AQA AS Level Physics Paper 2, June 2025: Question 5

1 mark · Medium difficulty · Multiple Choice

Calculate the new frequency of the first harmonic of a vibrating wire when an additional mass of 2M is added to the hanging mass M.

Practise this question

Question

Diagram showing a horizontal wire vibrating between a fixed support on the left and a frictionless pulley on the right. Over the pulley, the wire hangs vertically supporting a mass hanger with mass M. The text states the fundamental frequency is 420 Hz, and asks for the new frequency when an additional mass of 2M is added, offering four multiple choice options: A (210 Hz), B (590 Hz), C (730 Hz), and D (840 Hz).

Mark scheme

Show the mark scheme Mark scheme table indicating Question 5 has the correct key C, corresponding to the answer 730 Hz, mapped to assessment objective AO2.

How to answer it

First Harmonic Frequency & String Tension

📌 What this question tests

This question evaluates your understanding of stationary waves on strings (AQA Physics topic 3.3.1.2), specifically the mathematical relationship between the fundamental frequency (first harmonic) and string tension ( f ∝ √T ), combined with careful reading of changes in mass ( "additional mass" ).

Question 05 • Multiple Choice (1 Mark)

First Harmonic Frequency with Increased Mass

Assessment Objective: AO2 (Apply knowledge and understanding of scientific concepts)

✅ Correct Answer

Option C: 730 Hz

Mark Scheme: Key C earns 1 mark.

💡 Key Knowledge

  • First Harmonic Formula:
    f = (1 / 2L) × √(T / μ)
  • Where:
    • f = fundamental frequency (Hz)
    • L = vibrating length of wire (m)
    • T = tension in the wire (N) = m_total × g
    • μ = mass per unit length (kg m⁻¹)
  • Since L and μ remain constant:
    f ∝ √T

📐 Step-by-Step Calculation

  1. Determine the initial state:
    Initial mass suspended = M
    Initial tension = T₁ = M × g
    Initial frequency = f₁ = 420 Hz
  2. Determine the final tension:
    An additional mass of 2M is added to M .
    Total new mass = M + 2M = 3M
    New tension = T₂ = 3M × g = 3 × T₁
  3. Apply the proportionality relationship:
    f₂ / f₁ = √(T₂ / T₁)
    f₂ / 420 = √(3T₁ / T₁) = √3
  4. Calculate the new frequency:
    f₂ = 420 × √3 ≈ 420 × 1.73205 = 727.46 Hz
    To 2 significant figures (matching the data provided): f₂ = 730 Hz

🧠 Exam Technique & Strategy

  • Look for keyword triggers: The word "additional" is a classic physics exam trap. It tells you to add to the existing quantity rather than replace it.
  • Scaling factor method: For proportional reasoning questions, you don't need values for length L or mass per unit length μ . Just find the multiplying factor: here tension increases by a factor of 3, so frequency increases by a factor of √3 .
  • Estimate to eliminate quickly: √3 is approximately 1.73. Since 1.73 × 400 ≈ 700 , the answer must be around 700 Hz, making C the obvious choice without detailed calculator work.

❌ Common Errors & Distractor Traps

  • Selecting B (590 Hz): Assuming the new mass is simply 2M instead of 3M .
    420 × √2 ≈ 594 Hz → 590 Hz . A very common slip caused by misreading "additional mass".
  • Selecting D (840 Hz): Forgetting the square root and assuming direct proportionality between f and T ( 420 × 2 = 840 Hz ).
  • Selecting A (210 Hz): Inverting the relationship or halving the frequency ( 420 / 2 = 210 Hz ), mistaking tension with wavelength or length.

Topics

Physics · Required Practicals · 3.3 Waves · AS practicals (1–6)

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.