Sound Waves and Hearing

Sound as longitudinal waves; echoes; ultrasound applications; human hearing range

# Sound Waves and Hearing — GCSE Physics

Sound is one of the most familiar types of wave. From music to speech to medical ultrasound, sound waves play a crucial role in everyday life. Unlike light, sound needs a medium to travel through — it cannot travel through a vacuum.


1. Sound as a Longitudinal Wave

Sound is a longitudinal wave — the particles of the medium vibrate parallel to the direction of energy transfer.

When a speaker vibrates:

  1. It pushes air particles together → compression (high pressure)
  2. It pulls back, leaving a gap → rarefaction (low pressure)
  3. These compressions and rarefactions travel outwards through the air

The particles themselves don't travel — they oscillate back and forth, passing energy from one to the next.


2. Speed of Sound

Sound travels at different speeds in different media:

Medium Speed (m/s)
Air (20°C) ~340
Water ~1500
Steel ~5000
Vacuum 0 (cannot travel)

Sound is fastest in solids (particles close together, transfer energy quickly) and slowest in gases (particles far apart).

Measuring the Speed of Sound

Method 1 — Echo method:

  1. Stand a measured distance dd from a large wall
  2. Clap and time how long (tt) it takes to hear the echo
  3. v=2d/tv = 2d/t (sound travels there and back)

Method 2 — Two microphones:

  1. Place two microphones a known distance apart
  2. Make a sharp sound near one end
  3. An electronic timer measures the time difference
  4. v=d/tv = d/t

3. Frequency, Pitch, and Loudness

Wave Property Perception
Frequency Pitch (high frequency = high pitch)
Amplitude Loudness (large amplitude = loud)

Human Hearing Range

Humans can typically hear sounds between 20 Hz and 20,000 Hz (20 kHz).

  • Below 20 Hz: infrasound (elephants, earthquakes)
  • Above 20,000 Hz: ultrasound (bats, dog whistles, medical imaging)

Hearing range decreases with age — older people may not hear above ~16,000 Hz.


4. Ultrasound

Ultrasound is sound with a frequency above 20,000 Hz — beyond human hearing.

Applications

Medical imaging (pregnancy scans):

  • Ultrasound pulses are sent into the body
  • They reflect off boundaries between different tissues
  • The time delay and intensity of reflections build up an image
  • Safe — no ionising radiation (unlike X-rays)

Industrial quality control:

  • Ultrasound detects cracks or flaws inside metal structures
  • Pulses reflect off internal defects

Distance measurement (sonar):

  • Used by ships to measure ocean depth
  • Pulse sent down, echo detected d=v×t2d = \frac{v \times t}{2}

Cleaning:

  • Ultrasonic cleaners vibrate dirt off delicate objects (jewellery, surgical instruments)

5. Echoes and Reverberation

Echo: A reflection of sound off a hard surface that is heard as a separate, distinct sound. Requires the reflecting surface to be at least ~17 m away (for a time delay > 0.1 s).

Reverberation: When sound reflects off many surfaces and the reflections overlap, creating a prolonged sound. Common in large halls.

  • Too much reverberation → muddy sound, poor clarity
  • Too little reverberation → dry, unnatural sound
  • Solutions: soft furnishings, acoustic panels absorb sound and reduce reverberation

6. How We Hear

  1. Sound waves enter the ear canal
  2. They cause the eardrum to vibrate
  3. Vibrations pass through three small bones (hammer, anvil, stirrup) — these amplify the signal
  4. Vibrations reach the cochlea (fluid-filled spiral)
  5. Hair cells in the cochlea convert vibrations to electrical signals
  6. Signals travel along the auditory nerve to the brain

Hearing Damage

  • Prolonged exposure to loud sounds (>85 dB) can damage hair cells permanently
  • Hearing loss is irreversible
  • Prevention: ear protection, volume limits on devices

Worked Example: Ultrasound Depth

Problem

Question: A ship sends an ultrasound pulse to the seabed. The echo returns in 0.4 s. The speed of sound in water is 1500 m/s. Calculate the depth.

d=v×t2=1500×0.42=300 md = \frac{v \times t}{2} = \frac{1500 \times 0.4}{2} = 300 \text{ m}

Solution

Worked Example: Speed of Sound

Problem

Question: A student claps 50 m from a wall and hears the echo 0.3 s later. Calculate the speed of sound.

v=2dt=2×500.3=333 m/sv = \frac{2d}{t} = \frac{2 \times 50}{0.3} = 333 \text{ m/s}

Solution

Worked Example: Wavelength

Problem

Question: A sound of frequency 680 Hz travels at 340 m/s in air. Calculate the wavelength.

λ=vf=340680=0.5 m\lambda = \frac{v}{f} = \frac{340}{680} = 0.5 \text{ m}


Solution

8. Practice Questions

    1. Explain why sound cannot travel through a vacuum. (2 marks)
    1. A bat emits ultrasound at 50,000 Hz. The speed of sound is 340 m/s. Calculate the wavelength. (2 marks)
    1. Describe how ultrasound is used to produce an image of an unborn baby. (4 marks)
    1. Explain the difference between an echo and reverberation. (2 marks)
    1. A student stands 85 m from a cliff. They clap and hear the echo 0.5 s later. Calculate the speed of sound. (2 marks)

    Answers

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Summary

  • Sound is a longitudinal wave — compressions and rarefactions
  • Needs a medium — cannot travel through a vacuum
  • Speed: fastest in solids, slowest in gases (~340 m/s in air)
  • Pitch ∝ frequency; loudness ∝ amplitude
  • Human hearing: 20 Hz to 20 kHz; above 20 kHz = ultrasound
  • Ultrasound applications: medical imaging, sonar, cleaning
  • Echoes: d=vt/2d = vt/2

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