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To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from your distance of 10.0 m , it sounds like only an average whisper of 20.0 dB . So you decide to move closer to give the conversation a sound level of 60.0 dB instead. How close should you come

Sagot :

Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

            β = 10 log [tex]\frac{I}{I_o}[/tex]

let's write this expression for our case

           β₁ = 10 log \frac{I_1}{I_o}

           β₂ = 10 log \frac{I_2}{I_o}

           

          β₂ -β₁ = 10 ( [tex]log \frac{I_2}{I_o} - log \frac{I_1}{I_o}[/tex])

          β₂ - β₁ = 10 [tex]log \frac{I_2}{I_1}[/tex]

          log \frac{I_2}{I_1} = [tex]\frac{60 - 20}{10}[/tex] = 3

           [tex]\frac{I_2}{I_1}[/tex] = 10³

           I₂ = 10³ I₁

having the relationship between the intensities, we can use the definition of intensity which is the power per unit area

           I = P / A

           P = I A

the area is of a sphere

          A = 4π r²

           

the power of the sound does not change, so we can write it for the two points

          P =  I₁ A₁ =  I₂ A₂

          I₁ r₁² = I₂ r₂²

we substitute the ratio of intensities

          I₁ r₁² = (10³ I₁ ) r₂²

         r₁² = 10³ r₂²

         

         r₂ = r₁ / √10³

         

we calculate

          r₂ = [tex]\frac{10.0}{\sqrt{10^3} }[/tex]

          r₂ = 0.316 m

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