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}
- v — the speed of sound in the material, in m/s
- t — the total there-and-back echo time, in seconds
- the \div 2 is there because the sound covers the distance twice
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:
-
Prenatal scans (sonograms). A probe sends ultrasound pulses into the body;
the echoes from the boundaries inside are timed and turned into a picture of the baby. Because
it is just sound — no ionising radiation like an X-ray — it is considered
safe for the unborn baby.
-
Medical imaging generally. The same machines image a beating heart, muscles,
blood flow and organs, all from reflected ultrasound.
-
Industrial flaw detection. Fire ultrasound into a metal casting, an aircraft
wing or a weld: an unseen crack or air bubble inside is a boundary, so it
sends back an early echo — revealing the flaw before the part fails.
-
Sonar and echo-sounding. A ship pings a pulse down into the water and times
the echo from the seabed to measure the depth, or spots shoals of fish and
submarines the same way.
-
Echolocation in animals. Bats and dolphins do it naturally — squeaking or
clicking and listening for the echoes to "see" their world in the dark or the murk.
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.
-
Don't forget to HALVE. The echo time is for the round trip — there
and back — so the surface is only \tfrac{v t}{2} away, not
v t. Miss the \div 2 and every distance
comes out twice too big. (If a question asks for the total distance the sound
travelled, then it is v t — read carefully which one is
wanted.)
-
Ultrasound is high-frequency SOUND — not radiation. It is not an
electromagnetic wave, not light, not X-rays. That is exactly why a baby scan is safe: there
is no ionising radiation, just harmless vibrations.
-
Ultrasound needs a material to travel through. Being sound, it cannot cross
empty space or a vacuum. That is why a scanning probe is pressed on with a blob of gel — to
squeeze out the air and give the sound an unbroken path into the body.
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.