In a crowded Delhi Metro coach, a passenger puts on noise-cancelling earphones and presses a button. The rumble of the wheels and the hum of the air conditioner almost vanish, although nothing blocks the ear more than before. The earphones did not stop the noise. They added a second sound to it.
Where exactly is the first quiet spot?
The last lesson gave the rule for a quiet spot. The path difference must be half a wavelength, or one and a half, and so on. The rule says what must happen at the spot. It does not yet say where the spot is. For that we need the geometry of the two paths. The next example works it out completely.
The speed of sound in this lesson is \(343\,\text{m/s}\), the value for air at about \(20\,°\text{C}\). The last lesson used the same value.
Two speakers are \(3.0\,\text{m}\) apart and emit a \(572\,\text{Hz}\) tone in phase. A listener walks along a line parallel to the speakers, \(8.0\,\text{m}\) away. Where is the first quiet spot (destructive interference), measured from the centre line? Use \(v_{\text{sound}} = 343\,\text{m/s}\).
First, find the wavelength: \(\lambda = v/f = 343/572 \approx 0.600\,\text{m}\). Put \(S_1\) at \(y = +1.5\,\text{m}\) and \(S_2\) at \(y = -1.5\,\text{m}\). The listener stands at a distance \(y\) from the centre line, so \(L_1 = \sqrt{8.0^2 + (y - 1.5)^2}\) and \(L_2 = \sqrt{8.0^2 + (y + 1.5)^2}\).
For the first minimum, \(n = 0\): \(|L_1 - L_2| = \lambda/2 = 0.300\,\text{m}\). The squares of the paths give an exact relation: \(L_2^2 - L_1^2 = (y + 1.5)^2 - (y - 1.5)^2 = 6.0\,y\). Since \(L_2^2 - L_1^2 = (L_2 - L_1)(L_2 + L_1)\), we get \(y = \Delta L\,(L_1 + L_2)/6.0\) exactly.
The relation still contains \(y\) on the right, inside \(L_1 + L_2\), so solve it by a short numerical search. Guess \(y = 0.80\,\text{m}\): then \(L_1 = 8.031\,\text{m}\), \(L_2 = 8.324\,\text{m}\), and \(y = 0.300 \times 16.355/6.0 = 0.818\,\text{m}\). Repeat with \(y = 0.818\,\text{m}\): \(L_1 = 8.029\,\text{m}\), \(L_2 = 8.329\,\text{m}\), and \(y = 0.300 \times 16.358/6.0 = 0.818\,\text{m}\) again.
The first quiet spot is \(y \approx 0.82\,\text{m}\) from the centre line, on either side of it. Check: there \(L_2 - L_1 = 8.329 - 8.029 = 0.300\,\text{m} = \lambda/2\). The small-angle estimate of the next section gives \(0.80\,\text{m}\), about \(2\,\%\) low.
This example shows how interference creates regions of loud and quiet sound in space, at places that can be calculated. The quiet spot is less than a metre from the loud centre line. A listener who takes two steps sideways passes from the loudest place to the quietest.
Walk the line yourself in the figure below. Move the listener with the slider and watch the path difference.
On the centre line, \(y = 0\), the two paths are equal, \(8.14\,\text{m}\) each, and the spot is loud. At \(y = 0.82\,\text{m}\) the readout shows \(0.50\,\lambda\): the first quiet spot. At \(y = 1.66\,\text{m}\) the path difference is one whole wavelength, and the spot is loud again. At \(y = 2.55\,\text{m}\) it is \(1.5\,\lambda\), and the spot is quiet again. Loud and quiet spots alternate along the line.
Two speakers separated by \(2.0\,\text{m}\) emit \(343\,\text{Hz}\) sound waves in phase. If the speed of sound is \(343\,\text{m/s}\), what is the wavelength, in metres?
Two speakers play the same tone in phase. At a point P the path difference is \(0.40\,\text{m}\). For what frequency is P the first quiet spot (\(n = 0\))? Take \(v = 343\,\text{m/s}\). Give the answer in Hz.
A quick estimate when the listener is far away
The exact relation of the example, \(\Delta L = 2yd/(L_1 + L_2)\), holds for any speaker separation \(d\). It becomes simple when the listening line is far away. Then \(y\) and \(d\) are small compared with the distance \(L\) to the line, and each path is close to \(L\).
\[ L_2^2 - L_1^2 = \left(y + \tfrac{d}{2}\right)^2 - \left(y - \tfrac{d}{2}\right)^2 = 2yd \]
Why. Pythagoras for each path, with the speakers at \(\pm d/2\) and the line at a distance \(L\). The \(L^2\) terms cancel.
\[ (L_2 - L_1)(L_2 + L_1) = 2yd \]
Why. the difference of two squares. This line is exact.
\[ \Delta L = \dfrac{2yd}{L_1 + L_2} \]
Why. divide by \(L_1 + L_2\). This is the relation used in the example.
\[ \Delta L \approx \dfrac{yd}{L} \]
Why. when \(y\) and \(d\) are much smaller than \(L\), each path is nearly \(L\), so \(L_1 + L_2 \approx 2L\).
\[ y_n \approx \left(n + \tfrac12\right)\dfrac{\lambda L}{d} \]
Read it. set \(\Delta L = (n + \tfrac12)\lambda\), the condition for a quiet spot, and solve for \(y\).
For the example, \(y_0 \approx (0.600 \times 8.0)/(2 \times 3.0) = 0.80\,\text{m}\). The exact answer is \(0.82\,\text{m}\). The estimate is low because \(L_1 + L_2 = 16.36\,\text{m}\) there, a little more than \(2L = 16.0\,\text{m}\). The same formula spaces the loud spots about \(\lambda L/d = 1.60\,\text{m}\) apart; the figure found \(1.66\,\text{m}\).
The points where \(|L_1 - L_2|\) keeps one value form a hyperbola whose two foci are the speakers. So every quiet line across the room is a hyperbola. For the first quiet line, the half-difference is \(a = \lambda/4\), which is \(0.150\,\text{m}\). Then \(b^2 = 1.5^2 - a^2\), which is \(2.2275\,\text{m}^2\). On the listening line \(8.0\,\text{m}\) away, \[ y = a\sqrt{1 + 8.0^2/b^2}. \] This is \(0.150 \times 5.453 = 0.818\,\text{m}\), the answer of the search. Far from the speakers each hyperbola runs almost straight, which is why the small-angle estimate works.
The two speakers of the example play a tone of twice the frequency, \(1144\,\text{Hz}\), with nothing else changed. What happens to the first quiet spot on the listening line?
Interference at work
The interference of sound waves has many practical applications:
- Noise-cancelling headphones generate sound waves that interfere destructively with the noise around the listener.
- Concert halls are designed with care to avoid zones of destructive interference, where the music sounds muted.
- Ultrasonic imaging uses patterns of constructive and destructive interference to make detailed pictures.
A noise-cancelling earphone carries a small microphone that listens to the noise outside. A circuit makes the exact inverse of that pressure wave, and the earphone's speaker plays it. At the eardrum, the noise and this "anti-noise" arrive crest on trough, at every instant. Their pressure variations add to nearly zero.
The method works best for steady, low sounds, such as the drone of an engine. A low sound has a long wavelength, so a small error in timing is a small fraction of a cycle. A high, sudden sound, such as a voice or a clap, changes too fast for the circuit to follow. That is why the earphones remove the rumble of the Metro but not the announcements.
In 1934 the German physician Paul Lueg applied for a patent on "silencing sound oscillations", which was granted in the United States in 1936. His idea was to pick up a noise with a microphone and play it back inverted through a loudspeaker, so that the two cancel. The electronics of his time were too slow to make it work well. About fifty years later, fast circuits turned his idea into the headphones worn by pilots, and then by passengers.
In a concert hall, the sound reaches each seat directly from the stage and also after reflections from the walls and the ceiling. These waves travel different distances, so they interfere. At a few seats, a note may arrive crest on trough and sound muted; such a seat is called a dead spot. Architects shape the walls, and break up flat surfaces, so that no seat sits in a steady quiet zone.
The probe of an ultrasound scanner holds a row of tiny sources of ultrasound, above the range of human hearing. Each source is fired a tiny fraction of a microsecond after its neighbour. The delays are chosen so that all the waves arrive in phase at one chosen point inside the body. There they interfere constructively, and everywhere else they partly cancel. By changing the delays, the scanner sweeps this bright spot through the body and builds a detailed picture from the echoes.
For two identical speakers emitting the same frequency, there are certain points in space where you would hear nothing (complete silence) because of destructive interference.
In a real room the silence at a quiet spot is never quite complete. The two paths differ in length, so the two waves arrive with slightly different amplitudes. Walls also send in reflected sound. A quiet spot is therefore a place of strong but partial cancellation, and ideal silence needs equal amplitudes at the point.
Why do noise-cancelling headphones remove the steady rumble of an engine better than a sudden shout?
Now back to the Metro coach. The earphones did not block the rumble. They added its inverse, and at the eardrum the two pressure waves cancel, as the anti-noise figure shows. The first quiet spot of the two speakers was found to the centimetre as well: \(0.82\,\text{m}\) from the centre line, by a short numerical search.
- Using the separation of the speakers as the wavelength.The wavelength is \(v/f\). The separation \(d\) only enters the geometry of the two paths.
- Stopping at \(|L_1 - L_2| = \lambda/2\) and never finding the position.Write \(L_1\) and \(L_2\) in terms of \(y\), use \(L_2^2 - L_1^2 = 2yd\), and solve for \(y\), exactly or by the small-angle estimate.
- Using \(y \approx \lambda L/(2d)\) when the listener is close to the speakers.The estimate needs \(L\) much larger than \(d\) and \(y\). Close to the speakers, use the exact relation.