AQA A-Level Physics Paper 1, June 2025: Question 31
1 mark · Medium difficulty · Multiple Choice
Calculate the mass of an object attached to a vibrating spring-mass system at resonance given the spring constant and driving frequency.
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Question text
31 A spring with a spring constant of 50 N m−1 is attached to a rigid horizontal bar.
An object of mass m is attached to the bottom of the spring.
When the bar is made to move vertically up and down with a frequency of 2.6 Hz, the
mass–spring system undergoes resonance.
Ignore the effects of damping.
What is m?
[1 mark]
A 190 g
B 370 g
C 490 g
D 590 g
Mark scheme
Show the mark scheme
31 A 190 g AO2
How to answer it
Resonance of a Vertical Mass–Spring System
This question assesses your understanding of forced vibrations and resonance in simple harmonic motion (SHM). Specifically, it requires you to recognise the resonance condition (driving frequency = natural frequency), apply the formula for the time period of a mass-spring system, rearrange it to solve for unknown mass m, and correctly handle unit conversions from kilograms to grams.
Determining Unknown Mass at Resonance
AQA A-Level Physics • Further Mechanics & Thermal Physics (Periodic Motion)
✅ Correct Answer
A — 190 g
Awarded for correctly calculating m ≈ 0.187 kg and rounding to 2 significant figures (190 g).
💡 Key Knowledge
- Condition for Resonance: Occurs when the periodic driving frequency ( f = 2.6 Hz ) equals the natural frequency ( f₀ ) of the oscillating system.
- Time Period Formula: T = 2π√(m / k)
- Frequency Relation: Since f = 1 / T : f = (1 / 2π) × √(k / m)
📐 Step-by-Step Calculation
- Identify given values:
Spring constant, k = 50 N m⁻¹
Resonant frequency, f = 2.6 Hz - Calculate the angular frequency (ω):
ω = 2πf = 2 × π × 2.6 = 16.336 rad s⁻¹ - Rearrange the natural frequency equation for mass (m):
Since ω = √(k / m) , squaring both sides yields ω² = k / m
m = k / ω² = k / (2πf)² - Substitute the numbers:
m = 50 / (16.336)² = 50 / 266.87 = 0.18736 kg - Convert kilograms to grams:
m = 0.18736 × 1000 g = 187.4 g ≈ 190 g (2 s.f.)
→ Matches Option A.
🧠 Exam Technique
- Spot the trigger word: The word "resonance" is your clue that the driving frequency equals the system's natural frequency ( fdriving = f0 ).
- Use the Data Sheet efficiently: The formula is given as T = 2π√(m/k) . You can either find T = 1 / 2.6 = 0.3846 s first, or jump directly using ω = 2πf .
- Check units immediately: The answers are listed in grams ( g ), but SI standard equations give mass in kilograms ( kg ). Always multiply by 1000 before picking an option.
❌ Common Errors & Distractors
- Forgetting to square (2π): Entering 50 / 2π(2.6)² on a calculator instead of 50 / (2π × 2.6)² leads to an incorrect denominator of 42.47 , yielding m ≈ 1.18 kg .
- Inverting the ratio: Confusing m/k with k/m gives m ≈ 5.3 kg .
- Using the pendulum formula: Accidental retrieval of T = 2π√(l/g) instead of the mass-spring equation.
- Distractor values: Option B ( 370 g ) is roughly double the correct answer, commonly obtained if the student forgets the factor of 2 or squares π incorrectly.
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.