Changes in Decibels over a distance

[rusty65]

Homework Statement

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.

Homework 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

The Attempt at a Solution

My attempt at a solution is poor, I know, but this is the first problem like this I've 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.

 

[anathema]

Your solution for part 1 is almost correct. The only mistake you made was in converting from watts to decibels. The correct conversion is:

dB = 10log(P/Po)

where P is the power in watts and Po is the reference power of 1.0*10^-12 watts. So the correct conversion for 300000 watts is:

dB = 10log(300000/1.0*10^-12) = 10log(3*10^17) = 10(17.477) = 174.77 dB

This is a much louder sound level than what you got before, which makes sense since the jet plane is emitting a lot of sound energy.

To solve parts 2 and 3, you can use the same method you used for part 1, but you will need to take into account the distance and the air absorption rate. For part 2, the distance is 5.00 m and for part 3, the distance is 7.50 m. You will also need to use the inverse law of intensities to account for the distance. So for part 2, your equation would look like:

I1*(1)^2 = I2*(5.00)^2

And for part 3, your equation would look like:

I1*(1)^2 = I2*(7.50)^2

Once you have calculated the new intensity values for parts 2 and 3, you can plug them into the decibel equation to find the sound level at those distances.