Why use ear canal area in sound intensity calculation

[coconut62]

Homework Statement

Please refer to the image. Question (a).

Homework Equations

I= P/(4∏r^2)
P=E/t

The Attempt at a Solution

After a few attempts, I found the way to get the answer:

(6.3 x 10^-6) x (1.5 x 10^-4) x 60 = 5.67 x 10^-8

But I don't understand why the area of the ear canal is used, instead of the area which the energy carried by the wave passes through. From what I understand, intensity depends on the distance from the source, so how come we don't need to take into account the distance of 5.0m?

 

[SammyS]

Homework Statement

Please refer to the image. Question (a).

Homework Equations

I= P/(4∏r^2)
P=E/t

The Attempt at a Solution

After a few attempts, I found the way to get the answer:

(6.3 x 10^-6) x (1.5 x 10^-4) x 60 = 5.67 x 10^-8

But I don't understand why the area of the ear canal is used, instead of the area which the energy carried by the wave passes through. From what I understand, intensity depends on the distance from the source, so how come we don't need to take into account the distance of 5.0m?

The sound intensity at 5 m from the sound source is given. The ear canal is also at 5 m from the source. Therefore the distance of 5 m is taken into account.

 

[coconut62]

Thank you.