Question

At a distance of 14.0 m from a point source, the intensity level is measured to be 80 dB. At what distance from the source will the intensity level be 40 dB

Answer 1

Answer:

$d_2=19.8m$

Explanation:

The problem can be solved easily, because the relationship between intensities $I_1$ and $I_2$ at distances $d_1$ and $d_2$ respectively can be written as:

$\frac{I_2}{I_1} =\frac{d_1^2}{d_2^2}$

Where:

$I_1=Intensity\hspace{3}1\\I_2=Intensity\hspace{3}2\\d_1=Distance\hspace{3}1\\d_2=Distance\hspace{3}2$

In this case:

$I_1=80dB\\I_2=40dB\\d_1=14m\\d_2=?$

Solving for $d_2$

$d_2=\sqrt{\frac{d_1^2 *I_1}{I_2} } =\sqrt{\frac{14^2*80}{40} } =\sqrt{392} =19.79898987m\approx19.8m$

Answer 2

Answer:

1400m

Explanation:

For spherical waves we can use the following relationship between distance and intensity:

$\frac{I_{1}}{I_{2}}=\frac{r_{2}^2}{r_{1}^2}$

Where $I_{1}$ and $I_{2}$ are the first and second intensity, in $W/m^2$. And $r_{1}$ is the first distance:  $r_{1}=14m$ , and $r_{2}$ the distace we want to find.

Clearing the previous equation or $r_{2}$

$r_{2}^2=\frac{I_{1}r_{1}^2}{I_{2}} \\r_{2}=\sqrt{\frac{I_{1}r_{1}^2}{I_{2}}} \\r_{2}=r_{1}\sqrt{\frac{I_{1}}{I_{2}}}$.

This is what we will be using to find the answer, but fist we must convert the quantities 80dB and 40dB to $W/m^2$

We will call the quantities in dB $\beta_{1}$ and $\beta _{2}$:

$\beta_{1}=80dB$

$\beta_{2}=40dB$

We will use the following to find the corresponding intensities $I_{1}$ and $I_{2}$:

$\beta_{1}=10log\frac{I_{1}}{I_{o}}$

$\beta_{2}=10log\frac{I_{2}}{I_{o}}$

for both of these: $I_{0}=10^{-12}W/m^2$, and it is the minimum intensity that a human being perceives.

Thus, for $\beta_{1}=80dB$ we have:

$80dB=10log\frac{I_{1}}{I_{o}}$

$8dB=log\frac{I_{1}}{I_{o}}$

And to eliminate the logarithm, we use its inverse operation, exponentiation.

$10^8=10^{log\frac{I_{1}}{I_{0}} }$

$10^8=\frac{I_{1}}{I_{0}}$

$I_{0}(10^8)=I_{1}$

replacing $I_{0}=10^{-12}W/m^2$:

$10^{-12}W/m^2(10^8)=I_{1}$

$1x10^{-4}W/m^2=I_{1}$

and similarly for $I_{2}$

$40dB=10log\frac{I_{2}}{I_{o}}$

$4dB=log\frac{I_{2}}{I_{o}}$

$10^4=10^{log\frac{I_{2}}{I_{0}} }$

$10^4=\frac{I_{2}}{I_{0}}$

$I_{0}(10^4)=I_{2}$

replacing $I_{0}=10^{-12}W/m^2$:

$10^{-12}W/m^2(10^4)=I_{2}$

$1x10^{-8}W/m^2=I_{2}$

Now we substitute in the equation we had found for $r_{2}$

$r_{2}=r_{1}\sqrt{\frac{I_{1}}{I_{2}}}\\r_{2}=(14m)\sqrt{\frac{1x10^{-4}W/m^2}{1x10^{-8}W/m^2}}\\r_{2}=(14m)\sqrt{10000}\\ r_{2}=(14m)(100)\\r_{2}=1400m$