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

Answer:

The distance is $r_2 = 0.24 \ m$

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

      $\beta = 10dB log_{10}[\frac{d}{d_o} ]$

The intensity is  also mathematically represented as

      $d = \frac{P}{A}$

So

    $\beta = 10dB * log_{10}[\frac{P}{A* d_o} ]$

=>   $\frac{\beta}{10} = log_{10} [\frac{P}{A (l_o)} ]$

From the logarithm definition

=>    $\frac{P}{A * d_o} = 10^{\frac{\beta}{10} }$

=>      $P = A (d_o ) [10^{\frac{\beta }{ 10} } ]$

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

               $P_1 = A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]$

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

               $P_2 = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]$

Generally  power of the wave is constant at both positions  so  

    $A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ] = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]$

      $4 \pi r_1 ^2 [10^{\frac{\beta_1 }{ 10} } ] = 4 \pi r_2 ^2 [10^{\frac{\beta_2 }{ 10} } ]$

        $r_2 = \sqrt{r_1 ^2 [\frac{10^{\frac{\beta_1}{10} }}{ 10^{\frac{\beta_2}{10} }} ]}$

       substituting value

        $r_2 = \sqrt{ 24^2 [\frac{10^{\frac{ 40}{10} }}{10^{\frac{80}{10} }} ]}$

        $r_2 = 0.24 \ m$