How to Calculate Decibel Levels at Different Distances from a Sound Source?

[ninaw21]

Homework Statement

A car horn sounds and 15 m away the sound level is recorded as 85 dB.
Calculate the decibel level 25 m away from the horn.

Homework Equations

β = (10dB)logI/I_o

β = (20dB)logP/P_o

The Attempt at a Solution

I = P/A @ 15m
A = 4∏r²
A = 2827
I = 15/2827 = 0.005W/m

β = 10dBlog0.005/10^-12
= 87dB

I = P/A @ 25m
A = 4∏r²
A = 7854
I = 25/7854 = 0.003W/m

β = 10dBlog0.003/10^-12
= 85dB

I don't know if I'm even heading in the right direction with this one!

 

[I like Serena]

Hi ninaw21!

How did you get that P would be 15? It isn't true. P is unknown (as yet).

Please note that you are using two different versions of P here.
The P in your relevant equation is actually the sound pressure, usually denoted with a small p.
The P in the equation I = P/A is the power of the source.


You do have that at 15 m: β = (10dB)logI₁₅/Io = 85 dB
and furthermore that I₁₅ = P/A at 15 m.
From this you can calculate what I₂₅/I₁₅ is. Do you know how?

 

[ninaw21]

Thank you! No I don't know how to calculate it can you explain it please?

 

[I like Serena]

You already wrote that A = 4∏r².

So I₁₅ = P / 4∏15².
And I₂₅ = P / 4∏25².
In both cases P is the (unknown) power of the source.

Can you divide I₂₅ by I₁₅?

 

[ninaw21]

I divided I25 by I15 and got a decibel level of 3dB..Is that not very low?

 

[ninaw21]

do i add it to the original decibel level?

 

[I like Serena]

Hmm, suppose you add 3dB to the original decibel level... then the sound would be louder at 25 m? That can't be right...

Anyway, if I try to find the dB level, I do not get 3 dB...

Let's do one thing at a time.
What is I25/I15?

After that, what is the corresponding decibel level, which is 10log(I25/I15)?

 

[ninaw21]

I25/I15 = 2.78

10log2.78 = 4dB

 

[I like Serena]

Let's see...

I25 / I15 = (P / 4∏25²) / (P / 4∏15²) = 15² / 25².

erm... no that is not 2.78...