Intensity of isotropic sound waves

[Zdub311]

Homework Statement

A loudspeaker emits sound isotropically with a power of 100db. Find the intensity in w/m at a distance of 20 from the source

Homework Equations

Intensity=power/area, For sound wave... power=ro*a*2pi^2*f^2*v*A^2, and I=2pi^2*ro*f^2*v*A^2.

db=10log_10_(I/I_0_)

Also, the teacher wants us to use calculus in the answers, but in our book i found some easier equations for the question, I just don't if that's what he's looking for with how to solve it. Anyways, they are...

I=P/4pir^2,

The Attempt at a Solution

I've tried solving it using each of these equations...

 

[mrttrn]

My first answer.

To found the sound level we use β(in dB)=10*log(I/I₀) formula. I₀ is treshold of hearing which is 1*10^-10 W/m². We find the intesity I=10^-22 W/m². And the intesity decreases as the inverse square of the distance so the answer should be I/20² which is 2.5*10^-25 w/m². This is my first answer in this forum so if there is a mistake, forgive me :D

 

[az_lender]

Homework Statement

A loudspeaker emits sound isotropically with a power of 100db. Find the intensity in w/m at a distance of 20 from the source

Homework Equations

Intensity=power/area, For sound wave... power=ro*a*2pi^2*f^2*v*A^2, and I=2pi^2*ro*f^2*v*A^2.

db=10log_10_(I/I_0_)

I=P/4pir^2,

The threshold of human hearing is defined in multiple sources as 10^-12 W/m², not 10^-10.

dB = 10 log[I/(10^-12W/m²)],
so in your problem,
10¹⁰ = I/(10^-12 W/m<sup>2, and
I = 10-2 W/m2.

I agree that the sound intensity will attenuate with the square of the distance, but the 100 dB at the loudspeaker has no meaning IMO without a specification of how far the loudspeaker surface is from the source. Furthermore, "a distance of 20" means nothing without units. And finally, it's hard to believe that you were asked to calculate intensity "in w/m". ? Maybe your instructor is a moron.</sup>