Specification: AQA A-level Physics (7408) 3.3.1.1 | Paper: 1 | Data and formulae booklet: given in the exam
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📌 KEY DEFINITIONS – learn them word for word
Progressive wave: an oscillation that travels through a medium (or a vacuum), transferring energy without transferring matter.
Transverse and longitudinal: in a transverse wave the oscillations are perpendicular to the direction of energy transfer (e.g. all electromagnetic waves); in a longitudinal wave the oscillations are parallel to the direction of energy transfer, producing compressions and rarefactions (e.g. sound).
Amplitude, wavelength, period and frequency: amplitude is the maximum displacement from the equilibrium position; wavelength λ is the minimum distance between two points oscillating in phase; period T is the time for one complete oscillation; frequency f is the number of complete oscillations per second.
The equation
wave speed = frequency × wavelength c = fλ and frequency = 1 ÷ period f = 1 ÷ T
| Quantity | Symbol | Unit |
|---|---|---|
| wave speed | c | metres per second (m s⁻¹) |
| frequency | f | hertz (Hz), equivalent to s⁻¹ |
| wavelength | λ | metres (m) |
| period | T | seconds (s) |
Rearranged: f = c ÷ λ | λ = c ÷ f | T = 1 ÷ f | Convert kHz (× 103), MHz (× 106), ms (× 10−3) and nm (× 10−9) before you substitute. When a wave enters a new medium its frequency stays the same; the speed and wavelength change.
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) A transverse wave on a rope has a frequency of 50 Hz and a wavelength of 0.40 m. Calculate the speed of the wave.
Show solution
F: c = fλ
I: c = 50 × 0.40
F: c is already the subject
A: 20 m s⁻¹
Q2 (Core) A particle on a progressive wave takes 0.025 s to make one complete oscillation. Calculate the frequency of the wave.
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F: f = 1 ÷ T
I: f = 1 ÷ 0.025
F: f is already the subject
A: 40 Hz
Q3 (Core) A radio station broadcasts at a frequency of 97.6 MHz. Calculate the wavelength of the radio waves, to 3 significant figures. (c = 3.00 × 108 m s⁻¹)
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Convert first: 97.6 MHz = 9.76 × 107 Hz
F: c = fλ
I: 3.00 × 108 = 9.76 × 107 × λ
F: λ = 3.00 × 108 ÷ 9.76 × 107
A: 3.0737… m = 3.07 m (3 s.f.)
Q4 (Core) A green laser emits light of wavelength 532 nm. Calculate the frequency of the light, to 3 significant figures. (c = 3.00 × 108 m s⁻¹)
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Convert first: 532 nm = 5.32 × 10−7 m
F: c = fλ
I: 3.00 × 108 = f × 5.32 × 10−7
F: f = 3.00 × 108 ÷ 5.32 × 10−7
A: 5.639… × 1014 Hz = 5.64 × 1014 Hz (3 s.f.)
Watch out: 1 nm = 10−9 m, not 10−6 m. Visible light frequencies are always around 1014 Hz, so use that as a sense check.
Q5 (Exam) A microphone connected to an oscilloscope picks up a sound wave. One complete cycle of the trace spans 4.0 horizontal divisions and the time-base is set to 0.50 ms per division. Calculate the frequency and the wavelength of the sound. (speed of sound in air = 340 m s⁻¹)
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Step 1 – period from the trace
T = 4.0 divisions × 0.50 ms per division = 2.0 ms = 2.0 × 10−3 s
Step 2 – frequency
F: f = 1 ÷ T
I: f = 1 ÷ 2.0 × 10−3
A: 500 Hz
Step 3 – wavelength
F: c = fλ
I: 340 = 500 × λ
F: λ = 340 ÷ 500
A: 0.68 m
Watch out: measure a full cycle (peak to next peak), not peak to trough, and multiply divisions by the time-base before taking 1 ÷ T.
Q6 (Exam) A medical ultrasound scanner uses a frequency of 3.5 MHz. The speed of ultrasound in soft tissue is 1540 m s⁻¹. Calculate the wavelength of the ultrasound in the tissue, in mm, to 2 significant figures.
Show solution
Convert first: 3.5 MHz = 3.5 × 106 Hz
F: c = fλ
I: 1540 = 3.5 × 106 × λ
F: λ = 1540 ÷ 3.5 × 106
A: 4.4 × 10−4 m = 0.44 mm (2 s.f.)
Q7 (Exam) In a ripple tank, the distance from the 1st crest to the 9th crest is 12.0 cm. The dipper makes 40 complete oscillations in 8.0 s. Calculate the speed of the water waves in cm s⁻¹.
Show solution
Step 1 – wavelength
1st to 9th crest is 8 wavelengths, so λ = 12.0 ÷ 8 = 1.5 cm = 0.015 m
Step 2 – frequency
f = 40 ÷ 8.0 = 5.0 Hz
Step 3 – speed
F: c = fλ
I: c = 5.0 × 0.015
F: c is already the subject
A: 0.075 m s⁻¹ = 7.5 cm s⁻¹
Watch out: 9 crests contain only 8 gaps. Dividing by 9 is a classic counting error.
Q8 (Exam) A ship’s sonar sends a 25 kHz pulse straight down to the sea bed, 600 m below. The echo returns 0.80 s after the pulse is sent. Calculate the speed of sound in sea water and the wavelength of the sonar waves.
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Step 1 – speed from the echo
the pulse travels down and back, so distance = 2 × 600 = 1200 m
speed = distance ÷ time = 1200 ÷ 0.80 = 1500 m s⁻¹
Step 2 – wavelength
Convert first: 25 kHz = 25 000 Hz
F: c = fλ
I: 1500 = 25 000 × λ
F: λ = 1500 ÷ 25 000
A: 0.060 m = 6.0 cm
Watch out: an echo time is for the round trip. Forgetting the factor of 2 halves the speed.
Q9 (Stretch) A tuning fork vibrating at 512 Hz produces a sound wave that passes from air into water. The speed of sound is 340 m s⁻¹ in air and 1480 m s⁻¹ in water. (a) State the period of the wave in both media, to 3 significant figures. (b) Calculate the wavelength in each medium, to 3 significant figures. (c) Calculate how many times longer the wavelength is in water.
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Step 1 – (a) period
the frequency is set by the source, so it is 512 Hz in both media
T = 1 ÷ 512 = 1.953… × 10−3 s = 1.95 ms (3 s.f.) in both
Step 2 – (b) wavelengths
F: λ = c ÷ f
I: air: λ = 340 ÷ 512; water: λ = 1480 ÷ 512
A: air: 0.6640… m = 0.664 m; water: 2.890… m = 2.89 m (3 s.f.)
Step 3 – (c) ratio
λ is proportional to c when f is fixed, so ratio = 1480 ÷ 340 = 4.35 (3 s.f.)
Watch out: frequency never changes when a wave crosses a boundary. It is the speed and wavelength that change.
Q10 (Stretch) A helium-neon laser emits red light of wavelength 633 nm in a vacuum. The light enters a glass block, 5.0 mm thick, in which it travels at 2.00 × 108 m s⁻¹. (a) Calculate the frequency of the light, to 3 significant figures. (b) Calculate its wavelength inside the glass, to 3 significant figures. (c) Calculate the number of complete wavelengths that fit across the thickness of the block, to 2 significant figures. (c = 3.00 × 108 m s⁻¹)
Show solution
Convert first: 633 nm = 6.33 × 10−7 m; 5.0 mm = 5.0 × 10−3 m
Step 1 – (a) frequency (from vacuum values)
F: c = fλ
I: 3.00 × 108 = f × 6.33 × 10−7
F: f = 3.00 × 108 ÷ 6.33 × 10−7
A: 4.739… × 1014 Hz = 4.74 × 1014 Hz (3 s.f.)
Step 2 – (b) wavelength in glass (same frequency)
λ = 2.00 × 108 ÷ 4.739… × 1014 = 4.22 × 10−7 m = 422 nm (3 s.f.)
Step 3 – (c) number of wavelengths
N = 5.0 × 10−3 ÷ 4.22 × 10−7 = 11 848…
A: 1.2 × 104 wavelengths (2 s.f.)
Watch out: use the vacuum speed with the vacuum wavelength to find f, then the glass speed for the glass wavelength. Mixing them gives the wrong frequency.
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