# Changes in Decibels over a distance

1. May 29, 2012

### rusty65

1. The problem statement, all variables and given/known data

A jet plane emits 3.0×105 J of sound energy per second.
1) What is the sound level 21 m away?
2) Air absorbs sound at a rate of about 7.0 ; calculate what the sound level will be 5.00.
3) Calculate what the sound level will be 7.50 away from this jet plane, taking into account air absorption.

2. Relevant equations

J/s = W
inverse law of intensities: I_1*(r_1)^2=I_2*(r_2)^2
dB = 10log(I/I_o)
I_o = 1.0*10^-12

3. The attempt at a solution

My attempt at a solution is poor, I know, but this is the first problem like this ive come across.

I had no idea how to relate Joules per second to decibels, so I did the 1:1 conversion to Watts. Still having no idea how to relate watts to decibels, I used an online watt to decibel conversion engine. This gave me 300000 W = 54.8 dB, so I set off running with it.

From there I plugged 54.8 into the decibel formula:

54.8 = 10log(I/I_o) ---> 5.48 = log(I/I_o) ---> 10^5.48 = I/I_o
10^5.48 = 301995.172 ---> 301995.172 = I/I_o
I_o = 1.0*10^-12 ---> 301995.172*(1.0*10^-12) = I
I = 3.01995 * 10^-7

Using this intensity as I_1, and giving this initial intensity a distance of 1 from the jet, I plugged it into the inverse law of intensities:

I_1*(r_1)^2 = I_2*(r_2)^2
(3.01995*10^-7)*(1)^2 ---> 3.01995*10^-7 = I_2*(r_2)^2
3.01996*10^-7 = I_2*(21)^2 ---> (3.01995*10^-7)/441 = I_2
I_2 = 6.847959*10^-10

Then, I plugged this intensity value into the decibel equation, thinking it would give me the decibel level at 21 m:

dB = 10log((6.847959*10^-10)/(1.0*10^-12))

After calculating this, I got a value of 28.121 dB for part 1.
Unfortunately, masteringphysics has told me this is incorrect, and so I am out of ideas as to how to solve part 1, also rendering me incapable of solving parts 2 and 3.

Any help would be greatly appreciated.