# How can you focus sound?

I saw this TED talk and I am curious as to how the sound is focused on the general level. Can anyone explain this or does anyone have any good articles?

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It's worth noting that what he is doing is not sound focusing. He is creating audible sound from high frequency unaudible sound through acoustic non-linear process. An optical similar phenomenon called four-wave mixing does the same. –  Bernardo Kyotoku Nov 24 '10 at 11:30

I don't think anyone here has really answered your question. In this case, the sound is "focused" using phased arrays. The face of the audio spotlight has multiple transducers:

The same signal is output from each of them, but delayed slightly by different amounts, so that the wavefronts all reach the same point in front of the device at the same time. This "virtual focus" is called beamforming.

This is how modern radars focus their beams, too. Instead of spinning a satellite dish around, they have lots of little elements that don't move, but the signals are delayed to produce different beam shapes.

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That's actually diffraction, not focusing. –  ptomato Dec 1 '10 at 9:02
Diffraction? Are you thinking of a two-slit experiment? –  endolith Dec 1 '10 at 14:36
@ptomato- the individual beams diffract as any beam would... but the effect of introducing a variable phase delay between the beams to create a new wavefront is analogous to a wavefront passing through a lens, acquiring phase shift according to the length of each pass –  Pete Dec 10 '10 at 5:20
each pass (typo)--> each path –  Pete Dec 11 '10 at 4:25

Sound is a type of wave, so it has all the wave properties similar to other waves such as light waves. For light waves, you can use a lens to focus the light. A lens has higher refractive index, or lower light speed than the environment. The same is true for sound wave, so what you need is to make a high refractive region [1].

The air surrounding us can be approximated by the ideal gas, so the speed of sound is [2]

$c=\sqrt{\gamma\frac{P}{\rho}}$

where $\gamma$ is the adiabatic index, $p$ is pressure of the air, $\rho$ is density of the air

Here, we want to create a region with high refractive, or equivalently low sound speed. There are few way to achieve this, one is to decrease the pressure, another way is to decrease the temperature (by the ideal gas law $PV=NRT$). However, in both cases, you either need a hard container or a refrigerator near it to keep it cold.

On the other hand, increasing the density can be easily done by using a heavy gas such as carbon dioxide. You just need to fill the gas in a balloon and it can act as a very simple acoustic lens. Note that the size of the balloon or other container must be large compared with the wavelength. There are also other methods to focus sound without using lens. [3]

As said before, the same mechanism can be applied for other wave, for example, a water wave. In a shallow water tank, adding a lens shape obstacle at the bottom can converge water wave because water wave move slowly at the shallow region. This experiment can be easily performed in one's home.

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I think building reflectors would be easier than lenses –  endolith Apr 18 '12 at 21:27

On the general level, you focus sound the same way you focus light -- either by reflecting it from a parabolic surface, or letting it pass through an acoustic lens. An acoustic lens is just like an optical lens in that it consists of a material with a different propagation speed of sound, with varying thickness. See the Wikipedia article on acoustic mirrors.

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You may also focus light using diffractive lenses –  belisarius Nov 24 '10 at 12:26
As with sound, but I wanted to keep the answer simple ;-) –  ptomato Nov 24 '10 at 14:59
@belisarius: en.wikipedia.org/wiki/Zone_plate –  endolith Apr 19 '12 at 1:45
A CO${}_2$-filled balloon is a crude acoustic lens. Face a friend and talk to him. Then put a CO${}_2$-filled balloon between his head and yours. His voice will be louder. –  garyp Mar 19 at 17:32
These ‘audio spotlights’ work by emitting ultrasound at two different frequencies; it is the short wavelength of the ultrasound that causes the beam to be so directed. The two waves interfere and produce sum and different tones at frequencies of $f_1+f_2$ and $f_1-f_2$; if the ultrasound frequencies are, say, $f_1=45\,$kHz and $f_2=44\,$kHz, the difference tone will be at $1$kHz which is in the audible range for humans.