Due to a point isotropic sonic source, loudness at a point is L = 40 dB. If density of air is `r = 15//11 kg m^(-3) ` and velocity of sound in air, u `= 330 ms^(-1)` , calculate.
(i) pressure oscillation amplitude at the point of observation, and
(ii) ratio oscillation amplitude of particle of medium to wavelength of sonic waves.

Explanation

Due to propagation of longitudinal waves, pressure of the medium varies with time. Maximum change in pressure is given by
`(DeltaP)_("max") = (gammaP aomega)/(u) " ". . . (i) `
Speed of sound waves in a gas is `u = sqrt((gammaP)/(rho)) `
`gamma P = rho u ^(2) " ". . . (ii) `
Loudness at point is given by L = 10 `log_(10) (I)/(I_(0)) dB`
Where `I_(0) = 10 ^(-12) Wm^(-2)`
`:.` Intensity of sound waves at the point is `I = I_(0)`
anti `log_(10) ((L)/(10))`
or, `I = 1 xx 10^(-s) W m^(-2)`
But intensity `I - = 2 pi^(2) n^(2) a^(2) rho u = (1)/(2) (4 pi^(2) n^(2) a^(2))rho u`
`:. 4 pi^(2) n^(2) a^(2) = (2I)/(rhou)`
But ` 4 pi^(2) n^(2) a^(2) = (a omega)^(2) , "hence", a omega = sqrt((2I)/(rho u))" ". . . (iii)`
Substituting `gamma P = rho u ^(2)`
and `a omega = sqrt((2I)/(rho u)) ` in equation (i),
`(Delta P)_("max") = sqrt(2I rho u) = 3 xx 10^(-3) Nm^(-2)`
Substituting w = 2pn in equation (iii),

`2 pi na = sqrt((2I)/(rho u)) " ""But" n = (u)/(lambda)` `(a)/(lambda) = (1)/(2 pi u) sqrt((2I)/(rhou)) = (10^(-6))/(99 pi)`