AQA A-Level Physics Paper 1, June 2022: Question 31

1 mark · Medium difficulty · Multiple Choice

Calculate the maximum speed of a particle performing simple harmonic motion given its time period and amplitude.

Practise this question

Question

Multiple choice question 31 asks for the maximum speed of a particle performing simple harmonic motion with a time period of 1.4 s and an amplitude of 12 mm. Four options are provided: A, 8.6 mm s^{-1}; B, 27 mm s^{-1}; C, 54 mm s^{-1}; D, 110 mm s^{-1}, each with an oval selection box.
Question text

31 A particle performs simple harmonic motion with a time period of 1.4 s and an amplitude of

12 mm.

What is the maximum speed of the particle?

[1 mark]

A 8.6 mm s−1

B 27 mm s−1

C 54 mm s−1

D 110 mm s−1

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer is C (AO2), corresponding to 54 mm s^{-1}.

31 C (AO2) 54 mm s−1

How to answer it

Calculating Maximum Speed in Simple Harmonic Motion

What this question tests

This question assesses your ability to recall and apply the equation relating maximum speed, angular frequency, and amplitude in Simple Harmonic Motion (SHM). It tests unit consistency (converting milliunits) and formula manipulation under multiple-choice exam conditions.

Question 3.1

Exam Breakdown & Solution

✅ Correct Answer

C ( 54 mm s⁻¹ )

Awarded 1 mark for AO2 (Application of knowledge)

💡 Key Knowledge

  • The defining formula for maximum speed in SHM is: v_max = Aω
  • Angular frequency is linked to time period by: ω = 2π / T
  • Combining these gives: v_max = 2πA / T

🧠 Exam Technique

In multiple-choice calculation questions, always write out the formula before substituting values. Check the unit of the final options; since the answers are given in mm s⁻¹ , you can keep the amplitude in mm to save time and avoid unnecessary power-of-ten errors.

📐 Step-by-Step Calculation

  1. Identify given values:
    Amplitude, A = 12 mm
    Time period, T = 1.4 s
  2. Select the correct equation:
    v_max = (2 × π × A) / T
  3. Substitute the values:
    v_max = (2 × π × 12) / 1.4
  4. Evaluate:
    v_max = 75.398... / 1.4 = 53.85... mm s⁻¹
  5. Round appropriately:
    To 2 significant figures, this gives 54 mm s⁻¹ .

❌ Common Errors & Traps

  • Forgetting the factor of 2: Using v = πA / T instead of 2πA / T (often mixing up angular frequency with frequency).
  • Unnecessary conversions: Converting 12 mm to 0.012 m and then forgetting to convert back to mm s⁻¹ , resulting in an answer of 0.054 m s⁻¹ which doesn't match options unless carefully re-scaled.
  • Calculator errors: Forgetting brackets around the numerator (2 * π * 12) when dividing by 1.4 .

Topics

Physics · 3.6 Further mechanics and thermal physics (A-level only)

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.