Question

The sound of normal breathing is not very loud, with an intensity of about 11 dB at a distance of 1 m away from the face of the breather.
Note that in this problem sound intensity in decibels is denoted β; intensity in W/m² is denoted I.
Given that a person with normal hearing canbarely detect a sound with intensity of 1 dB at a frequency of 1kHz (the sensitivity of the human ear peaking near 1 kHz), how faraway could this person detect another person breathingnormally?
Express the distance $D_{detect}$ in meters.

Answer 1

Answer:

 r₂ = 1,648 m

Explanation:

The expression for the sound level in decibels is

         β = 10 log I / I₀

Let's apply this expression for the colon

Initial

         β₁ = 11 dB

         β₁ = 10 log I₁ / I₀

Final

         β₂ = 1 dB

         β₂₂ = 10 log I₂ / I₀

         I₁ / I₀ = exp ( β₁ / 10)

         I₂ / I₀ = exp β₂ / 10

         I₁ / I₂ = exp ( (β₁ -β₂) / 10)           (1)

The expression for intensity is the power emitted per unit area

             I = P / A

We apply  

           I₁ = P / A₁

           I₂ = P / A₂

           I₁ A₁ = I₂ A₂

           I₁ / I₂ = A₂ / A₁

The area of ​​a sphere is

             A = 4π r²

            I₁ / I₂ = r₂² / r₁²

We substitute in the equation with decibels, equation 1

             r₂² / r₁² = exp (β₁ -β₂) / 10

They give us the initial distance

              r₁ = 1 m

Let's calculate

             r₂² = r₁² exp ((β₁ - β₂) / 10)

             r₂² = 1 exp ((11-1) / 10)

             r₂ = √ 2,718

             r₂ = 1,648 m