AQA A-Level Physics Paper 1, June 2024: Question 15

1 mark · Medium difficulty · Multiple Choice

Determine the wavelength of sound emitted by a loudspeaker above a water-filled tube when the first and next intensity maxima are observed at air column lengths L1 and L2.

Practise this question

Question

A multiple-choice question with a diagram showing a vertical tube filled partially with water, with a loudspeaker at the top and a tap at the bottom. Dimensions L1 and L2 indicate lengths of the air column from the top to the water level at two different stages. Four options A, B, C, and D are given as expressions involving L1 and L2.
Question text

15 A loudspeaker producing a single-frequency sound is mounted above a tube filled with

water. A tap at the bottom of the tube is opened to allow the water to run out.

A student observes the change in loudness of the sound emitted by the tube as the water

runs out.

When the length of the column of air in the tube reaches L1, the loudness is at its first

maximum.

The next maximum is reached when the length of the column of air is L2.

What is the wavelength of the sound emitted by the loudspeaker?

[1 mark]

A L2

B 2L1

C L2 − L1

D 2(L2 − L1)

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer is option D, which is 2(L2 - L1).

15 D 2(L2 − L1) AO1

How to answer it

Stationary Waves in Closed Air Columns

What this question tests:

This question assesses your understanding of stationary waves formed in a closed-end air column (closed at the water surface, open at the loudspeaker end). It tests your knowledge of resonance lengths, node-antinode positioning, and how successive harmonic resonance positions relate directly to the wavelength of the sound wave.

Question 15: Multiple Choice Solution

Determining Wavelength from Successive Resonance Lengths

Correct Option: D — 2(L₂ - L₁)

✅ Correct Answer Breakdown

For a tube closed at one end and open at the other:

  • First resonance length: L₁ = λ / 4
  • Next (second) resonance length: L₂ = 3λ / 4
  • The difference in length between consecutive resonances is: L₂ - L₁ = (3λ / 4) - (λ / 4) = 2λ / 4 = λ / 2
  • Rearranging for wavelength: λ = 2(L₂ - L₁)

💡 Key Knowledge

  • Boundary conditions: A closed tube always has a displacement node at the closed water surface and a displacement antinode at the open end.
  • Resonance spacing: The distance between any two adjacent resonant lengths in a closed (or open) pipe is always equal to half a wavelength ( λ / 2 ).

🧠 Exam Technique

When dealing with consecutive resonance positions in pipes, do not try to work out absolute node positions from scratch if you get confused. Remember the golden rule: the difference between two adjacent resonance lengths is always λ / 2 .

❌ Common Errors

  • Choosing option B ( 2L₁ ) by assuming the first length equals half a wavelength instead of a quarter.
  • Choosing option C ( L₂ - L₁ ) and forgetting to multiply by 2 to convert the half-wavelength difference into a full wavelength ( λ ).
Mark Scheme Allocation: 1 mark for selecting D (AO1 - Demonstration of knowledge).

Topics

Physics · 3.3 Waves

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.