AQA A-Level Physics Paper 1, November 2020: Question 18

Stationary Waves on a Fixed Rope

What this question tests

This question tests your understanding of the properties of stationary (standing) waves on strings or ropes fixed at both ends. You need to recall and apply relationships between string length ($L$), wavelength ($\lambda$), harmonics, and the positions of nodes and antinodes.

Question 18 - Multiple Choice

Exam Breakdown & Analysis

✅ Correct Answer: D

Statement D is not correct, making it the right choice to select. For the 5th harmonic, there are actually 6 nodes (including the two fixed ends at the boundaries).

💡 Key Knowledge

First harmonic ($n=1$): $\lambda = 2L$. Nodes = 2, Antinodes = 1.
Nodes ($n$th harmonic): There are always $n + 1$ nodes for a string fixed at both ends.
Antinodes ($n$th harmonic): There are $n$ antinodes.
Midpoint behavior: Even-numbered harmonics always have a node at the exact midpoint, whereas odd-numbered harmonics have an antinode at the midpoint.

🧠 Exam Technique

When faced with a "which statement is not correct" multiple-choice question, evaluate statements systematically from A to D. Do not stop at the first one that looks plausible—verify each option until you find the definitive contradiction.

❌ Common Errors

A frequent student mistake is confusing the number of harmonics or loops with the number of nodes. Students often assume the $n$-th harmonic has $n$ nodes, forgetting that fixed ends require a node at both boundaries ($n+1$ nodes total).

Detailed Evaluation of All Options

Option A Evaluation (True)

For the first harmonic on a rope of length $L = 1.0\text{ m}$ fixed at both ends, the wavelength is given by $\lambda = 2L = 2 \times 1.0\text{ m} = 2.0\text{ m}$.
Verdict: Correct statement.

Option B Evaluation (True)

Even-numbered harmonics (e.g., $n = 2, 4, 6$) feature an even number of loops, placing a node directly at the centre ($L/2$).
Verdict: Correct statement.

Option C Evaluation (True)

Let's check if $\lambda = 0.4\text{ m}$ fits the harmonic formula $\lambda = \frac{2L}{n}$:
$0.4 = \frac{2(1.0)}{n} \implies n = \frac{2.0}{0.4} = 5$
Since $n = 5$ is a valid integer (the 5th harmonic), this wavelength can be set up.
Verdict: Correct statement.

Option D Evaluation (False - The Answer!)

For the 5th harmonic ($n = 5$), the number of nodes is $n + 1 = 5 + 1 = 6$ nodes (not 5). The statement claims there are 5 nodes, which is incorrect.
Verdict: NOT correct (Correct option to choose).

Mark Scheme Note: 1 mark is awarded for selecting option D. No working out is required for multiple-choice questions, but quick mental checks of formulas ($\lambda = 2L/n$ and Nodes = $n+1$) ensure absolute accuracy.