# Part1What are is the resultant noise from Â 5 gunshots at same timeeach 125 dB a distance of 1 meter from each otherThe audience of the sound is exactly 4 meter from nearest gun shotÂ (Distance...

Part1

What are is the resultant noise from Â 5 gunshots at same time

each 125 dB a distance of 1 meter from each other

The audience of the sound is exactly 4 meter from nearest gun shotÂ

(Distance of audience from Ist gun is 4m, from 2nd gun is 5 m, 3rd Gun is 6 m..respectively)

Â

Part 2

How can you theoritically & mathematically proveÂ

Multiple impulse noise/impact noise from same or different sources Â cannot be considered as continious noise

### 2 Answers | Add Yours

For some reasons after posting, two words are missing from the above text. Please read:

**This means the wave front is a sphere, and since the surface of a sphere is `S =4*pi*R^2` the intensity of sound will vary inversely proportional to `R^2` .**

The sound is a wave and from the point where it has been produced it propagates in all directions uniformly. This means the wave front is a sphere, and since the surface of a sphere is ` ` the intensity of sound will vary inversely proportional to ` ` .

First we need to find the intensity of the sound at the point where it has been emitted. Since the dB level of the sound intensity is defined as as:

`dB = 10*log(I/I_0)` where `I_0 =10^-12 W/m^2` is the reference.

For all 5 sounds each having 125 dB the intensity at `0m` distance is

`125 =10*log(I/10^-12) rArr I = 10^12.5*10^-12 =10^0.5 = sqrt(10) W/m^2`

Thus for the first sound the intensity at a distance of 4 m is

`I(4m) = (I(0 m))/4^2 =sqrt(10)/16`

For the second sound the intensity at a distance of 5 m is

`I(5m) =(I(0m))/5^2 =sqrt(10)/25`

and so on for the rest of sounds:

`I(6m) = sqrt(10)/36` , `I(7m) = sqrt(10)/49` ,`I(8) = sqrt(10)/64`

The total intensity of the sound heard by the audience is simply the sum of the 5 sound intensities (the sounds are produced at the same moment):

`I = sqrt(10)*(1/16 +1/25+1/36+1/49+1/64) =0.166*sqrt(10)`

The audience will hear a total sound having the level (see above definition):

`L =10*log((0.166*sqrt(10))/10^-12) =117.2 dB`

**Answer: The resultant noise heard by the audience is 117.2 dB**

For the part two the mathematical demonstration is the following. A single impulse noise can be modeled by the Dirac function.The continuous noise can be modeled by the Heaviside (or step) function. In mathematics it is shown that the Dirac function is the first derivative of the Heaviside function. It follows that only an infinite sum of Dirac functions are making a whole Heaviside function, or in other words one will need an infinite impulse noises to make a continuous sound.

**Sources:**