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Speed Of Sound & Echo

Key idea: O Level echo questions: sound reflection, why distance is 2d, and how to find speed of sound or distance using v = 2d/t.

  • SEC G3 Physics 2027
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Learning objectives

  • Describe wave generation by vibrating sources, ropes and springs
  • Describe ripple-tank waves using wavefronts
  • Explain that waves transfer energy
  • Explain that wave energy transfer does not transfer matter
  • Use amplitude, frequency and wavelength to describe wave motion
  • Define and use wave speed and period and interpret wave graphs
  • Recall and apply wave speed = frequency × wavelength
  • Compare transverse and longitudinal waves and give examples
  • Explain sound production by vibration and the need for a medium
  • Describe sound using compressions and rarefactions
  • Relate sound loudness to amplitude and pitch to frequency
  • Explain reflected-sound echoes and use them to measure distance
  • Explain ultrasound use in sonar and soft-tissue scanning
  • Use the normal, angle of incidence and angle of reflection
  • Apply the law of reflection in constructions, measurements and calculations
  • Use the normal, angle of incidence and angle of refraction
  • Apply sin i divided by sin r as a constant for a fixed pair of media
  • Define refractive index as vacuum light speed divided by medium light speed
  • Explain the critical angle
  • Explain the conditions for total internal reflection
  • Apply total internal reflection to optical fibres and state advantages
  • Describe how a thin converging lens acts on a light beam
  • Define the focal length of a converging lens
  • Construct real and virtual image ray diagrams for a thin converging lens
  • Describe lens images as real or virtual, magnified or diminished, and upright or inverted

1. Definition

A. Speed of sound (in air)

At room temperature, the speed of sound in air is about 340 m s⁻¹.

B. Echo

An echo is a distinct sound heard due to the reflection of a sound wave from a surface.

2. Key Ideas

  • Sound can be reflected by large, hard surfaces (walls, cliffs).
  • In echo questions, sound travels to the wall and back, so the total distance is 2d.
  • Key relationship: v = distance/time
  • Echo formula: v = 2d/t or d = vt/2
  • In the repeated-clap method, adjust the rhythm until each clap coincides with the previous echo, then time N clap intervals: v = 2Nd/tₜₒₜₐₗ

3. Detailed Explanations

A. How an echo forms

  1. A sound is produced (clap/shout).
  2. The sound wave travels to a reflecting surface.
  3. The wave is reflected and travels back to the listener.
  4. If the reflected sound arrives late enough, it is heard as a separate sound (an echo).
Sound pulse and echo pathA sound pulse travels from a source and receiver to a reflecting wall and back, covering twice the one-way distance.Source / listenerWallOutgoing pulseEcho returnd2d = vt, so d = vt / 2
The measured delay is for the complete outward-and-return path: total distance = 2d.

B. Measuring distance using an echo

If the time delay between the original sound and the echo is t, then the sound has travelled to the surface and back, so:

2d = vt ⇒ d = vt/2

C. Practical: measuring the speed of sound using echoes

  1. Measure the one-way distance d to a large wall/cliff.
  2. Clap repeatedly and adjust the rhythm until each clap coincides with the echo of the previous clap.
  3. Time N clap intervals in one continuous measurement to obtain tₜₒₜₐₗ.
  4. Use:

v = 2Nd/tₜₒₜₐₗ

Why time many echoes?

Each clap interval is one sound round trip, 2d. Timing many intervals in one continuous measurement makes the total time larger, so reaction time is a smaller percentage uncertainty.

4. Common Mistakes

  • Forgetting the factor of 2 (using d = vt instead of d = vt/2).
  • Using the one-way distance when the question gives a round-trip time (or vice versa).
  • Mixing units (cm and m, ms and s).

5. Exam Tips

  • Start with a labelled sketch or a sentence: “sound travels to the surface and back”.
  • Write the formula with symbols before substituting:
    • v = 2d/t or d = vt/2
  • For the repeated-clap method, state that each clap is synchronised with the previous echo and many intervals are timed together.

6. Worked Examples

Modelled example 1

Distance from an echo

Core

Problem

A student stands d metres from a wall. The echo is heard 0.40 s after the clap. Take v = 340 m s⁻¹. Find d.
Study the worked solution

  1. Interpret the measured time

    Method

    Use the delay for travel to the wall and back.

    Reason

    The echo is received only after the reflected sound completes a round trip.

    Working

    Round-trip distance = vt = (340)(0.40) = 136 m.

  2. Find the one-way distance

    Method

    Divide the round-trip distance by two.

    Reason

    The outward and return legs each have length d.

    Working

    d = vt/2 = 136/2 = 68 m

Guided practice 2

Speed of sound from the repeated-clap method

About 5 min

Problem

A student stands 50 m from a wall and claps so each clap coincides with the previous echo. The total time for 50 clap intervals is 15.0 s. Find the speed of sound.

Count round trips, not claps alone

Unit: m/s

Hints

Hint 1: one interval
Each clap interval corresponds to one 100 m round trip.
Hint 2: all intervals
For 50 intervals, use 2Nd = 2(50)(50).
View solution step by step

  1. Find the aggregate distance

    Method

    Multiply one round trip by the 50 timed intervals.

    Reason

    Each new clap coincides with the preceding clap’s returning echo.

    Working

    2Nd = 2(50)(50) = 5000 m

  2. Calculate the speed

    Method

    Divide total distance by the continuous timing.

    Reason

    Timing many intervals reduces the percentage effect of reaction-time uncertainty.

    Working

    v = 5000/15.0 = 333 m s⁻¹

Common misconception 3

Minimum distance idea (echo heard separately)

Find and correct the mistake

Learner response

An echo is heard separately only if the delay is at least 0.10 s. Take v = 340 m s⁻¹. A student calculates the minimum wall distance as (340)(0.10) = 34 m. Locate the error.

Separate total path from wall distance

Unit: m

View solution step by step

  1. Interpret 34 metres

    Method

    Recognise 34 m as the total distance travelled.

    Reason

    The delay ends when the reflected sound returns to the listener.

    Working

    vt = (340)(0.10) = 2d

  2. Find the minimum wall distance

    Method

    Halve the total path length.

    Reason

    The wall is one equal leg of the round trip.

    Working

    d = 34/2 = 17 m

Examiner practice 4

Echo time from distance

3 marks

Examination question

A student stands 85 m from a cliff. Take v = 340 m s⁻¹. Find the time between the shout and the echo. [3 marks]

Show path length, equation and result

View solution step by step

  1. Find the travel distance

    1 mark

    Method

    Double the one-way cliff distance.

    Reason

    The sound travels to the cliff and back.

    Working

    2d = 2(85) = 170 m

  2. Calculate the delay

    2 marks

    Method

    Divide the round-trip distance by speed.

    Reason

    The measured interval covers the complete echo path.

    Working

    t = 2d/v = 170/340 = 0.50 s

Challenge 5

Finding distance (given speed and time)

Minimal support

Changed-speed transfer

An echo is heard 0.60 s after a clap. In these conditions, take v = 330 m s⁻¹. Find the distance to the wall.

Use the supplied speed and justify the factor of two

Hints

Hint 1: use the stated conditions
Do not replace 330 m s⁻¹ with a memorised value.
Hint 2: interpret the delay
The measured 0.60 s is for 2d.
View solution step by step

  1. Calculate the one-way distance

    Method

    Multiply speed by round-trip time, then halve.

    Reason

    The echo delay includes equal outward and return legs.

    Working

    d = vt/2 = (330)(0.60)/2 = 99 m

7. Mind Stretchers

Mind stretcher 1: Soft vs hard surfacesExtension

Why do curtains and carpets reduce echoes in a room?

Show Answer

Soft, porous materials absorb more sound energy and reflect less, so the reflected sound is weaker and echoes are reduced.

Mind stretcher 2: Echo sounder ideaExtension

An echo sounder sends a sound pulse into water and measures the time for the echo to return. Why must the distance formula still include a factor of 2?

Show Answer

The pulse travels from the source to the seabed (or fish) and then back to the receiver. The measured time is for the round trip, so the total distance is twice the one-way distance.

8. Practice and next step

For every echo calculation, sketch the outward and return path before using d = vt/2. Continue to ultrasound to apply the same timing logic to sonar and scanning.

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027