Where is the wall?

The idea

Because reflected sound travels to the surface and back, the distance of a reflecting surface is (v × t) / 2, where v is the speed of sound and t the time until the echo returns.

Clap in a corridor

You clap in an empty corridor and hear the echo 0.5 s later. Sound travels at 340 m s⁻¹.

What do you think the distance to the wall is?

• 85 m
• 170 m
• 340 m

Time to distance

Choose the time after which the echo returns. Sound travels at 340 m s⁻¹.

Half of there and back

The sound travels to the surface and back, so the distance it covers is v × t. The surface is only half this far from you.

Reflected sound travels to the surface and back, so the distance of the reflecting surface is (v × t) / 2, where v is the speed of sound and t is the time until the echo returns.

Notes

Distance of the reflecting surface = (v × t) / 2, because the sound goes to the surface and back.

Check yourself

You clap in an empty corridor and hear an echo after 0.5 s. The speed of sound is 340 m s⁻¹. How far is the wall, in m?

Answer: 85 m

Distance = (340 × 0.5) ÷ 2 = 85 m.

An echo returns after 2 s. The speed of sound is 340 m s⁻¹. How far is the reflecting surface, in m?

Answer: 340 m

Distance = (340 × 2) ÷ 2 = 340 m.

Why do we divide v × t by 2 when finding the distance of the reflecting surface?

Which formula gives the distance of the reflecting surface?

• The sound slows down by half. The sound does not slow down. The time includes the trip to the surface and the trip back.
• The wall absorbs half of the sound. The division by 2 is because the sound covers the distance to the surface twice, there and back.
• The sound travels to the surface and back — correct. Yes!
• (v × t) / 2 — correct. Yes!
• v × t. This is the distance travelled there and back. The surface is half this distance away.
• (v × t) × 2. The sound covers the distance twice, so the surface is half of v × t away, not double.