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Echoes and sound measurements

Key idea: An echo travels to a reflecting surface and back. For a source and listener together, distance to the surface is sound speed times echo delay divided by two.

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Shout towards a cliff and you hear your voice come back a moment later. The sound has travelled to the cliff, reflected and travelled back, and the delay tells you how far away the cliff is. The same idea, timing a reflected pulse, lets ships measure the depth of the sea and lets you measure the speed of sound with nothing more than a wall and a stopwatch.

Echoes and measuring distance

An echo is a sound reflected from a surface and heard after the original sound. Hard, flat surfaces such as walls, cliffs and tall buildings reflect sound well. Soft materials such as curtains and cushions absorb much of it.

If you know the speed of sound, an echo lets you measure how far away a surface is. The sound travels to the surface and back, so the total distance it travels is twice the distance to the surface:

2d = vt d = vt/2

Here t is the time from making the sound to hearing the echo and v is the speed of sound. This works when the source and the listener are in the same place and the surface is not moving.

Sound pulse and echo pathTwo direction arrows represent a pulse travelling from a source and receiver at the same position to a stationary reflector and back along the same path. The arrows are drawn apart for clarity. Each leg has length d; the measured delay covers both legs, so the total path is 2d.Time the full journey from emission to echo arrival.Source / listenerWallOutgoing pulseReturning echodArrows are drawn apart; both legs follow the same path. Total path: 2d.
Sound pulse and echo pathTwo direction arrows represent a pulse travelling from a source and receiver at the same position to a stationary reflector and back along the same path. The arrows are drawn apart for clarity. Each leg has length d; the measured delay covers both legs, so the total path is 2d.Time the full journey from emission to echo arrival.Source / listenerWallOutgoing pulseReturning echodArrows are drawn apart; both legs follow the same path. Total path: 2d.
For a stationary reflector and a sender and receiver together, with constant speed along the path, 2d = vt. Therefore d = vt/2. The two arrows are schematic and separated for clarity.

Worked example 1

Finding the distance to a wall

Problem

A student claps and hears the echo from a wall 0.40 s later. The speed of sound is 340 m/s. How far away is the wall?

Show full solution

  1. Find the full journey length

    Method

    Multiply sound speed by the echo delay.

    Reason

    The 0.40 s covers travel to the wall and back to the student.

    Working

    Answer

    vt = 340 m/s × 0.40 s = 136 m.

  2. State the one-way distance

    Working

    Answer

    The wall is 136 m ÷ 2 = 68 m away.

Check your understanding 1 To halve or not?

A detector across a field receives a pulse directly from its source 0.20 s after it is sent out. Sound speed is 340 m/s. A learner calculates 340 × 0.20 ÷ 2 = 34 m for the distance. Decide which path the sound took and find the correct distance.

The sound's path
Unit: m
Show answer

The detector is across the field, so this sound travelled one way only; it is not an echo. The distance is 340 × 0.20 = 68 m. Before you halve, check whether the time includes a journey there and back.

Check your understanding 2 Between two cliffs

A hiker stands between two parallel cliffs and shouts. The echo from one cliff arrives after 1.0 s and the echo from the other after 1.5 s. The speed of sound is 340 m/s. How far apart are the cliffs?

Unit: m
Unit: m
Show the steps Optional working
Unit: m
Unit: m
Unit: m
Show answer

Nearer cliff: d₁ = 340 × 1.0 ÷ 2 = 170 m. Further cliff: d₂ = 340 × 1.5 ÷ 2 = 255 m. She stands between them, so the cliffs are 170 + 255 = 425 m apart.

Measuring the speed of sound with echoes

A single echo from a nearby wall arrives too quickly to time accurately by hand: your reaction time is a large part of it. Instead, a student claps steadily so that each clap coincides with the echo of the one before, and times many intervals at once.

Worked example 2

Timing many echo intervals

Problem

A student claps just as the echo of each previous clap returns from a wall. She times 20 complete intervals, from the first clap to the 21st: 10.0 s. Take sound speed as 340 m/s. Find the distance to the wall and explain why timing many intervals reduces the effect of stopwatch reaction time.

Show full solution

  1. Find one echo delay

    Method

    Divide the total time by the number of complete intervals.

    Reason

    Each interval covers one journey to the wall and back. There are 20 gaps between 21 claps.

    Working

    Answer

    t = 10.0 s ÷ 20 = 0.50 s.

  2. Relate time to wall distance

    Method

    Use d = vt/2 with the time for one interval.

    Reason

    The source and listener are together, so the two legs are equal.

    Working

    Answer

    d = 340 m/s × 0.50 s ÷ 2 = 85 m.

  3. Give the distance and timing benefit

    Working

    Answer

    The wall is 85 m away. The reaction-time error in starting and stopping the watch is a much smaller fraction of 10.0 s than of a single 0.50 s interval.

Twenty complete intervals need 21 claps: start the stopwatch on the first and stop it on the 21st. Repeat the measurement and take the mean. If the distance to the wall is measured instead, the same timing gives the speed of sound from v = 2d/t.

Check your understanding 3 Counting the intervals

A learner times from the first clap to the 20th and divides the result by 20. How many intervals were timed, and what should change?

Unit: intervals
Corrected method
Show answer

There are 19 complete intervals between those 20 claps. Divide by 19, or time from the first clap to the 21st if you want 20 intervals. Count the gaps, not the claps.

Check your understanding 4 Speed of sound from echoes

A wall is 51 m away. Fifteen complete echo intervals take 4.5 s. Find the time for one interval, the distance the sound travels in that time and the speed of sound.

Unit: s
Unit: m
Show the steps Optional working
Unit: s
Unit: m
Unit: m/s
Show answer

t = 4.5/15 = 0.30 s, so v = 2d/t = 2 × 51/0.30 = 340 m/s. Fifteen intervals need 16 claps.

Check your understanding 5 Work out the actual reflected path

A source is 34 m from a wall and a detector is 17 m from the same wall, along the same perpendicular line on the same side. Sound speed is 340 m/s. Find the time from emission until the reflected pulse reaches the detector. Does half of the full reflected path give the source’s distance to the wall? Sketch and label the two legs on paper first.

Unit: m
Show the steps Optional working
Unit: m
Unit: s
Does half the reflected path give the source’s distance to the wall?
Show answer

Self-review the sketch: draw wall, detector and source on one line, with 34 m from source to wall and 17 m from wall to detector. The reflected path is 34 + 17 = 51 m, so its delay from emission is 51/340 = 0.15 s. Half of 51 m is 25.5 m, which is neither stated wall distance: the legs are unequal. The rule d = vt/2 gives the wall distance when source and receiver are together, not for this separated arrangement. The direct pulse travels only 17 m and arrives earlier; the question asks for the reflected pulse.

Common mistakes

  • Halving every travel time. Halve only for a journey there and back; a sound received one way uses d = vt.
  • Using d = vt/2 when the sender and receiver are in different places. Work out the actual path first.
  • Timing the gap between two echoes instead of from the sound to its echo.
  • Treating 20 claps as 20 intervals. There are 19 gaps between the first and the 20th clap.

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