How Loud Is a Firework Explosion at 4300 Meters Distance?

[songoku]

Homework Statement

A firework charge is detonated many metres above the ground. At a distance d₁=550 m from the explosion, the acoustic pressure reaches a maximum of ΔP_max = 10 Pa. Assume the speed is constant at 343 m/s throughout the atmosphere over the region considered and the ground absorbs all the sound falling on it. Assume that the density of air 1.2 kgm^-3. What is the sound level at a distance of d₂ = 4.30 x 10³ from the explosion?

Homework Equations

Don't know

The Attempt at a Solution

I don't even know what the relevant equations that should be used to solve this question. Please help me to start, I don't have clue

 

[tiny-tim]

hi songoku!

A firework charge is detonated many metres above the ground. At a distance d₁=550 m from the explosion, the acoustic pressure reaches a maximum of ΔP_max = 10 Pa. Assume the speed is constant at 343 m/s throughout the atmosphere over the region considered and the ground absorbs all the sound falling on it. Assume that the density of air 1.2 kgm^-3. What is the sound level at a distance of d₂ = 4.30 x 10³ from the explosion?

i'll guess that you're supposed to assume that the height is negligible, and that there's no reflection (ie, the ground isn't there)

 

[songoku]

hi songoku!


i'll guess that you're supposed to assume that the height is negligible, and that there's no reflection (ie, the ground isn't there)

hi tiny-tim

Sorry still not know how to proceed. Acoustic pressure is the difference between the total pressure and atmospheric pressure, so the total pressure at that point is almost the same as atmospheric pressure.

I don't understand how to combine all the information given; I have pressure, density, speed. What is the relation between them?

Should I use TI₂=TI₁ + 10.log (r₁/r₂) to find the sound level?
TI₁ = 10 log (I/I₀) and I = power / area and I don't have the information to find all the variables needed. I even don't know whether I am on the right track or not

 

[tiny-tim]

why so complicated?

see http://en.wikipedia.org/wiki/Sound_pressure#Distance_law ​

 

[songoku]

why so complicated?

see http://en.wikipedia.org/wiki/Sound_pressure#Distance_law ​

hi tiny-tim

Oh I never know that formula...

P₁ = 10 + 1 x 10⁵ = 100010 Pa
r₁ = 550 m
r₂ = 4.3 x 10³ m

So P₂ = 127.9 Pa

sound level = 10 log (P / P_ref)²

Do I use P₁ as the P_ref?

 

[tiny-tim]

hi songoku!

(just got up :zzz:)

from that wikipedia: The distance law for the sound pressure p in 3D is inverse-proportional to the distance r of a punctual sound source

so you apply it directly to the 10 Pa

Do I use P₁ as the P_ref?

sorry, no idea

 

[songoku]

hi tiny-tim

hi songoku!

(just got up :zzz:)

from that wikipedia: The distance law for the sound pressure p in 3D is inverse-proportional to the distance r of a punctual sound source

so you apply it directly to the 10 Pa

What does it mean by "punctual sound source"? Does it mean the pressure at that point or just the difference of pressure from atmospheric pressure?

sorry, no idea

Do you have idea of alternative equation that can be used?

 

[tiny-tim]

What does it mean by "punctual sound source"?

it means a point source (as opposed to a spread-out source) ​

"punctual" is a word meaning "at the correct time" which some idiots are trying to re-define as meaning "related to a point"

 

[songoku]

it means a point source (as opposed to a spread-out source) ​

"punctual" is a word meaning "at the correct time" which some idiots are trying to re-define as meaning "related to a point"

Ok. Thanks for the help