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

Step 1. Given data:

$$\begin{gathered} \left({\mathrm{r}}_{\mathrm{A}}\right)=5\mathrm{m}; \\ \left({\mathrm{r}}_{\mathrm{B}}\right)=50\mathrm{m}; \\ \left({\mathrm{\beta }}_{\mathrm{A}}\right)=40\mathrm{dB} \end{gathered}$$

${\mathrm{\beta }}_{\mathrm{A}}$ is the sound level at $A$

$r_A$ is the distance from the point $A$

$r_B$ is the distance from the point $B$

Step 2: Formula used,

Sound level, $\mathrm{\beta }=10{\log }_{10}\left(\frac{\mathrm{l}}{{\mathrm{l}}_{\mathrm{o}}}\right)$ (equation i)

$I$ is the intensity of sound.

Sound level at $A$, ${\mathrm{\beta }}_{\mathrm{A}}=10{\log }_{10}\left(\frac{{\mathrm{l}}_{\mathrm{A}}}{{\mathrm{l}}_{\mathrm{o}}}\right)$

Sound level at $B$, ${\mathrm{\beta }}_{\mathrm{B}}=10{\log }_{10}\left(\frac{{\mathrm{l}}_{\mathrm{B}}}{{\mathrm{l}}_{\mathrm{o}}}\right)$

$I$ is the intensity of sound from the point source

$I_A$ is the intensity of sound at the point $A$

$I_B$ is the intensity of sound at the point $B$

Step 3: The sound level at points A and B

As per the given statement,

${\mathrm{\beta }}_{\mathrm{A}}=10{\log }_{10}\left(\frac{{\mathrm{l}}_{\mathrm{A}}}{{\mathrm{l}}_{\mathrm{o}}}\right)$ (equation ii)

from equation (i) and equation (ii), we get:

$\frac{{\mathrm{l}}_{\mathrm{A}}}{{\mathrm{I}}_{\mathrm{o}}}=10\left(\frac{{\mathrm{\beta }}_{\mathrm{A}}}{10}\right)$ equation a

Again,

${\mathrm{\beta }}_{\mathrm{B}}=10{\log }_{10}\left(\frac{{\mathrm{l}}_{\mathrm{B}}}{{\mathrm{l}}_{\mathrm{o}}}\right)$ equation iii)

$\frac{{\mathrm{l}}_{\mathrm{B}}}{{\mathrm{I}}_{\mathrm{o}}}=10\left(\frac{{\mathrm{\beta }}_{\mathrm{B}}}{10}\right)$ (equation b)

Step 4: Dividing the equation a and b

From (a) and (b)

$\frac{{\mathrm{I}}_{\mathrm{A}}}{{\mathrm{I}}_{\mathrm{B}}}={10}^{\frac{\left({\mathrm{\beta }}_{\mathrm{A}}-{\mathrm{\beta }}_{\mathrm{B}}\right)}{10}}$ (equation c)

Step 5: Relating the intensities of sound at points A and B with the distance

$\frac{I_A}{I_B}={\left(\frac{r_B}{r_A}\right)}^2={\left(\frac{50}{5}\right)}^2=100$

$\frac{{\mathrm{I}}_{\mathrm{A}}}{{\mathrm{I}}_{\mathrm{B}}}=100$ (equation iv)

Step 6: calculating the sound level at point B

From equations (c) and (iv),

$$\begin{gathered} \frac{{\mathrm{I}}_{\mathrm{A}}}{{\mathrm{I}}_{\mathrm{B}}}={10}^{\frac{\left({\mathrm{\beta }}_{\mathrm{A}}-{\mathrm{\beta }}_{\mathrm{B}}\right)}{10}}=100 \\ {10}^{\frac{\left({\mathrm{\beta }}_{\mathrm{A}}-{\mathrm{\beta }}_{\mathrm{B}}\right)}{10}}={10}^2 \\ \frac{\left({\mathrm{\beta }}_{\mathrm{A}}-{\mathrm{\beta }}_{\mathrm{B}}\right)}{10}=2 \\ \left({\mathrm{\beta }}_{\mathrm{A}}-{\mathrm{\beta }}_{\mathrm{B}}\right)=20 \\ \left(40-{\mathrm{\beta }}_{\mathrm{B}}\right)=20 \\ {\mathrm{\beta }}_{\mathrm{B}}=40-20=20 \end{gathered}$$

Therefore, the sound level of a point $50m$ away from the point source is $20\mathrm{dB}$.