AQA A-Level Physics Paper 1, June 2025: Question 14
1 mark · Medium difficulty · Multiple Choice
Identify which quantity produces a straight-line graph through the origin when plotted against the first harmonic frequency of a stationary wave on a wire.
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Question text
14 In an investigation of stationary waves on different wires, the frequency f of the
first harmonic is measured.
Which quantity produces a straight-line graph through the origin when plotted against f?
In each case all other variables are kept constant.
[1 mark]
A
diameter of the wire
B (mass per unit length of the wire)2
C length of the wire
D (tension in the wire)2
Mark scheme
Show the mark scheme
14 A AO1
diameter of the wire
How to answer it
Stationary Waves: First Harmonic & Graphical Proportionality
This question assesses your ability to relate physical quantities in the first harmonic frequency equation for stationary waves on a string. Specifically, it tests your ability to break down composite quantities (such as mass per unit length, μ) in terms of fundamental dimensions (such as wire diameter, d) and deduce direct proportionality (graphs of the form y = mx that pass through the origin).
First Harmonic Relationships & Graphs
Identifying the quantity directly proportional to fundamental frequency
✅ Correct Answer
A: 1 / diameter of the wire
💡 Key Knowledge
- First harmonic formula:
f = (1 / 2L) × √(T / μ) - Mass per unit length (μ):
μ = mass / length = (ρ × volume) / L = ρ × A - Cross-sectional area of a wire:
A = πd² / 4 , so μ = (ρπd²) / 4 - Straight line through the origin: Requires direct proportionality ( y ∝ x , or y = mx where the y-intercept is zero).
📐 Step-by-Step Algebraic Derivation
- Write the base equation:
f = (1 / 2L) × √(T / μ) - Substitute μ in terms of diameter (d) and density (ρ):
Since μ = ρ × A = ρ × (πd² / 4) , take the square root:
√μ = √[(ρπ / 4) × d²] = d × √(ρπ / 4) - Substitute √μ back into the frequency formula:
f = [1 / (2L)] × [√T / (d × √(ρπ / 4))] - Group all constant terms together:
Since L, T, and ρ are kept constant:
f = (constant) × (1 / d)
Rearranging gives: (1 / d) = (constant) × f - Conclusion:
Because (1 / d) ∝ f , plotting 1 / diameter against f produces a straight-line graph through the origin ( y = mx ).
🧠 Exam Technique & Eliminating Distractors
- Option B: (μ)²
Since f ∝ 1 / √μ , squaring gives f² ∝ 1 / μ , which means μ² ∝ 1 / f⁴ . This is an inverse relationship, producing a curve, not a straight line. - Option C: length (L)
Since f ∝ 1 / L , L ∝ 1 / f . A plot of L against f yields a reciprocal curve (hyperbola), not a straight line. - Option D: T²
Since f ∝ √T , squaring both sides gives T ∝ f² , so T² ∝ f⁴ . A plot of T² against f produces a steep curve starting at the origin, not a straight line.
❌ Common Traps & Misconceptions
- Forgetting the square root on μ: Students often mistakenly write f ∝ 1/μ instead of f ∝ 1/√μ , leading to wrong guesses.
- Confusing diameter with cross-sectional area: Remember that √A ∝ d . The square root in the formula cancels the squared term of diameter ( √(d²) = d ), leaving a clean 1 / d relationship.
- Confusing "straight line" with "straight line through origin": Even if a relationship is linear, it must satisfy y = mx (zero y-intercept) to pass through the origin. Here, direct proportionality guarantees this.
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 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.