Question

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

Answer 1

r₂ = 0.316 m

Step-by-step explanation:

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

β = 10 log
`(I)/(I_o)`

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 (
`log (I_2)/(I_o) - log (I_1)/(I_o)`)

β₂ - β₁ = 10
`log (I_2)/(I_1)`

log \frac{I_2}{I_1} =
`(60 - 20)/(10)` = 3

`(I_2)/(I_1)` = 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 =
`(10.0)/(√(10^3) )`

r₂ = 0.316 m