Answer:
[tex]28.11m[/tex] far from the speaker the intensity drops to 85 dB.
Explanation:
In the equation for the Decibel scale
[tex](1). \: \:\beta =10 log(\dfrac{I}{I_0})[/tex]
The ratio of the intensities can be written as
[tex]$ \frac{I}{I_0} = \dfrac{\frac{P}{A} }{\frac{P}{A_0} } $[/tex]
[tex]\dfrac{I}{I_0} = \dfrac{A_0}{A}.[/tex]
And since
[tex]A = 4\pi r^2[/tex]
and
[tex]A_0 = 4 \pi r_0^2[/tex],
[tex]\dfrac{A_0}{A} = \dfrac{4 \pi r_0^2}{4 \pi r^2} = \dfrac{r_0^2}{r^2}[/tex]
meaning
[tex]\dfrac{I}{I_0} = \dfrac{r_0^2}{r^2}.[/tex]
Putting this into equation (1), we get:
[tex]\boxed{ (2).\: \: \beta = 10log(\dfrac{r_0^2}{r^2})}[/tex]
Now, if the intensity is 100 dB when the distance is 5 meters, we have:
[tex]100dB=10 log(\dfrac{r_02}{(5m)^2})[/tex]
[tex]10= log(\dfrac{r_0^2}{25})[/tex]
by taking both sides to the exponent:
[tex]10^{10}= \dfrac{r_0^2}{25}[/tex]
[tex]r^2 = 25 *10^{10}\\r = 5 *10^5[/tex]
Now equation (2) becomes
[tex]\beta = 10log(\dfrac{25*10^{10}}{r^2})[/tex]
when the intensity level is 85 dB we have
[tex]85 = 10log(\dfrac{25*10^{10}}{r^2})[/tex]
[tex]8.5 = log(\dfrac{25*10^{10}}{r^2})[/tex]
take both sides to exponents and we get:
[tex]10^{8.5} =10^{ log(\dfrac{25*10^{10}}{r^2})}[/tex]
[tex]10^{8.5} =\dfrac{25*10^{10}}{r^2}[/tex]
[tex]r^2 = \dfrac{25*10^{10}}{10^{8.5}}[/tex]
[tex]\boxed{r = 28.11m}[/tex]
Thus, [tex]28.11m[/tex] far from the speaker the intensity drops to 85 dB.
