Question

a straight road goes nearby a residential building, 50 m from the road . How much farther away would the house be placed so that the sound intensity level would have been 5 dB less. The traffic is unchanged

Answer 1

I have to use these equations

\[I=\frac{ P }{ 4\pi r }\]

\[\beta=10*\log(\frac{ I }{ I_0 })\]

Answer 2

?

Answer 3

@Michele_Laino

Answer 4

I did get a tip from my professor that I should count as the intensity decreases with 1/r, where r is the distance from the road...

Answer 5

we have to find the value I_1 such that:
\[{\beta _1} = 10 \times \log \left( {\frac{{{I_1}}}{{{I_0}}}} \right) = \beta - 5 = 10 \times \log \left( {\frac{I}{{{I_0}}}} \right) - 5\]

Answer 6

or:
\[\Large 10 \times \log \left( {\frac{I}{{{I_1}}}} \right) = 5\]
where:
\[\Large I\]
is the initial intensity. So we have to know what is P, in your previous formula

Answer 7

yea.. but how?

Answer 8

do you know the power P?

Answer 9

No, I no nothing else than what Ive written in the question:S

Answer 10

please wait, I try to solve your question

Answer 11

yes

Answer 12

I think that we can use this formula:
\[\Large \frac{I}{{{I_1}}} = {\left( {\frac{r}{{{r_1}}}} \right)^2}\]
where:
\[r = 50\;meters\]
and
\[{{r_1}}\] is the requested distance
I think that the right formula is:
\[\Large I = \frac{P}{{4\pi {r^2}}}\]

Answer 13

so, what is :
\[\large \frac{I}{{{I_1}}} = ...?\]

Answer 14

what does \[\frac{ I }{ I_1 } =(\frac{ r }{ r_1 })^2\]

comes from?

Answer 15

@Michele_Laino

Answer 16

since the power is the same, so we can write:
\[\Large 4\pi {r^2}I = 4\pi r_1^2{I_1}\]

Answer 17

namely the power of the emitted sound doesn't change, since the traffic is unchanged

Answer 18

I dont understand the algebra how you get it to \[\frac{ I }{ I_1} =(\frac{ r }{ r_1})^2\]

Answer 19

\[\frac{ I }{ I_1 }=(\frac{ r_1 }{ r })^2\]

Answer 20

@Michele_Laino

Answer 21

I get it to the last I wrote

Answer 22

yes! Sorry I have made an error, the right formula is:
\[\Large \frac{{{I_1}}}{I} = {\left( {\frac{r}{{{r_1}}}} \right)^2}\]

Answer 23

So \[\frac{ I_1 }{ (-5) }=(\frac{ 50 }{ r })^2\]

@Michele_Laino

Answer 24

no, we have to use the condition above, so we get:
\[\Large \frac{I}{{{I_1}}} = \sqrt {10} = {\left( {\frac{{{r_1}}}{{50}}} \right)^2}\]
@pate16

Answer 25

what, where does\[\sqrt{10}\]
comes from?

Answer 26

im sorry for the questions.. @Michele_Laino

Answer 27

no worries :)
it comes from the condition that the new intensity has to be 5 dB less than the initial intensity

Answer 28

and now you did write \[\frac{ r_1 }{ r }\] again?

Answer 29

yes! and r=50 meters

Answer 30

yah but you corrected it and said \[\frac{ r }{ r_1 }\] ?

Answer 31

as you well commented, we have:
\[\frac{I}{{{I_1}}} = {\left( {\frac{{{r_1}}}{r}} \right)^2}\]

Answer 32

sorry sorry!!!! but can you explain the squareroot 10 again?

Answer 33

so, substituting r = 50 meters, we have:
\[\frac{I}{{{I_1}}} = {\left( {\frac{{{r_1}}}{{50}}} \right)^2}\]

Answer 34

your problem requests, that the new intensity I, has to be such that the subsequent quantity:
\[{\beta _1} = 10 \times \log \left( {\frac{{{I_1}}}{{{I_0}}}} \right)\]
is 5 less than this quantity:

\[\beta = 10 \times \log \left( {\frac{I}{{{I_0}}}} \right)\]
where:

\[I\] is the initial intensity, whereas \[{{I_1}}\] is the new intensity, and \[{{I_0}}\] is a refernce intensity, whose value is a standard value

Answer 35

so we can write:
\[\large 10 \times \log \left( {\frac{{{I_1}}}{{{I_0}}}} \right) = 10 \times \log \left( {\frac{I}{{{I_0}}}} \right) - 5\]

Answer 36

or, using the properties of logarithm:
\[\large \begin{gathered}
10 \times \log \left( {\frac{I}{{{I_0}}}} \right) - 10 \times \log \left( {\frac{{{I_1}}}{{{I_0}}}} \right) = 5 \hfill \\
\hfill \\
\log \left( {\frac{I}{{{I_1}}}} \right) = \frac{1}{2} \hfill \\
\hfill \\
\frac{I}{{{I_1}}} = {10^{1/2}} = \sqrt {10} \hfill \\
\end{gathered} \]

Answer 37

yeah im with you now

Answer 38

\[\sqrt{10}=(\frac{ r1 }{ 50 })^2\]

Answer 39

solve for r?

Answer 40

@Michele_Laino 88.91 m?

Answer 41

that's right! @pate16

Answer 42

damn it your good michele!!

Answer 43

thanks! :)

Answer 44

Thanks alot!!!!