Question

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

Answer 1

Let ${\beta }_A$ be the sound level at a point 5 m (= r₁) away from the point source and ${\beta }_B$ be the sound level at a distance of 50 m (= r₂) away from the point source.
∴​ ${\beta }_A$ = 40 dB
Sound level is given by:
$\beta =10{\log }_{10}\left(\frac{I}{I_0}\right)$
According to the question,
$$\begin{gathered} \mathit{}{\beta }_{\mathrm{A}}=10{\log }_{10}\left(\frac{I_A}{I_0}\right). \\ \Rightarrow \frac{I_{\mathrm{A}}}{I_0}={10}^{\left(\frac{{\beta }_{\mathrm{A}}}{10}\right)}.....\left(1\right) \\ {\beta }_B=10{\log }_{10}\left(\frac{I_B}{I_o}\right) \\ \Rightarrow \frac{{\mathrm{I}}_{\mathrm{B}}}{{\mathrm{I}}_0}={10}^{\left(\frac{{\beta }_{\mathrm{B}}}{10}\right)}.....\left(2\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}: \\ \frac{I_{\mathrm{A}}}{I_{\mathrm{B}}}={10}^{\left(\frac{{\beta }_{\mathrm{A}}-{\beta }_{\mathrm{B}}}{10}\right)}....\left(3\right) \\ \mathrm{Also}, \\ \frac{I_{\mathrm{A}}}{I_{\mathrm{B}}}=\frac{r_{\mathrm{B}}^2}{r_{\mathrm{A}}^2}={\left(\frac{50}{5}\right)}^2={10}^2.....\left(4\right) \\ From\left(3\right)\mathrm{and}\left(4\right),\mathrm{we}\mathrm{get}: \\ {10}^2={10}^{\left(\frac{{\beta }_{\mathrm{A}}-{\beta }_{\mathrm{B}}}{10}\right)} \\ \Rightarrow \frac{{\beta }_A-{\beta }_B}{10}=2 \\ \Rightarrow {\beta }_{\mathrm{A}}-{\beta }_{\mathrm{B}}=20 \\ \Rightarrow \mathit{}{\beta }_{\mathrm{B}}=40-20=20\mathrm{dB} \end{gathered}$$

Thus, the sound level of a point 50 m away from the point source is 20 dB.