OCR GCSE Physics A: Sound waves at boundaries and hearing (J249) – Revision Notes

These notes condense statements P5.1f to P5.1i of the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026). They are Higher Tier only. For full explanations and longer worked examples, read the study guide. Test yourself with the practice questions, tick topics off on the checklist, see the OCR GCSE Physics hub and find weak spots with the free diagnostics.

Key facts at a glance

Statement	Key point	Tier
P5.1f	Crossing a boundary: frequency fixed, speed and wavelength change together	Higher Tier only
P5.1g	Reflection, transmission and absorption at an interface; ultrasound and sonar use echoes	Higher Tier only
P5.1h	Sound in air and vibrations in solids convert both ways; the ear is the main example	Higher Tier only
P5.1i	Conversions work over a limited frequency range; human range is roughly 20 Hz to 20 kHz; it narrows with age	Higher Tier only

P5.1f: sound moving between media

Rules

• Frequency stays the same. The source sets it.
• Speed changes with the material. Usually slowest in gases, faster in liquids, fastest in solids.
• Wavelength changes in proportion to speed: λ₂ / λ₁ = v₂ / v₁.

Equations

Quantity	Equation	Units
Wave speed	v = f λ	m/s, Hz, m
Wavelength	λ = v / f	m
Frequency	f = v / λ	Hz

Method in steps

1. Write down the frequency and convert kHz or MHz to Hz.
2. Find the wavelength in the first medium: λ = v / f.
3. Keep the same f. Use the new speed to find the new wavelength.
4. Check: did the wavelength change by the same factor as the speed?

Worked reminder. A 640 Hz note passes from air (340 m/s) into water (1480 m/s). In air, λ = 340 / 640 = 0.53 m. In water, λ = 1480 / 640 = 2.3 m. Both values are given to 2 significant figures. The frequency is 640 Hz in both.

Another direction. A note has wavelength 2.5 m in water, where the speed is 1480 m/s. Frequency = 1480 / 2.5 = 592 Hz. In air, λ = 340 / 592 = 0.57 m.

P5.1g: interfaces

Effect	Meaning	Where the energy goes
Reflection	Wave bounces back from the boundary (an echo)	Remains wave energy in the first medium
Transmission	Wave carries on into the second medium	Wave energy in the second medium, with new speed and wavelength
Absorption	Wave is taken in by a material	Thermal store of the material, which warms slightly

• Usually all three happen at once, in different amounts.
• A big difference between the two materials gives a strong reflection.
• Soft, porous materials absorb sound and reduce echoes.

Ultrasound. Frequency above the upper limit of human hearing. Pulses go into the body, echoes come back from tissue boundaries, and the timing builds an image. Gel on the skin removes the air gap that would reflect nearly all of the ultrasound.

Sonar. A pulse is sent through water and the echo is timed. Depth or distance = (speed × time) / 2. The halving is for the return journey.

Worked reminder. Echo from the seabed after 0.68 s, speed in water 1500 m/s. Distance there and back = 1500 × 0.68 = 1020 m. Depth = 510 m.

P5.1h: converting sound and vibrations in solids

Sound (a pressure wave in air) makes a solid surface vibrate. A vibrating solid pushes air and makes sound. Both directions occur.

Examples to quote

• Loudspeaker cone: vibrating solid produces sound in air.
• Microphone diaphragm: sound in air makes a thin solid sheet vibrate.
• Tuning fork on a table: the table vibrates and radiates the sound more loudly.
• The ear.

Chain in the ear (learn in order)

1. Pinna and ear canal collect and guide the sound.
2. Eardrum vibrates.
3. Three small bones (ossicles) pass the vibration on.
4. Cochlea (fluid-filled): the vibration becomes a wave in liquid.
5. Hair cells produce electrical signals.
6. Auditory nerve carries them to the brain.

Only a simple structure of the ear is expected, so you do not need to name the individual ossicles.

P5.1i: limited range and human hearing

Why the range is limited

• Every vibrating part has mass and stiffness.
• At very high frequency the part cannot change direction fast enough to keep in step, so its vibration is tiny.
• At very low frequency the response is weak and the part does not follow the pressure changes efficiently.
• Outside the working range, the vibration is too small to be passed on and no signal is produced.

Human hearing

• Typical young person: roughly 20 Hz to 20 kHz.
• Below 20 Hz: infrasound. Above 20 kHz: ultrasound.
• Other species have different ranges because their ears are built differently.
• With age, the upper limit falls first. Hair cell damage and long-term exposure to loud noise contribute.

Worked reminder. Wavelength of a 20 kHz sound in air at 340 m/s: 340 / 20 000 = 0.017 m. A 20 Hz sound: 340 / 20 = 17 m.

Model sentences to learn

Explain why the wavelength changes but the frequency does not. Each compression arriving at the boundary starts one compression in the second medium, so the number per second is the same. The speed is different in the new material, and since v = f λ with f fixed, the wavelength changes in proportion to the speed.

Describe what happens when ultrasound meets a boundary between two tissues. Some of the ultrasound is reflected back to the probe, some is transmitted into the second tissue where its speed and wavelength change, and a small amount is absorbed, which warms the tissue slightly.

Describe how the ear converts a sound wave into a signal. Sound in air makes the eardrum vibrate. The ossicles pass the vibration to the cochlea, where it becomes a wave in fluid. Hair cells produce electrical signals that travel along the auditory nerve.

Explain why we cannot hear a 40 kHz sound. The parts of the ear that must vibrate have mass and stiffness, so they cannot follow such rapid changes. Their vibration is too small to stimulate the hair cells and no signal reaches the brain.

Ultrasound wavelength reminder

Ultrasound of frequency 3.5 MHz in tissue where the speed is 1540 m/s: convert 3.5 MHz to 3 500 000 Hz, then λ = 1540 / 3 500 000 = 4.4 × 10⁻⁴ m, which is 0.44 mm.

Must-know distinctions

Idea	Do not confuse with
Frequency is unchanged at a boundary	Wavelength, which changes
Absorption (energy goes to the thermal store)	Transmission (wave carries on)
Ultrasound (above 20 kHz)	Infrasound (below 20 Hz)
Echo distance = half of speed × time	Speed × time (which is the round trip)
Eardrum: air wave to solid vibration	Cochlea: vibration to liquid wave, then nerve signals

Quick self-test

1. Which quantity stays constant when sound passes from air into water?
2. A 500 Hz note travels in air at 340 m/s. Find its wavelength.
3. The same note enters a liquid where its wavelength is 2.9 m. Find the speed in the liquid.
4. Name the three things that can happen to a wave at an interface.
5. Why is gel used in an ultrasound scan?
6. A sonar echo returns after 0.20 s. The speed is 1500 m/s. How deep is the object?
7. Put in order: cochlea, eardrum, ossicles, auditory nerve.
8. Why is a loudspeaker cone an example of P5.1h?
9. Give the approximate frequency range of human hearing.
10. Which end of the range is lost first with age?

Answers

1. Frequency.
2. 340 / 500 = 0.68 m.
3. v = 500 × 2.9 = 1450 m/s.
4. Reflection, transmission, absorption.
5. It removes the air gap, which would reflect almost all of the ultrasound at the skin and stop it entering the body.
6. 1500 × 0.20 / 2 = 150 m.
7. Eardrum, ossicles, cochlea, auditory nerve.
8. A vibrating solid converts to a sound wave in air.
9. Roughly 20 Hz to 20 kHz.
10. The high-frequency end, so the upper limit falls.

Where marks are usually lost

• Stating that frequency changes at the boundary. Examiners expect frequency to be unchanged.
• Forgetting to convert kHz or MHz to Hz before using v = f λ.
• Forgetting to halve the time or distance for echoes. A single-leg answer is a clear sign it was missed.
• Reporting the wavelength in the new medium using the old speed.
• Mixing up absorbed and transmitted, or leaving out where absorbed energy goes.
• Listing ear parts without saying what each does to the wave. A strong answer states the change of form at each stage.
• Giving “the ear is not sensitive enough” as the reason for the limited range, instead of referring to mass, stiffness or inertia.
• Writing the hearing range without units, or as 20 to 20 000 only.

Official syllabus

OCR, GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), first assessment 2018. Statements P5.1f to P5.1i.