Define ultrasound as sound with a frequency higher than 20 kHz

3.4 Sound

Learning Objectives

• Explain how a vibrating source produces sound and why the wave is longitudinal.
• State the audible frequency range for humans (20 Hz – 20 kHz) and define ultrasound as sound with a frequency \(f > 20\;\text{kHz}\).
• Describe the relationship between amplitude, frequency, loudness and pitch.
• Recall the speed of sound in air (330 – 350 m s⁻¹) and compare it with the speeds in water and steel (supplementary data).
• Explain echo as a reflection of sound and use it to determine distances.
• Identify practical uses of ultrasound.
• Apply the wave relationship \(v = f\lambda\) to calculate wavelength.

How Sound Is Produced

• A vibrating source (e.g. a tuning‑fork, a loud‑speaker diaphragm) repeatedly compresses and expands the surrounding particles.
• These periodic compressions and rarefactions travel outward as a wave.

Why Sound Needs a Material Medium

Sound is a mechanical disturbance; it cannot travel through a vacuum because there are no particles to transmit the compressions and rarefactions. A solid, liquid or gas is therefore required.

Why Sound Is Longitudinal

• In a longitudinal wave the particle displacement is parallel to the direction of wave propagation.
• Each particle pushes its neighbour, creating a series of compressions (high pressure) and rarefactions (low pressure).

Amplitude, Frequency, Loudness and Pitch

Property	What It Represents	Effect on Perception
Amplitude (A)	Maximum displacement of particles (m)	Greater amplitude → louder sound (higher intensity)
Frequency (f)	Number of cycles per second (Hz)	Greater frequency → higher pitch

Speed of Sound – Core Values

For the IGCSE syllabus the following speeds are required:

Medium	Speed \(v\) (m s⁻¹)
Air (20 °C, 1 atm)	≈ 330 – 350 m s⁻¹ (commonly quoted as 340 m s⁻¹)

Supplementary data (optional for extended syllabus)

Medium	Speed \(v\) (m s⁻¹)
Water (fresh, 20 °C)	≈ 1 500 m s⁻¹
Steel	≈ 5 000 m s⁻¹

Sound travels faster in media that are less compressible and more rigid. Gases are highly compressible, so the speed is lowest; liquids are less compressible, and solids are the least compressible, giving the highest speeds.

Quantitative illustration – “faster in solids”

Time to travel 1 m:

• Air: \(t = \dfrac{1\;\text{m}}{340\;\text{m s}^{-1}} \approx 2.9\times10^{-3}\;\text{s}\)
• Steel: \(t = \dfrac{1\;\text{m}}{5\,000\;\text{m s}^{-1}} = 2.0\times10^{-4}\;\text{s}\)

The wave reaches the same point about 15 times faster in steel than in air.

Echo – Reflection of Sound

• An echo occurs when a sound wave reflects from a surface and returns to the source.
• The round‑trip distance can be found from the measured time \(t\): \[ d = \frac{v\,t}{2} \] where \(v\) is the speed of sound in the medium.

Frequency Ranges

Type of Sound	Frequency Range	Typical Examples
Infrasound	< 20 Hz	Earthquakes, volcanic eruptions, large explosions
Audible Sound	20 Hz – 20 kHz	Speech, musical instruments, everyday noises
Ultrasound	\(> 20\;\text{kHz}\)	Medical imaging, industrial cleaning, sonar

Ultrasound

Sound with a frequency \(f > 20\;\text{kHz}\) is called ultrasound. It is inaudible to humans but offers several practical advantages:

• Short wavelength (\(\lambda = v/f\)) → higher resolution in imaging.
• Can penetrate many solids and liquids that visible light cannot.
• Can be focused to a small spot, enabling precise cleaning, cutting, or medical diagnostics.

Compression & Rarefaction

During one complete cycle of a longitudinal sound wave the particles of the medium undergo:

1. Compression – particles are pushed together, creating a region of higher pressure.
2. Rarefaction – particles are pulled apart, creating a region of lower pressure.

Key Formula

The fundamental wave relationship is:

\[ v = f\lambda \]

where

• \(v\) = speed of sound (m s⁻¹)
• \(f\) = frequency (Hz)
• \(\lambda\) = wavelength (m)

Worked Example – Wavelength of Ultrasound in Air

Calculate the wavelength of a 40 kHz ultrasound wave travelling in air (use \(v = 340\;\text{m s}^{-1}\)).

1. Write the rearranged formula: \(\lambda = \dfrac{v}{f}\).
2. Insert the values: \(\lambda = \dfrac{340\;\text{m s}^{-1}}{40\,000\;\text{Hz}} = 8.5\times10^{-3}\;\text{m}\).
3. Result: \(\lambda = 8.5\;\text{mm}\). The short wavelength explains the high resolution of medical ultrasound.

Practical Investigation – Measuring the Speed of Sound (Echo Method)

1. Place a loudspeaker at one end of a straight tube and a microphone at the opposite end.
2. Generate a short pulse (e.g., with an electronic pulse generator).
3. Record the time \(t\) between the emitted pulse and the echo received after reflecting from the far end.
4. Measure the distance \(L\) between speaker and reflector.
5. Calculate the speed using \(v = \dfrac{2L}{t}\) (the factor 2 accounts for the round‑trip).

This experiment reinforces the concepts of wave speed, frequency, wavelength and echo.

Suggested Diagram