Soundproofing
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To study the sound attenuation introduced by a wall, we adopt the following summary model :
In a pipe of cross section \(S\), a harmonic incident plane progressive sound wave of angular frequency \(\omega\) arrives on a piston of surface \(S\), of thickness \(e\) and of density \(\mu\), free to move around \(x = 0\).
Note \(c\) the speed of sound in air and \(\mu_0\) density of air at rest.
It seeks the form of velocity field :
\({v_1}(x < 0,t) = {A_1}{e^{j(\omega t - kx)}} + {B_1}{e^{j(\omega t + kx)}}\)
And :
\({v_1}(x > e,t) = {A_2}{e^{j(\omega t - kx + ke)}}\)
Question
Justify these expressions and write the corresponding overpressures \(p_1(x,t)\).
Solution
The piston created a reflected wave and, using acoustic impedances :
\({p_1}(x < 0,t) = {\mu _0}c{A_1}{e^{j(\omega t - kx)}} - {\mu _0}c{B_1}{e^{j(\omega t + kx)}}\;\;\;and\;\;\;{p_1}(x > e,t) = {\mu _0}c{A_2}{e^{j(\omega t - kx + ke)}}\)
Question
Write the boundary conditions on the non-deformable piston and deduce that :
\(\frac{{{{\underline A }_2}}}{{{A_1}}} = \frac{1}{{1 + \frac{{j\omega \;\mu e}}{{2{\mu _0}c}}}}\)
Solution
The continuity of speeds gives :
\(A_1+B_1=A_2\)
The Newton's second law applied to the piston gives (the piston speed is for example given by \(v_1(x=e,t)\)) :
\(j\omega (\mu Se){A_2} = S{p_1}(0,t) - S{p_1}(e,t)\)
Is :
\(j\omega \mu e{A_2} = {\mu _0}c({A_1} - {B_1} - {A_2})\)
We deduce then :
\(\frac{{{{\underline A }_2}}}{{{A_1}}} = \frac{1}{{1 + \frac{{j\omega \;\mu e}}{{2{\mu _0}c}}}}\)
Question
Deduce the factor of transmittance in power \(T\) of the wall.
We give :
\({\mu _0} = 1,3\;kg.{m^{ - 3}},\;\mu = {2.10^3}\;kg.{m^{ - 3}},\;c = 340\;m.{s^{ - 1}}\)
What should be the minimum thickness of the wall if you want an attenuation of at least \(- 40\) decibels for \(f=1\;kHz\) then for \(f=100\;Hz\) ?
Solution
The factor of transmittance in power of the wall \(T\) here :
\(T = {\left| {\frac{{{A_2}}}{{{A_1}}}} \right|^2} = \frac{1}{{1 + \frac{{{\mu ^2}{e^2}{\omega ^2}}}{{4\mu _0^2{c^2}}}}}\)
We wish attenuation \(- 40\; dB\) (at least). Therefore, \(T<10^{-4}\).
We obtain \(e>7\;mm\) at \(1\;kHz\) and \(e>7\;cm\) at \(100\;Hz\).