Question

In a test of a subsonic jet, the jet flies overhead at an altitude of room. The sound intensity on the
ground as the jet passes overhead is $60{\text{ }}dB$. At what altitude should the jet plane fly so that
the ground noise is not greater than $120{\text{ }}dB$?

Answer 1

1. Loudness is given by the formula $L = 10 \times {\log _{10}}\left( {\dfrac{I}{{{I_0}}}} \right)$. It is measured in decibels.
2. The intensity I is given by $I = \dfrac{{{I_0}}}{{{r^2}}}$

Complete step by step Solution:

The relationship between loudness & the intensity is given by
$L = 10\log \left( {\dfrac{I}{{{I_0}}}} \right)$
Where ‘L is the sound intensity level $\left( {loudness} \right),$I is the sound’s intensity & I0 is the intensity of the reference ${I_0} = 1pw/{m^2}$
By using this you can solve the problem.
Given attitude height $= {\text{ }}100{\text{ }}m$
The sound intensity on the ground as the jet passes by is $160{\text{ }}dB$. Hence from the formula
$L = 10\log \left( {\dfrac{I}{{{I_0}}}} \right)$
$= 10\log \left( {\dfrac{{{I_0}}}{{{r^2}{I_0}}}} \right) = 10\log \left( {\dfrac{1}{{{r^2}}}} \right)$
$L = 160\,\,dB,\,\,r = 100\,m$
$160\,\,dB = 10\,\log \left( {\dfrac{1}{{{{\left( {\log } \right)}^2}}}} \right)$ ……….(1)
Let the destitute be x meter for the sound intensity of $120{\text{ }}dB.$
So, $120 = 10{\log _{10}}\left( {\dfrac{1}{{{x^2}}}} \right)$ ………(2)
Subtraction equation (2) form (1)
$\dfrac{{160 - 120}}{{10}} = {\log _{10}}\left( {\dfrac{1}{{10000}}} \right) - {\log _{10}}\left( {\dfrac{1}{{{x^2}}}} \right)$
$4 = {\log _{10}}\left( {\dfrac{{{x^2}}}{{10000}}} \right)$
$\Rightarrow {x^2} = {\left( {10000} \right)^2}$
${x^2} = 10000 = 10\,\,km$
$\therefore$ the jet should pass $10{\text{ }}km$ above the ground so that noise 9 not greater than $120{\text{ }}dB.$

Note:
1. Intensity is also given as power used per unit surface area. $\dfrac{\rho }{{4\pi {r^2}}}$
2. Loudness also depends on the pressure. If the pressure increases, pitch increases, so loudness increases.