91影视

1. Frequency is $2000\text{鈥塇锄}$.
2. The distance ${\mathrm{r}}_2$ is, $30.0\text{鈥尘}$.
3. Intensity at distance ${\mathrm{r}}_1=6.10\text{鈥尘}$ is $0.960\text{鈥尘奥}/{\text{m}}^{\text{2}}$.

Intensity is inversely proportional to the radius. To find displacement amplitude, use the formula for intensity in terms of displacement amplitude. To find the amplitude of pressure, we can use the formula for pressure amplitude in terms of displacement amplitude.

Formula:

The intensity of the wave,

$\mathbf{I}\mathbf{=}\frac{\mathbf{P}}{\mathbf{4}{\mathbf{蟺谤}}^{\mathbf{2}}}$ 鈥�(颈)

The displacement amplitude of the wave,

${\mathbf{S}}_{\mathbf{w}}\mathbf{=}\sqrt{\frac{\mathbf{2}\mathbf{I}}{{\mathbf{蚁惫蝇}}^{\mathbf{2}}}}$ 鈥�(颈颈)

The pressure amplitude of the wave,

$\mathbf{螖辫}\mathbf{=}\mathbf{蚁惫蝇}\mathbf{脳}{\mathbf{s}}_{\mathbf{w}}$ 鈥�(颈颈颈)

The angular frequency of the wave,

$\mathbf{蝇}\mathbf{=}\mathbf{2}\mathbf{蟺蹿}$ 鈥�(颈惫)

From equation (i), we can see that the intensity of the wave is inversely proportional to the square of the distance.

$\mathrm{I}鈭�\frac{1}{{\mathrm{r}}^2}$ 鈥�(补)

We know from given data that,

${\mathrm{I}}_1=0.960脳{10}^{鈭�3}\text{鈥墂}/{\text{m}}^{\text{2}}\text{at}{\mathrm{r}}_1=6.1\text{鈥尘}$

$I_2$is intensity at${\mathrm{r}}_2=30.0\text{鈥尘}$

Hence, from equation (a), we can get

$\begin{array}{c}\frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}={\left(\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}\right)}^2\\ \frac{{\mathrm{I}}_2}{0.960\text{鈥尘奥}/{\text{m}}^{\text{2}}}={\left(\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}\right)}^2\\ {\mathrm{I}}_2={\left(\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}\right)}^2脳0.960脳{10}^{鈭�3}\text{鈥塛}/{\text{m}}^{\text{2}}\\ ={\left(\frac{6.10\text{鈥尘}}{30\text{鈥尘}}\right)}^2脳0.960脳{10}^{鈭�3}\text{鈥塛}/{\text{m}}^{\text{2}}\\ {\mathrm{I}}_2=3.97脳{10}^{鈭�5}\text{鈥�}\text{W}/{\text{m}}^{\text{2}}\end{array}$

Hence, the value of the intensity at 3.00 m is, $3.97脳{10}^{鈭�5}\text{鈥�}\text{W}/{\text{m}}^{\text{2}}$.

Angular frequency of the wave using equation (iv) is given as:

$\begin{array}{c}\mathrm{蝇}=2\mathrm{蟺}脳2000\text{鈥塇锄}\\ =4000\mathrm{蟺}\text{鈥塺补诲}/\text{s}\\ \mathrm{蝇}=12560\text{鈥塺补诲}/\text{s}\end{array}$

The density of air is, $\mathrm{蚁}=1.21\text{鈥塳驳}/{\text{m}}^{\text{3}}$and velocity of sound is $\mathrm{v}=343\text{鈥尘}/\text{s}$. Using this value of angular frequency in equation (ii), we get the displacement amplitude as:

$\begin{array}{c}{\mathrm{S}}_{\mathrm{w}}=\sqrt{\frac{2脳0.960脳{10}^{鈭�3}\text{鈥塛}/{\text{m}}^{\text{2}}}{1.21\text{鈥塳驳}/{\text{m}}^{\text{3}}脳{(12560\text{鈥塺补诲}/\text{s})}^2脳343\text{鈥尘}/\text{s}}}\\ =1.71245脳{10}^{鈭�7}\text{鈥尘}\\ {\mathrm{S}}_{\mathrm{w}}鈮�1.71脳{10}^{鈭�7}\text{鈥尘}\end{array}$

Hence, the value of displacement is, $1.71脳{10}^{鈭�7}\text{鈥尘}$.

Using equation (iii), the value of pressure amplitude at 6.10 m is given as:

$\begin{array}{c}\mathrm{螖辫}=1.21\text{鈥塳驳}/{\text{m}}^{\text{3}}脳343\text{鈥尘}/\text{s}脳12560\text{鈥塺补诲}/\text{s}脳1.71245脳{10}^{鈭�7}\text{鈥尘}\\ =0.89\text{鈥塒补}\end{array}$

Hence, the value of pressure amplitude is, $0.89\text{鈥塒补}$.