The sound intensity is $\mathbf{0}\mathbf{.0080}\mathbf{\text{\hspace{0.17em}W}}\mathbf{/}{\mathbf{\text{m}}}^{\mathbf{\text{2}}}$at a distance of $\mathbf{10}\mathbf{\text{\hspace{0.17em}m}}$from an isotropic point source of sound. (a) What is the power of the source? (b) What is the sound intensity from the source? (c) What is the sound level$\mathbf{10}\mathbf{\text{\hspace{0.17em}m}}$ from the source?

1. Sound intensity, $I=0.0080\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$.
2. Radius, $r=10\text{\hspace{0.17em}m}$.

Using the formula for power in terms of intensity we can get its value. Then, using the relation between intensity and distance, we can find the sound intensity at 5.0 m from the source. Inserting the value of intensity at 10 m in the formula for sound level, we can find its value at that distance.

Formulae:

$\mathit{I}\mathbf{=}\frac{\mathbf{P}}{\mathbf{4}\mathbf{\pi }{\mathbf{r}}^{\mathbf{2}}}$ …(1)

$\mathit{\beta }\mathbf{=}\mathbf{\left(}\mathbf{10}\mathbf{\text{\hspace{0.17em}dB}}\mathbf{\right)}\mathit{l}\mathit{o}\mathit{g}\frac{\mathbf{I}}{{\mathbf{I}}_{\mathbf{0}}}$ (2)

Intensity of waves is

$I=\frac{P}{4\pi r^2}$

Substitute all the value in the above equation.

$\begin{array}{c}0.0080\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}=\frac{P}{4\times 3.14\times {\left(10\text{\hspace{0.17em}m}\right)}^2}\\ P=10.048\text{\hspace{0.17em}W}\\ P\simeq 10\text{\hspace{0.17em}W}\end{array}$

Hence the power of source is, $10\text{\hspace{0.17em}W}$.

Using the above value of power, we can find the intensity for $r=5.0\text{\hspace{0.17em}m}$ as

$I=\frac{P}{4\pi r^2}$

Substitute all the value in the above equation.

$\begin{array}{c}I=\frac{10\text{\hspace{0.17em}W}}{4\times 3.14\times {\left(5.0\text{\hspace{0.17em}m}\right)}^2}\\ I=0.03187\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}\\ I=0.032\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}\end{array}$

Hence the sound intensity is, $0.032\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$.

The standard reference intensity is, $I_0=1.0\times {10}^{-12}\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$.

We know the equation for sound level

$\beta =\left(10\text{\hspace{0.17em}dB}\right)\log \frac{I}{I_0}$

Substitute all the value in the above equation.

With $I=0.0080\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$ and $I_0=1.0\times {10}^{-12}\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$.

$\begin{array}{c}\beta =\left(10\text{\hspace{0.17em}dB}\right)\log \frac{0.0080\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}}{1.0\times {10}^{-12}\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}}\\ \beta =99\text{\hspace{0.17em}dB}\end{array}$

Hence the sound level is, $99\text{\hspace{0.17em}dB}$.