Party hearing. As the number of people at a party increases, you must raise your voice for a listener to hear you against the background noise of the other partygoers. However, once you reach the level of yelling, the only way you can be heard is if you move closer to your listener, into the listener’s “personal space.” Model the situation by replacing you with an isotropic point source of fixed power $\mathbf{P}$ and replacing your listener with a point that absorbs part of your sound waves. These points are initially separated by${\mathbf{r}}_i\mathbf{=}\mathbf{}\mathbf{\text{1.20\hspace{0.17em}m}}$ . If the background noise increases by$\mathbf{\Delta \beta }\mathbf{=}\mathbf{}\mathbf{5}\mathbf{dB}$ , the sound level at your listener must also increase. What separation${\mathbf{r}}_f\mathbf{}$ is then required?

Difference in sound level: $\text{Δ}\mathrm{\beta }=5\text{dB}$

Distance of first source of sound is, ${\mathrm{r}}_{\mathrm{i}}=1.20\text{\hspace{0.17em}m}$

If power is constant, then, intensity of sound is inversely proportional to the square of distance of the point from the source of sound.

$\mathbf{I}\mathbf{}\propto \frac{1}{{\mathbf{r}}^2}$

The difference in sound level is given as- $\text{Δ}\mathrm{\beta }=10\log \left(\frac{{\mathrm{I}}_{\mathrm{f}}}{{\mathrm{I}}_{\mathrm{i}}}\right)$

where,I is intensity andr is radius.

Since, power is constant, so, intensity of sound will be inversely proportional to the square of distance of the point from the source of sound. Thus,

$\mathrm{I}\mathrm{\alpha }\frac{1}{{\mathrm{r}}^2}$

$\frac{{\mathrm{I}}_{\mathrm{f}}}{{\mathrm{I}}_{\mathrm{i}}}={\left(\frac{{\mathrm{r}}_{\mathrm{i}}}{{\mathrm{r}}_{\mathrm{f}}}\right)}^2$

So, difference in sound level will be-

$\begin{array}{c}\text{Δ}\mathrm{\beta }=10\text{dB}\log \left(\frac{{\mathrm{I}}_{\mathrm{f}}}{{\mathrm{I}}_{\mathrm{i}}}\right)\\ =10\log {\left(\frac{{\mathrm{r}}_{\mathrm{i}}}{{\mathrm{r}}_{\mathrm{f}}}\right)}^2\end{array}$

For the given values, we have-

$\begin{array}{c}\left(5\text{dB}\right)=\left(10\text{\hspace{0.17em}dB}\right)\log {\left(\frac{1.20\text{\hspace{0.17em}m}}{{\mathrm{r}}_{\mathrm{f}}}\right)}^2\\ {\left(\frac{1.20\text{\hspace{0.17em}m}}{{\mathrm{r}}_{\mathrm{f}}}\right)}^2={10}^{0.5}\\ {\left(\frac{1.20\text{\hspace{0.17em}m}}{{\mathrm{r}}_{\mathrm{f}}}\right)}^2=3.16227\\ {\mathrm{r}}_{\mathrm{f}}=\frac{1.20\text{\hspace{0.17em}m}}{1.7783}\end{array}$

On further solving-

${\mathrm{r}}_{\mathrm{f}}=0.67\text{\hspace{0.17em}m}$

Hence, distance of the other source of sound is,${\mathrm{r}}_{\mathrm{f}}=0.6\text{7\hspace{0.17em}m}$ .