91影视

Difference in sound level: $\text{螖}\mathrm{尾}=5\text{dB}$

Distance of first source of sound is, ${\mathrm{r}}_{\mathrm{i}}=1.20\text{鈥尘}$

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

$\mathbf{I}\mathbf{}\text{鈥夆赌�}鈭�\text{鈥夆赌�}\frac{1}{{\mathbf{r}}^2}$

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

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,

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

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

So, difference in sound level will be-

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

For the given values, we have-

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

On further solving-

${\mathrm{r}}_{\mathrm{f}}=0.67\text{鈥尘}$

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