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Sound, ultrasound and infrasound

Sound, ultrasound and infrasound

4.3.1 Sound waves, vibrations and the human ear

From sound to hearing

Definition

Sound wave

A longitudinal mechanical wave produced by a vibrating source and carried through a medium as a series of compressions and rarefactions.

  1. A vibrating source pushes and pulls nearby particles in a medium.
  2. The particles oscillate parallel to the direction of energy transfer, producing alternating compressions and rarefactions.
  3. The particles transfer the disturbance to neighbouring particles, so energy travels through the medium without the medium moving overall.
  4. Sound cannot travel through a vacuum because there are no particles to pass on the vibration.

Sound and solid vibrations

Definition

Frequency

The number of complete waves or oscillations passing a point each second.

  1. When sound reaches a solid surface, changing pressure exerts an alternating force on it and can make it vibrate.
  2. A vibrating solid can also push and pull the surrounding medium, converting its vibration into a sound wave.
  3. The frequency of the sound wave matches the frequency of the source vibration.
  4. A larger vibration amplitude usually produces a sound wave with a larger amplitude and a greater intensity.

The human ear

  1. The pinna collects sound and channels it through the ear canal.
  2. Pressure variations make the eardrum vibrate at the sound frequency.
  3. The three ossicles transfer the vibration through the middle ear and increase the pressure applied to the oval window.
  4. Movement at the oval window produces pressure waves in the fluid of the cochlea.
  5. Different regions of the cochlea respond most strongly to different frequencies.
  6. Sensory hair cells bend and convert the mechanical disturbance into electrical impulses.
  7. The auditory nerve carries these impulses to the brain, where they are interpreted as sound.

A limited frequency range

  1. A structure responds effectively only over a limited frequency range because its mass, stiffness and damping affect how easily it can vibrate.
  2. At some frequencies the response amplitude is large, while very low or very high frequencies produce a weaker response.
  3. The combined response of the eardrum, ossicles, cochlea and hair cells limits normal human hearing to roughly 20 Hz20\ \text{Hz}20 Hz to 20 000 Hz20\,000\ \text{Hz}20000 Hz.
  4. The upper limit usually falls with age and after exposure to intense sound.
Example

Linking pitch to vibration

  • A loudspeaker cone completes 750750750 vibrations in 0.50 s0.50\ \text{s}0.50 s.
  • Its frequency is f=7500.50=1500 Hzf=\dfrac{750}{0.50}=1500\ \text{Hz}f=0.50750​=1500 Hz.
  • The sound wave and the eardrum vibrate at 1500 Hz1500\ \text{Hz}1500 Hz, so the frequency is within the normal human hearing range.
Exam technique
  • For an ear question, follow the energy transfer in order: sound wave, eardrum, ossicles, oval window, cochlear fluid, hair cells, auditory nerve.
  • Use the word converts only when the form of the disturbance changes, such as a mechanical vibration becoming an electrical impulse.
Common Mistake
  • Do not say that air particles travel from the source to the ear; they oscillate about fixed positions.
  • Do not confuse frequency with amplitude: frequency affects pitch, whereas amplitude is linked to loudness.
Self review
  • How does a vibrating source produce a sound wave?
  • Why can sound not travel through a vacuum?
  • What sequence transfers a vibration from the eardrum to the cochlea?
  • How do hair cells contribute to hearing?
  • Why does the ear work over only a limited frequency range?

4.3.2 Ultrasound and infrasound

Frequencies beyond hearing

Definition

Ultrasound

Sound with a frequency greater than 20 000 hertz, above the upper limit of normal human hearing.

Definition

Infrasound

Sound with a frequency less than 20 hertz, below the lower limit of normal human hearing.

  1. Sound from about 20 Hz20\ \text{Hz}20 Hz to 20 000 Hz20\,000\ \text{Hz}20000 Hz lies within the typical hearing range of a young person.
  2. The limits describe frequency, not loudness, so an intense ultrasound wave remains inaudible to humans.
  3. Ultrasound and infrasound are mechanical waves and therefore require a medium.

Wave properties

  1. For every sound wave, speed, frequency and wavelength are linked by v=fλv=f\lambdav=fλ.
  2. At a fixed sound speed, ultrasound has a shorter wavelength than audible sound, while infrasound has a longer wavelength.
  3. Short wavelengths can detect small details, which is useful for imaging and flaw detection.
  4. Long wavelengths can travel around large obstacles and may propagate over very long distances.

Producing and detecting ultrasound

  1. A rapidly alternating potential difference can make a piezoelectric crystal change shape and vibrate.
  2. The vibrating crystal produces ultrasound at the driving frequency.
  3. Returning ultrasound makes the crystal vibrate and produces an alternating potential difference, so the same type of transducer can act as a detector.

Sources of infrasound

  1. Large, slow vibrations can produce infrasound, including earthquakes, volcanic activity, ocean waves and large machinery.
  2. Some animals can produce or detect frequencies outside the human hearing range.
  3. Detecting a wave does not mean a person hears it; electronic sensors can convert it into a measurable electrical signal.
Example

Comparing wavelengths

  • Sound travels at 340 m s−1340\ \text{m s}^{-1}340 m s−1 in air.
  • For ultrasound at 40 000 Hz40\,000\ \text{Hz}40000 Hz, λ=34040 000=8.5×10−3 m\lambda=\dfrac{340}{40\,000}=8.5\times10^{-3}\ \text{m}λ=40000340​=8.5×10−3 m.
  • For infrasound at 10 Hz10\ \text{Hz}10 Hz, λ=34010=34 m\lambda=\dfrac{340}{10}=34\ \text{m}λ=10340​=34 m.
  • The infrasound wavelength is much larger than the ultrasound wavelength.
Exam technique
  • Classify a sound by comparing its frequency with the two numerical boundaries and include the unit hertz.
  • When explaining a use, link frequency to wavelength and then to the required effect, such as resolving small structures.
Common Mistake
  • Do not write that ultrasound is sound above the loudness threshold; ultrasound is defined only by frequency.
  • Use 20 000 Hz20\,000\ \text{Hz}20000 Hz, not 20 000 kHz20\,000\ \text{kHz}20000 kHz.
Self review
  • What frequency range defines ultrasound?
  • What frequency range defines infrasound?
  • Why does ultrasound have a short wavelength in a given medium?
  • How can a piezoelectric crystal produce and detect ultrasound?
  • Name two natural sources of infrasound.

4.3.3 Uses of ultrasound and infrasound

Uses of high and low frequencies

  1. Ultrasound and infrasound are useful because their wavelengths, propagation and reflection differ from those of audible sound.
  2. A complete explanation links the wave property to the way information is obtained.

Sonar

Definition

Sonar

A technique that sends ultrasound pulses through water and uses the time for their echoes to return to find the distance to an object or the depth of water.

  1. A transmitter sends an ultrasound pulse through water.
  2. The pulse reflects from the seabed, a shoal of fish or another boundary and returns to a detector.
  3. The measured time is for the outward and return journeys, so the one-way distance is d=vt2d=\dfrac{vt}{2}d=2vt​.
  4. Short pulses and short wavelengths help distinguish nearby objects and provide better detail.
Example

Finding sea depth

  • A sonar pulse returns from the seabed after 0.080 s0.080\ \text{s}0.080 s, and sound travels through seawater at 1500 m s−11500\ \text{m s}^{-1}1500 m s−1.
  • The total travelled distance is vt=1500×0.080=120 mvt=1500\times0.080=120\ \text{m}vt=1500×0.080=120 m.
  • The water depth is d=1202=60 md=\dfrac{120}{2}=60\ \text{m}d=2120​=60 m.

Foetal scanning

  1. A transducer sends pulses of ultrasound into the body through coupling gel, which reduces reflection at the skin.
  2. At boundaries between tissues, some ultrasound is reflected while the rest continues deeper.
  3. The time delay gives the depth of a boundary, and the echo strength helps distinguish different tissues.
  4. A computer combines many echoes to form an image of the foetus.
  5. Ultrasound is non-ionising, so it does not carry the ionisation risk associated with X-rays, although exposure is still kept as low as needed.

Exploring the Earth

  1. Earthquakes produce seismic waves with very low frequencies, including infrasound and vibrations that pass through the Earth.
  2. Seismic waves change speed, refract and reflect at boundaries between layers.
  3. Detectors around the Earth record arrival times and wave paths.
  4. Differences in speed and the presence of shadow zones provide evidence for internal layers, including the mantle and core.
  5. Different seismic wave types behave differently in solids and liquids, which helps identify whether a layer is solid or liquid.
Exam technique
  • For an echo calculation, state that the measured time covers two journeys before dividing by 222.
  • For a scanning explanation, link each echo's time delay to depth and its amplitude to the nature of the boundary.
  • For the Earth's core, refer to changes in speed, refraction, reflection and detected arrival patterns.
Common Mistake
  • Do not describe a foetal scan as an X-ray image; it is formed from reflected ultrasound pulses.
  • Do not forget the factor of 222 in pulse-echo distance calculations.
Self review
  • How does sonar determine the distance to an object?
  • Why is coupling gel used in an ultrasound scan?
  • What information is obtained from echo time and echo amplitude?
  • How do seismic-wave paths provide evidence about the Earth's interior?
  • Why must pulse-echo travel time be divided by two?

4.3.4 Sound waves passing between media

Sound crossing a boundary

Definition

Medium

A substance whose particles pass on the oscillations of a mechanical wave, such as sound, from one place to another.

  1. When sound crosses from one medium into another, its speed usually changes because particle spacing, stiffness and density affect how quickly vibrations are transferred.
  2. The source remains unchanged, so the frequency is unchanged at the boundary.
  3. The wavelength changes because v=fλv=f\lambdav=fλ.

Linking speed and wavelength

  1. If sound speed increases while frequency stays constant, wavelength increases.
  2. If sound speed decreases while frequency stays constant, wavelength decreases.
  3. The ratio is v2v1=λ2λ1\dfrac{v_2}{v_1}=\dfrac{\lambda_2}{\lambda_1}v1​v2​​=λ1​λ2​​ because fff is the same on both sides.
  4. The period T=1fT=\dfrac{1}{f}T=f1​ also remains unchanged.

Direction and transmission

  1. Sound arriving at an angle may refract because one side of the wavefront changes speed first.
  2. If sound speeds up, it bends away from the normal; if it slows down, it bends towards the normal.
  3. Sound arriving along the normal changes speed and wavelength without changing direction.
  4. At the same boundary, some sound energy may be reflected and some may be absorbed, so the transmitted amplitude can be smaller.

Sound in different materials

  1. Sound generally travels faster in solids than in liquids and faster in liquids than in gases because the restoring forces between particles are stronger in more rigid media.
  2. Density alone is not a complete rule because wave speed depends on both density and stiffness.
  3. A change in wave speed does not mean the sound source has changed pitch because pitch is linked to frequency, which remains constant.
Example

Sound entering water

  • A sound wave of frequency 500 Hz500\ \text{Hz}500 Hz travels at 340 m s−1340\ \text{m s}^{-1}340 m s−1 in air and 1500 m s−11500\ \text{m s}^{-1}1500 m s−1 in water.
  • In air, λ1=340500=0.68 m\lambda_1=\dfrac{340}{500}=0.68\ \text{m}λ1​=500340​=0.68 m.
  • In water, λ2=1500500=3.0 m\lambda_2=\dfrac{1500}{500}=3.0\ \text{m}λ2​=5001500​=3.0 m.
  • The frequency stays at 500 Hz500\ \text{Hz}500 Hz while speed and wavelength both increase.
Exam technique
  • For a three-quantity question, state each result explicitly: speed changes, frequency stays constant and wavelength changes in the same direction as speed.
  • Support the relationship with v=fλv=f\lambdav=fλ and keep units consistent.
Common Mistake
  • Do not state that frequency changes because the medium changes; the source sets frequency.
  • Do not use density alone to predict sound speed because stiffness also affects the result.
Self review
  • Which sound-wave quantity remains unchanged at a boundary?
  • What happens to wavelength when sound speed increases?
  • Why can sound refract when it enters a new medium at an angle?
  • Why is density alone insufficient to predict sound speed?
  • How are v2/v1v_2/v_1v2​/v1​ and λ2/λ1\lambda_2/\lambda_1λ2​/λ1​ related?

Recap questions

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A loudspeaker is switched on inside a chamber, and the air is pumped out until the sound outside can no longer be heard. Why does the sound stop reaching you?

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Longitudinal sound wave in air showing compressions, rarefactions, wavelength, particle motion, and direction of travel

Sound is produced by vibrations. In air it travels as a longitudinal wave, so particles vibrate back and forth parallel to the direction the wave moves.

Sound needs a medium such as air, water or metal, so it is a mechanical wave and cannot travel through a vacuum. Crowded regions are compressions and spread-out regions are rarefactions, while the particles only oscillate about fixed positions, so energy is transferred but matter is not.

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Which of the following frequencies is classified as ultrasound?

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What causes a sound wave?

4.3 Sound, ultrasound and infrasound Revision Guide

  1. GCSE
  2. /Physics
  3. /4.3 Sound, ultrasound and infrasound

Revision notes for Edexcel GCSE Physics 4.3 Sound, ultrasound and infrasound. Open the guide for explanations and worked examples. Written against the Edexcel GCSE Physics (1PH0) specification, so the content matches what's examinable rather than general Physics background.