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FractionRush AQA GCSE Physics 4.6
AQA GCSE Physics 4.6

Sound Waves & Ultrasound

Sound as longitudinal wave · Speed in media · Echoes · Ultrasound in medicine and sonar

Learning Objectives

🌊 Describe sound as a longitudinal wave and explain compression and rarefaction
⚡ Use the wave speed equation v = fλ to solve problems involving sound
🏗️ Explain how the speed of sound varies in solids, liquids and gases
📡 Use the echo equation to calculate distances and speeds
🔬 Describe the properties of ultrasound and its uses in medical imaging
🚢 Explain how sonar uses ultrasound to map the seabed and detect objects

🌊 Sound as a Longitudinal Wave

Sound is a mechanical, longitudinal wave. Unlike transverse waves (where the oscillation is perpendicular to the direction of travel), in a longitudinal wave the particles oscillate parallel to the direction of energy transfer.

Longitudinal wave: A wave in which the particles of the medium vibrate back and forth along the same direction as the wave travels.

As a sound wave travels through a medium, it creates alternating regions of compression (where particles are pushed closer together, increasing pressure) and rarefaction (where particles are pulled further apart, decreasing pressure). These regions move outward from the source, carrying energy — but the particles themselves only oscillate about fixed positions.

Sound CANNOT travel through a vacuum because there are no particles to carry the compressions and rarefactions. This is famously demonstrated by the bell-in-a-jar experiment, where the sound fades as air is pumped out.

The frequency of a sound wave determines its pitch — a higher frequency gives a higher pitch. The amplitude determines loudness — a larger amplitude means more energy and a louder sound. The human ear can detect sounds in the approximate range 20 Hz to 20 000 Hz.

Ultrasound: Sound with a frequency above 20 000 Hz (20 kHz) — above the upper limit of human hearing.

The wave equation links speed, frequency and wavelength for all waves including sound:

v = f × λ
v = wave speed (m/s)  |  f = frequency (Hz)  |  λ = wavelength (m)
SymbolQuantitySI Unit
vWave speedmetres per second (m/s)
fFrequencyhertz (Hz)
λWavelengthmetres (m)
AAmplitudemetres (m)
TPeriodseconds (s)

🏗️ Speed of Sound in Different Media

The speed of sound depends on the density and elasticity of the medium through which it travels. Sound travels fastest in solids, slower in liquids, and slowest in gases. This is because particles in a solid are tightly packed and strongly bonded, so they can pass on vibrations more efficiently.

MediumApproximate Speed of Sound
Air (20°C)~343 m/s
Water~1480 m/s
Steel~5100 m/s
Concrete~3400 m/s
Glass~5000 m/s

In general: vsolid > vliquid > vgas. The speed of sound in air also increases with temperature — warmer air has more energetic molecules that can transmit the pressure variations more rapidly.

The speed of sound in air is approximately 330–340 m/s at room temperature. For GCSE calculations, use 340 m/s unless given otherwise.

Because sound travels much faster in solids, you can hear a distant train through the rails long before you hear it through the air. Similarly, whales use the ocean as an efficient medium for long-distance communication because sound travels about four times faster in water than in air.

It is important to note that the speed of a wave depends only on the medium — changing the frequency of the source does not change the wave speed in a given medium. Instead, changing frequency changes the wavelength (since v = fλ and v is constant).

📡 Echoes and Distance Calculations

An echo is a reflected sound wave. When sound hits a hard surface (such as a cliff, wall, or the seabed), it reflects back towards the source. By measuring the time between transmitting a pulse of sound and receiving its echo, we can calculate how far away the reflecting surface is.

Echo: A reflected sound wave that returns to the original source after bouncing off a surface.

The key equation uses the fact that the total distance travelled by the sound equals twice the distance to the reflector (there and back):

d = v × t ÷ 2
d = distance to reflecting surface (m)  |  v = speed of sound (m/s)  |  t = total time for echo (s)

Alternatively written as: 2d = v × t, so d = vt / 2

This principle is used in several real-world technologies:

Always divide the total echo time by 2 when calculating distance — the sound has to travel to the object AND back.

🔬 Ultrasound — Properties and Uses

Ultrasound is sound with a frequency above 20 000 Hz (20 kHz). It shares all the properties of sound: it is a mechanical longitudinal wave, it requires a medium, and it travels faster in denser materials. However, its very high frequency gives it useful properties for technology.

Why use ultrasound rather than audible sound?

Medical Imaging (Ultrasound Scanning)

In medicine, a transducer placed on the skin emits pulses of ultrasound. These pulses travel through body tissues and are partially reflected at boundaries between tissues of different densities (e.g. between muscle and fat, or between tissue and bone). The reflected pulses (echoes) are detected by the same transducer and converted into electrical signals. A computer processes the time delays and amplitudes of the echoes to build up a 2D or 3D image of internal structures.

Ultrasound scanning is routinely used to image unborn babies (prenatal scanning) because it is safe — it does not use ionising radiation. It is also used to image soft tissues such as tendons, muscles and organs.

A coupling gel is applied to the skin to prevent air gaps (which would cause total reflection of the ultrasound beam before it enters the body).

SONAR (Sound Navigation And Ranging)

SONAR uses ultrasound pulses emitted from a ship or submarine. The pulses travel through water, reflect off the seabed or other objects, and return to a receiver. By measuring the time delay and knowing the speed of sound in seawater (~1500 m/s), the depth or distance to objects can be calculated using d = vt/2.

SONAR is used to:

The same echo-location principle is used biologically by bats (in air, ~40–100 kHz) and dolphins (in water, up to 150 kHz) to navigate and hunt in the dark or murky water.

📐 Summary of Key Equations

Wave speed:   v = f × λ
Echo distance:   d = (v × t) ÷ 2
Period and frequency:   T = 1 ÷ f
ConceptKey fact
Type of waveLongitudinal (compressions & rarefactions)
Speed in mediaSolid > Liquid > Gas
Speed in air~340 m/s
Speed in water~1500 m/s
Ultrasoundf > 20 000 Hz
Ultrasound advantageNon-ionising, high resolution, safe
Echo equationd = vt/2 (divide by 2 — two-way trip)
Example 1: A sound wave travels through air at 340 m/s with a frequency of 850 Hz. Calculate the wavelength of the sound wave.
1 Write down the wave equation and identify known quantities.
v = f × λ  |  v = 340 m/s  |  f = 850 Hz  |  λ = ?
2 Rearrange for wavelength.
λ = v ÷ f
3 Substitute values.
λ = 340 ÷ 850
4 Calculate.
λ = 0.4 m
λ = 0.40 m
Example 2: A ship uses SONAR to find the depth of the ocean. An ultrasound pulse is emitted and the echo returns 0.24 seconds later. The speed of sound in seawater is 1500 m/s. Calculate the depth of the ocean at that point.
1 Identify the echo equation and known quantities.
d = (v × t) ÷ 2  |  v = 1500 m/s  |  t = 0.24 s  |  d = ?
2 Note: we divide by 2 because the sound travels DOWN and then back UP — the total distance is 2d, not d.
3 Substitute values.
d = (1500 × 0.24) ÷ 2
4 Calculate numerator first.
1500 × 0.24 = 360 m (total distance)
5 Divide by 2.
d = 360 ÷ 2 = 180 m
Depth = 180 m
Example 3: An ultrasound scanner operating at 3.5 MHz is used in a medical scan. The speed of ultrasound in soft tissue is 1540 m/s. Calculate (a) the wavelength of the ultrasound in tissue, and (b) the period of the wave.
1 Convert frequency to Hz.
f = 3.5 MHz = 3.5 × 10⁶ Hz = 3 500 000 Hz
2 Part (a): find wavelength using v = f × λ.
λ = v ÷ f = 1540 ÷ 3 500 000
3 Calculate λ.
λ = 4.4 × 10⁻⁴ m = 0.00044 m (0.44 mm)
4 Part (b): find period using T = 1 ÷ f.
T = 1 ÷ 3 500 000
5 Calculate T.
T = 2.86 × 10⁻⁷ s
(a) λ = 4.4 × 10⁻⁴ m  |  (b) T = 2.86 × 10⁻⁷ s
Example 4: A student stands 68 m from a large wall and claps once. She hears the echo 0.4 s later. Calculate the speed of sound in air from this data.
1 The sound travels to the wall and back — total distance = 2 × 68 = 136 m.
2 Use the speed equation: v = total distance ÷ time.
v = 136 ÷ 0.4
3 Calculate.
v = 340 m/s
Speed of sound = 340 m/s

Question 1: Which of the following correctly describes how sound travels through air?

Question 2: A sound wave has a frequency of 440 Hz and a wavelength of 0.775 m. What is the speed of this sound wave?

Question 3: A submarine detects an echo from the seabed 0.36 s after sending an ultrasound pulse. The speed of sound in seawater is 1500 m/s. What is the depth of the seabed below the submarine? Enter your answer in metres.

Question 4: Why is ultrasound preferred over X-rays for imaging an unborn baby?

Question 5: A sound wave travels through steel at 5100 m/s. Its frequency is 1700 Hz. Calculate the wavelength of the sound in steel. Enter your answer in metres.

Challenge 1: A fishing vessel uses SONAR to locate a shoal of fish. An ultrasound pulse is transmitted and an echo is detected 0.12 s later from the fish and another echo 0.30 s later from the seabed. The speed of sound in seawater is 1500 m/s.

(a) Calculate the depth of the shoal of fish below the ship.

(b) Calculate the depth of the seabed.

(c) Calculate the distance between the shoal of fish and the seabed.

Challenge 2: An ultrasound medical scanner uses a frequency of 5.0 MHz. The speed of sound in muscle tissue is 1580 m/s and in fat is 1450 m/s.

(a) Calculate the wavelength of ultrasound in muscle tissue.

(b) Calculate the wavelength of ultrasound in fat.

(c) Explain why ultrasound is partially reflected at the boundary between muscle and fat tissue, and why this is useful in medical imaging.

Challenge 3: A student measures the speed of sound using an echo method. She stands 85 m from a building and measures the time between clapping her hands and hearing the echo. She repeats the measurement five times and records: 0.50 s, 0.48 s, 0.51 s, 0.49 s, 0.50 s.

(a) Calculate the mean time for the echo.

(b) Use the mean time to calculate the speed of sound.

(c) Suggest two sources of error in this experiment and explain how each could be reduced.

Challenge 4 (Extended): A bat emits ultrasound pulses at a frequency of 80 kHz and detects an echo from a moth 0.006 s after emission. The speed of sound in air is 340 m/s.

(a) Calculate the distance between the bat and the moth.

(b) Calculate the wavelength of the bat's ultrasound pulse in air.

(c) A second bat emits pulses at 40 kHz instead of 80 kHz. Explain how this affects the wavelength of the sound and discuss whether this would make it better or worse at detecting small insects.