Question

# A certain loudspeaker system emits sound isotropically with a frequency of $2000 \mathrm{~Hz}$ and an intensity of $0.990 \mathrm{~mW} / \mathrm{m}^{2}$ at a distance of $6.10 \mathrm{~m}$. Assume that there are no reflections. (a) What is the intensity at $30.0 \mathrm{~m}$ ? At $6.10 \mathrm{~m}$. what are (b) the displacement amplitude and (c) the pressure amplitude?

Solution

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$\textbf{Given:}$

$f = 2000 \: \: \mathrm{Hz}$

$I_o = 0.960 \: \: \mathrm{\frac{mW}{m^2}}$

$r_o = 6.10 \: \: \mathrm{m}$

$r = 30.0 \: \: \mathrm{m}$

$\textbf{A:}$

We know that the sound intensity is defined as:

\begin{align} I &= \dfrac{P}{A} \end{align}

And for an isotropic source of sound, the area can be defined as:

\begin{align} A &= 4 \pi r^2 \end{align}

Plugging in Equation 2 to 1, we get:

\begin{align} I &= \dfrac{P}{4 \pi r^2} \end{align}

Then we could get the quotient of the two intensities at different distance as shown below:

\begin{align*} &\dfrac{I}{I_o} = \dfrac{P}{4 \pi r^2} \cdot \dfrac{4 \pi r_o^2}{P} \\\\ &\boxed{\dfrac{I}{I_o} = \dfrac{r_o^2}{r^2} } \end{align*}

Solving for $I$ and substituting all known values, we get:

\begin{align*} I &= I_o \dfrac{r_o^2}{r^2} \\\\ &= 0.960 \times 10^{-3} \cdot \dfrac{6.10^2}{30^2} \\\\ &= \boxed{3.969 \times 10^{-5} \: \: \mathrm{\dfrac{W}{m^2}}} \end{align*}