91Ó°ÊÓ

The intensity of the sound is given as $\mathrm{β}=10\log \left(\frac{\mathrm{l}}{{\mathrm{l}}_0}\right)$.Substitute the values in the above equation we get,

$$\begin{gathered} 5.3\mathrm{db}=10\log \left(\frac{\mathrm{l}}{{10}^{−12}\mathrm{W}/{\mathrm{m}}^2}\right) \\ 5.3\mathrm{db}=\log \left(\frac{\mathrm{l}}{{10}^{−12}\mathrm{W}/{\mathrm{m}}^2}\right) \\ \mathrm{Now}{,10}^{\log \left(\mathrm{x}\right)}=\mathrm{x} \\ {10}^{5.3}={10}^{\log }\left(\frac{\mathrm{l}}{{10}^{−12}\mathrm{W}/{\mathrm{m}}^2}\right) \\ \frac{1}{{10}^{−12}\mathrm{W}/{\mathrm{m}}^2}={10}^{5.3} \\ \mathrm{I}=2×{10}^{−7}\mathrm{W}/{\mathrm{m}}^2 \end{gathered}$$

Therefore, the intensity of the sound is$2×{10}^{-7}\mathrm{W}/{\mathrm{m}}^2$

Formula is$\frac{l_2}{l_1}=\frac{r_1^2}{r_2^2}$

On Rearrange ${\mathrm{r}}_2={\mathrm{r}}_1\sqrt{\frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}}$

RI = 3 m which is the distance between the source and the point A, and is the distance between the source and the point at which the intensity is one-fourth of $2×{10}^{-7}\mathrm{W}/{\mathrm{m}}^2$what it was at A, meaning that${\mathrm{l}}_2=\frac{2×{10}^{-7}\mathrm{W}/{\mathrm{m}}^2}{4}=0.5×{10}^{-7}\mathrm{W}/{\mathrm{m}}^2$

Substitute the values in equation ${\mathrm{r}}_2={\mathrm{r}}_1\sqrt{\frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}}$we get,

$$\begin{gathered} {\mathrm{r}}_2=3\mathrm{m}×\sqrt{\frac{2×{10}^{-7}\mathrm{W}/{\mathrm{m}}^2}{0.5×{10}^{-7}\mathrm{W}/{\mathrm{m}}^2}} \\ =6\mathrm{m} \end{gathered}$$

Use the relation that describes the difference between the sound intensity levels for two different

intensities, which is given by${\mathrm{β}}_2-{\mathrm{β}}_1=10\log \left(\frac{{\mathrm{l}}_2}{{\mathrm{l}}_1}\right)$

$$\begin{gathered} \frac{{\mathrm{β}}_1}{4}−{\mathrm{β}}_1=10\log \left(\frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}\right) \\ \frac{53\mathrm{db}}{4}−53\mathrm{db}=10\log \left(\frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}\right) \\ −39.8\mathrm{db}=10\log \left(\frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}\right) \\ −3.98=\log \left(\frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}\right) \\ \frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}={10}^{−3.98} \\ {\mathrm{I}}_2=10.4×{10}^{−5}{\mathrm{I}}_1 \\ {\mathrm{I}}_2=10.4×{10}^{−5}×2×{10}^{−7}\mathrm{W}/{\mathrm{m}}^2 \\ {\mathrm{I}}_2=2.1×{10}^{−11}\mathrm{W}/{\mathrm{m}}^2 \\ {\mathrm{I}}_2=10.4×{10}^{−5}{\mathrm{I}}_1 \end{gathered}$$

Therefore, the intensity is$2.1×{10}^{-11}\mathrm{W}/{\mathrm{m}}^2$

Where is the intensity of the sound wave at the point A and we have already calculated it in part (a), so, replace ${\mathrm{l}}_1$with $2.1×{10}^{-11}\mathrm{W}/{\mathrm{m}}^2$to get ${\mathrm{l}}_2$

$$\begin{gathered} {\mathrm{I}}_2=10.4×{10}^{−5}{\mathrm{I}}_1 \\ {\mathrm{I}}_2=10.4×{10}^{−5}×2×{10}^{−7}\mathrm{W}/{\mathrm{m}}^2 \\ {\mathrm{I}}_2=2.1×{10}^{−11}\mathrm{W}/{\mathrm{m}}^2 \end{gathered}$$

To determine the distance ${\mathrm{r}}_2$of that point

$$\begin{gathered} {\mathrm{r}}_2=3\mathrm{m}×\sqrt{\frac{2×{10}^{-7}\mathrm{W}/{\mathrm{m}}^2}{2.1×{10}^{-11}\mathrm{W}/{\mathrm{m}}^2}} \\ =292\mathrm{m} \end{gathered}$$

The distance ${\mathrm{r}}_2$is 292 m.

As we can see from (2), the intensity obeys the inverse square law but sound intensity level does not