PhysicsICSE Class 10 · Reflection of Sound and Echoes

Echo Numericals: Using v = 2d/t Step by Step (ICSE Class 10)

Q: What is the formula for echo numericals?

v = 2d/t, where d is the distance of the reflector and t the time after which the echo is heard. The 2 appears because sound goes to the reflector and back.

Q: Why must the reflector be at least 17 m away to hear an echo?

The ear keeps a sound for about 0.1 s. In 0.1 s sound in air at 340 m/s travels 34 m, which is 17 m there and 17 m back.

Q: How is the third echo formed between two cliffs?

Sound reflected from one cliff travels to the other cliff, is reflected again and returns. Its time is the sum of the first two echo times.

Updated 15 October 2026 · Fact-checked

In every echo problem the sound goes to the reflector and comes back, covering a distance 2d in time t, so v = 2d/t. Rearranged, d = vt/2 and t = 2d/v. Convert minutes to seconds and km to metres first, and remember that a distinct echo needs at least 0.1 s.

The sound covers d going and d returning, so the total path is 2d; always use v = 2d/t and check that t is at least 0.1 s.

Illustration drawn for QuestionBuddy.

Understand Echo Numericals (v = 2d/t)

The one idea. An echo is reflected sound. Between the moment the sound is made and the moment its echo is heard, the sound travels to the reflector and back. If the reflector is a distance $d$ away and the echo is heard after time $t$ :

$v = \frac{2 d}{t} .$

Every echo numerical is this equation, rearranged:

To find	Use
distance of the reflector	$d = \frac{v t}{2}$
speed of sound	$v = \frac{2 d}{t}$
time for the echo	$t = \frac{2 d}{v}$

The factor 2 is the most common source of lost marks. A quick check: the one-way time is $\frac{t}{2}$ , so $d = v \times \frac{t}{2}$ .

Is the echo distinct? Our ear keeps the sensation of a sound for about $0.1$ s. The echo is heard separately only if $t \geq 0.1$ s. In air at $340$ m/s this needs $d \geq \frac{340 \times 0.1}{2} = 17$ m. In general

$d_{m i n} = \frac{v \times 0.1}{2},$

so the minimum distance is proportional to the speed of sound: larger in water or steel, a little larger in warm air. When two media are compared, $d_{1} : d_{2} = v_{1} : v_{2}$ .

Two reflectors (person between two cliffs). If the cliffs are $d_{1}$ (nearer) and $d_{2}$ away:

first echo: $t_{1} = \frac{2 d _{1}}{v}$ ;
second echo: $t_{2} = \frac{2 d _{2}}{v}$ ;
third echo (sound reflected from one cliff to the other and back to the person): $t_{3} = \frac{2 ( d _{1} + d _{2} )}{v} = t_{1} + t_{2}$ ;
distance between the cliffs: $d_{1} + d_{2} = \frac{v ( t _{1} + t _{2} )}{2}$ .

If one cliff is closer than $d_{m i n}$ , its echo merges with the original sound; only the farther cliff gives a distinct first echo.

Clapping in rhythm (finding the speed of sound). A person claps so that each clap coincides with the echo of the previous one. Then the time between claps is the echo time. If $n$ claps are made in total time $T$ (timing from the first clap to the last), there are $n - 1$ intervals:

$t = \frac{T}{n - 1}, v = \frac{2 d}{t} .$

SONAR and ultrasound. The same formula works under water (sound in sea water about $1500$ m/s) and inside the body (take the value given, about $1540$ m/s in soft tissue). Only the speed changes.

Units checklist. Distance in metres, time in seconds, speed in m/s. $1$ min $= 60$ s; $1$ km $= 1000$ m; $1$ ms $= 1 0^{- 3}$ s. Give the final answer with its unit.

Echo Numericals (v = 2d/t): formulas and key facts

Echo equation: $v = \frac{2 d}{t}$; 2d is the round trip to the reflector and back.
Distance of the reflector: $d = \frac{v t}{2}$; Use the full echo time t, then halve.
Minimum distance for a distinct echo: $d_{m i n} = \frac{v \times 0.1}{2}$; 17 m in air at 340 m/s; proportional to v.
Third echo between two cliffs: $t_{3} = \frac{2 ( d _{1} + d _{2} )}{v} = t_{1} + t_{2}$; Sound reflected from both cliffs.
Echo method checklist: Round trip distance is 2d
Units metres, seconds, m/s
Distinct echo t at least 0.1 s
Two media minimum distances in the ratio of the speeds
Clap rhythm interval equals total time over (n − 1); Run through this before writing the answer.

Round trip	distance is 2d
Units	metres, seconds, m/s
Distinct echo	t at least 0.1 s
Two media	minimum distances in the ratio of the speeds
Clap rhythm	interval equals total time over (n − 1)

How to solve Echo Numericals (v = 2d/t) questions

Follow the same five steps for every echo numerical.

1List the data and convert: minutes to seconds, km to m, ms to s.
2Decide what the time is: the full echo time (there and back), or a one-way time.
3Write $v = \frac{2 d}{t}$ and rearrange for the unknown before substituting.
4Substitute with units and calculate; give the answer with its unit.
5Check: is $t \geq 0.1$ s (is the echo distinct)? For two cliffs, is each cliff farther than $d_{m i n}$ ?

Echo Numericals (v = 2d/t) 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.

Shouting across a valley

Suppose you are trekking and shout towards a rock face across a valley. You hear your echo 2.5 s later. The air is mild and sound travels at 344 m/s.

How far away is the rock face?

Try it first, then show the solutionSolution

The sound goes there and back: $d = \frac{v t}{2}$ .
$d = \frac{344 \times 2.5}{2} = 430$ m.

Answer: About 430 m.

Memory hook: Halve the time, then multiply by the speed.

Shouting into a well

Imagine you shout into a dry well and hear the echo from the bottom 0.3 s later. (Speed of sound 340 m/s)

How deep is the well, and would you hear a separate echo from a well only 15 m deep?

Try it first, then show the solutionSolution

Depth $= \frac{340 \times 0.3}{2} = 51$ m.
For 15 m: $t = \frac{2 \times 15}{340} \approx 0.088$ s, which is less than 0.1 s.

Answer: 51 m; no separate echo from a 15 m well, since it returns in under 0.1 s.

Memory hook: Under 17 m in air, the echo hides inside your own voice.

Tricks to remember Echo Numericals (v = 2d/t)

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

Two-way traffic
'Echo is a return ticket': the sound pays for 2d, never d.
Use it when: Remembering the 2 in v = 2d/t.
Seventeen in air
340 × 0.1 ÷ 2 = 17: 'seventeen to hear it again'.
Use it when: Minimum echo distance in air.
Fence posts
n claps have n − 1 gaps, like n fence posts with n − 1 spaces between them.
Use it when: Clap-rhythm problems.
Third echo adds up
Between two cliffs, the third echo time is the sum of the first two.
Use it when: Two-reflector problems.

Echo Numericals (v = 2d/t): solved examples

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

Example 1: Distance from echo time · 1 mark · MCQBoard level

A boy hears the echo of his shout from a building 0.75 s after shouting. If sound travels at 336 m/s, the building is:
(a) 63 m away
(b) 126 m away
(c) 252 m away
(d) 448 m away

Try it first, then show the solutionSolution

$d = \frac{v t}{2} = \frac{336 \times 0.75}{2} = 126$ m.

Answer: (b) 126 m away

Example 2: Speed of sound from an echo · 2 marks · Short answerBoard level

A girl stands 198 m from a high wall and fires a toy cap gun. She hears the echo 1.2 s later. Calculate the speed of sound in air.

Try it first, then show the solutionSolution

Total distance $= 2 \times 198 = 396$ m. (1)
$v = \frac{2 d}{t} = \frac{396}{1.2} = 330$ m/s. (1)

Answer: 330 m/s

Example 3: Echo inside the body · 2 marks · Short answerHarder board level

In an ultrasound scan, a pulse is reflected from a kidney stone and the echo is received 0.13 ms after the pulse is sent. Taking the speed of sound in body tissue as 1540 m/s, find the depth of the stone below the probe in cm.

Try it first, then show the solutionSolution

$t = 0.13 \times 1 0^{- 3}$ s. (½)
$d = \frac{v t}{2} = \frac{1540 \times 0.13 \times 1 0 ^{- 3}}{2} \approx 0.100$ m $= 10$ cm. (1½)

Answer: About 10 cm

Example 4: Between two cliffs · 3 marks · Short answerHarder board level

A man fires a gun while standing somewhere between two parallel cliffs. The two echoes reach him 1.2 s and 2.0 s after the shot. The speed of sound is 340 m/s. Find (a) his distance from each cliff, (b) the distance between the cliffs, (c) the time at which he hears the third echo.

Try it first, then show the solutionSolution

(a) $d_{1} = \frac{340 \times 1.2}{2} = 204$ m; $d_{2} = \frac{340 \times 2.0}{2} = 340$ m. (1)
(b) $d_{1} + d_{2} = 204 + 340 = 544$ m. (1)
(c) $t_{3} = \frac{2 \times 544}{340} = 3.2$ s (check: $1.2 + 2.0 = 3.2$ s). (1)

Answer: (a) 204 m and 340 m (b) 544 m (c) 3.2 s after firing

Example 5: Clapping in step with the echo · 3 marks · Short answerHarder board level

Standing 60 m from the boundary wall of her school, a student claps in time with the echoes: every clap is made just as the echo of the previous clap arrives. She makes 31 claps in 11.25 s, timing from the first clap to the last. Find (a) the time between successive claps, (b) the speed of sound.

Try it first, then show the solutionSolution

(a) 31 claps have 30 intervals: $t = \frac{11.25}{30} = 0.375$ s. (1½)
(b) Each interval is the echo time: $v = \frac{2 \times 60}{0.375} = 320$ m/s. (1½)

Answer: (a) 0.375 s (b) 320 m/s

Example 6: Walking towards the wall · 3 marks · Short answerHOTS / competency

A person standing in front of a wall hears the echo of a clap after 0.6 s. She walks 34 m straight towards the wall and claps again. (Speed of sound 340 m/s) (a) How far from the wall was she at first? (b) After how long does she hear the new echo? (c) How much closer can she go and still hear a distinct echo?

Try it first, then show the solutionSolution

(a) $d = \frac{340 \times 0.6}{2} = 102$ m. (1)
(b) New distance $= 102 - 34 = 68$ m; $t = \frac{2 \times 68}{340} = 0.4$ s. (1)
(c) She must stay at least $d_{m i n} = \frac{340 \times 0.1}{2} = 17$ m away, so she can go a further $68 - 17 = 51$ m closer. (1)

Answer: (a) 102 m (b) 0.4 s (c) 51 m more

Example 7: Case study: a temple courtyard · 4 marks · Case-basedHOTS / competency

A guide claps in a temple courtyard. The courtyard wall facing her is 119 m away and the wall behind her is only 12 m away. The speed of sound is 340 m/s.
(a) After how long is the echo from the far wall heard?
(b) Does she hear a distinct echo from the wall behind her? Justify with a calculation.
(c) A visitor says 'there must be a third echo at the sum of the two times'. Find this time and say whether the visitor's echo would be heard separately.
(d) If the air warms up and sound travels faster, how does the answer to (a) change?

Try it first, then show the solutionSolution

(a) $t = \frac{2 \times 119}{340} = 0.7$ s. (1)
(b) $t = \frac{2 \times 12}{340} \approx 0.071$ s $< 0.1$ s, so no: it merges with the original clap (12 m is less than 17 m). (1)
(c) $t_{3} = \frac{2 ( 119 + 12 )}{340} \approx 0.771$ s. It comes only about 0.071 s after the 0.7 s echo, which is less than 0.1 s, so it merges with that echo and is not heard as a separate sound. (1)
(d) $t = \frac{2 d}{v}$ , so a larger $v$ gives a slightly shorter time. (1)

Answer: (a) 0.7 s (b) No, 0.071 s < 0.1 s (c) About 0.771 s; merges with the first echo (d) Slightly shorter

Where students lose marks in Echo Numericals (v = 2d/t)

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

Mistake: Forgetting the round trip
The echo time covers the path to the reflector and back.
Wrong working
$d = v t = 1500 \times 3.2 = 4800$ m for the sea depth.
Right working
$d = \frac{v t}{2} = 2400$ m.
Do this: Write $v = \frac{2 d}{t}$ as your first line every time.
Usually costs: CISCE examiners reported that candidates did not account for the reflection in the 2025 Q7(ii) submarine problem.
Mistake: Not converting minutes to seconds
Speed is in m/s, so time must be in seconds.
Wrong working
Taking 1 min 4 s as 1.04 s or 104 s.
Right working
1 min 4 s $= 60 + 4 = 64$ s.
Do this: Convert on a separate line before substituting.
Usually costs: Reported by CISCE examiners for 2025 Q7(ii), along with wrong units.
Mistake: Using a reflector that is too close
A reflector nearer than $d_{m i n}$ (17 m in air at 340 m/s) gives no distinct echo.
Wrong working
Working out an echo time from a wall 12 m away as if it would be heard.
Right working
Check $t = \frac{2 \times 12}{340} \approx 0.071$ s $< 0.1$ s: no distinct echo; use the farther wall.
Do this: Compare every distance with $d_{m i n}$ before using it.
Usually costs: Reported by CISCE examiners for 2024 Q7(i).
Mistake: Not giving a comparison as a ratio
A 'compare x and y' question needs x : y in simplest form, with 0.1 s used for both media.
Wrong working
'x = 16.5, y = 74.25' with no ratio, or dividing speeds at random.
Right working
$x : y = \frac{330 \times 0.1}{2} : \frac{1485 \times 0.1}{2} = 2 : 9$ .
Do this: Finish with 'x : y = …'.
Usually costs: Reported by CISCE examiners for 2024 Q2(vii).
Mistake: Counting claps instead of intervals
Timing from the first clap to the last, n claps make n − 1 intervals.
Wrong working
$t = \frac{11.25}{31}$ .
Right working
$t = \frac{11.25}{30} = 0.375$ s.
Do this: Draw three claps: you see two gaps.
Usually costs: The method mark.

Exam tips

Start every answer with $v = \frac{2 d}{t}$ ; it earns the formula mark even if arithmetic slips.
When the question gives a diagram with two reflectors, check both distances against 17 m (or $d_{m i n}$ for the given speed) before calculating.
Give large distances in both m and km if the question mentions km.
For SONAR questions, name the waves as ultrasonic before calculating.

Practice questions on Echo Numericals (v = 2d/t)

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
A wall is 68 m from a person. If sound travels at 340 m/s, the echo of a clap is heard after:
(a) 0.2 s
(b) 0.4 s
(c) 0.8 s
(d) 5 s
Hint
$t = \frac{2 d}{v}$ .
Answer
(b) 0.4 s
Q2 · 1 mark · Very short answerBoard level
On a warm day sound travels at 350 m/s. Find the minimum distance of a reflector for an echo to be heard.
Hint
Use 0.1 s.
Answer
$d_{m i n} = \frac{350 \times 0.1}{2} = 17.5$ m.
Q3 · 2 marks · Short answerBoard level
The echo of a firecracker from a hill is heard 3 s after the explosion. Find the distance of the hill. (Speed of sound 330 m/s)
Hint
Halve the total path.
Answer
$d = \frac{330 \times 3}{2} = 495$ m.
Q4 · 2 marks · Short answerHarder board level
An echo sounder on a research vessel times a pulse at 3.2 s from leaving the hull to returning from the ocean floor. Taking sound to travel at 1500 m/s in sea water, find the depth of the ocean there.
Hint
Same formula, speed in water.
Answer
$d = \frac{1500 \times 3.2}{2} = 2400$ m (2.4 km).
Q5 · 2 marks · Short answerHarder board level
In air sound covers 330 m each second; in water it covers 1485 m each second. If $x$ and $y$ are the minimum distances for a distinct echo in air and in water, find $x : y$ .
Hint
Use 0.1 s in both; common factors cancel.
Answer
$x = \frac{330 \times 0.1}{2} = 16.5$ m, $y = \frac{1485 \times 0.1}{2} = 74.25$ m; $x : y = 330 : 1485 = 2 : 9$ .
Q6 · 3 marks · Short answerHarder board level
Two cliffs face each other. A girl between them, 170 m from the nearer cliff and 510 m from the farther one, claps once. (Speed of sound 340 m/s) Find the times of the first, second and third echoes.
Hint
Third echo: total path $2 (d_{1} + d_{2})$ .
Answer
$t_{1} = \frac{2 \times 170}{340} = 1$ s; $t_{2} = \frac{2 \times 510}{340} = 3$ s; $t_{3} = \frac{2 \times 680}{340} = 4$ s.
Q7 · 2 marks · Short answerHarder board level
A rifle is fired 660 m from a cliff and the echo is heard 4 s later. Find the speed of sound. Would the echo be heard if the rifle were fired 15 m from the cliff?
Hint
Then check $d_{m i n}$ at this speed.
Answer
$v = \frac{2 \times 660}{4} = 330$ m/s. At 15 m: $d_{m i n} = \frac{330 \times 0.1}{2} = 16.5$ m; 15 m is less, so no distinct echo.
Q8 · 2 marks · Short answerHarder board level
A submarine's SONAR receives an echo from an iceberg 1 min 4 s after the pulse is sent. Speed of sound in sea water is 1500 m/s. Find the distance of the iceberg in km.
Hint
Change the time to seconds first.
Answer
$t = 64$ s; $d = \frac{1500 \times 64}{2} = 48000$ m $= 48$ km.
Q9 · 3 marks · Short answerHOTS / competency
In an ultrasound scan of an organ, echoes from its front and back surfaces return 0.05 ms and 0.11 ms after a pulse is sent. Taking the speed of sound in tissue as 1540 m/s, find (a) the depth of each surface below the probe, (b) the thickness of the organ, in cm.
Hint
Each surface: $d = \frac{v t}{2}$ .
Answer
(a) Front: $\frac{1540 \times 0.05 \times 1 0 ^{- 3}}{2} = 0.0385$ m $= 3.85$ cm; back: $\frac{1540 \times 0.11 \times 1 0 ^{- 3}}{2} \approx 0.0847$ m $= 8.47$ cm. (b) Thickness $= 8.47 - 3.85 = 4.62$ cm.

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.Echo formula?Show answer
$v = \frac{2 d}{t}$ .
2.Minimum time gap for a distinct echo?Show answer
0.1 s.
3.Minimum echo distance in air at 340 m/s?Show answer
17 m.
4.Third echo time between two cliffs?Show answer
$t_{1} + t_{2}$ .
5.How many intervals in 21 claps?Show answer
20.

Echo Numericals (v = 2d/t): frequently asked questions

What is the formula for echo numericals?

v = 2d/t, where d is the distance of the reflector and t the time after which the echo is heard. The 2 appears because sound goes to the reflector and back.

Why must the reflector be at least 17 m away to hear an echo?

The ear keeps a sound for about 0.1 s. In 0.1 s sound in air at 340 m/s travels 34 m, which is 17 m there and 17 m back.

How is the third echo formed between two cliffs?

Sound reflected from one cliff travels to the other cliff, is reflected again and returns. Its time is the sum of the first two echo times.