AQA AS Level Physics Paper 2, June 2022: Question 3
10 marks · Medium difficulty · Short Answer
Calculate the average output power of a battery, describe differences between microwaves and sound waves, state conditions for coherence, calculate path difference from interference setup, and determine wave frequency.
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Question text
03 A student buys a portable loudspeaker that is powered by its own internal battery.
The battery in the loudspeaker is initially uncharged.
03.1 The battery is connected to a charger that maintains a constant potential difference
of 5.0 V across the battery. It takes 2.6 hours for the battery to become fully charged.
The average current in the battery during this time is 2.0 A.
The battery is disconnected from the charger.
The fully-charged battery operates the loudspeaker for 12 hours before it is
completely discharged.
Calculate the average output power of the battery during these 12 hours.
[2 marks]
average output power = W
03.2 A mobile phone transmits data to the loudspeaker using microwaves. The data are
processed at the loudspeaker to produce sound waves.
Microwaves and sound waves travel at different speeds.
Describe two other differences between microwaves and sound waves.
[2 marks]
03.3 A second loudspeaker receives the same data from the mobile phone. The two
loudspeakers act as coherent sources of sound waves.
State the two conditions required for the sources to be coherent.
[2 marks]
Figure 5 shows two loudspeakers A and B that act as coherent point sources of
sound of a single frequency.
Figure 5
C is the midpoint between A and B.
Distances OA and OB are equal.
OP is perpendicular to CO.
The student uses a sound-level meter to measure the intensity of the sound. The
meter detects a maximum intensity at O.
The student moves the meter along OP. The intensity decreases and reaches a first
minimum at M. The intensity then increases as the meter moves towards P.
The student records the following distances:
AB = 1.80 m
CO = 8.00 m 15
OM = 1.24 m.
03.4 Show that the difference between the path lengths AM and BM is
approximately 0.3 m.
[2 marks]
03.5 The speed of sound is 340 m s−1.
Determine the frequency of the sound waves.
[2 marks]
frequency = Hz
Mark scheme
Show the mark scheme
Question Answers Additional Comments/Guidance Mark AO
03.1 attempts to calculate energy stored during 2.6 hr period Correctly rounded answer gains both marks. 2 2 ×
(Calculator value is = 2.16666667) AO2
OR
12 hr For 1✓ stored energy = 93.6 kJ
attempts to calculate average output power during
period using their energy stored 1✓ For 1✓ condone use of t in hours. (2.6 hr = 9360 s;
12 hr = 43200 s)
If no other mark given, award 1 mark for calculating
2.2 (W) 2✓
charge transfer during 2.6 hr period as 18.7 kC
03.2 Max 2 from: ✓ ✓ Apply list principle. Do not allow reference to 2 2 ×
applications e.g. cooking food. AO1
For first point, allow weak descriptions in terms of
microwaves are transverse; sound are longitudinal;
parallel and perpendicular oscillations/vibrations
microwaves have higher frequency (than sound); with direction of energy transfer.
microwaves can be polarised but sound can’t;
microwaves can travel through a vacuum but sound
can’t/requires a medium OR sound are mechanical waves but
microwaves are EM waves
– SICS – – JUNE 2022
1403.3 For 1✓ do not accept “in phase” or fixed path 2 2 ×
difference. AO1
fixed/constant phase difference 1✓
For 2✓ condone “same wavelength”.
same frequency 2✓
Ignore reference to other features e.g. amplitude or
type of wave.
22 2 No credit for using double-slit equation.
03.4 evaluates AM from AM = 8.00 + 0.34 2 2 ×
Expect 8.01 (m) for AM and 8.28 (m) for BM AO2
OR evaluates BM from BM2 = 8.002 + 2.142 ✓
8.28 − 8.01 = 0.27 (m) ✓
No credit for using double-slit equation.
03.5 statement that path difference = λ/2 OR uses wavelength = 2 2 2 ×
× their 03.4 answer 1✓ For 1✓ expect to see 0.54 or 0.60 m for wavelength AO2
For 2✓ expect 570 Hz (from 0.3 m) OR 630 Hz
(from 0.27 m) OR 620 Hz (from 0.274 m).
evaluates (Hz) 2✓
correct λ If no other mark given, allow 1130 Hz or 1260 Hz.
for 1 mark.
Total 10
How to answer it
Portable Loudspeaker, Microwaves, and Interference Study Guide
What this question tests
This multi-part exam question bridges foundational concepts in electricity (current, charge, energy, and power calculations) and waves (electromagnetic vs. mechanical waves, wave properties, coherence, path difference, and interference minima).
Battery Power and Energy Transfer
✅ Correct Answer
2.2 W (Exact calculator value: 2.166... W )
💡 Key Knowledge
- Charge Q = I × t
- Energy E = V × Q = V × I × t
- Power P = E / t
📐 Step-by-Step Calculation
- Find charge transferred during charging: Q = 2.0 A × (2.6 × 3600 s) = 18720 C
- Find total energy stored: E = 5.0 V × 18720 C = 93600 J
- Calculate average output power over 12 hours (43200 s): P = 93600 J / (12 × 3600 s) = 2.166... W
- Round to appropriate sig figs: 2.2 W (matching input data of 2.0 A and 5.0 V).
❌ Common Errors & Exam Technique
Time unit traps: Forgetting to convert hours into seconds ( × 3600 ) when calculating energy, or mixing up the 2.6-hour charging time with the 12-hour discharging time. Always double-check your time multipliers!
Comparing Microwaves and Sound Waves
✅ Correct Answers (Any 2)
- Microwaves are transverse; sound waves are longitudinal.
- Microwaves have a much higher frequency than sound waves.
- Microwaves can be polarised; sound waves cannot.
- Microwaves are electromagnetic waves and can travel through a vacuum; sound waves are mechanical waves and require a medium.
💡 Key Knowledge
Be precise with wave classifications. Avoid everyday applications like "microwaves are used for cooking"—the examiner wants fundamental physical properties.
Conditions for Coherence
✅ Correct Answers
- Fixed / constant phase difference
- Same frequency (or same wavelength)
❌ Common Errors
Strict terminology: Do not write "in phase" as a standalone condition (sources must maintain a constant phase difference, which could be zero or any fixed value). Also, do not confuse path difference with phase difference.
Path Difference Calculation
✅ Correct Answer
Path difference = 8.28 m - 8.01 m = 0.27 m (or approximately 0.3 m as shown)
🧠 Exam Technique & Geometry
Do not use the double-slit fringe spacing equation ( w = λD / s ) here, as this is a geometric path-length problem involving point sources, not a standard double-slit interference setup.
📐 Step-by-Step Calculation
- Identify coordinates from Figure 5: Distance from central axis to speaker A/B is half of 1.80 m = 0.90 m .
- Point M is 1.24 m away from central line O.
- For distance AM: 1.24 - 0.90 = 0.34 m vertical distance from source A. Using Pythagoras: AM = √(8.00² + 0.34²) = 8.0072 m ≈ 8.01 m
- For distance BM: 1.24 + 0.90 = 2.14 m vertical distance from source B. Using Pythagoras: BM = √(8.00² + 2.14²) = 8.2816 m ≈ 8.28 m
- Subtract path lengths: 8.28 m - 8.01 m = 0.27 m (rounds to approximately 0.3 m).
Determining Frequency from Interference Minima
✅ Correct Answer
Frequency = 630 Hz (using 0.27 m) or 570 Hz (using 0.3 m)
💡 Key Knowledge
Point M is stated to be the first minimum of intensity. For destructive interference from two coherent sources in antiphase/out of phase, the path difference equals an odd multiple of half-wavelengths, or specifically for the first minimum from identical sources, path difference = λ / 2 .
📐 Step-by-Step Calculation
- Relate path difference to wavelength: Path difference = λ / 2 , meaning wavelength λ = 2 × path difference .
- Using the more precise path difference ( 0.27 m ): λ = 2 × 0.27 = 0.54 m .
- Apply wave equation c = f × λ rearranged for frequency: f = c / λ = 340 / 0.54 = 629.6 Hz ≈ 630 Hz .
- *(Note: If using the rounded 0.3 m path difference, λ = 0.6 m giving 570 Hz; both are fully credited).*
Topics
Physics · 3.4 Mechanics and materials · 3.3 Waves · 3.5 Electricity
Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.