Question

You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.0dB . So you decide to move closer to give the conversation a sound level of 80.0dB instead. How close should you come?

Answer 1

The distance is
`r_2  =  0.24 \  m`

Step-by-step explanation:

From the question we are told that

The distance from the conversation is
`r_1    =  24.0 \ m`

The intensity of the sound at your position is
`\beta _1 =  40 dB`

The intensity at the sound at the new position is
`\beta_2 =  80.0dB`

Generally the intensity in decibel is is mathematically represented as

`\beta  =  10dB log_(10)[(d)/(d_o) ]`

The intensity is also mathematically represented as

`d =  (P)/(A)`

So

`\beta  =  10dB *  log_(10)[(P)/(A* d_o) ]`

=>
`(\beta)/(10)  =  log_(10) [(P)/(A (l_o)) ]`

From the logarithm definition

=>
`(P)/(A  *  d_o)  =  10^{(\beta)/(10) }`

=>
`P =  A (d_o ) [10^{(\beta )/( 10) } ]`

Here P is the power of the sound wave

and A is the cross-sectional area of the sound wave which is generally in spherical form

Now the power of the sound wave at the first position is mathematically represented as

`P_1 =  A_1 (d_o ) [10^{(\beta_1 )/( 10) } ]`

Now the power of the sound wave at the second position is mathematically represented as

`P_2 =  A_2 (d_o ) [10^{(\beta_2 )/( 10) } ]`

Generally power of the wave is constant at both positions so

`A_1 (d_o ) [10^{(\beta_1 )/( 10) } ]  = A_2 (d_o ) [10^{(\beta_2 )/( 10) } ]`

`4 \pi r_1 ^2   [10^{(\beta_1 )/( 10) } ]  = 4 \pi r_2 ^2   [10^{(\beta_2 )/( 10) } ]`

`r_2 =  \sqrt{r_1 ^2 [\frac{10^{(\beta_1)/(10) }}{ 10^{(\beta_2)/(10) }} ]}`

substituting value

`r_2 =   \sqrt{ 24^2 [\frac{10^{( 40)/(10) }}{10^{(80)/(10) }} ]}`

`r_2  =  0.24 \  m`