A firework charge is detonated many metres above the ground. At a distance of 400 m from the explosion, the acoustic pressure reaches a maximum of `10.0(N)/(m^2)`. Assume that the speed of sound is constant at 343`(m)/(s)` throughout the atmosphere over the region considered, the ground absorbs all the sound falling on it, and the air absorbs sound energy at the rate of 7.00 `(dB)/(km)`. What is the sound level (in decibels) at 4.00 km from the axplosion?

Explanation

At a distance of 4 km, an explosion sould be audible, but probably not extremely loud. So based on the data given we might expect the sound level to be somewhere between 40 and 80 dB. From the sound pressure data given in the problem, we can find the intensity, which is used to find the sound level in dB. The sound intensity will decrease with increased distance from the source and from the absorption of the sound by the air.
At distance of 400 m from the explosion, `triangleP_(max)=10.0Pa` At this point the intensity is
`I=(triangleP_(max)^(2))/(2rhov)=((10.0(N)/(m))^2)/(2(1.20(kg)/(m^3)(343(m)/(s)))=0.121(W)/(m^2)`
From the inverse square law, we can calculating the intensity and decibel level (due to distance alone) 4km away
`I'=I((400m)^2)/((4000m)^2)=1.21xx10^-3(W)/(m^2)`
and `beta=10log((I')/(I_0))=10log((1.21xx10^-3(W)/(m^2))/(1.00xx10^(-12)(W)/(m^2)))=90.8dB`
At a distance of 4 km from the explosion, absorption from the air will decrease the dound level by an additional amount,
`trianglebeta=(7.00(dB)/(km))(3.60km)=25.2dB`
so at 4 km the sound level will be
`beta_f=beta-trianglebeta=90.8dB-25.2dB=65.6dB`
this sound level falls within our expected range. Evidently, this explosioin is rather loud (about the same as a vacuum cleaner) even at a distance of 4 km from the source. It is interesting to note that the distance and absorption effect each reduces the sound level by about the same amount `(-25dB)`, If the explosion were at ground level, the sound level would be further reduced by reflection and absorption form abstacles between the source and observer, and the calculation could be much more complicated.