Question

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?

Short answer

1. Power of source is $10\text{\hspace{0.17em}W}$.
2. The sound intensity 5.0 m from the source is $0.032\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$.
3. The sound level 10 m from the source is $99\text{\hspace{0.17em}dB}$.

Step-by-step solution

The given data

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

Understanding the concept of power and intensity

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:

1.Intensity,

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

2. Sound level,

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

Step 3(a): Determination of power of source

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, $10\text{\hspace{0.17em}W}$.

Step 4(b): Determination of sound intensity

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, $0.032\text{\hspace{0.17em}W}/{\text{m}}^{\text{2}}$.

Step 5(c): Determination of sound level

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, $99\text{\hspace{0.17em}dB}$.