K326 / K327 / 2027
General properties of waves overview

Lesson 5 of 5

Echoes and distance

  • I can explain reflected sound echoes and their use in measuring distances.Syllabus

    Syllabus K326 / K327, 9(i). Explain reflected sound echoes and their use in measuring distances.

An echo is reflected sound received after travelling to a surface and back. The measured delay includes the complete sound path.

For approximately constant speed, distance travelled = speed x time. Before calculating, identify where the sound is emitted, where it reflects and where it is received.

Draw the outward and return paths

Place a sound source and receiver together, a distance d from a reflecting surface. A short pulse travels distance d to the surface and distance d back. If the sound speed is approximately uniform and the paths match, the total distance is 2d.

2d = vt   so   d = vt/2t is the delay from emission to reception of the echo. v is the sound speed. d is the one-way distance from the source/receiver to the reflector.

Count both parts of the echo's path

First example: the source and receiver are together. Sound travels at the supplied 340 m/s and returns after 0.40 s.

An echo travels 68 metres out and 68 metres backA combined sound source and receiver is on the left and a reflector is 68 metres to its right. The outward arrow points right and the returning echo arrow points left. They are vertically separated only for readability: the example uses the same outward and return path. The total path is 136 metres in 0.40 seconds at 340 metres per second. The reflector distance is half the total path.SourceandreceiverReflectorOutward: 68 mReturn: 68 mTotal path: 136 m in 0.40 s

Separate recorded example: subtract the timestamps

Emission at 20 milliseconds and return at 220 milliseconds give a 200 millisecond intervalThis separate example is a timeline, not the 0.40 second path example above. An emission event is at 20 milliseconds and the corresponding echo arrives at 220 milliseconds. The elapsed time is 200 milliseconds, or 0.200 seconds. At 340 metres per second this interval gives a reflector distance of 34.0 metres. Vertical event marks identify timestamps; their heights do not represent a sound amplitude.Emitted20 msEcho received220 ms060120180240Recorded time / ms220 - 20 = 200 ms
The sound path goes to the reflector and back. The 0.40 s path example and the separate timestamp example both use a round-trip delay, but the timestamp example first requires subtraction.

Worked example

An echo returns after 0.40 s

Use the supplied sound speed 340 m/s. The source and receiver are together, and the reflector remains stationary.

  1. Total sound path: vt = 340 x 0.40 = 136 m.
  2. One-way reflector distance: d = 136/2 = 68 m.

The 136 m is not the reflector distance. It is the sum of the outward and return distances.

A reception time is not automatically a delay

In a separate recording, a pulse is emitted at 20 ms and its echo is received at 220 ms. The timing origin occurred before emission.

Reading recorded times

Subtract, convert, then use the full path

  1. Delay: t = 220 - 20 = 200 ms.
  2. Convert: 200 ms = 0.200 s, since 1 ms = 0.001 s.
  3. Total path: 340 x 0.200 = 68.0 m.
  4. Reflector distance: d = 68.0/2 = 34.0 m.

If the reflector distance were independently measured as 34.0 m, the same record could instead determine speed: v = 2d/t = 68.0/0.200 = 340 m/s.

Try it step by step

Work it out in steps: Total path vt in m, then Distance to the reflector in m.

  1. Problem 1. A student claps and hears the echo from a wall 0.60 s later. Take the speed of sound as 340 m/s. How far away is the wall?
  2. Problem 2. An echo returns from a cliff 0.30 s after a shout. Take the speed of sound as 340 m/s. How far away is the cliff?
  3. Problem 3. Thunder-like echo tests: a sound returns from a hillside 1.5 s after it is made. Take the speed of sound as 340 m/s. How far away is the hillside?
Show the answer

Problem 1.

Total path vt in m: vt = 340 x 0.60 = 204 m (there and back)

Distance to the reflector in m: d = 204 / 2 = 102 m, which is 100 m to 2 sf

Problem 2.

Total path vt in m: vt = 340 x 0.30 = 102 m

Distance to the reflector in m: d = 102 / 2 = 51 m (2 sf)

Problem 3.

Total path vt in m: vt = 340 x 1.5 = 510 m

Distance to the reflector in m: d = 510 / 2 = 255 m, which is 260 m to 2 sf

Measure the intended echo

A microphone with electronic timing or a recorded signal is useful for a short interval. Identify the emission event and the echo of that same pulse. Use a time scale with enough resolution, and account for any stated delay introduced by the instrument.

Several surfaces can produce different echoes. Select the return from the intended reflector rather than pairing the emission with an unrelated later pulse. If determining speed, measure the distance to the actual reflecting surface.

Repeating a human stopwatch measurement does not remove reaction-time limitations or guarantee that a very short echo interval has been resolved. An electronic record addresses timing; a clearer reflector arrangement addresses competing echoes. Choose the improvement that matches the problem.

The path determines whether to divide by two. A one-way transmission from a source to a separate receiver uses its one-way path. Do not divide every sound distance by two merely because sound is involved.

Check your understanding

A source and receiver are together. A pulse is emitted at 20 ms and its echo arrives at 220 ms. With sound speed 340 m/s and the same outward and return path, how far away is the reflector?

  1. 37.4 m.
  2. 68.0 m.
  3. 34.0 m.
Show the answer

34.0 m.

The delay is 220 - 20 = 200 ms = 0.200 s. The total sound path is 340 x 0.200 = 68.0 m. This is outward plus return, so the one-way distance is 34.0 m.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

A student stands 85 m in front of a tall wall and claps once. She hears the echo 0.50 s after the clap. Calculate the speed of sound in air. [3]

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