AQA A-Level Physics Paper 2, June 2025: Question 24

1 mark · Medium difficulty · Multiple Choice

Determine the frequency of an ac signal displayed on an oscilloscope screen given a time-base setting of 5 μs div⁻¹.

Practise this question

Question

An oscilloscope screen grid displaying a sinusoidal waveform spanning 10 horizontal divisions. The waveform shows approximately 1.7 cycles across the display, with one full cycle covering approximately 5.9 horizontal divisions. The text states that the time base is set to 5 microseconds per division and asks to find the frequency of the signal from options A (34 kHz), B (37 kHz), C (42 kHz), and D (83 kHz).

Mark scheme

Show the mark scheme Mark scheme table indicating for question 24 that the correct option is A (34 kHz), targeted at assessment objective AO2.

How to answer it

Determining Signal Frequency from an Oscilloscope Trace

📋 What This Question Tests
  • Oscilloscope Graticule Reading: Determining the number of horizontal divisions representing one complete waveform cycle (the time period, T ).
  • Time-base Calibration: Converting horizontal divisions to time in seconds using the time-base setting ( 5 µs div⁻¹ ).
  • Frequency Calculation: Applying the wave equation f = 1 / T with correct unit conversions ( µs to s and Hz to kHz ).

Question 24 (Multiple Choice)

Cathode-Ray Oscilloscope (CRO) • 1 Mark • AO2

✅ Correct Answer

A — 34 kHz

Mark Scheme: Award 1 mark for option A.

💡 Key Knowledge

  • Period ( T ): The time taken to complete one full cycle (e.g., peak to peak, or zero-crossing to matching zero-crossing).
  • Subdivisions: Each major division on the graticule is divided into 5 small subdivisions; therefore, each small tick = 0.2 divisions .
  • Time-base factor: 1 µs = 1 × 10⁻⁶ s .
  • Relationship: f = 1 / T

📐 Step-by-Step Calculation

  1. Measure the horizontal length of one complete wave:
    Starting from the upward zero-crossing on the left edge ( x = 0 ):
    • The trace completes one full sine wave cycle and crosses upward again just before the 6th vertical gridline.
    • Careful reading shows it crosses at approximately 5.9 divisions (or measuring peak-to-peak: from x ≈ 1.4 div to x ≈ 7.3 div , giving 7.3 − 1.4 = 5.9 div ).
  2. Calculate the time period ( T ):
    T = divisions × time-base
    T = 5.9 div × 5 µs div⁻¹ = 29.5 µs = 29.5 × 10⁻⁶ s
  3. Calculate the frequency ( f ):
    f = 1 / T = 1 / (29.5 × 10⁻⁶ s) ≈ 33 898 Hz
  4. Convert to kHz:
    f = 33 898 / 1000 ≈ 34 kHz (Matches Option A)

🧠 Exam Technique

  • Pick distinct points: Peak-to-peak is often easiest to identify, but checking where the curve crosses the central axis gives a quick secondary verification.
  • Count subdivisions precisely: A common source of hesitation in MCQ options (such as choosing between 34 kHz and 37 kHz) comes from misreading 5.9 divisions as 5.4 divisions. Verify against multiple consecutive features.
  • Use powers of 10 safely: Input 1 / (29.5 × 10⁻⁶) into your calculator with brackets to prevent order-of-operation errors.

❌ Common Distractors & Traps

  • Option D (83 kHz): Measuring half a cycle ( ~2.4 div to ~2.9 div ) instead of a full cycle, resulting in roughly double the true frequency.
  • Option B (37 kHz): Miscounting divisions as 5.4 div ( T = 27 µs → f ≈ 37 kHz ).
  • Option C (42 kHz): Misreading the period as roughly 4.8 div ( T = 24 µs → f ≈ 41.7 kHz ).
  • Unit conversion error: Forgetting that 1 µs = 10⁻⁶ s , leading to wrong decimal placement.

Topics

Physics · Practical skills · 3.7 Fields and their consequences (A-level only) · Data analysis

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