Q23E - University Physics with Modern Physics

Question

At point A, 3.0 m from a small source of sound that is emitting uniformly in all directions, the sound intensity level is 53 db. (a) What is the intensity of the sound at A? (b) How far from the source must you go so that the intensity is one-fourth of what it was at A? (c) How far must you go so that the sound intensity level is one-fourth of what it was at A? (d) Does intensity obey the inverse-square law? What about sound intensity level?

Step-by-Step Solution

Answer

a) The intensity of the sound is $2\times {10}^{-7} \mathrm{W}/{\mathrm{m}}^2$ b) The ratio of the intensities is 6m c) The distance ${\mathrm{r}}_2$ is 292 m d) The intensity obeys the inverse square law but sound intensity level does not

STEP 1 Concept of the intensity of the sound

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

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

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

STEP 2 Concept of the ratio of the intensities

 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 $2\times {10}^{-7}\mathrm{W}/{\mathrm{m}}^2$ what it was at A, meaning that ${\mathrm{l}}_2=\frac{2\times {10}^{-7}\mathrm{W}/{\mathrm{m}}^2}{4}=0.5\times {10}^{-7}\mathrm{W}/{\mathrm{m}}^2$

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

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

STEP 3 Find the point at which the sound intensity level is one-fourth

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

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

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

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

STEP 4 Calculate the value of r

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 $2.1\times {10}^{-11}\mathrm{W}/{\mathrm{m}}^2$ to get ${\mathrm{l}}_2$

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

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

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

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