Explanation:
Sound intensity level [tex]\beta[/tex] is a logarithmic quantity and is defined as:
[tex]\beta=(10dB) log(\frac{I}{I_{o}})[/tex]
Where:
[tex]I[/tex] is the sound intensity
[tex]I_{o}=10^{-12}W/m^2[/tex] is the reference sound intensity
In this case we are given a sound intensity [tex]\beta_{1}=170dB[/tex] (during takeoff at a distance [tex]r_{1}=34m[/tex]):
[tex]\beta_{1}=(10dB) log(\frac{I_{1}}{I_{o}})=170dB[/tex] (1)
Now, from (1) we have to find [tex]I_{1}[/tex].
Dividing both sides of the equation by [tex]10dB[/tex]:
[tex]\frac{10dB}{10dB} log(\frac{I_{1}}{I_{o}})=\frac{170dB}{10dB}[/tex] (2)
[tex]log(\frac{I_{1}}{I_{o}})=17[/tex] (3)
Applying logarithm properties on both sides:
[tex]\frac{I_{1}}{I_{o}}=10^{17}[/tex] (4)
Isolating [tex]I_{1}[/tex]:
[tex]I_{1}=10^{17}I_{o}[/tex] (5)
[tex]I_{1}=10^{17}(10^{-12}W/m^2)[/tex] (6)
[tex]I_{1}=100000W/m^2[/tex] (7) This is the sound intensity at a distance of 34 m
On the other hand, we know the sound intensity[tex]I[/tex] is inversely proportional to the square of the distance [tex]r[/tex]:
[tex]I\propto\frac{1}{r^2}[/tex]
This means:
[tex]I_{1}r_{1}^2=I_{2}r_{2}^2[/tex] (7)
Where [tex]r_{2}=1km=1000m[/tex]
Finding [tex]I_{2}[/tex] from (7):
[tex]I_{2}=\frac{I_{1}r_{1}^2}{r_{2}^2}[/tex] (8)
[tex]I_{2}=\frac{(100000W/m^2)(34m)^2}{(1000m)^2}[/tex] (9)
[tex]I_{2}=115.6W/m^2[/tex] (9) This is the sound intensity at a distance of 1km
Now, knowing the value of [tex]I_{2}[/tex] we will be able to find the sound intensity level [tex]\beta_{2}[/tex] at a distance of 1km:
[tex]\beta_{2}=(10dB) log(\frac{I_{2}}{I_{o}})[/tex] (10)
[tex]\beta_{2}=(10dB) log(\frac{115.6W/m^2}{100000W/m^2})[/tex] (11)
Finally:
[tex]\beta_{2}=140.629dB[/tex] (12) This is the sound intensity level at a distance of 1 km.
