PhysicsICSE Class 9 · Infrasonic, Sonic and Ultrasonic Sound

Ultrasound: Properties, Uses and SONAR Echo Calculations (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

Ultrasound is sound of frequency above 20 kHz. Its short wavelength lets it travel as a narrow beam and reflect from small objects, so it is used for medical imaging, SONAR (depth = Vt/2 because the pulse goes out and back), cleaning, flaw detection and by bats and dolphins for echolocation.

The pulse travels down and back, a path of 2d, so the depth is only half of V × t.

Illustration drawn for QuestionBuddy.

Understand Ultrasound: Properties, Uses and SONAR Echo Calculations

Ultrasound. Sound waves of frequency above 20 kHz (20 000 Hz) are ultrasonic; the normal human ear cannot hear them. Dogs hear up to about 45 kHz; bats and dolphins produce and hear well above that.

Why it is useful. From $V = f λ$ , a high frequency gives a short wavelength. In air at 340 m s⁻¹, 40 kHz gives $λ = \frac{340}{40 000} = 8.5$ mm. A short wavelength means

the waves can travel as a narrow beam with little spreading;
they are reflected by small objects;
fine details can be resolved (details smaller than about one wavelength are not seen).

Echo and SONAR. An echo is sound reflected back from a surface. In SONAR (Sound Navigation And Ranging) a ship sends an ultrasonic pulse downward and measures the time $t$ for the echo to return. The pulse goes to the sea bed and back, so the distance travelled is $2 d$ :

$d = \frac{V \times t}{2} .$

SONAR finds the depth of the sea, locates submarines, shoals of fish and wrecks.

Uses of ultrasound.

*Medical imaging (ultrasonography):* echoes from organs and from an unborn baby form an image; no ionising radiation as in X-rays.
*Cleaning:* in a liquid bath, vibrations loosen dirt from small parts, jewellery and surgical instruments.
*Detecting cracks or flaws* in metal blocks and castings.
*Echolocation* by bats and dolphins; *dog whistles* above human hearing.

Echo problems. Always identify one-way and two-way distances: for an echo the sound covers the distance twice. If the speed in the medium is not the same as in air, use the given value (for example about 1500 m s⁻¹ in sea water).

Changing medium. The frequency of the ultrasound stays the same when it passes into the body or water; only $V$ and $λ$ change.

Ultrasound: Properties, Uses and SONAR Echo Calculations: formulas and key facts

Ultrasonic range: $f > 20 kHz$; Above human hearing.
Echo ranging (SONAR): $d = \frac{V \times t}{2}$; t = time for out-and-back journey.
Wavelength: $λ = \frac{V}{f}$; High f gives small λ.

How to solve Ultrasound: Properties, Uses and SONAR Echo Calculations questions

Sketch out and back journey first.

1Convert kHz to Hz and ms to s.
2For echoes use $2 d = V t$ .
3Use the speed given for the medium (air, sea water or tissue).
4For a moving ship or target, find two distances and divide the change by the time between them.
5Use $λ = V / f$ for the size of detail that can be resolved.

Ultrasound: Properties, Uses and SONAR Echo Calculations in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A fishing boat’s echo sounder

Suppose a fishing boat’s echo sounder receives an echo from the sea bed 1.6 s after it sends a pulse, and the speed of ultrasound in sea water is 1500 m s⁻¹.

How deep is the water?

Try it first, then show the solutionSolution

$d = \frac{V t}{2} = \frac{1500 \times 1.6}{2} = 1200$ m.

Answer: 1200 m.

Memory hook: Out and back: halve it.

Tricks to remember Ultrasound: Properties, Uses and SONAR Echo Calculations

Short memory aids. Say each one aloud once, then use it on a question the same day.

Out and back, so halve
SONAR depth = Vt ÷ 2.
Use it when: Echo numericals.
High f, small λ
Short waves make narrow beams and show small details.
Use it when: Why ultrasound?
Ultra means over 20 kHz
Ultra = above, Infra = below.
Use it when: Ranges.

Ultrasound: Properties, Uses and SONAR Echo Calculations: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Why SONAR uses ultrasound · 1 mark · MCQBoard level

SONAR uses ultrasonic waves rather than audible sound mainly because ultrasound:
(a) travels faster
(b) has a short wavelength and forms a narrow beam
(c) cannot be absorbed by water
(d) is louder

Try it first, then show the solutionSolution

Short wavelength means a narrow beam and reflection from small objects.

Answer: (b)

Example 2: Depth of the sea · 2 marks · Short answerBoard level

A ship receives the echo of an ultrasonic pulse from the sea bed 1.6 s after sending it. The speed of ultrasound in sea water is 1500 m s⁻¹. Find the depth.

Try it first, then show the solutionSolution

The pulse covers $2 d$ in 1.6 s: $2 d = 1500 \times 1.6 = 2400$ m. (1)
$d = 1200$ m. (1)

Answer: 1200 m

Example 3: Smallest detail · 3 marks · Short answerHarder board level

Ultrasound of frequency 5 MHz travels in soft tissue at 1540 m s⁻¹. Find its wavelength and state the order of size of the smallest detail it can resolve.

Try it first, then show the solutionSolution

$λ = \frac{V}{f} = \frac{1540}{5 \times 1 0 ^{6}} = 3.08 \times 1 0^{- 4}$ m. (1½)
This is about 0.31 mm, so details of about this size (a fraction of a millimetre) can be resolved. (1½)

Answer: λ ≈ 0.31 mm; details of about 0.3 mm

Example 4: A moving submarine · 3 marks · Short answerHarder board level

A ship sends SONAR pulses to a submarine on its line of travel. The echo returns after 0.8 s; 10 s later, a second pulse returns after 0.6 s. Taking the speed of ultrasound as 1500 m s⁻¹, find the two distances and the speed with which the distance is decreasing.

Try it first, then show the solutionSolution

$d_{1} = \frac{1500 \times 0.8}{2} = 600$ m. (1)
$d_{2} = \frac{1500 \times 0.6}{2} = 450$ m. (1)
Closing speed $= \frac{600 - 450}{10} = 15$ m s⁻¹. (1)

Answer: 600 m; 450 m; 15 m s⁻¹

Example 5: A bat in a cave · 4 marks · Case-basedHOTS / competency

A bat emits a squeak of frequency 68 kHz in a cave where the speed of sound is 340 m s⁻¹.
(a) Find the wavelength.
(b) Could it detect an insect 2 mm across? Explain.
(c) It hears the echo from a wall 0.06 s after the squeak. Find the distance of the wall.
(d) One second later, after flying 5 m towards the wall, find the new echo time.

Try it first, then show the solutionSolution

(a) $λ = \frac{340}{68 000} = 5$ mm. (1)
(b) Only poorly: details smaller than the wavelength (5 mm) are not reflected well; a frequency of 170 kHz ( $λ = 2$ mm) would be better. (1)
(c) $d = \frac{340 \times 0.06}{2} = 10.2$ m. (1)
(d) New distance $5.2$ m; echo time $= \frac{2 \times 5.2}{340} \approx 0.031$ s. (1)

Answer: (a) 5 mm (b) poorly, λ larger than insect (c) 10.2 m (d) 0.031 s

Where students lose marks in Ultrasound: Properties, Uses and SONAR Echo Calculations

The slips examiners see most often, and what to write instead.

Mistake: Echo distance not halved
The pulse goes out and back.
Wrong working
d = 1500 × 1.6 = 2400 m
Right working
d = 1500 × 1.6 ÷ 2 = 1200 m
Do this: Write "2d = Vt".
Usually costs: The answer.
Mistake: kHz not converted
V = fλ needs Hz.
Wrong working
λ = 340/40 = 8.5 m
Right working
λ = 340/40 000 = 8.5 mm
Do this: 1 kHz = 1000 Hz.
Usually costs: The answer.
Mistake: Ultrasonic confused with supersonic
Ultrasonic describes frequency; supersonic describes speed.
Wrong working
An ultrasonic aircraft.
Right working
A supersonic aircraft.
Do this: Frequency vs speed.
Usually costs: 1 mark.

Exam tips

Give the reason (short wavelength, narrow beam) with each use.
Show "2d = Vt" in echo questions.
Convert kHz and MHz to hertz before substituting.
Say that frequency stays the same when ultrasound changes medium.

Practice questions on Ultrasound: Properties, Uses and SONAR Echo Calculations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which of these frequencies is ultrasonic?
(a) 15 Hz
(b) 800 Hz
(c) 18 kHz
(d) 40 kHz
Hint
Above 20 kHz.
Answer
(d) 40 kHz
Q2 · 1 mark · Very short answerBoard level
What does SONAR stand for?
Hint
Expand the acronym.
Answer
Sound Navigation And Ranging.
Q3 · 2 marks · Short answerHarder board level
Find the wavelength in air (340 m s⁻¹) of ultrasound of frequency 40 kHz.
Hint
λ = V/f.
Answer
$λ = \frac{340}{40 000} = 8.5 \times 1 0^{- 3}$ m $= 8.5$ mm.
Q4 · 3 marks · Short answerHarder board level
Ultrasound of frequency 25 kHz is sent first through air (340 m s⁻¹) and then through water (1500 m s⁻¹). Find its wavelength in each and the ratio of the wavelengths.
Hint
Frequency unchanged.
Answer
Air: $\frac{340}{25 000} = 1.36$ cm. Water: $\frac{1500}{25 000} = 6$ cm. Ratio $= 4.4$ .
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): Ultrasound is preferred to X-rays for scanning an unborn baby.
Reason (R): Ultrasound does not involve the ionising radiation of X-rays.
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Hint
Safety.
Answer
(a)
Q6 · 3 marks · Short answerHOTS / competency
A jet flies at Mach 2.5 where the speed of sound is 320 m s⁻¹. Find its speed in m s⁻¹ and km h⁻¹ and the distance it covers in 90 s. Is it correct to call it an "ultrasonic jet"?
Hint
Mach = speed ÷ speed of sound.
Answer
Speed $= 2.5 \times 320 = 800$ m s⁻¹ $= 2880$ km h⁻¹; distance $= 800 \times 90 = 72$ km. No: the correct term is supersonic (speed); ultrasonic refers to the frequency of a sound.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Frequency range of ultrasound?Show answer
Above 20 kHz.
2.SONAR depth formula?Show answer
d = Vt/2.
3.Why does ultrasound form a narrow beam?Show answer
Its wavelength is very short.
4.Name an animal using echolocation.Show answer
Bat or dolphin.
5.Use of ultrasound in medicine?Show answer
Imaging organs and an unborn baby (ultrasonography).

Ultrasound: Properties, Uses and SONAR Echo Calculations: frequently asked questions

What are the uses of ultrasound?

Medical imaging, SONAR, cleaning of small parts, detecting flaws in metals, and echolocation by bats and dolphins.

How does SONAR measure depth?

It times the echo of an ultrasonic pulse; depth = (speed × time)/2 because the pulse goes down and comes back.

Why is ultrasound safer than X-rays for imaging?

Ultrasound is a mechanical wave and does not involve the ionising radiation of X-rays.