Question

You are trying to overhear a juicy conversation, but from your distance of 20.0 m , it sounds like only an average whisper of 30.0 dB . So you decide to move closer to give the conversation a sound level of 80.0 dB instead?

Answer 1

r₂ = 0.2 m

Step-by-step explanation:

given,

distance = 20 m

sound of average whisper = 30 dB

distance moved closer = ?

new frequency = 80 dB

using formula

`\beta = 10 log((I_1)/(I_0))`

I₀ = 10⁻¹² W/m²

now,

`30 = 10 log((I_1)/(10^(-12)))`

`(I_1)/(10^(-12))= 10^3`

`I_1= 10^(-8)\ W/m^2`

to hear the whisper sound = 80 dB

`80 = 10 log((I_2)/(10^(-12)))`

`(I_2)/(10^(-12))= 10^8`

`I_2= 10^(-4)\ W/m^2`

we know intensity of sound is inversely proportional to square of distances

`(I_1)/(I_2)=(r_2^2)/(r_1^2)`

`(10^(-8))/(10^(-4))=(r_2^2)/(20^2)`

`10^(-4)=(r_2^2)/(20^2)`

r₂ = 0.2 m