

In 3 (0,0) must be in centre of line.

4 there is no ecf from 1 or 2 .

03.2 Accept on either quadrant 1 AO1

their max displacement

P at (judge by eye)

�2

Expect to see 3.54 x 10-2 m AND 2.1 m s-2

Total 5

How to answer it

SHM Graphs: Acceleration-Displacement & Kinetic Energy

WHAT THIS QUESTION TESTS

This question assesses your ability to link the mathematical definitions of Simple Harmonic Motion (SHM) to graphical representations and energy relationships:

  • Calculating angular frequency ( ω ), amplitude ( A ), and maximum acceleration ( amax ) from period ( T ) and maximum velocity ( vmax ).
  • Understanding the fundamental defining equation: a = −ω²x and plotting it accurately across all four quadrants.
  • Applying quadratic energy considerations in SHM: calculating displacement when kinetic energy is at a specified fraction of its maximum value.
QUESTION 03.1 [4 MARKS]

Acceleration against Displacement Graph

Plotting the characteristic SHM line with proper axis scales and units

📐 Step-by-Step Calculation

Step 1: Calculate angular frequency (ω)
ω = 2π / T = 2π / 0.81 s = 7.757 rad s⁻¹
Step 2: Find the amplitude (maximum displacement, A)
Using vmax = ωA :
A = vmax / ω = 0.39 / 7.757 = 0.0503 m ≈ 0.05 m
Step 3: Calculate maximum acceleration (amax)
Using amax = ω²A = ω × vmax :
amax = 7.757 × 0.39 = 3.03 m s⁻² ≈ 3.0 m s⁻²
Step 4: Determine gradient
Gradient = −ω² = −(7.757)² ≈ −60.2 s⁻²

✅ Correct Graph Specification

  • Axes & Units:
    • Horizontal: displacement / m (or x / m )
    • Vertical: acceleration / m s⁻² (or a / m s⁻² )
  • Scales: Origin (0, 0) must be in the exact centre of the grid. Linear scales showing up to at least ±0.05 m and ±3.0 m s⁻² .
  • Graph Line:
    • A single continuous, ruler-drawn straight line passing directly through (0, 0) .
    • Negative gradient spanning across all 4 quadrants.
    • Ends cleanly at coordinates (+0.05, −3.0) and (−0.05, +3.0) .

💡 Key Knowledge

The defining condition for SHM is:

a ∝ −x  ⇒  a = −ω²x

  • Acceleration is directly proportional to displacement and always directed towards the equilibrium position (hence the negative sign).
  • On an a–x graph, this always yields a straight line through the origin with a negative gradient equal to −ω² .

❌ Common Errors & Examiner Traps

  • Positive gradient: Plotting a line with a positive slope ( a = +ω²x ) loses the third mark.
  • Missing units: Writing just "displacement" or "acceleration" on the axes without m or m s⁻² .
  • Line not centred: Drawing the line only in the positive quadrant, or not having (0,0) at the midpoint of the line.
  • Extending past limits: Drawing arrows or extending the line beyond the amplitudes ( ±0.05 m ). In SHM, the object cannot exist beyond amplitude A !
Mark Breakdown (4 Marks):
• 1✓: Calculating ω = 2π / T OR finding A = 0.05 m (or A = V / ω ).
• 2✓: Using a = −ω²x OR calculating amax = 3.0 m s⁻² .
• 3✓: Straight line through origin with negative gradient featuring in all four quadrants with (0, 0) at the centre.
• 4✓: Line correctly terminates at (0.05, −3.0) and (−0.05, +3.0) (no ECF allowed for this mark).
QUESTION 03.2 [1 MARK]

Locating 50% Maximum Kinetic Energy

Determining displacement from quadratic energy relationships

📐 Step-by-Step Derivation

Step 1: Write the kinetic energy expression
Ek = ½mω²(A² − x²)
Maximum kinetic energy occurs at x = 0 :
Ek,max = ½mω²A²
Step 2: Set Ek to 50% of Ek,max
½mω²(A² − x²) = 0.5 × [½mω²A²]
Cancel common factors ½mω² :
A² − x² = 0.5 A²
x² = 0.5 A² = A² / 2
Step 3: Solve for x and corresponding a
x = A / √2 = 0.0503 / 1.414 = 0.0354 m (3.54 × 10⁻² m)
Corresponding acceleration:
|a| = amax / √2 = 3.03 / 1.414 = 2.14 m s⁻² ≈ 2.1 m s⁻²

🧠 Exam Technique & What to Mark

  • Mark point P clearly with a dot and cross or letter P on the line drawn in 03.1.
  • Acceptable on either quadrant:
    • At x ≈ +0.035 m , a ≈ −2.1 m s⁻²
    • OR at x ≈ −0.035 m , a ≈ +2.1 m s⁻²
  • Notice that 1 / √2 ≈ 0.707 , meaning point P is located at roughly 71% of the maximum displacement, well over halfway along the axis!

❌ The #1 Discriminator Mistake: The "Linear" Energy Trap

A very common misconception is assuming that 50% kinetic energy occurs at half the maximum displacement ( x = 0.5 A = 0.025 m ).

Why this is wrong: Energy in SHM depends on the square of displacement ( x² ), not displacement linearly. If x = 0.5 A , then x² = 0.25 A² , which means potential energy is 25% and kinetic energy is actually 75% of maximum!

Mark Breakdown (1 Mark):
• 1✓: Label P placed on the line at (their max displacement) / √2 (judged by eye). Expected values: x = ±3.54 × 10⁻² m and a = ∓2.1 m s⁻² .

Topics

Physics · Practical skills · 3.6 Further mechanics and thermal physics (A-level only) · Data analysis

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.