AQA AS Level Physics Paper 2, June 2025: Question 16

1 mark · Medium difficulty · Multiple Choice

Determine the harmonic number N given the change in distance between adjacent nodes when the string vibrates at the (N+1)th harmonic.

Practise this question

Question

Question 16 presents a scenario where a string vibrates at the Nth harmonic with a separation x between adjacent nodes of 0.60 m. When vibrating at the (N + 1)th harmonic, x decreases by 0.12 m. It asks for the value of N, offering options A (3), B (4), C (5), and D (6).

Mark scheme

Show the mark scheme Mark scheme for question 16 indicates the correct answer is option B, corresponding to a value of 4, assessed under AO2.

How to answer it

Harmonics and Node Separation on a Stretched String

📌 What this question tests

This question assesses your understanding of stationary waves on strings: specifically, the relationship between harmonic number (N), the separation between adjacent nodes (x = λ/2), and the total fixed length of the string (L). It tests your ability to set up simultaneous algebraic expressions for different vibrating modes and solve for an unknown integer.

Question 16 (Multiple Choice)

AQA AS Physics — Waves & Stationary Waves

✅ Correct Answer

B — 4

The harmonic number is N = 4 .

Mark Scheme: Option B (1 mark, AO2 - Applying knowledge of physics).

💡 Key Knowledge

  • On a stationary wave, the distance between adjacent nodes is half a wavelength ( x = λ/2 ), representing one single loop.
  • For a string of fixed length L fixed at both ends, the N th harmonic has N loops.
  • Therefore, total length: L = N × x .
  • As harmonic order increases, node spacing decreases because x = L / N .

📐 Step-by-Step Solution

  1. Define total string length for the Nth harmonic:
    At the N th harmonic, the node separation is x₁ = 0.60 m .
    Since the string consists of N identical loops of length x₁ :
    L = N × 0.60   — (Equation 1)
  2. Find node separation for the (N + 1)th harmonic:
    The node separation decreases by 0.12 m :
    x₂ = 0.60 - 0.12 = 0.48 m
  3. Define total string length for the (N + 1)th harmonic:
    The string length is fixed, so:
    L = (N + 1) × 0.48   — (Equation 2)
  4. Equate and solve for N:
    Since the length L has not changed:
    0.60 N = 0.48(N + 1)
    0.60 N = 0.48 N + 0.48
    0.60 N - 0.48 N = 0.48
    0.12 N = 0.48
    N = 0.48 / 0.12 = 4

🧠 Exam Technique: Fast Elimination

In multiple-choice questions, you can test the options directly if setting up algebra feels slow:

  • If N = 3 : L = 3 × 0.60 = 1.80 m . For N = 4 : x = 1.80 / 4 = 0.45 m (decrease = 0.15 m ≠ 0.12 m).
  • If N = 4 : L = 4 × 0.60 = 2.40 m . For N = 5 : x = 2.40 / 5 = 0.48 m (decrease = 0.60 - 0.48 = 0.12 m). Matches perfectly!

❌ Common Traps & Misconceptions

  • Confusing node separation with full wavelength: Thinking x = λ instead of x = λ/2 . While this gives the same numerical value for N here due to cancellation, it causes errors in structured questions requiring string length or speed.
  • Misreading "decreases by": Setting the new node spacing as 0.12 m instead of subtracting it from 0.60 m ( 0.60 - 0.12 = 0.48 m ).
  • Confusing harmonic number with number of nodes: An N th harmonic has N loops and (N + 1) nodes. Remember that the question asks for N (the harmonic order), not total node count.

Topics

Physics · 3.3 Waves

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.