# Equivalent noise from gunshots

What are is the resultant noise from 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 shot

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

The sound power level dB is

$L = 10*log(P/P0)$

where P is the sound power at the specified distance (1 m) and $P0 =10^{-12} W$ (0 dB)

therefore at a distance of 1 m the power of the sound (in W) is

$125 =10*log(P/P0)$

$10^{12.5} =P/P0$

$P = P0*10^12.5 =10^0.5 = \sqrt{10} W$

The power of the sound is inversely proportional to the area of the wave front (which is spherical).

It means

$P(4)/P(1) = A(1)/A(4) = 1*1/(4*4) =1/16$,

$P(4) =(1/16)*\sqrt{10}$     (4 meters distance from first sound)

$P(5)/P(1) = 1/25$,
$P(5) = 1/25*\sqrt{10}$      (5 meters distance from second sound)

$P(6)/P(1) =1/364$,     $P(6) = 1/36*\sqrt(10)$

$P(7)/P(1) = 1/49$,    $P(7) = 1/49*\sqrt(10)$

$P(8)/P(1) = 1/64$,    $P(8) = 1/64*\sqrt(10)$

The total sound power at 4 m distance from first sound is
$P = (1/16 +1/25+ 1/36 +1/49 +1/64)*\sqrt(10) =0.1663*\sqrt(10)$

The dB value of this sound is

$L = 10*log(0.1663*10^0.5/10^{-12}) = 10*log(0.1663*10^{12.5}) =117.209 dB$