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Hope the graphics will help me explaining my question better. Let's say the box would be a room in the fourth floor, and the sound source would come from cars in the street. It is clear that in case A the sound would enter the room and would be clearly heard from the human ear.

What I am interested to inspect is case B. The red line there represents a sound-proof shield. The shield would block the direct sound geometrically as you see. In front of the flat there would be no nearby building that could bounce the sound back to the room from the other direction. So, there is just air. I know sound is not a bullet that travels in straight line, so my question is: Would there be a significant decrease in the sound heard inside the room if a sound-proof shield would be placed as in the picture in case B? Can we calculate a rough percentage?

  • $\begingroup$ It may be good to realize that sound is a longitudinal air pressure wave. That is to say, sound is air moving back and forth. But just as a fan cannot push air straight forward, so can't sound. $\endgroup$ – MSalters Jun 11 '15 at 14:30

The situation you are describing is an example of Fresnel diffraction (or near-field diffraction).

In general, when a wave propagates every point of the wave front can be thought of as its own source of waves traveling in all directions (called Huygens construction). It turns out that neighboring point sources along an infinite straight wave front reinforce the "forward" direction only, but if you put an obstacle in the way you can see this diffraction.

The mathematics needed is simplified when you look at the effect of this diffraction "far away" (far compared to the wavelength of the wave). In the case of sound, a frequency of 55 Hz (low end of the range of sounds you hear) has a wavelength of about 6 m, so on the scale of your drawing diffraction would occur.

This explains why you can hear the thumping bass of a loud car stereo before the car turns the corner, and only make out the song when the car is in sight.

The calculation of relative sound level into the room as drawn is tricky - it involves an integral that is usually evaluated using a graphical technique called the Cornu spiral, and strongly depends on dimensions and frequency. But as a rule of thumb, "high frequencies travel straighter". And "sound barriers" do work (somewhat) to reduce nuisance noise (for example the noise of cars speeding along a highway).

If you want to estimate the attenuation, you will find this link has some helpful equations and graphs.


There is a problem with the link given: it defines the Fresnel number as

$$N = \frac{2d}{\lambda}$$

But has a confusing definition of $d$. To make things work, you need to set the straight line distance from source to receiver to $D$ (not $d$). If you do that, then

$$d = A + B - D$$

and the Fresnel number is

$$N = 2\;\frac{A+B-D}{\lambda}$$

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  • $\begingroup$ So, according the Fresnel-Barrier graph in the link you provided, we're talking about an attenuation of somewhere between 5 and 25 dB, correct? $\endgroup$ – multigoodverse Jun 11 '15 at 12:32
  • $\begingroup$ @ArditS. It depends on the Fresnel number. The plot shows the result for values between 0.01 and 100, which is why they show attenuation a from 5 to 25 dB, but you should calculate it (the Fresnel number) for your geometry and the wavelength of the sound of interest (from the $1/\lambda$ relationship you can see that longer wavelengths attenuate less). Of course the real situation is further complicated because of the window - but we are ignoring that for now. $\endgroup$ – Floris Jun 11 '15 at 13:07
  • $\begingroup$ Got it! In my case, d is 20 meters, and traffic noise frequency is around 3000 Hz, so a wavelength of 0.1 m. The Fresnel number would be: N = 2d/λ = 2*20 / 0.1m = 400 With a Fresnel number of 400, the attenuation would be near 30 dB then. Thanks! $\endgroup$ – multigoodverse Jun 11 '15 at 13:30
  • $\begingroup$ Note - $d$ is the difference between the direct line from source to receiver, and the indirect line going around the obstacle. For a reasonable setup, I don't think it is as much as 20 m. Look at the diagram in the link. If you can't figure it out I will update the answer... $\endgroup$ – Floris Jun 11 '15 at 13:35
  • $\begingroup$ I just re-read the link and realize that VERY UNHELPFULLY they use the symbol $d$ twice - once for the total distance, and once again in the expression $d = A+B-d$. The $d$ on the left hand side of that equation really ought to be a different symbol than on the right. Call distance between source and receiver $D$, then $d = A+B-D$. And use $d$ to calculate the Fresnel number. $\endgroup$ – Floris Jun 11 '15 at 13:38

You will benefit by finding some tutorials on wave theory. In brief, assuming a spherical wavefront from the emitter, you are correct there's no direct path to the receiver. However, the edge of yourabsorber there causes diffraction (Huygen's principle), so thatsome of the sound wave (energy) will make its way to the receiver. You can see a demo of this, e.g., at mike-willis tutorial .

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