AQA AS Level Physics Paper 1, June 2024: Question 3

11 marks · Medium difficulty · Short Answer

Calculate the wave speed and mass of a guitar string vibrating at given frequencies and tensions, draw its third harmonic, and deduce how touching the string's midpoint affects the harmonics present.

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Question

Four figures related to a guitar string experiment. Figure 3 shows a guitar string fixed at X and Y with length 648 mm. Figure 4 is a frequency spectrum graph showing relative amplitudes of harmonics f1, f2, f3, f4, and f5 up to 800 Hz. Figure 5 is an outline of the guitar for drawing the third harmonic. Figure 6 shows the string touched at its midpoint Z, dividing it into two 324 mm sections.

Question text

03 Figure 3 shows a guitar with only one of its strings attached.

The string is fixed at X and Y.

The string is plucked and vibrates freely between X and Y.

The distance XY is 648 mm.

Figure 3

03.1 The frequency of the first harmonic is 147 Hz.

Calculate the speed of the wave travelling in the string.

[2 marks]

speed of wave = m s−1

03.2 The tension in the string is 71 N.

Calculate the mass of the string between10 X and Y.

[3 marks]

The sound produced by the guitar is analysed.

The sound is the superposition of the first harmonic f1 with harmonics f2, f3, f4 and f5

of the stationary waves that exist on the string.

Figure 4 shows the frequencies of these harmonics and their relative amplitudes.

Figure 4

mass = kg

03.3 Draw, on Figure 5, the stationary wave that produces the harmonic f3.

Label the positions of all nodes N and all antinodes A.

[3 marks]

Figure 5

03.4 The string is vibrating freely.

The player then touches the string lightly at its midpoint Z as shown in Figure 6. This

prevents the string from vibrating at Z.

The sections XZ and ZY of the string continue to vibrate.

*10* Figure 6

The sound produced by the guitar is analysed.

Deduce, with reference to frequency, how the harmonics present in this sound

compare with the harmonics present in Figure 4.

[3 marks]

Mark scheme

Show the mark scheme

Question Answers Additional Comments/Guidance Mark AO

03.1 λ = 2 × 0.648 Allow 1296 (mm) or 1.296 (m) or 129.6 (cm) 2 1 × AO1

seen. 1 × AO2

Use of v = f λ

Condone one error in their substitution where

λ and f have been substituted and v would be

the subject:

Allow

(v=) 0.648 x 147 (forgets to double L)

0.648

(v=) x 147 (halves L)

Do not allow:

(v=) 648 x 147 (POT error and forgets to

double L)

NOR

(v=) x 147 (POT error and halves L)

(v =) 191 (m s−1)

Calculator display= 190.512

190 (ms-1) correct to 2 sf

03.2 1 T Condone one error where f, l and T have been 3 1 × AO1

Use of f = substituted.

2l μ 2 × AO2

𝜇𝜇 would be subject of a correctly rearranged

𝑇𝑇 𝑇𝑇

Use of v = � expression (𝜇𝜇=) 2 2

𝜇𝜇 4𝑙𝑙 𝑓𝑓

(𝜇𝜇 =) 1.956 × 10-3 (kg m-1)

Their l:

must be seen in MP1:

condone a POT error (if already penalised in

Use of MP1 or 03.1)

m = their 𝜇𝜇 × 𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑙𝑙

𝑇𝑇

allow ecf from 03.1 where v = � seen

𝜇𝜇

𝑇𝑇 71

MP1 𝜇𝜇= 2 or 𝜇𝜇= 2

𝑣𝑣 (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑒𝑒𝑒𝑒 𝑡𝑡𝑡𝑡 03.1)

MP2 (m=) their ecf 𝜇𝜇 × 0.648

MP3 ecf answer

(m =) 1.3 × 10−3 (kg)

Calculator display= 1.267618831 × 10-3 (kg)

03.3 At least one NAN envelope with its nodes N and antinodes A 3 1 × AO1

labelled.

1 × AO2

1 × AO3

The positions of all nodes N and all antinodes A labelled

3 (NAN) envelopes drawn

All 3 envelopes drawn same dimensions and all nodes N and Must not have any obvious differences in

antinodes A labelled correctly height and width by eye.

MP3: Do not allow unequal width and unequal

height.

Penalise A labelled twice at one antinode in

MP3.

03.4 MP1 ECF from 03.3 where third harmonic drawn 3 AO3

with node at midpoint.

Node at the midpoint

MP1 node at midpoint (ecf)

MP2

MP2 idea that one of the even harmonics,

Idea that stationary wave can only exist if one of its nodes

would have an antinode here and can’t exist

coincides with midpoint.

(ecf)

14 OR

Idea odd harmonic(s) require(s) an antinode to exist at this

idea that one of the odd harmonics can exist,

point and therefore cannot exist.

would have a node here (ecf)

MP3 f1, f3 and f5 all exist. (ecf)

Idea that the frequency f2 / (~)300 Hz can exist (when string is

Alternative MP2

touched lightly at midpoint)

OR Determines longest wavelength that can form

stationary wave between X and Z (or

Idea that the frequency f4 / (~)600 Hz can exist (when string is equivalent) to arrive at 294 Hz (allow ecf from

touched lightly at midpoint) (about 600 Hz) speed 03.1)

Determines next longest that can form

stationary wave between X and Z (or

equivalent) to arrive at 588 Hz (allow ecf from

speed 03.1)

How to answer it

Stationary Waves on a Guitar String

AQA AS-Level Physics • Waves & Stationary Waves

What this question tests

This question assesses your understanding of stationary waves on strings, wave equations ( v = fλ ), mass per unit length calculations ( μ = T / v² ), sketching harmonic profiles with correct nodes and antinodes, and applying boundary conditions when a vibrating string is damped or shortened.

Question 03.1

Calculate the speed of the wave travelling in the string (First harmonic frequency = 147 Hz, Length = 648 mm)

✅ Correct Answer

Speed (v) = 191 m s⁻¹ (Accept 190.5 m s⁻¹)

📐 Step-by-Step Calculation

Determine wavelength (λ): For the first harmonic, length L = λ / 2 , so λ = 2 × 0.648 m = 1.296 m .
Apply wave equation: v = fλ
Substitute values: v = 147 Hz × 1.296 m = 190.512 m s⁻¹
Round appropriately: 191 m s⁻¹ (to 2 significant figures).

❌ Common Errors

Forgetting to double the length L to find the wavelength of the first harmonic.
Power of ten (POT) errors when converting millimeters to meters ( 648 mm = 0.648 m ).

Question 03.2

Calculate the mass of the string between X and Y given a tension of 71 N

✅ Correct Answer

Mass (m) = 1.3 × 10⁻³ kg (Accept 1.27 × 10⁻³ kg)

💡 Key Knowledge

The speed of a wave on a stretched string is related to tension T and mass per unit length μ by:

v = √(T / μ) or μ = T / v²

📐 Step-by-Step Calculation

Rearrange for mass per unit length: μ = T / v² = 71 / (190.512)² = 1.956 × 10⁻³ kg m⁻¹
Calculate total mass: m = μ × L
Substitute values: m = 1.956 × 10⁻³ × 0.648 = 1.268 × 10⁻³ kg
Final value: 1.3 × 10⁻³ kg

Question 03.3

Draw on Figure 5 the stationary wave that produces the third harmonic (f₃) and label nodes (N) and antinodes (A)

✅ Correct Answer

Three complete "loops" (envelope sections) drawn between fixed ends X and Y, with nodes N at every zero-amplitude point and antinodes A at the center of every loop.

🧠 Exam Technique & Marking Points

MP1: Correct positions of nodes and antinodes labelled. Ends X and Y must be nodes.
MP2: Exactly 3 distinct loops (NAN envelopes) drawn.
MP3: Consistent dimensions—all loops must have equal height and width by eye. Do not double-label antinodes in the same loop.

Question 03.4

Deduce how the harmonics present when touched lightly at its midpoint Z compare with Figure 4

✅ Correct Answer

Only odd harmonics ( f₁, f₃, f₅ ) remain present; all even harmonics ( f₂, f₄ ) disappear.

💡 Key Knowledge & Examiner Commentary

Touching the string at its midpoint (Z) forces a node to exist at that exact position.
Even harmonics ( f₂, f₄ ) naturally possess an antinode at the midpoint, so they cannot vibrate when touched there and are suppressed.
Odd harmonics ( f₁, f₃, f₅ ) naturally possess a node at the midpoint, meaning their vibration is unaffected, so they continue to exist.

Topics

Physics · Required Practicals · 3.3 Waves · AS practicals (1–6)

Question and mark scheme from the AQA AS Level Physics examination, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.