← Sound, the Doppler effect, and resonance

Echo ranging

Maren shouts across a canyon. Three seconds later, the canyon wall sends her shout back.

The echo's delay is a measurement. With the speed of sound and a stopwatch, Maren can find how far away the wall is. The catch: the sound made a round trip.

Maren's shout reflects off the far wall, the bouncing back you met at reflection, and returns to her as an echo.

The echo returns 3 s after the shout. Sound travels through the air at 340 m/s.

Marencanyon wallout: shout, 340 m/sback: echo, 340 m/s?round trip: 3 s
The timed 3 s covers the out journey and the back journey together.

In 3 s, the sound covers: 340 × 3 = 1020 m.

But that 1020 m is the whole round trip: out to the wall, and back again.

The wall sits at half the round trip: 1020 ÷ 2 = 510 m.

Marencanyon wallout: shout, 340 m/sback: echo, 340 m/s510 mround trip: 340 × 3 = 1020 m, so one way: 1020 ÷ 2 = 510 m
Half the round trip is the distance to the wall: 510 m.

The distance the sound covers comes from: distance = speed × time.

For an echo, halve the result, because the timed journey is there and back.

Halve first or halve last, the answer agrees: half of 3 s is 1.5 s, and 340 × 1.5 = 510 m.

Noll stands in a field facing a long brick wall. His clap's echo returns after 4 s. Sound travels at 340 m/s in air. Calculate the distance to the wall.

Write down the values:

speed = 340 m/s, round-trip time = 4 s, distance to the wall = ?

Write down the equation:

distance = speed × time

Substitute in the values, calculate the round trip, and halve it:

round trip = 340 × 4 = 1360

1360 ÷ 2 = 680

The wall is 680 m away.

A research ship's sonar pings the seabed, and the echo returns after 0.6 s. Sound travels at 1500 m/s in water. Calculate the depth of the seabed.

Write down the values:

speed = 1500 m/s, round-trip time = 0.6 s, depth = ?

Write down the equation:

distance = speed × time

Substitute in the values, calculate the round trip, and halve it:

round trip = 1500 × 0.6 = 900

900 ÷ 2 = 450

The seabed is 450 m below the ship.

An echo returns 5 s after a shout. Enter the time the sound took to reach the cliff, in s.

A ship's sonar ping echoes off the seabed after 0.8 s. Sound travels at 1500 m/s in water. Enter the depth of the seabed in m.

Ottoline claps in a canyon, and the echo returns after 1.8 s. Sound travels at 340 m/s in air. Enter the distance to the canyon wall in m.

Your turn

Pim shouts at a cliff, and the echo returns after 1.3 s. Sound travels at 340 m/s in air. Enter the distance to the cliff in m.

At a trailhead, a ranger blows a whistle at a rock face across a valley, and the echo takes 7 s to return. Sound travels at 340 m/s in air. Enter the distance from the ranger to the rock face in m.

A rail engineer taps a long steel rail, and an instrument times the echo returning from a crack: 0.4 s. Sound travels at 5000 m/s in steel. Enter the distance to the crack in m.