AQA AS Level Physics Paper 1, November 2020: Question 6

✅ Correct Answer

• Antiphase / completely out of phase / π radians out of phase (Allow ½ cycle or 180° out of phase).
• Similar amplitudes of vibration (or similar displacements at any instant).
• Same period or frequency, and move with the same speed.

❌ Common Errors

Writing vague statements like "opposite amplitudes" or "when P is positive, Q is negative" will be treated as neutral. You must explicitly state they are in antiphase or out of phase.

📐 Step-by-Step Calculation

1. Determine wavelength (λ): From Figure 9, the distance between P and Q is 0.55 m, which contains 2 complete loops (one full wavelength). Therefore, λ = 0.55 / 2 = 0.275 m (or use v = fλ directly recognising the distance spans 2λ).
2. Apply wave equation: v = f × λ
3. Substitute values: v = 250 Hz × 0.275 m = 68.75 m s⁻¹
4. Round appropriately: 69 m s⁻¹ (Allowing range 67.5 to 70.0 m s⁻¹).

🧠 Exam Technique

Always show your substitution clearly. Even if your wavelength extraction has a minor slip, stating v = fλ can secure method marks.

💡 Key Knowledge

A stationary wave is formed by the superposition of two progressive waves travelling in opposite directions. For a stable stationary wave pattern to form, these two component waves must have:

• Same speed
• Moving in opposite directions
• Same frequency / same wavelength / similar amplitudes

❌ Common Errors

Examiners note that generic answers like "progressive", "transverse", or "transfer energy" are insufficient because progressive waves transfer energy, whereas stationary waves store energy.

✅ Correct Answer

• A completely horizontal line drawn straight across from P to Q (representing destructive interference where displacement is zero everywhere at this specific instant).

🧠 Exam Technique

Recognise that when two identical waves travelling in opposite directions are exactly out of phase during superposition, destructive interference results in a flat line of zero displacement momentarily.

✅ Correct Answer

• Mark an A at each end of the string (at the fixed boundaries if applicable to the harmonic context, or mark all 5 antinodes evenly spaced by eye along the horizontal line).

❌ Common Errors

Careless placement of antinodes that do not align with the mathematical peaks/loops of the underlying harmonic structure will lose the second mark.

📐 Step-by-Step Deduction

1. Recall wave speed equation for a string: v = √(T / μ)
2. Analyze change in μ: If mass per unit length increases by a factor of 9 (9μ), the wave speed v decreases by a factor of √9 = 3.
3. Relate speed to frequency and wavelength: Since frequency f is constant, wavelength λ must also decrease by a factor of 3 ( λ_new = λ_old / 3 ).
4. Determine harmonic: A shorter wavelength means more loops fit into the fixed length L . Since λ becomes 3 times smaller, the number of loops multiplies by 3, making it the third harmonic (3 symmetrical loops).

🧠 Top-Level Examiner Insight

To score full marks, top responses explicitly state that the wave speed reduces by a factor of 3, which reduces the wavelength by 3, meaning three times as many loops fit into the string length, confirming the third harmonic.