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