Question

Two atmospheric sound sources $\mathbf{A}\mathbf{}\mathbf{\text{and}}\mathbf{}\mathbf{B}\mathbf{}$ emit isotropically at constant power. The sound levels $\mathit{\beta }$of their emissions are plotted in Figure versus the radial distance $\mathit{r}$from the sources . The vertical axis scale is set by ${\mathbf{\beta }}_{\mathbf{1}}\mathbf{=}\mathbf{\text{85.0dB}}$and ${\mathbf{\beta }}_2\mathbf{=}\mathbf{\text{65.0dB}}$.

What are (a) the ratio of the larger power to the smaller power and

(b) the sound level difference at $\mathbf{r}\mathbf{=}\mathbf{}\mathbf{\text{10m}}$?

Short answer

1. The ratio of larger to lower power is$3.2$.
2. The sound level difference is $\text{5.0dB}$.

Step-by-step solution

Step 1: Given

• Intensity level ,${\mathrm{\beta }}_1=8\text{5dB}$
• Other intensity level ,${\mathrm{\beta }}_2=65\text{dB}$
• Distance ,$\mathrm{r}=10\text{m}$

Determining the concept

Write the sound level in terms of intensity. Also, write the relation between intensity, power, and radius. Substituting the intensity in terms of power in the formula for sound intensity, and using the data from the given graph, find the ratio of the power. Also, see the difference between the intensity values at any given radius from the graph and compare it to the value asked in the problem.

The expression for the intensity in terms of power is given by,

$\mathrm{I}=\frac{\mathrm{P}}{4{\mathrm{\pi r}}^2}$

Here $P$is the power, $I$is the intensity,$\mathrm{r}$ is the radius.

The expression for the sound level is given by,

${\mathrm{\beta }}_1-{\mathrm{\beta }}_2=10\log \left(\frac{{\mathrm{I}}_1}{{\mathrm{I}}_2}\right)$

Here, ${\mathrm{\beta }}_1$,${\mathrm{\beta }}_2$ are the sound level, $I_1$, $I_2$are the intensities.

(a) Determining the ratio of larger power to lower power

Sound level can be written as,

$\mathrm{\beta }=10\log \left(\frac{\mathrm{I}}{{\mathrm{I}}_0}\right)$

Since,

${\mathrm{\beta }}_1-{\mathrm{\beta }}_2=10\log \left(\frac{{\mathrm{I}}_1}{{\mathrm{I}}_0}\right)-10\log \left(\frac{{\mathrm{I}}_2}{{\mathrm{I}}_0}\right)$

Now,

$\mathrm{I}=\frac{\mathrm{P}}{4{\mathrm{\pi r}}^2}$

${\mathrm{\beta }}_1-{\mathrm{\beta }}_2=10\log \left(\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}\right)$

At any given radius r, the difference between sound levels, $\mathrm{\Delta \beta }={\mathrm{\beta }}_1-{\mathrm{\beta }}_2$is $\text{5.0dB}$, therefore,

$\begin{array}{c}5.0=10\log \left(\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}\right)\\ 0.5=\log \left(\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}\right)\\ \frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}={10}^{0.5}\\ =3.16\\ \approx 3.2\end{array}$

Hence, the ratio of larger to lower power is$3.2$ .

(b) Determine the sound level difference 

From the graph, it can be seen that the sound level difference is constant between the two sources for all the values of $\mathrm{r}$ . Therefore, for role="math" localid="1661410281447" $r=10m$, it would be equal to $\text{5.0dB}$.

Hence, the sound level difference is $\text{5.0dB}$.

Therefore, using the relationship between power, intensity, and sound level, the ratio can be found. Also, the sound level at any other radius can be found using the data from the graph.