Lesson focus: Echoes are reflected sound waves whose return time can reveal distance, structure or movement.
Explain echoes and ultrasoundSound with a frequency above 20,000 Hz (20 kHz), too high for humans to hear. applications and calculate one-way distance from a round-trip echoA sound you hear again because it bounced off a surface, such as a wall or cliff. time.
Build on earlier learning
Detecting soundThe ear detects reflected sound as an echo. This lesson uses the delay between emission and detection as quantitative evidence.
You will be successful when you can…
- Define echo, reverberation and ultrasound.
- Explain why an echo arrives after the original sound.
- Calculate range using a round-trip time.
- Evaluate sonar and medical-ultrasound uses without claiming they are identical.
Words that carry the science
- echo
- A reflected sound detected separately from the original sound.
- reflection
- A wave changing direction at a boundary and returning into its original medium.
- reverberation
- Many closely spaced reflections that persist after the source sound.
- ultrasound
- Sound with frequency above the upper limit of human hearing, conventionally above 20 kHz.
- sonar
- A system that emits sound and analyses returning echoes to locate objects or boundaries.
- range
- The one-way distance from the emitter to a target.
Turn reflected sound into ranging evidence
01An echo is delayed reflection
When sound meets a boundary, some energy may reflectionWhen light or sound bounces off a surface, like light from a mirror.. A separate echoA sound you hear again because it bounced off a surface, such as a wall or cliff. is heard when the reflected wave returns after a sufficient delay.
Many rapid overlapping reflections produce reverberationWhen lots of echoes blur together so a sound seems to last longer, like in a big empty hall.. Soft, porous materials can absorb more sound and reduce strong reflections in theatres or recording spaces.
02Round-trip time must be halved
An emitted pulse travels to the target and back. The measured echo time therefore covers twice the range. Calculate total distance with distance = speed × time, then divide by two.
A diagram of the path prevents a common factor-of-two error. Use the speed in the actual medium, because sound travels much faster in water than in air.
03Ultrasound has several evidence-rich uses
ultrasoundSound with a frequency above 20,000 Hz (20 kHz), too high for humans to hear. has frequency above 20 kHz. sonarA way of finding objects underwater by sending out sound and timing the echoes. uses echoes in water to estimate depth, map the seabed or detect objects. Bats and dolphins use biological echolocation.
Medical ultrasoundSound with a frequency above 20,000 Hz (20 kHz), too high for humans to hear. sends pulses into the body and detects reflections from boundaries between tissues. A computer combines many return times and signal strengths into an image. This differs from sonar in context and processing, even though both use reflected sound.
Measuring distance with an echo
A sound pulse travels to the reflector and back. Distance = speed × time ÷ 2, because the sound goes there and back.
Controls
Speed of sound: about 340 m/s in air and 1500 m/s in sea water. The animation runs slower than real time.
Calculate sea depth from sonar
A sonar echo returns after 0.80 s. Sound speed in the water is 1500 m/s. Calculate the water depth.
- Write total distance = speed × time.
- Calculate 1500 × 0.80 = 1200 m for the outward and return path.
- Recognise that this path equals twice the depth.
- Divide 1200 m by 2.
How to make the reasoning visible: The water depth is 600 m.
Interpret four echo measurements
Every listed time is the total outward-and-return time.
| Context | Sound speed / m/s | Echo time / s | One-way range |
|---|---|---|---|
| Wall in air | 340 | 0.20 | 34 m |
| Bat prey in air | 340 | 0.030 | 5.1 m |
| Seabed in water | 1500 | 1.60 | 1200 m |
| Submerged object | 1500 | 0.040 | 30 m |
Show how the 5.1 m bat-prey range is obtained.
Why would using 340 m/s for the seabed row give a poor result?
Check your understanding
Answer from memory, then check against the suggested answer.
What is an echo?
What frequency range is called ultrasound?
Why divide an echo path by two?
A submarine receives an echo 2.4 s after emitting a pulse. Sound speed is 1500 m/s. Calculate the range.
Two sonar echoes return at the same time but one is weaker. Explain why equal range does not guarantee identical targets.
Exam Question Practice
Exam Question Practice
Mark points shown after marking
A boat sends a sound pulse down to the seabed. The echo returns 0.60 s later. The speed of sound in water is 1500 m/s.
Calculate the depth of the seabed. Show your working.
(3)Explain why you divide by two.
(1)Total for this question: 4 marks
Command word: For calculate and explainA command word pair: work out the number answer, then say what it means or why your method works., show the total path before halving it and state why the halving is required.