Sound And Ultrasound

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P23: Sound and Ultrasound

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Sound waves, ultrasound and their applications

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๐Ÿ“‹ Key Definitions

Sound waves: Longitudinal waves produced by vibrating objects. They require a medium (solid, liquid or gas) to travel — they cannot travel through a vacuum.
Ultrasound: Sound waves with a frequency above 20,000 Hz (20 kHz). Humans cannot hear ultrasound.
Infrasound: Sound waves with a frequency below 20 Hz. Humans cannot hear infrasound.
Echo: A reflected sound wave. The time delay between a sound and its echo can be used to calculate distance.

๐Ÿ“ Properties of Sound

How Sound Travels

Range of Human Hearing

The human ear can detect sound frequencies between 20 Hz and 20,000 Hz.

Category Frequency range Can humans hear it?
Infrasound Below 20 Hz No
Audible sound 20 Hz – 20,000 Hz Yes
Ultrasound Above 20,000 Hz No

Sound and the Ear

How we hear: Sound waves enter the ear canal → eardrum vibrates → vibrations pass through small bones (ossicles) to the cochlea → cochlea converts vibrations to electrical signals → auditory nerve carries signals to the brain.

Human hearing is limited by the size and shape of the eardrum and the structure of the inner ear.

๐Ÿ“Š Uses of Ultrasound

Medical Scanning (Pre-natal Scanning)

Principle: Ultrasound waves are directed into the body. When they reach a boundary between different tissues (e.g. fluid and bone), some of the wave is reflected back and some passes through. The time taken for the reflections (echoes) to return is used to build up an image. Ultrasound is non-ionising — safer than X-rays for imaging soft tissue and during pregnancy.

Sonar (Depth Finding)

Principle: Ships use sonar to find the depth of the sea. An ultrasound pulse is sent from the ship into the water. It reflects off the sea floor and returns to the ship as an echo. The depth is calculated from the time taken.

Industrial Uses

๐Ÿ“ Echo Calculations

Distance = speed × time ÷ 2

Where:
distance = distance to the reflecting surface (m)
speed = speed of sound in the medium (m/s)
time = total time for the sound to travel there AND back (s)

You divide by 2 because the sound travels to the surface AND back.
Why divide by 2? The time measured is for the sound to travel to the object and back again. The actual distance to the object is only half of the total distance the sound has travelled.

๐Ÿงฎ Worked Examples

Example 1: Calculating depth using sonar

A ship sends out a sonar pulse. The echo returns after 0.8 seconds. The speed of sound in water is 1500 m/s. Calculate the depth of the sea.

Solution:

Total distance = speed × time = 1500 × 0.8 = 1200 m

Depth = total distance ÷ 2 = 1200 ÷ 2 = 600 m

Example 2: Calculating distance from an echo in air

A person claps their hands and hears an echo from a wall 0.3 seconds later. The speed of sound in air is 330 m/s. How far away is the wall?

Solution:

Total distance = 330 × 0.3 = 99 m

Distance to wall = 99 ÷ 2 = 49.5 m

Example 3: Medical ultrasound timing

In an ultrasound scan, a pulse takes 4 × 10⁻⁵ s to return from a boundary. The speed of ultrasound in tissue is 1500 m/s. How far below the skin is the boundary?

Solution:

Total distance = 1500 × 4 × 10⁻⁵ = 0.06 m

Depth = 0.06 ÷ 2 = 0.03 m = 3 cm

Example 4: Calculating time for an echo

A ship is floating above a sea floor at a depth of 3000 m. The speed of sound in water is 1500 m/s. How long will it take for the sonar echo to return?

Solution:

Total distance (there and back) = 3000 × 2 = 6000 m

Time = distance ÷ speed = 6000 ÷ 1500 = 4 s

โ“ Practice Questions

Q1: Foundation State the range of frequencies that humans can hear.

Q2: Foundation Explain why sound cannot travel through a vacuum.

Q3: Higher A sonar pulse returns after 1.2 s. The speed of sound in water is 1500 m/s. Calculate the depth of the water.

Q4: Higher Describe how ultrasound is used in medical scanning. Explain why ultrasound is safer than X-rays for pre-natal scanning.

Q5: Foundation A person shouts at a cliff and hears the echo 2.0 s later. The speed of sound in air is 330 m/s. How far away is the cliff?

โœ… Answers

  1. Q1: Humans can hear frequencies between 20 Hz and 20,000 Hz (20 kHz).
  2. Q2: Sound is a longitudinal wave that requires particles to vibrate and transfer energy. In a vacuum there are no particles, so there is no medium for the sound to travel through.
  3. Q3: Total distance = 1500 × 1.2 = 1800 m. Depth = 1800 ÷ 2 = 900 m
  4. Q4: An ultrasound transducer sends high-frequency sound waves into the body. When the waves reach a boundary between different tissues, some are reflected back as echoes. The time taken for echoes to return gives the distance to each boundary, building up an image. Ultrasound is non-ionising, so it does not damage cells or DNA, making it much safer than X-rays for scanning a developing foetus.
  5. Q5: Total distance = 330 × 2.0 = 660 m. Distance to cliff = 660 ÷ 2 = 330 m

๐ŸŽฏ Exam Tips

๐Ÿ”ข Maths Skills

Mathematical Skills

Echo calculations use distance = speed ร— time รท 2. The division by 2 is critical because sound travels to the object AND back. Always check that the time given is the total echo time, not a one-way time. Convert all units consistently (ms โ†’ s, km โ†’ m).
Maths Example

A bat emits an ultrasound pulse and detects the echo 0.06 s later. The speed of sound in air is 330 m/s. How far away is the insect? Total distance = 330 ร— 0.06 = 19.8 m. Distance to insect = 19.8 รท 2 = 9.9 m.

โš ๏ธ Common Misconceptions

Watch Out!

1. Wrong: Sound can travel through a vacuum because light can Correct: Sound is a mechanical, longitudinal wave that requires particles to vibrate โ€” it cannot travel through a vacuum

2. Wrong: Forgetting to divide by 2 in echo calculations Correct: The time measured is for the sound to travel there AND back, so distance to object = (speed ร— time) รท 2

3. Wrong: Ultrasound and infrasound are just quiet sounds that humans struggle to hear Correct: Ultrasound (above 20,000 Hz) and infrasound (below 20 Hz) are completely inaudible to humans โ€” the ear cannot detect these frequencies at any volume

โœ๏ธ 6-Mark Question

Extended Answer

6 marks: Explain how ultrasound is used in medical scanning during pregnancy. Compare the safety of ultrasound with X-rays for this purpose.

An ultrasound transducer is placed on the skin and sends high-frequency sound waves (above 20,000 Hz) into the body. When the ultrasound reaches a boundary between different tissues (e.g. between fluid and bone, or between the foetus and amniotic fluid), some of the wave is reflected back as an echo and some continues through. The time taken for each echo to return is measured, and since the speed of ultrasound in tissue is known, the distance to each boundary can be calculated (distance = speed ร— time รท 2). A computer processes these times to build up an image of the foetus. Ultrasound is much safer than X-rays for pre-natal scanning because it is non-ionising โ€” it does not carry enough energy to damage cells or DNA. X-rays are ionising and could damage the developing foetus, potentially causing mutations or cancer. Ultrasound can be used repeatedly without harmful effects.

Mark scheme: 1 mark โ€” ultrasound is above 20,000 Hz, 1 mark โ€” waves reflect at tissue boundaries producing echoes, 1 mark โ€” time of echo gives distance to boundary, 1 mark โ€” computer builds image from echoes, 1 mark โ€” ultrasound is non-ionising (safe), 1 mark โ€” X-rays are ionising and can damage DNA/cells in a developing foetus

๐Ÿ“Š AO3: Analyse & Evaluate

Analysis and Evaluation

A ship uses sonar to measure sea depth. The speed of sound in water is 1500 m/s. Three readings are taken: 2.0 s, 2.2 s, and 1.8 s. The sonar manufacturer states the device has a timing uncertainty of ยฑ0.05 s.

(a) Calculate the mean depth of the sea at this location.

(b) Calculate the maximum and minimum possible depths using the timing uncertainty.

(c) The ship moves to shallower water and the echo time decreases. Explain why a shorter echo time means shallower water, and suggest why the readings might become less reliable in shallow water.

Answers: (a) Mean time = (2.0 + 2.2 + 1.8)/3 = 2.0 s. Mean depth = (1500 ร— 2.0) รท 2 = 1500 m. (b) Max depth: time = 2.0 + 0.05 = 2.05 s, depth = (1500 ร— 2.05) รท 2 = 1537.5 m. Min depth: time = 2.0 โˆ’ 0.05 = 1.95 s, depth = (1500 ร— 1.95) รท 2 = 1462.5 m. (c) Shorter echo time means the sound has less distance to travel there and back, so the sea floor is closer (shallower). In shallow water, the echo returns so quickly that the timing uncertainty (ยฑ0.05 s) becomes a larger percentage of the total time, making depth measurements less reliable.

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