AQA A-Level Physics Paper 1, June 2023: Question 14

1 mark · Medium difficulty · Multiple Choice

Determine the wavelength and frequency of a wave on a string of length 1.2 m vibrating at its second harmonic, given a displacement-time graph.

Practise this question

Question

A multiple-choice question showing a displacement-time graph for a point on a string of length 1.2 m vibrating at its second harmonic. The graph plots displacement in cm from 0 to 2.0 against time in ms from 0 to 6, showing one complete sine wave cycle with a period of 6 ms. A table below lists four options A, B, C, and D for wavelength in m (0.6 or 1.2) and frequency in kHz (0.17 or 0.34).
Question text

14 A string with a length of 1.2 m vibrates at its second harmonic.

The diagram shows the displacement–time graph for a point on the string.

What are the wavelength and frequency of the wave on the string?

[1 mark]

Wavelength / m Frequency / kHz

A 0.6 0.17

B 0.6 0.34

C 1.2 0.17

D 1.2 0.34

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer is option C, with a wavelength of 1.2 m and a frequency of 0.17 kHz.

14 C 1.2 0.17

How to answer it

Wavelength and Frequency of Stationary Waves

What this question tests

This question assesses your understanding of stationary waves on strings, specifically relating harmonic numbers and physical string length to wavelength, as well as extracting time periods and frequencies from displacement-time graphs.

Question 1.4 — Multiple Choice

Determining Wavelength and Frequency

✅ Correct Answer: C

Wavelength: 1.2 m

Frequency: 0.17 kHz

💡 Key Knowledge

  • For the second harmonic ($n = 2$), the length of the string equals one full wavelength: L = λ .
  • A displacement-time graph shows the motion of a single point over time, giving the time period T of the oscillation, not the wavelength.
  • Frequency is the reciprocal of the time period: f = 1 / T .

🧠 Exam Technique

  • Step 1 (Wavelength): Read the string length given in the stem ( 1.2 m ). Recall that for the second harmonic, 2 loops fit onto the string, meaning L = λ . Therefore, λ = 1.2 m . This immediately eliminates options A and B!
  • Step 2 (Time Period): Look at the displacement-time graph. One full wave cycle completes in 6 ms ( 6 × 10⁻³ s ).
  • Step 3 (Frequency Calculation): Calculate f = 1 / (6 × 10⁻³ s) = 166.7 Hz . Convert to kilohertz by dividing by 1000: 0.167 kHz , which rounds to 0.17 kHz .

❌ Common Errors

  • Harmonic confusion: Assuming L = λ / 2 (which is true only for the first harmonic / fundamental mode). This leads to an incorrect wavelength of 0.6 m .
  • Graph misinterpretation: Confusing a displacement-time graph with a displacement-position (snapshot) graph and attempting to read wavelength off the time axis.
  • Unit conversion traps: Forgetting to convert milliseconds ( ms ) to seconds ( s ) or Hertz ( Hz ) to kilohertz ( kHz ).
Examiner Insight: Distractor B is a common trap for students who correctly identify the frequency but incorrectly apply the fundamental harmonic formula ( λ = 2L / n ) to find wavelength. Always sketch the wave profile for the stated harmonic to verify the relationship between L and λ !

Topics

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

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.