91影视

1. Initial distance${\mathrm{r}}_2=10\text{鈥塳尘or}10000\text{鈥尘}$
2. Frequency is$\mathrm{f}=100\text{鈥塇锄}$

We use the concept of sound intensity. Using the equation of sound intensity related to power and area, we can write the equations of intensities for ${\mathbf{r}}_{\mathbf{1}}$and${\mathbf{r}}_{\mathbf{2}}$. We are given,${\mathbf{r}}_{\mathbf{2}}$ so by taking the ratio, we find the distance${\mathbf{r}}_{\mathbf{1}}$.

Formulae:

The intensity of a wave,

$\mathbf{I}\mathbf{=}\frac{\mathbf{P}}{\mathbf{A}}$ 鈥�(颈)

The surface area of the sphere,

$\mathbf{A}\mathbf{=}\mathbf{4}{\mathbf{蟺谤}}^{\mathbf{2}}$ 鈥�(颈颈)

We know sound waves travel spherically, so the area at which waves is focused is given using equation (ii).The power of the source is the same for any distance. Hence, using equation (i), we can write the intensity at surface of radius$r_1$and${\mathrm{r}}_2$as:

$\begin{array}{l}{\mathrm{I}}_1=\frac{\mathrm{P}}{4{\mathrm{蟺谤}}_1^2}\\ {\mathrm{I}}_2=\frac{\mathrm{P}}{4{\mathrm{蟺谤}}_2^2}\end{array}$

Dividing$I_2$by$I_1$, we get

$\begin{array}{c}\frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}=\frac{\left(\frac{\mathrm{P}}{4{\mathrm{蟺谤}}_2^2}\right)}{\left(\frac{\mathrm{P}}{4{\mathrm{蟺谤}}_1^2}\right)}\\ \frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}=\frac{4{\mathrm{蟺谤}}_1^2}{4{\mathrm{蟺谤}}_2^2}\end{array}$

$\frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}=\frac{{\mathrm{r}}_1^2}{{\mathrm{r}}_2^2}$ 鈥�(补)

We know that intensity ratio between barely audible and painful threshold is:

$\frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}={10}^{鈭�12}$

Substitute all the value in the equation (a).

$\begin{array}{c}{10}^{鈭�12}=\frac{{\left({\mathrm{r}}_1\right)}^2}{{\left(10000\text{鈥尘}\right)}^2}\\ {\mathrm{r}}_1^2={10}^{鈭�12}脳{10}^8{\text{鈥尘}}^{\text{2}}\\ ={10}^{鈭�4}{\text{鈥尘}}^{\text{2}}\\ {\mathrm{r}}_1=\sqrt{{10}^{鈭�4}{\text{鈥尘}}^{\text{2}}}\\ ={10}^{鈭�2}\text{鈥尘}\\ =0.01\text{鈥尘}\\ {\mathrm{r}}_1=1\text{鈥塩尘}\end{array}$.

Hence, at $1\text{鈥塩尘}$it would begin to cause pain.