AQA A-Level Physics Paper 1, November 2020: Question 30

Simple Pendulum Period Scaling

What this question tests

This question assesses your understanding of simple harmonic motion (SHM) formulas, specifically how the period ($T$) of a simple pendulum scales proportionally with its length ($L$). You must be able to manipulate square root relationships and set up simultaneous proportional equations to solve for an unknown initial quantity.

Question 30 (Multiple Choice)

Exam Breakdown & Solution Guide

✅ Correct Answer: B (0.60 m)

Option B is the correct original length of the pendulum.

💡 Key Knowledge

The formula for the period of a simple pendulum is: T = 2π√(L/g)
Since $2$ and $π$ and gravitational field strength $g$ are constant, T ∝ √L .
If the period doubles ( 2T ), the square root of the length must also double.

🧠 Exam Technique

For scaling and proportionality multiple-choice questions, set up a ratio equation comparing initial and final states rather than substituting numbers blindly. This prevents algebraic slips.

❌ Common Errors

A common trap is assuming linear scaling (e.g., halving or dividing the increase by 2 linearly, leading to incorrect choices like 0.90 m). Students forget that squaring both sides is required when dealing with square-root relationships.

📐 Step-by-Step Calculation

Write down the proportionality rule:
T ∝ √L
Set up the equation for the changed state:
Let the original length be L and the new length be L + 1.8 .
Since the period is doubled: 2T ∝ √(L + 1.8)
Form a ratio of final to initial states:
(2T / T) = √( (L + 1.8) / L )
2 = √( (L + 1.8) / L )
Square both sides to eliminate the square root:
2² = (L + 1.8) / L
4 = (L + 1.8) / L
Rearrange and solve for L :
4L = L + 1.8
3L = 1.8
L = 1.8 / 3 = 0.60 m

Mark Scheme Note: 1 mark awarded for selecting B.

AQA A-Level Physics Paper 1, November 2020: Question 30

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