Specification: AQA A-level Physics (7408) 3.3.1.1 | Paper: 1 | Data and formulae booklet: not given – learn φ = 2πd/λ
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📌 KEY DEFINITIONS – learn them word for word
Phase difference: how far one point on a wave is ahead of or behind another point in its cycle, measured as an angle in radians or degrees. One full cycle is 2π rad = 360°.
In phase and antiphase: points are in phase when their phase difference is 0, 2π, 4π… rad (separated by a whole number of wavelengths); they are in antiphase when their phase difference is π, 3π, 5π… rad (separated by an odd number of half wavelengths).
Path difference: the difference in distance travelled by two waves from their sources to a point, measured in metres (or in wavelengths). It is a distance, not an angle.
The equation
phase difference = 2π × separation ÷ wavelength φ = 2πd ÷ λ (in radians) or φ = 360° × d ÷ λ and, for a time difference, φ = 2πΔt ÷ T
| Quantity | Symbol | Unit |
|---|---|---|
| phase difference | φ | radians (rad) or degrees (°) |
| separation of the two points along the direction of travel | d | metres (m) |
| wavelength | λ | metres (m) |
| time difference between the two oscillations | Δt | seconds (s) |
| period | T | seconds (s) |
Rearranged: d = φλ ÷ 2π | λ = 2πd ÷ φ | Δt = φT ÷ 2π | Work out d ÷ λ first: it is the fraction of a cycle, and a whole number of cycles can be removed because the points are back in step. To convert, π rad = 180°.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) Two points on a progressive wave are 0.20 m apart in the direction of travel. The wavelength is 0.80 m. Calculate the phase difference between the points in radians.
Show solution
F: φ = 2πd ÷ λ
I: φ = 2π × 0.20 ÷ 0.80
F: φ is already the subject
A: π/2 rad (1.57 rad, 3 s.f.)
Q2 (Core) Two points on a wave of wavelength 1.2 m are 0.40 m apart. Calculate their phase difference in degrees, and give it in radians to 3 significant figures.
Show solution
F: φ = 360° × d ÷ λ
I: φ = 360° × 0.40 ÷ 1.2
F: φ is already the subject
A: 120° = 2π/3 rad = 2.09 rad (3 s.f.)
Q3 (Core) Two points on a stretched spring oscillate with a phase difference of π/3 rad. The wavelength of the wave is 90 cm. Calculate the minimum distance between the points, in m.
Show solution
Convert first: 90 cm = 0.90 m
F: φ = 2πd ÷ λ
I: π/3 = 2π × d ÷ 0.90
F: d = (π/3) × 0.90 ÷ 2π = 0.90 ÷ 6
A: 0.15 m
Q4 (Core) A sound wave of frequency 5.0 kHz travels through air at 340 m s⁻¹. Calculate the phase difference between two points 1.7 cm apart along the direction of travel.
Show solution
Convert first: 5.0 kHz = 5000 Hz; 1.7 cm = 0.017 m
Step 1 – wavelength
λ = c ÷ f = 340 ÷ 5000 = 0.068 m
Step 2 – phase difference
F: φ = 2πd ÷ λ
I: φ = 2π × 0.017 ÷ 0.068
F: φ is already the subject
A: π/2 rad (90°)
Q5 (Exam) A wave has a wavelength of 0.60 m. Points B, C and D are 0.30 m, 0.90 m and 1.20 m from point A along the direction of travel. State whether each point is in phase or in antiphase with A, and calculate the phase difference between A and a point E 0.45 m from A.
Show solution
F: d ÷ λ gives the separation in wavelengths
I: B: 0.30 ÷ 0.60 = 0.5; C: 0.90 ÷ 0.60 = 1.5; D: 1.20 ÷ 0.60 = 2.0
F: odd number of half wavelengths → antiphase; whole number of wavelengths → in phase
A: B and C are in antiphase with A; D is in phase with A
Point E: 0.45 ÷ 0.60 = 0.75, so φ = 2π × 0.75 = 3π/2 rad (270°)
Watch out: in phase means a whole number of wavelengths apart, not just ‘one wavelength’. D is 2λ from A and is still in phase.
Q6 (Exam) Two signals of frequency 2.0 kHz are displayed on a dual-beam oscilloscope. The peaks of trace Y occur 0.10 ms after the peaks of trace X. Calculate the phase difference between the signals in radians, to 3 significant figures, and in degrees.
Show solution
Convert first: 2.0 kHz = 2000 Hz; 0.10 ms = 1.0 × 10−4 s
Step 1 – period
T = 1 ÷ f = 1 ÷ 2000 = 5.0 × 10−4 s
Step 2 – phase difference
F: φ = 2πΔt ÷ T
I: φ = 2π × 1.0 × 10−4 ÷ 5.0 × 10−4 = 0.4π
F: φ is already the subject
A: 1.256… rad = 1.26 rad (3 s.f.) = 72°; Y lags X
Q7 (Exam) Two loudspeakers connected to the same signal generator emit sound in phase at 680 Hz. The speed of sound is 340 m s⁻¹. A listener stands 3.00 m from one speaker and 3.75 m from the other. Calculate the path difference and the phase difference at the listener, and state what she hears.
Show solution
Step 1 – wavelength
λ = c ÷ f = 340 ÷ 680 = 0.50 m
Step 2 – path difference
path difference = 3.75 − 3.00 = 0.75 m = 0.75 ÷ 0.50 = 1.5λ
Step 3 – phase difference
F: φ = 2πd ÷ λ
I: φ = 2π × 0.75 ÷ 0.50 = 3π rad, which is the same as π rad (antiphase)
A: path difference 0.75 m; the waves arrive in antiphase and interfere destructively, so she hears a quiet sound (a minimum)
Watch out: examiners report students mixing up path difference and phase difference. Path difference is a distance in metres (here 1.5λ); phase difference is an angle in radians (here π rad). Do not write ‘phase difference = λ/2’.
Q8 (Exam) A wave of frequency 12 Hz travels at 3.6 m s⁻¹ along a slinky spring. Points P and Q are 0.21 m apart along the spring, with the wave reaching P first. Calculate the phase difference between P and Q in radians, to 3 significant figures, and in degrees.
Show solution
Step 1 – wavelength
λ = c ÷ f = 3.6 ÷ 12 = 0.30 m
Step 2 – phase difference
F: φ = 2πd ÷ λ
I: φ = 2π × 0.21 ÷ 0.30 = 2π × 0.70 = 1.4π
F: φ is already the subject
A: 4.398… rad = 4.40 rad (3 s.f.) = 252° (Q lags P)
Note: going the other way round the cycle, this is equivalent to Q leading P by 2π − 1.4π = 0.6π = 1.88 rad (108°)
Q9 (Stretch) To measure the speed of sound, a loudspeaker driven at 400 Hz faces two microphones connected to a dual-beam oscilloscope. With the microphones side by side, the traces are in phase. One microphone is moved directly away from the loudspeaker; the traces come back into phase 4 times as it moves through 3.40 m. (a) Calculate the wavelength and the speed of sound. (b) State how far the microphone had moved when the traces were first in antiphase. (c) Calculate the phase difference between the traces when the microphones are 1.00 m apart, in radians to 3 significant figures.
Show solution
Step 1 – (a) wavelength and speed
back in phase each time the separation increases by one wavelength, so λ = 3.40 ÷ 4 = 0.85 m
F: c = fλ
I: c = 400 × 0.85
A: λ = 0.85 m, c = 340 m s⁻¹
Step 2 – (b) first antiphase
antiphase first occurs at d = λ/2 = 0.425 m
Step 3 – (c) phase difference at 1.00 m
d ÷ λ = 1.00 ÷ 0.85 = 1.176…, i.e. 1 whole cycle plus 0.176… of a cycle
φ = 2π × 0.176… = 1.11 rad (3 s.f.)
Watch out: counting several wavelengths and dividing reduces the percentage uncertainty in λ compared with measuring just one.
Q10 (Stretch) A transverse wave of frequency 50 Hz and amplitude 3.0 cm travels at 20 m s⁻¹ along a rope from point P towards point Q, which is 0.10 m further along. (a) Calculate the phase difference between P and Q. (b) Calculate the time between P reaching maximum displacement and Q reaching maximum displacement, and show that it gives the same phase difference. (c) A point R, between 0.90 m and 1.20 m from P, oscillates in antiphase with P. Find its distance from P and state its displacement when P is at +3.0 cm.
Show solution
Step 1 – (a) wavelength and phase difference
λ = c ÷ f = 20 ÷ 50 = 0.40 m
F: φ = 2πd ÷ λ
I: φ = 2π × 0.10 ÷ 0.40
A: π/2 rad (90°), with Q lagging P
Step 2 – (b) time difference
Δt = d ÷ c = 0.10 ÷ 20 = 5.0 × 10−3 s = 5.0 ms; T = 1 ÷ 50 = 0.020 s
φ = 2πΔt ÷ T = 2π × 0.0050 ÷ 0.020 = π/2 rad ✓
Step 3 – (c) point R
antiphase points are at d = (n + ½)λ = 0.20 m, 0.60 m, 1.00 m, 1.40 m…, so R is 1.00 m from P
in antiphase, R is always at the opposite displacement to P, so R is at −3.0 cm
Watch out: being in antiphase does not mean being ‘half a wavelength away’ only. Any odd number of half wavelengths works, which is why R is at 2.5λ.
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