Under the huge dome of the Gol Gumbaz in Vijayapura, Karnataka, visitors climb to the gallery that runs round its base. A single clap there comes back again and again, a string of fading replies. Yet in an ordinary classroom, a clap brings no reply at all, although its walls reflect sound too.
What makes an echo
The last lesson showed that a rigid wall reflects almost all of the sound that reaches it. When the reflected sound reaches your ear distinctly after the original, you hear it as a separate sound. That separate, repeated sound is an echo.
Suppose you stand at a distance \(d\) from a large wall or a cliff and clap once. The sound travels to the wall and back, a total distance of \(2d\). At a speed \(v\), the echo reaches you after a time \[ t = \dfrac{2d}{v}. \] The factor 2 is the whole of the difficulty. The sound makes a round trip, so the time measures twice the distance.
The ear needs a tenth of a second
The ear does not separate two sounds that arrive very close together. The sensation of a sound lasts about \(0.1\,\text{s}\) after the sound itself stops. This is the persistence of hearing. A reflection that returns sooner than this blends with the original and only makes it sound longer. A reflection that returns \(0.1\,\text{s}\) or more later is heard as a separate echo.
So the echo time must be at least \(0.1\,\text{s}\). With the speed of sound in air taken as \(340\,\text{m/s}\), the least distance to the reflector is \[ d_{\min} = \dfrac{v \times 0.1\,\text{s}}{2} = \dfrac{340 \times 0.1}{2} = 17\,\text{m}. \] A classroom wall is closer than this, so its reflection merges with the clap. That is why a classroom gives no echo.
Try the figure above at three distances. At \(d = 8.5\,\text{m}\) the echo returns after \(0.050\,\text{s}\) and merges with the clap. At \(d = 17\,\text{m}\) it returns after exactly \(0.100\,\text{s}\), just heard apart. At \(d = 34\,\text{m}\) it returns after \(0.200\,\text{s}\), a clear echo.
The persistence of hearing is about \(0.1\,\text{s}\), and sound travels at \(340\,\text{m/s}\). How near to a cliff can a person stand and still hear the echo of a clap, in m?
A trekker in the Western Ghats shouts towards a cliff across a valley and hears the echo \(1.5\,\text{s}\) later. Take the speed of sound as \(340\,\text{m/s}\). (a) How far away is the cliff? (b) Would the trekker still hear an echo from a cliff \(10\,\text{m}\) away?
(a) The sound makes a round trip, so \(d = vt/2 = 340 \times 1.5/2 = 255\,\text{m}\).
(b) The echo would return after \(t = 2d/v = 20/340 \approx 0.059\,\text{s}\). That is less than \(0.1\,\text{s}\), so the reflection merges with the shout, and no separate echo is heard.
A student claps at one end of a hall \(12\,\text{m}\) long, facing the far wall. Take the speed of sound as \(340\,\text{m/s}\). Does she hear a separate echo from that wall?
A student stands \(85\,\text{m}\) from a tall wall and claps. She adjusts her rhythm until each clap coincides with the echo of the one before. A friend counts \(20\) intervals between claps in \(10\,\text{s}\). Find the speed of sound.
Each interval equals the echo time, so \(t = 10/20 = 0.5\,\text{s}\).
The sound covers \(2d = 170\,\text{m}\) in that time, so \(v = 2d/t = 170/0.5 = 340\,\text{m/s}\).
Echo sounding and SONAR
The relation \(d = vt/2\) turns an echo into a measuring tape. A ship finds the depth of the sea this way. A device under the hull sends a short pulse of sound straight down. The pulse reflects from the sea bed, and a receiver on the hull records the time of the echo. This is echo sounding.
Two things differ from a clap in air. First, the speed of sound in sea water is much larger, about \(1500\,\text{m/s}\). Second, the pulse is usually ultrasound, of frequency above \(20\,\text{kHz}\). Its short wavelength spreads less and reflects well even from small objects. The same method, used to find submarines, shoals of fish and wrecks, is called SONAR, for sound navigation and ranging.
A research ship off the coast of Chennai sends an ultrasonic pulse straight down. The echo from the sea bed returns after \(1.2\,\text{s}\). The speed of sound in sea water is \(1500\,\text{m/s}\). How deep is the sea there?
The echo time is \(t = 1.2\,\text{s}\), and the speed in sea water is \(v = 1500\,\text{m/s}\).
The pulse goes down and comes back, so \(d = vt/2\).
\(d = 1500 \times 1.2/2 = 900\,\text{m}\).
The sea is \(900\,\text{m}\) deep below the ship.
A fishing boat's echo sounder receives the echo from a shoal of fish \(0.04\,\text{s}\) after sending the pulse. The speed of sound in sea water is \(1500\,\text{m/s}\). How deep is the shoal, in m?
Bats that hunt at night send out short cries of ultrasound, far above the range of human hearing, and listen for the echoes. An insect \(1.7\,\text{m}\) away returns an echo after \(2 \times 1.7/340 = 0.01\,\text{s}\). From the delay a bat judges the distance, and from the change in the echo it judges the direction and the motion. Dolphins find fish in murky water the same way.
Reverberation in a hall
In a large hall the sound does not return once. It reflects from the walls, the floor and the ceiling, again and again. Each reflection loses a little energy to the surface it strikes. The reflections arrive so close together that the ear cannot separate them. Instead, the sound seems to linger and fade slowly after the source stops. This is reverberation.
The reverberation time is the time taken for the sound level to fall by \(60\,\text{dB}\), to a millionth of its intensity, after the source stops. A long reverberation time makes speech blurred, because each word is still sounding when the next arrives. A very short one makes music sound dry and dead. Good lecture halls aim at about one second, and concert halls at about two.
Soft materials absorb sound instead of reflecting it: curtains, carpets, cushioned seats and people. More absorbing area means less energy returned at each reflection, so the sound dies faster. That is why an empty hall sounds louder and more ringing than the same hall full of people. Designers choose the surfaces so that the hall rings for just the right time.
Wallace Sabine found an approximate rule for the reverberation time, \(T \approx 0.161\,V/A\). Here \(V\) is the volume of the hall in m³, and \(A\) the total absorbing area in m². For a hall of \(5000\,\text{m}^3\) with \(400\,\text{m}^2\) of absorbing area, \(T \approx 0.161 \times 5000/400 = 2.0\,\text{s}\). Doubling the absorbing area to \(800\,\text{m}^2\) halves it to \(1.0\,\text{s}\), as in Fig 5.1.
This links back to the concert halls of the second lesson. There, reflections from the walls interfered with the direct sound and could leave a dead spot. Here, the same reflections decide how long the hall rings. An architect must manage both: the reflections must fill the hall evenly, and they must die away at the right rate.
In 1895 Harvard University asked a young physics lecturer, Wallace Sabine, to cure a new lecture room where speech could not be understood. Working at night, he carried seat cushions in and out of halls and timed how long a note from an organ pipe remained audible. He found that the time depended on the volume of the room and on its absorbing surfaces. In 1900 Boston's Symphony Hall opened, designed with his rule, and it is still praised for its sound.
A school's new assembly hall has bare concrete walls, and speeches in it sound blurred. Which change helps most?
You clap in front of a long school building across the playground and hear the echo about \(0.4\,\text{s}\) later. Estimate the distance to the building, in m, with sound at \(340\,\text{m/s}\).
Now return to the Gol Gumbaz. Its dome is so wide that a reflection from the far side returns more than a tenth of a second after the clap. So each reflection comes back as a separate echo, and the smooth, hard stone loses little energy at each one. A single clap therefore returns again and again. A classroom wall is closer than \(17\,\text{m}\), so its reflection merges with the clap instead.
- Forgetting the factor 2 in \(d = vt/2\).The echo time covers the trip to the reflector and back. Halve the product of speed and time.
- Using the speed of sound in air for SONAR.Sound in sea water travels at about \(1500\,\text{m/s}\), more than four times as fast as in air.
- Saying that a small room gives no reflection.A small room reflects sound well, but the reflections return within \(0.1\,\text{s}\) and blend with the original. They add reverberation, not an echo.