Question

The sound level at a point $5.0m$ away from a point source is $40dB$. What will be the level at a point $50m$ away from the source ?

Answer 1

We know that $\beta =10{\log }_{10}\left(\frac{I}{I_0}\right)$
${\beta }_A=10\log \frac{I_A}{I_O},$ ${\beta }_B=10\log \frac{I_B}{I_O}$
$\Rightarrow I_A/I_0={10}^{\left({\beta }_A/10\right)}$ $I_B/I_O={10}^{\left({\beta }_B/10\right)}$

$\Rightarrow \frac{I_A}{I_B}=\frac{r_B^2}{r_A^2}={\left(\frac{50}{5}\right)}^2$

$\Rightarrow {10}^{\frac{({\beta }_A-{\beta }_B)}{10}}={10}^2$

$\Rightarrow \frac{{\beta }_A-{\beta }_B}{10}=2$

$\Rightarrow {\beta }_A-{\beta }_B=20$

$\Rightarrow {\beta }_B=40+20=60dB$.