Sound Waves — CIE IGCSE Physics Revision Notes

What you'll learn

Sound waves form a fundamental part of the CIE IGCSE Physics specification, appearing in both Paper 2 (core) and Paper 4 (extended) examinations. This topic covers the production and properties of sound waves, how they travel through different media, and their practical applications including ultrasound. Expect questions on calculations involving wave speed, frequency and wavelength, as well as explanations of phenomena such as echoes and the limitations of human hearing.

Key terms and definitions

Longitudinal wave — a wave in which particles of the medium vibrate parallel to the direction of energy transfer, creating compressions and rarefactions.

Frequency — the number of complete waves passing a point per second, measured in hertz (Hz).

Amplitude — the maximum displacement of particles from their rest position, which determines the loudness of a sound.

Wavelength — the distance between two successive compressions (or rarefactions) in a sound wave, measured in metres.

Compression — a region in a longitudinal wave where particles are pushed closer together than their normal spacing.

Rarefaction — a region in a longitudinal wave where particles are spread further apart than their normal spacing.

Echo — a reflected sound wave that can be heard as a distinct repetition of the original sound.

Ultrasound — sound waves with frequencies above 20,000 Hz, beyond the upper limit of human hearing.

Core concepts

Nature of sound waves

Sound waves are longitudinal waves that require a medium (solid, liquid or gas) to travel through. Unlike transverse waves such as light, sound cannot travel through a vacuum because there are no particles to vibrate and transmit the energy.

When a sound source vibrates:

It creates regions of high pressure (compressions) where particles are pushed together
It creates regions of low pressure (rarefactions) where particles are pulled apart
These compressions and rarefactions travel outward from the source
The particles themselves oscillate back and forth around fixed positions rather than travelling with the wave

The bell jar experiment demonstrates that sound requires a medium:

Place an electric bell inside a sealed glass jar
Connect the bell to a power supply so it rings
Use a vacuum pump to gradually remove air from the jar
The sound becomes quieter as air is removed
When most air is removed, no sound can be heard even though the hammer is still striking the bell

This proves sound cannot travel through a vacuum, unlike electromagnetic waves.

Speed of sound in different media

The speed of sound varies significantly depending on the medium:

Air (20°C): approximately 340 m/s
Water: approximately 1500 m/s
Steel: approximately 5000 m/s

Sound travels faster in solids than liquids, and faster in liquids than gases. This occurs because particles in solids are closer together and more tightly bound, allowing vibrations to be transmitted more rapidly between neighbouring particles.

Temperature also affects sound speed in gases. As temperature increases, particles move faster and collide more frequently, transmitting sound more quickly. The speed of sound in air increases by approximately 0.6 m/s for every 1°C rise in temperature.

Wave equation and calculations

The fundamental wave equation applies to all waves, including sound:

wave speed = frequency × wavelength

or v = f × λ

Where:

v = wave speed (m/s)
f = frequency (Hz)
λ = wavelength (m)

This equation appears frequently in CIE IGCSE exam questions. You must be able to rearrange it:

f = v / λ
λ = v / f

Frequency, pitch and amplitude

Frequency determines the pitch of a sound:

High frequency = high pitch (e.g. whistle, bird song)
Low frequency = low pitch (e.g. bass drum, thunder)

The human ear can detect frequencies between approximately 20 Hz and 20,000 Hz. This range decreases with age, particularly at the upper limit.

Amplitude determines the loudness of a sound:

Large amplitude = loud sound (particles vibrate with greater displacement)
Small amplitude = quiet sound (particles vibrate with smaller displacement)

Oscilloscope traces provide visual representations:

Pitch is shown by the spacing between peaks (closer together = higher frequency)
Loudness is shown by the height of the peaks (taller = greater amplitude)

Reflection of sound and echoes

Sound waves obey the law of reflection: the angle of incidence equals the angle of reflection.

An echo occurs when a sound wave reflects off a hard surface and returns to the listener with sufficient delay to be heard as a separate sound. The human brain requires approximately 0.1 seconds between the original sound and its reflection to perceive them as distinct.

For an echo to be heard, the reflecting surface must be at least 17 m away:

Sound travels at 340 m/s in air
To create 0.1 s delay, sound must travel 34 m total (to surface and back)
Therefore minimum distance = 34 m ÷ 2 = 17 m

Echo calculations follow this principle:

Calculate total distance travelled: distance = speed × time
Divide by 2 to find distance to reflecting surface

Soft materials (carpets, curtains, foam) absorb sound energy rather than reflecting it, reducing echoes. Concert halls and theatres use acoustic design to control reflections and create optimal sound quality.

Ultrasound and its applications

Ultrasound waves have frequencies above 20,000 Hz. Although inaudible to humans, they have important practical applications:

Medical imaging (prenatal scans):

Ultrasound waves are transmitted into the body
Waves reflect at boundaries between different tissues (different densities)
A detector receives the reflected waves
Computer software calculates distances using time delays and wave speed
An image is built up showing internal structures
Safe for use during pregnancy (no ionising radiation)

Sonar and depth measurement:

Ships and submarines use ultrasound to measure water depth
A pulse is transmitted downward from the vessel
Reflection occurs at the seabed
Time for echo to return is measured
Depth = (speed × time) ÷ 2

Industrial quality control:

Ultrasound detects cracks and flaws inside metal structures
Waves reflect from internal defects
Used to test welds, pipelines and aircraft components without damaging them

Distance measurement formula: distance to reflector = (speed of sound × time for echo) ÷ 2

The division by 2 accounts for the sound travelling to the reflector and back again.

Hearing and the ear

The human ear detects sound through mechanical vibrations:

Sound waves enter the ear canal and cause the eardrum to vibrate
Three small bones (ossicles) amplify these vibrations
Vibrations pass to the cochlea, a fluid-filled organ
Hair cells in the cochlea convert vibrations to electrical impulses
The auditory nerve transmits signals to the brain

Hearing range decreases with age and exposure to loud sounds. Prolonged exposure to sounds above 85 dB can cause permanent hearing damage. Ear protection should be worn in noisy environments such as construction sites and concerts.

Worked examples

Example 1: Wave equation calculation

Question: A sound wave travelling through air has a frequency of 256 Hz and a wavelength of 1.3 m. Calculate the speed of sound in air. [2 marks]

Solution: Using v = f × λ v = 256 Hz × 1.3 m v = 332.8 m/s ≈ 333 m/s

Mark scheme: 1 mark for correct formula or substitution; 1 mark for correct answer with unit.

Example 2: Echo calculation

Question: A girl stands 85 m from a large wall and claps her hands. She hears an echo 0.5 s later. Calculate the speed of sound. [3 marks]

Solution: Total distance travelled by sound = 85 m × 2 = 170 m (Sound travels to the wall and back)

Speed = distance ÷ time Speed = 170 m ÷ 0.5 s Speed = 340 m/s

Mark scheme: 1 mark for recognising distance = 170 m (or 2 × 85); 1 mark for correct working; 1 mark for answer with unit.

Example 3: Ultrasound depth measurement

Question: A ship uses ultrasound to measure the depth of water beneath it. An ultrasound pulse is sent downward and the echo is detected 0.24 s later. The speed of sound in seawater is 1500 m/s. Calculate the depth of water. [3 marks]

Solution: Total distance travelled = speed × time Total distance = 1500 m/s × 0.24 s = 360 m

Depth of water = 360 m ÷ 2 = 180 m (The ultrasound travels down to the seabed and back up)

Mark scheme: 1 mark for calculating total distance (360 m); 1 mark for dividing by 2; 1 mark for correct final answer with unit.

Common mistakes and how to avoid them

Mistake: Stating that sound is a transverse wave. Correction: Sound is always a longitudinal wave in which particle vibrations are parallel to the direction of energy transfer.
Mistake: In echo calculations, forgetting to double the distance or divide the time by 2. Correction: Sound travels to the reflecting surface and back, covering twice the distance to the reflector. Always use: distance to object = (speed × time) ÷ 2.
Mistake: Confusing frequency with amplitude. Students often say "increasing amplitude makes sound higher pitched". Correction: Frequency determines pitch (high/low); amplitude determines loudness (loud/quiet).
Mistake: Using the wrong speed of sound value. Correction: Check whether the question specifies sound in air (≈340 m/s), water (≈1500 m/s) or another medium. Use the value given or stated in data sheets.
Mistake: Stating that ultrasound is "above the human hearing range" without giving a specific frequency value. Correction: For full marks, state that ultrasound has frequencies above 20,000 Hz or 20 kHz.
Mistake: In wave speed calculations, failing to convert units correctly (e.g. using cm instead of m). Correction: Ensure consistent SI units: metres for wavelength, hertz for frequency, metres per second for speed.

Exam technique for Sound waves

Describe/Explain questions about ultrasound applications require clear step-by-step accounts. For medical imaging: state that (1) ultrasound is transmitted into body, (2) reflections occur at boundaries, (3) detector receives echoes, (4) computer calculates distances from time delays, (5) image is produced. Each step typically earns one mark.
Calculate questions always require three elements: formula (or correct substitution), working, and answer with unit. Show every step clearly. In echo problems, explicitly state that you're dividing by 2 and explain why.
Comparison questions such as "Compare the speed of sound in air and water" require statements about both media. State which is faster and give approximate values (e.g. "Sound travels faster in water at approximately 1500 m/s compared to 340 m/s in air").
Drawing or interpreting oscilloscope traces requires understanding that horizontal spacing shows frequency (time period) and vertical height shows amplitude. High-pitched sounds have closely-spaced peaks; loud sounds have tall peaks.

Quick revision summary

Sound waves are longitudinal waves requiring a medium to travel. Key equation: v = f × λ links speed (≈340 m/s in air), frequency and wavelength. Frequency determines pitch; amplitude determines loudness. Human hearing range: 20–20,000 Hz. Echoes occur when sound reflects; minimum distance for echo is 17 m. Ultrasound (>20,000 Hz) applications include medical imaging, sonar and quality control. In calculations involving echoes or depth measurement, remember to divide total distance by 2 because sound travels to the reflector and back.