Echoes and Ultrasound

Stand at the mouth of a cave, or across a valley from a big cliff, and shout your name. A moment later your name comes flying back at you out of the rock — "…name… name…" You didn't shout twice. Nobody in the cliff shouted back. So what happened?

Sound is a wave, and when a wave runs into a hard, flat surface it bounces off — it reflects, just like a ball thrown at a wall. The reflected sound racing back to your ears is an echo. This page takes that one simple idea — sound reflects — and shows how it becomes a ruler for measuring distances you could never reach with a tape measure: the depth of the ocean, the crack hidden inside a steel bridge, even a baby before it is born.

An echo is a stopwatch for distance

Here is the clever part. Sound travels at a known, steady speed — about 340\ \text{m/s} in air (you met this on wave speed). So if you can time how long an echo takes to come back, you can work out how far away the reflecting surface is. Distance is just speed multiplied by time:

\text{distance} = \text{speed} \times \text{time} = v \times t.

But there is a catch that trips almost everyone up. The sound doesn't travel to the wall and stop — it travels all the way there and all the way back before you hear it. The time t you measured is the time for the round trip, which is twice the distance to the wall. So to find the distance to the wall itself, you must halve it:

If a sound reflects off a surface and the echo returns after a time t, the distance to that surface is

\text{distance} = \frac{v \times t}{2}

Worked example — how far is the cliff? You clap, and the echo from a cliff returns 2\ \text{s} later. Sound travels at 340\ \text{m/s}. How far away is the cliff?

\text{distance} = \frac{v \times t}{2} = \frac{340 \times 2}{2} = 340\ \text{m}.

The sound actually flew 680\ \text{m} in total (there and back) — but the cliff is 340\ \text{m} away. Forget the \div 2 and you'd say the cliff was twice as far as it really is.

Watch the pulse go and come back

Below, a source on the left sends out a short pulse of sound. It races across to the wall, reflects, and comes racing back. Drag the second slider to send the pulse on its round trip and watch the running total. Then move the wall further away and send it again — the echo takes longer to return, because the sound has further to travel. The readout shows the round-trip time t and the distance the formula recovers, d = \tfrac{v\,t}{2}.

Notice the pulse crosses the same gap twice. That is the whole reason for the \div 2: the clock runs for both legs of the journey, so half the timed distance is the one-way distance to the wall.

Ultrasound: sound too high to hear

Your ears can only hear sounds within a certain range of frequencies — from about 20\ \text{Hz} (a very low rumble) up to about 20\,000\ \text{Hz} = 20\ \text{kHz} (a very high whistle). Sound with a frequency above 20\ \text{kHz} — too high for any human to hear — is called ultrasound.

Ultrasound is still just sound: a wave of vibrations passed along through a material, exactly like the sound you can hear. It is not light, not radio, not X-rays — it carries no radiation and needs a material to travel through. What makes it so useful is that, like all sound, it reflects at a boundary between two different materials — where soft tissue meets bone, where metal meets an air-filled crack, where water meets the seabed. Send a pulse of ultrasound into an object, time the reflections that bounce back from each boundary inside it, and those echo times reveal what is hidden inside — without ever cutting it open.

Two reasons. First, a higher frequency means a shorter wavelength (remember v = f\lambda), and a shorter wavelength can pick out smaller details — you can "see" fine structures a low, long-wavelength sound would sail straight past. Second, ultrasound is above hearing, so it doesn't deafen anyone and doesn't get drowned out by the everyday noise around us. A doctor can flood the body with it and the room stays silent.

Where echoes and ultrasound are used

The same trick — send a pulse, time the echo, compute the distance with d = \tfrac{v t}{2} — turns up everywhere:

More worked examples

Example 1 — the depth of the sea. A ship's echo-sounder sends a pulse straight down and the echo returns from the seabed after 0.4\ \text{s}. Sound travels through seawater at v = 1500\ \text{m/s}. How deep is the water?

\text{depth} = \frac{v \times t}{2} = \frac{1500 \times 0.4}{2} = \frac{600}{2} = 300\ \text{m}.

The seabed is 300\ \text{m} below the ship.

Example 2 — timing the echo the other way round. A wall is 85\ \text{m} away and sound travels at 340\ \text{m/s}. How long after you clap will you hear the echo? Here the sound covers the distance twice, so rearrange the formula for t:

t = \frac{2 \times \text{distance}}{v} = \frac{2 \times 85}{340} = \frac{170}{340} = 0.5\ \text{s}.

Half a second later, the echo arrives.

Example 3 — a flaw inside metal. An ultrasound pulse fired into a steel block (where sound travels at v = 6000\ \text{m/s}) reflects off a hidden crack and returns in 0.0002\ \text{s}. How far below the surface is the crack?

\text{depth} = \frac{v \times t}{2} = \frac{6000 \times 0.0002}{2} = \frac{1.2}{2} = 0.6\ \text{m}.

The crack lies 0.6\ \text{m} in from the surface.

A bat hunts on the wing in total darkness and never bumps into a thing. Its secret is echolocation — living, breathing echo-timing. The bat lets out a stream of ultrasonic squeaks (far too high for us to hear) and listens for the faint echoes bouncing back off everything around it. From how long each echo takes it works out how far away an obstacle — or a tasty moth — is; from which ear hears it first, which direction it lies in. It builds a complete sound-picture of the night, precise enough to snatch an insect out of mid-air.

Dolphins do the very same thing underwater, clicking and listening to find fish in cloudy water where eyes are useless. Human sonar, prenatal scanners and crack-detectors are really just machines re-inventing a trick that bats and dolphins have been using for millions of years.