A point isotropic source generates sound oscillations with frequency `v=1.45 kHz`. At a distance `r_(0)=5.0 m` from the source the displacement amplitude of particles of the medium is equal to the `a_(0)=50 mu m,` and at the point `A` located at a distance `r=10.0 m` from the source the displacement ampitude is `eta=3.0 ` times less than `a_(0)` . Find `:`
`(a)` the damping coefficient `gamma` of the wave,
`(b0` the velocity oscillation amplitude of particles of the medium at the point `A`.

Explanation

`(a)` We known that the equation of a spherical wave in a homogeneous absorbing medium of wave dampint coefficient `gamma` is `:`
`xi=(a'_(0) e^(-gammar))/(r) cos ( omegat- kr)`
Thus particl's displacement amplitude equals
`(a'_(0) e^(-gammar))/(r)`
According to the conditions of the problem,
at `r=r_(0), a_(0)=(a'_(0)e^(-gammar_(0)))/( r) ....(1)`
and when ` r= r, (a_(0))/( eta)=(a'_(0)^(-gammar))/(r_(0)) ...(2)`
Thus from Eqns (1) and (2)
`e^(gamma(1-r_(0)))=eta(r_(0))/( r)`
or, `gamma(r-r_(0))=1n( etar_(0))-1nr`
or, `gamma=(1n eta+ 1 n r_(0)-1nr)/(r-r_(0))=(1n3+1n5-1n10)/( 5)=0.08m ^(-1)`
`(b)` As `x9=(a'_(0)e^(-gammar))/(r) cos ( omegat- kr )`
So, `( delta xi)/( deltat)-( a'_(0)e^(-gammar))/( r) omega sin ( omegat- kr )`
`((deltaxi)/( deltat))_(n)=(a'_(0)e^(- gammar))/( r) omega`
But at point `A,(a'_(0)e^(-gammar))/( r)=(a_(0))/( eta)`
So, `((deltaxi)/( deltat))_(m)(a_(0)omega)/(eta)=(a_(0)2pi)/( eta)=(50xx10^(-6))/( 3)xx2xx(22)/(7)xx1.45xx10^(3)=15 m//s`