Sound WavesAQA GCSE Physics: Revision notes

Sound is a longitudinal wave that requires a medium to travel through. This means:

• Sound waves cannot travel through a vacuum (unlike light)
• The particles in the medium vibrate parallel to the direction of wave travel
• Sound travels through solids, liquids, and gases by causing particles to compress and stretch in sequence
• The speed of sound varies depending on the medium:
  • Air: approximately 330 m/s
  • Water: approximately 1500 m/s
  • Steel: approximately 5000 m/s

Sound travels slower in less dense media and faster in denser media.

The human ear detects sound through a series of structures:

1. Sound waves enter the ear canal and cause the eardrum (tympanum) to vibrate
2. The eardrum passes vibrations to three tiny bones (ossicles) in the middle ear: the hammer, anvil, and stirrup
3. These bones amplify the vibrations and pass them to the cochlea in the inner ear
4. The cochlea contains fluid and hair cells that vibrate in response to the sound energy
5. Hair cells convert vibrations into electrical signals that are sent to the brain via the auditory nerve
6. The brain interprets these signals as sound

The ear is a very sensitive detector capable of responding to extremely small vibrations. Microphones work by converting sound vibrations directly into electrical signals using a similar principle – a diaphragm vibrates with the sound wave and generates a corresponding electrical output.

Sound waves have two key properties that relate to how we perceive them:

Loudness and Amplitude:

• Louder sounds have larger amplitudes (particles vibrate with greater displacement)
• Quieter sounds have smaller amplitudes
• Amplitude is measured as the maximum displacement of particles from their rest position

Pitch and Frequency:

• Higher frequency sounds have a higher pitch (more vibrations per second)
• Lower frequency sounds have a lower pitch
• Frequency is the number of complete waves passing a point per second
• Doubling the frequency of a sound makes it sound one octave higher

Human hearing range:

• The normal range of human hearing is approximately 20 Hz to 20 kHz (20,000 Hz)
• Sounds below 20 Hz are called infrasound
• Sounds above 20 kHz are called ultrasound

Ultrasound is sound with a frequency above 20 kHz, beyond the range of human hearing. Ultrasound waves have a shorter wavelength than audible sound, which allows them to pass through materials and be reflected back, making them useful for detection and imaging.

Key applications of ultrasound:

1. Pre-natal scanning (medical imaging)

   • Ultrasound is passed through the abdomen into the uterus
   • Different tissues reflect ultrasound by different amounts
   • A receiver detects the reflected waves and creates an image of the foetus
   • Ultrasound is safe for the foetus because it has lower energy than X-rays

2. Sonar (sound navigation and ranging)

   • Used by ships and submarines to detect objects underwater
   • An ultrasound pulse is sent out and reflects off objects (fish, submarines, sea bed)
   • The time taken for the echo to return is measured
   • The distance to the object can be calculated using d = vt

3. Other applications

   • Cleaning delicate items (jewellery, electronic components)
   • Breaking up kidney stones in medical treatment
   • Non-destructive testing to find cracks in metals and concrete

Why ultrasound is useful:

• High frequency means short wavelength, allowing detection of small objects
• Can penetrate soft tissue safely
• Reflects well from boundaries between different materials
• Non-ionising radiation (safer than X-rays)

The equation d = vt relates distance, speed, and time:

• d = distance (metres, m)
• v = wave speed (metres per second, m/s)
• t = time (seconds, s)

For echo calculations (sonar and ultrasound):

When calculating the distance to an object using echoes, you must remember that the sound travels to the object and back, so the total distance is twice the distance to the object:

1. Measure the total time taken for the echo to return (to and from the object)
2. Rearrange to find distance: d = v × t
3. This gives you the total distance travelled (there and back)
4. Divide by 2 to find the actual distance to the object

Formula for finding distance to object:

Distance to object = (v × t) ÷ 2

Worked example:

A ship uses sonar. An ultrasound pulse is sent downwards and the echo returns after 3.2 seconds. The speed of ultrasound in water is 1500 m/s. Calculate the depth of the sea bed.

1. Calculate total distance: d = vt = 1500 × 3.2 = 4800 m
2. Divide by 2 (because sound went down and came back up): Depth = 4800 ÷ 2 = 2400 m