Speed Of Sound & Echo

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/tquador 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).

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 Rightarrow 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

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

Example 1: Distance from an echoCore

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.

Show Answer

d = vt/2 = (340)(0.40)/2 = 68 m.

Example 2: Speed of sound from the repeated-clap methodCore

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.

Show Answer

Total distance travelled = 2Nd = 2(50)(50) = 5000 m.
v = 2Nd/tₜₒₜₐₗ = 5000/15.0 = 333 m s⁻¹.

Example 3: Minimum distance idea (echo heard separately)Core

An echo is heard separately only if the time delay is at least 0.10 s. Take v = 340 m s⁻¹. Estimate the minimum distance from a wall needed to hear an echo separately.

Show Answer

d = vt/2 = (340)(0.10)/2 = 17 m.

Example 4: Echo time from distanceCore

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

Show Answer

Total distance travelled = 2d = 170 m.
t = 2d/v = 170/340 = 0.50 s.

Example 5: Finding distance (given speed and time)Core

An echo is heard 0.60 s after a clap. Take v = 330 m s⁻¹. Find the distance to the wall.

Show Answer

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.