ScienceCBSE Class 9 · Sound Waves: Characteristics and Applications

Echo, Reverberation, SONAR, Echolocation and Ultrasound (Class 9)

Updated 17 October 2026 · Fact-checked

Sound reflects from hard surfaces by the same laws as light. An echo is a reflected sound heard separately from the original; because the ear retains a sound for about 0.1 s, the echo must come back at least 0.1 s later, so the reflector must be at least about 17 m away in air (340 m s⁻¹). The distance to a reflector is d = vt ÷ 2, where t is the round-trip time, and the same formula gives the depth in SONAR. Reverberation is the lingering of sound from repeated reflections in a large hall, and ultrasound (above 20 kHz) is used in imaging, cleaning, flaw detection and echolocation.

Sound covers the distance twice, there and back, so in 0.1 s it travels 34 m and the wall must be at least 17 m away.

Illustration drawn for QuestionBuddy.

Understand Echo, Reverberation, Sonar and Ultrasound

Reflection of sound. Sound reflects from solid or liquid surfaces and obeys the same laws as light: the incident and reflected directions make equal angles with the normal, and all three lie in one plane. Hard, smooth surfaces give strong echoes, soft surfaces such as curtains absorb sound, and rough surfaces scatter it.

Echo. An echo is a reflected sound heard separately from the original. The ear keeps a sound for about 0.1 s, so the echo must arrive at least 0.1 s later. In 0.1 s sound travels 340 × 0.1 = 34 m, and since it goes to the reflector and back, the reflector must be at least 34 ÷ 2 = 17 m away. The distance to a reflector is d = v × t ÷ 2, where t is the time between the original sound and the echo (the round trip).

Reverberation. In a big hall, repeated reflections make the sound linger after the source stops; the reflections arrive within about 0.05 s of each other and blur together. Too much reverberation makes speech unclear. Halls use sound-absorbing panels, upholstered seats, curtains and carpets to reduce it, and curved ceilings and stage walls to direct sound to the audience. The Whispering Gallery of Gol Gumbaz in Vijayapura, Karnataka, lets a whisper be heard many times across the dome.

Ultrasonic waves (above 20 kHz). Uses: ultrasonography (imaging organs and a baby in the womb without surgery); breaking kidney stones into small pieces; cleaning delicate machine parts; ultrasonic welding; finding cracks and defects inside metal blocks; locating objects by reflection. Infrasonic waves (below 20 Hz) detect earthquakes, volcanic eruptions and severe storms because they travel far.

Echolocation and SONAR. Echolocation is locating objects by reflected sound. Most bats send out short bursts of ultrasound and use the echoes from prey and obstacles; dolphins and whales do this too. SONAR (sound navigation and ranging) sends ultrasonic pulses into water from a ship and times the echo to find the distance, direction and speed of submarines, shoals of fish, shipwrecks or the sea bed. Depth = v × t ÷ 2 with v about 1500 m s⁻¹ in sea water. Car parking sensors use the same idea in air (about 40 kHz).

Why ultrasound is used for ranging. Ultrasonic waves have short wavelengths, so they can be sent as narrow beams that do not spread much, can be detected as echoes from small objects, and do not disturb people because they are above the audible range.

Echo, Reverberation, Sonar and Ultrasound: formulas and key facts

Echo and SONAR distance: $d = v \times t \div 2, w i t h tt h er o u n d - t r i pt im e$
Minimum distance for an echo in air: $d = 340 \times 0.1 \div 2 = 17 m$
Time of the round trip: $t = 2 d \div v$
Speeds to use: Air 340 m s⁻¹ | sea water about 1500 m s⁻¹
Echo versus reverberation: Echo: heard at least 0.1 s later | reverberation: reflections within about 0.05 s blend
Hearing range: Ultrasonic above 20 kHz | infrasonic below 20 Hz

How to solve Echo, Reverberation, Sonar and Ultrasound questions

Use this method for echo, SONAR and reverberation numericals.

1Identify the medium and the speed (340 m s⁻¹ for air, about 1500 m s⁻¹ for sea water) from the data.
2Decide whether the time given is the round trip (echo, SONAR) or one way (lightning).
3For distance from an echo use d = vt ÷ 2; for the time of an echo use t = 2d ÷ v.
4Compare the time with 0.1 s to decide if an echo can be heard separately.
5For two reflectors, find each distance separately and add or subtract as the geometry needs.
6State the answer with units and a one-line reason (for example 'because the sound goes there and back').

Echo, Reverberation, Sonar and Ultrasound 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.

Clapping near a cliff

Suppose a boy stands 51 m from a rocky cliff and claps once on a day when the speed of sound is 340 m s⁻¹.

When does he hear the echo, and will it be heard separately?

Try it first, then show the solutionSolution

The sound travels there and back, 2 × 51 = 102 m.
Time = 102 ÷ 340 = 0.3 s.
Since 0.3 s is more than 0.1 s, the echo is heard separately.

Answer: After 0.3 s, as a separate echo.

Memory hook: There and back: double the distance.

A bat finding an insect

Suppose a bat sends an ultrasonic pulse and hears the echo from a moth 0.02 s later, with the speed of sound in air 340 m s⁻¹.

Find how far the moth is from the bat.

Try it first, then show the solutionSolution

The echo time 0.02 s is the round trip.
Distance = v × t ÷ 2 = 340 × 0.02 ÷ 2 = 3.4 m.

Answer: 3.4 m.

Memory hook: Half the round trip.

Tricks to remember Echo, Reverberation, Sonar and Ultrasound

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

Divide by two
An echo goes there and back, so distance is v × t ÷ 2.
Use it when: Every echo and SONAR numerical.
0.1 and 17
Ear memory 0.1 s; at 340 m s⁻¹ the sound goes 34 m, so the wall must be 17 m away.
Use it when: Condition for hearing an echo.
Echo is a copy, reverberation is a smear
An echo is a separate repeat; reverberation is overlapping repeats that blur the sound.
Use it when: Distinguishing the two.
SONAR = sound navigation and ranging
A ship pings the sea and counts the time to the echo; depth is half of v × t.
Use it when: Explaining SONAR.

Echo, Reverberation, Sonar and Ultrasound: solved examples

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

Example 1: Conditions for an echo · 3 marks · Short answerBoard level

(a) State the condition for hearing an echo. (1)
(b) Show that the reflecting surface must be at least 17 m away when the speed of sound is 340 m s⁻¹. (1)
(c) A person claps 10 m from a wall. Will she hear an echo? (1)

Try it first, then show the solutionSolution

(a) The reflected sound must reach the ear at least 0.1 s after the original sound, because the ear retains a sound for about 0.1 s. (1)
(b) In 0.1 s sound travels 340 × 0.1 = 34 m; this is the round trip, so the minimum distance is 34 ÷ 2 = 17 m. (1)
(c) Time = 2 × 10 ÷ 340 = 0.059 s, which is less than 0.1 s, so no separate echo is heard. (1)

Answer: (a) at least 0.1 s later; (b) 34 ÷ 2 = 17 m; (c) no, t = 0.059 s.

Example 2: Depth of the sea by SONAR · 3 marks · Short answerBoard level

A ship sends an ultrasonic pulse towards the sea bed and receives the echo 0.8 s later. The speed of sound in sea water is 1500 m s⁻¹. Find (a) the depth, (b) the depth if the echo had returned after 1.2 s, and (c) why the formula has a division by 2.

Try it first, then show the solutionSolution

(a) Depth = v × t ÷ 2 = 1500 × 0.8 ÷ 2 = 600 m. (1)
(b) Depth = 1500 × 1.2 ÷ 2 = 900 m. (1)
(c) The measured time covers the journey down and back, so the depth is half the total distance. (1)

Answer: (a) 600 m; (b) 900 m; (c) the pulse goes down and back.

Example 3: Echoes between two cliffs · 5 marks · Long answerHarder board level

A boy stands in a valley between two cliffs and claps once. He hears the first echo after 2.0 s and the second after 3.5 s. The speed of sound is 340 m s⁻¹.
(a) Find his distance from each cliff. (2)
(b) Find the distance between the cliffs. (1)
(c) Find the time between the two echoes. (1)
(d) On a warmer day the speed is 344 m s⁻¹ and the echo times are unchanged. Find the new distances. (1)

Try it first, then show the solutionSolution

(a) Near cliff: d₁ = 340 × 2.0 ÷ 2 = 340 m. Far cliff: d₂ = 340 × 3.5 ÷ 2 = 595 m. (2)
(b) The boy is between the cliffs, so the distance = 340 + 595 = 935 m. (1)
(c) 3.5 − 2.0 = 1.5 s between the echoes. (1)
(d) d₁ = 344 × 2.0 ÷ 2 = 344 m; d₂ = 344 × 3.5 ÷ 2 = 602 m; the cliffs are 946 m apart. (1)

Answer: (a) 340 m and 595 m; (b) 935 m; (c) 1.5 s; (d) 344 m, 602 m (946 m apart).

Example 4: Tracking a submarine with SONAR · 5 marks · Long answerHarder board level

A ship sends an ultrasonic pulse and the echo from a submarine returns after 4.0 s. A second pulse 60 s later returns after 4.2 s. The speed of sound in sea water is 1500 m s⁻¹.
(a) Find the distance of the submarine at the first pulse. (1)
(b) Find its distance at the second pulse. (1)
(c) Find the speed at which it moves relative to the ship along the line of sight and say whether it is approaching or receding. (2)
(d) Why are ultrasonic waves used for SONAR? (1)

Try it first, then show the solutionSolution

(a) d₁ = 1500 × 4.0 ÷ 2 = 3000 m. (1)
(b) d₂ = 1500 × 4.2 ÷ 2 = 3150 m. (1)
(c) It moved 3150 − 3000 = 150 m in 60 s, speed = 150 ÷ 60 = 2.5 m s⁻¹; the distance is increasing, so it is receding. (2)
(d) Ultrasound can be sent as a narrow beam, is reflected by small objects, and does not disturb people because it is above the audible range. (1)

Answer: (a) 3000 m; (b) 3150 m; (c) 2.5 m s⁻¹ receding; (d) narrow beams, reflected by small objects.

Example 5: Echo or reverberation in a hall? · 4 marks · Case-basedHOTS / competency

Suppose a speaker in a hall (speed of sound 340 m s⁻¹) is 8 m from the back wall and also 20 m from a hard wall at the front.
(a) Find the time for sound to go to the back wall and return. (1)
(b) Find the time for the round trip to the front wall. (1)
(c) Which reflection is heard as a separate echo, and which merges as reverberation? (1)
(d) Suggest two ways of reducing the reverberation. (1)

Try it first, then show the solutionSolution

(a) t = 2 × 8 ÷ 340 = 16 ÷ 340 = 0.047 s. (1)
(b) t = 2 × 20 ÷ 340 = 40 ÷ 340 = 0.118 s. (1)
(c) The front wall's reflection returns after 0.118 s, more than 0.1 s, so it is heard as a separate echo; the back wall's reflection returns in 0.047 s, within about 0.05 s, so it blends with the original sound (reverberation). (1)
(d) Sound-absorbing panels, curtains, carpets or upholstered seats reduce reverberation. (1)

Answer: (a) 0.047 s; (b) 0.118 s; (c) front wall: echo; back wall: reverberation; (d) absorbing panels and curtains.

Where students lose marks in Echo, Reverberation, Sonar and Ultrasound

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

Mistake: Not dividing by 2 in echo and SONAR distance.
The time given is for the round trip.
Wrong working
Depth = 1500 × 0.8 = 1200 m
Right working
Depth = 1500 × 0.8 ÷ 2 = 600 m
Do this: Ask: does the time include going and returning?
Usually costs: 2 marks.
Mistake: Using 17 m for the minimum distance in any medium.
The value 17 m is for air at 340 m s⁻¹.
Wrong working
The reflector must be 17 m away in water as well.
Right working
Minimum distance = v × 0.1 ÷ 2, which is 75 m for v = 1500 m s⁻¹.
Do this: Recompute with the speed in the medium.
Usually costs: 1 mark.
Mistake: Confusing echo with reverberation.
Both come from reflections.
Wrong working
Reverberation is a clear repeat of the sound after 0.1 s.
Right working
That is an echo. Reverberation is the lingering blur from repeated reflections arriving within about 0.05 s.
Do this: Echo: separate. Reverberation: overlapping.
Usually costs: 1 mark.
Mistake: Writing that SONAR uses audible sound.
The word sound is taken at face value.
Wrong working
SONAR uses sounds of 1 kHz.
Right working
SONAR uses ultrasonic waves (above 20 kHz).
Do this: Ranging uses ultrasound.
Usually costs: 1 mark.
Mistake: Using the speed of sound in air for a sea-depth question.
340 m s⁻¹ is learned as 'the speed of sound'.
Wrong working
Depth = 340 × 0.8 ÷ 2 = 136 m
Right working
Use 1500 m s⁻¹ in sea water: 600 m
Do this: Pick the speed of the medium.
Usually costs: 2 marks.

Exam tips

Underline whether the time is a round trip; then write d = vt ÷ 2 (or t = 2d ÷ v).
Quote 0.1 s and 17 m for echoes in air, and the speed of sound you assume.
In SONAR answers mention ultrasonic pulse, reflection from the bottom and timing.
Define echo and reverberation separately and give one remedy for reverberation.
State the uses of ultrasound as specific examples (kidney stones, ultrasonography).

Practice questions on Echo, Reverberation, Sonar and Ultrasound

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

Q1 · 2 marks · Very short answerBoard level
Give two uses of ultrasonic waves and one use of infrasonic waves.
Hint
Think of medicine, cleaning and earthquakes.
Answer
Ultrasound: ultrasonography to image organs or a baby in the womb; breaking kidney stones; cleaning machine parts; detecting cracks in metal (any two) (1 mark). Infrasound: detecting earthquakes, volcanic eruptions or storms (1 mark).
Q2 · 3 marks · Short answerHarder board level
A bat emits an ultrasonic pulse and the echo from an insect returns in 0.02 s. The speed of sound in air is 340 m s⁻¹. (a) Find the distance of the insect. (b) How long would the echo take if the insect were 1.7 m away? (c) Name this method of locating objects.
Hint
The time given is for the round trip.
Answer
(a) d = 340 × 0.02 ÷ 2 = 3.4 m (1 mark). (b) t = 2 × 1.7 ÷ 340 = 0.01 s (1 mark). (c) Echolocation (1 mark).
Q3 · 2 marks · Assertion-reasonBoard level
Assertion (A): No echo can be heard if the reflecting wall is 10 m from the listener in air. Reason (R): The reflected sound returns within 0.1 s of the original sound. Choose: (a) A and R both true, R explains A; (b) A and R both true, R does not explain A; (c) A true, R false; (d) A false, R true.
Hint
Find the round-trip time for 10 m.
Answer
(a). The round trip is 20 m, taking 20 ÷ 340 = 0.059 s, less than 0.1 s, so the echo merges with the original sound and is not heard separately.
Q4 · 3 marks · Short answerHarder board level
A ship's SONAR measures an echo time of 1.2 s from the sea bed. A little later the time is 0.9 s. (Speed in sea water 1500 m s⁻¹.) (a) Find both depths. (b) By how much has the depth changed? (c) What does this say about the sea bed?
Hint
Depth = vt ÷ 2 each time.
Answer
(a) 1500 × 1.2 ÷ 2 = 900 m and 1500 × 0.9 ÷ 2 = 675 m (1 mark). (b) The depth fell by 900 − 675 = 225 m (1 mark). (c) The sea bed is rising, so the ship has moved over shallower water, for example towards a slope or an undersea ridge (1 mark).
Q5 · 4 marks · Case-basedHOTS / competency
A car parking sensor works with ultrasound of frequency 40 kHz in air (340 m s⁻¹). (a) Find the wavelength of the sound. (b) Find the time for the echo from a wall 0.5 m behind the car. (c) Find the time when the wall is 0.1 m away. (d) Explain why an ultrasonic sensor is better than an audible one.
Hint
λ = v ÷ ν.
Answer
(a) λ = 340 ÷ 40 000 = 0.0085 m = 8.5 mm (1 mark). (b) t = 2 × 0.5 ÷ 340 = 0.0029 s, about 2.9 ms (1 mark). (c) t = 2 × 0.1 ÷ 340 = 0.00059 s, about 0.59 ms (1 mark). (d) Ultrasound is above the audible range so it does not disturb people, and its short wavelength gives a narrow beam reflected by small objects (1 mark).

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.How long does the ear retain a sound?Show answer
About 0.1 s.
2.What is the minimum distance for an echo in air?Show answer
About 17 m.
3.What does SONAR stand for?Show answer
Sound navigation and ranging.
4.Name an animal that uses echolocation.Show answer
A bat (also dolphins and whales).
5.What is reverberation?Show answer
The lingering of sound in a large hall due to repeated reflections.
6.What is the formula for the depth found by SONAR?Show answer
Depth = v × t ÷ 2.

Echo, Reverberation, Sonar and Ultrasound: frequently asked questions

Why do we not hear echoes in a small room?

The walls are too close, so the reflected sound returns in less than 0.1 s and merges with the original sound.

Why do concert halls have curtains and carpets?

They absorb sound and reduce reverberation, making speech and music clearer.

Can sound reflect from water?

Yes. Sound reflects from solid and liquid surfaces, which is how SONAR works.

Why do bats use ultrasound rather than audible sound?

Ultrasound has a short wavelength, so it reflects from small prey and gives sharp echoes.

What laws does the reflection of sound obey?

The same as light: the angle of incidence equals the angle of reflection, and the incident ray, reflected ray and normal lie in one plane.