Suppose a certain jet plane creates an intensity level of 124 dB at a distance of 5.01 m. What intensity level does it create on the ground directly underneath it when flying at an altitude of 2.25 km?

Respuesta :

Answer:71 dB

Explanation:

Given

sound Level [tex]\beta _1=124 dB[/tex]

distance [tex]r_1=5.01 m[/tex]

From sound Intensity

[tex]\beta =10dB\log (\frac{I_1}{I_0})[/tex]

[tex]124=10dB\log (\frac{I_1}{I_0})[/tex]

[tex]12.4=\log (\frac{I_1}{I_0})[/tex]

[tex]I_1=(1\times 10^{-12})\times 10^{12.4}[/tex]

[tex]I_1=2.51 W/m^2[/tex]

we know Intensity [tex]I\propto ^\frac{1}{r^2}[/tex]

[tex]I_1r_1^2=I_2r_2^2[/tex]

[tex]I_2=I_1(\frac{r_1}{r_2})^2[/tex]

[tex]I_2=2.51\cdot (\frac{5.01}{2.25\times 10^3})^2[/tex]

[tex]I_2=1.24\times 10^{-5} W/m^2[/tex]

Sound level corresponding to [tex]I_2[/tex]

[tex]\beta _2=10\log (\frac{I_2}{I_0})[/tex]

[tex]\beta _2=10\log (\frac{1.24\times 10^{-5}}{1\times 10^{-12}})[/tex]

[tex]\beta _2=70.93\approx 71 dB[/tex]

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