Question

A certain loudspeaker system emits sound isotropically with a frequency of $\mathbf{2000}\mathbf{\text{\hspace{0.17em}Hz}}$ and an intensity ofrole="math" localid="1661500478873" $\mathbf{0.960}\mathbf{\text{\hspace{0.17em}mW}}\mathbf{/}{\mathbf{\text{m}}}^{\mathbf{\text{2}}}$ at a distance ofrole="math" localid="1661501289787" $\mathbf{6.10}\mathbf{}\mathbf{\text{\hspace{0.17em}m}}$ . Assume that there are no reflections. (a) What is the intensity at $\mathbf{30.0}\mathbf{\text{\hspace{0.17em}m}}$? At $\mathbf{6}\mathbf{.10}\mathbf{\text{\hspace{0.17em}m}}$, what are (b) the displacement amplitude and (c) the pressure amplitude?

Short answer

1. The intensity at $30.0\text{\hspace{0.17em}m}$ is, $3.97\times {10}^{-5}\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$.
2. The displacement at $6.10\text{\hspace{0.17em}m}$ is, $1.71\times {10}^{-7}\text{\hspace{0.17em}m}$.
3. The pressure amplitude at $6.10\text{\hspace{0.17em}m}$ is, $0.89\text{\hspace{0.17em}Pa}$.

Step-by-step solution

The given data

1. Frequency is $2000\text{\hspace{0.17em}Hz}$.
2. The distance ${\mathrm{r}}_2$ is, $30.0\text{\hspace{0.17em}m}$.
3. Intensity at distance ${\mathrm{r}}_1=6.10\text{\hspace{0.17em}m}$ is $0.960\text{\hspace{0.17em}mW}/{\text{m}}^{\text{2}}$.

Understanding the concept of intensity

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,

$\mathbf{I}\mathbf{=}\frac{\mathbf{P}}{\mathbf{4}{\mathbf{\pi r}}^{\mathbf{2}}}$ …(i)

The displacement amplitude of the wave,

${\mathbf{S}}_{\mathbf{w}}\mathbf{=}\sqrt{\frac{\mathbf{2}\mathbf{I}}{{\mathbf{\rho v\omega }}^{\mathbf{2}}}}$ …(ii)

The pressure amplitude of the wave,

$\mathbf{\Delta p}\mathbf{=}\mathbf{\rho v\omega }\mathbf{\times }{\mathbf{s}}_{\mathbf{w}}$ …(iii)

The angular frequency of the wave,

$\mathbf{\omega }\mathbf{=}\mathbf{2}\mathbf{\pi f}$ …(iv)

a) Calculation of intensity at 30.0 m

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

$\mathrm{I}\propto \frac{1}{{\mathrm{r}}^2}$ …(a)

We know from given data that,

${\mathrm{I}}_1=0.960\times {10}^{-3}\text{\hspace{0.17em}w}/{\text{m}}^{\text{2}}\text{at}{\mathrm{r}}_1=6.1\text{\hspace{0.17em}m}$

$I_2$is intensity at${\mathrm{r}}_2=30.0\text{\hspace{0.17em}m}$

Hence, from equation (a), we can get

$\begin{array}{c}\frac{{\mathrm{I}}_2}{{\mathrm{I}}_1}={\left(\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}\right)}^2\\ \frac{{\mathrm{I}}_2}{0.960\text{\hspace{0.17em}mW}/{\text{m}}^{\text{2}}}={\left(\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}\right)}^2\\ {\mathrm{I}}_2={\left(\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}\right)}^2\times 0.960\times {10}^{-3}\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}\\ ={\left(\frac{6.10\text{\hspace{0.17em}m}}{30\text{\hspace{0.17em}m}}\right)}^2\times 0.960\times {10}^{-3}\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}\\ {\mathrm{I}}_2=3.97\times {10}^{-5}\text{W}/{\text{m}}^{\text{2}}\end{array}$

Hence, the value of the intensity at 3.00 m is, $3.97\times {10}^{-5}\text{W}/{\text{m}}^{\text{2}}$.

b) Calculation of displacement at 6.10 m

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

$\begin{array}{c}\mathrm{\omega }=2\mathrm{\pi }\times 2000\text{\hspace{0.17em}Hz}\\ =4000\mathrm{\pi }\text{\hspace{0.17em}rad}/\text{s}\\ \mathrm{\omega }=12560\text{\hspace{0.17em}rad}/\text{s}\end{array}$

The density of air is, $\mathrm{\rho }=1.21\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}$and velocity of sound is $\mathrm{v}=343\text{\hspace{0.17em}m}/\text{s}$. Using this value of angular frequency in equation (ii), we get the displacement amplitude as:

$\begin{array}{c}{\mathrm{S}}_{\mathrm{w}}=\sqrt{\frac{2\times 0.960\times {10}^{-3}\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}}{1.21\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}\times {(12560\text{\hspace{0.17em}rad}/\text{s})}^2\times 343\text{\hspace{0.17em}m}/\text{s}}}\\ =1.71245\times {10}^{-7}\text{\hspace{0.17em}m}\\ {\mathrm{S}}_{\mathrm{w}}\approx 1.71\times {10}^{-7}\text{\hspace{0.17em}m}\end{array}$

Hence, the value of displacement is, $1.71\times {10}^{-7}\text{\hspace{0.17em}m}$.

c) Calculation of pressure amplitude at 6.10 m

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

$\begin{array}{c}\mathrm{\Delta p}=1.21\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}\times 343\text{\hspace{0.17em}m}/\text{s}\times 12560\text{\hspace{0.17em}rad}/\text{s}\times 1.71245\times {10}^{-7}\text{\hspace{0.17em}m}\\ =0.89\text{\hspace{0.17em}Pa}\end{array}$

Hence, the value of pressure amplitude is, $0.89\text{\hspace{0.17em}Pa}$.