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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.
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4.3.2 Ultrasound and infrasound Revision Guide

  1. GCSE
  2. /Physics
  3. /4.3.2 Ultrasound and infrasound

Revision notes for Edexcel GCSE Physics 4.3.2 Ultrasound and infrasound: explanations and worked examples.

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