AQA A-Level Physics Paper 1, June 2025: Question 32

1 mark · Medium difficulty · Multiple Choice

Calculate the difference in oscillation frequency of an identical simple pendulum on Earth and on the Moon.

Practise this question

Question

Question 32 asks: 'A simple pendulum of length 8.0 cm oscillates on Earth with a frequency f_1. An identical pendulum on the Moon oscillates with a frequency f_2. The acceleration due to gravity is 1.6 m s^-2 on the surface of the Moon. What is f_1 - f_2?' Followed by four multiple-choice options: A: 1.8 Hz, B: 1.1 Hz, C: 0.85 Hz, D: 0.35 Hz, with tick boxes.
Question text

32 A simple pendulum of length 8.0 cm oscillates on Earth with a frequency f1.

An identical pendulum on the Moon oscillates with a frequency f2.

The acceleration due to gravity is 1.6 m s−2 on the surface of the Moon.

What is f1 – f2?

[1 mark]

A 1.8 Hz

B 1.1 Hz

C 0.85 Hz

D 0.35 Hz

Mark scheme

Show the mark scheme Mark scheme table row for question 32 showing the key 'B', the answer '1.1 Hz', and the assessment objective 'AO2'.

32 B 1.1 Hz AO2

How to answer it

Simple Harmonic Motion: Pendulum Frequency on Earth vs Moon

📋 What this question tests

This question assesses your ability to apply the formula for the time period of a simple pendulum to determine oscillation frequency under different gravitational fields, manage unit conversions, and calculate differences between physical values.

  • Recall & Manipulation: Relating time period to frequency ( f = 1 / T ) using T = 2π√(L / g) .
  • Standard Constants: Recalling acceleration due to gravity on Earth ( g = 9.81 m s⁻² ) from the formula sheet.
  • Unit Conversion: Converting length from centimetres to metres ( 8.0 cm = 0.080 m ).
Question 32 (1 Mark)

Frequency Comparison: f₁ − f₂

Simple pendulum oscillation in two different gravitational environments

✅ Correct Answer

B: 1.1 Hz

Mark Scheme: 1 mark for identifying option B (AO2).

💡 Key Knowledge

  • Time period of a simple pendulum:
    T = 2π√(L / g)
  • Frequency is the reciprocal of period:
    f = 1 / T = (1 / 2π) × √(g / L)
  • Earth's surface gravity:
    g₁ = 9.81 m s⁻²
  • Moon's surface gravity (given):
    g₂ = 1.6 m s⁻²

📐 Step-by-Step Calculation

  1. Convert the length to SI units:
    L = 8.0 cm = 0.080 m
  2. Calculate the frequency on Earth (f₁):
    T₁ = 2π × √(0.080 / 9.81) = 2π × √(0.008155) ≈ 0.5674 s
    f₁ = 1 / T₁ = 1 / 0.5674 ≈ 1.762 Hz
  3. Calculate the frequency on the Moon (f₂):
    T₂ = 2π × √(0.080 / 1.6) = 2π × √(0.050) ≈ 1.405 s
    f₂ = 1 / T₂ = 1 / 1.405 ≈ 0.712 Hz
  4. Compute the difference (f₁ − f₂):
    f₁ − f₂ = 1.762 Hz − 0.712 Hz = 1.050 Hz
    Rounding to two significant figures gives 1.1 Hz.

🧠 Exam Technique & Time-Saver

You can combine the expressions to calculate the difference directly on your calculator in one line:

f₁ − f₂ = [√(9.81) − √(1.6)] / [2π × √(0.080)]

  • √(9.81) − √(1.6) ≈ 3.1321 − 1.2649 = 1.8672
  • 2π × √(0.080) ≈ 1.7772
  • 1.8672 / 1.7772 ≈ 1.051 Hz → 1.1 Hz

Factoring out 1 / (2π√L) saves substantial time and prevents intermediate rounding errors.

❌ Common Traps & Distractors

  • Option A (1.8 Hz): Trap for calculating only f₁ and forgetting to subtract f₂ .
  • Subtracting time periods (T₂ − T₁ ≈ 0.84 s): Leads directly to distractor C (0.85 Hz). Remember: 1 / (T₂ − T₁) ≠ (1 / T₁) − (1 / T₂) .
  • Unit conversion failure: Leaving L = 8.0 rather than 0.080 m introduces a factor of 10 error inside the square root ( √10 ≈ 3.16 ).
  • Inverting the ratio: Incorrectly using √(L / g) directly for frequency instead of √(g / L) leads to distractor D (0.35 Hz).

Topics

Physics · 3.6 Further mechanics and thermal physics (A-level only)

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.