A source of sound operates at 2.0 kHz, 20 W emitting sound uniformly in all direction. The speed of sound in air is `340 m s^(-1)` at a distance of air is `1.2kgm ^(-3)` (a) What is the intensity at a distance of 6.0 m from the source ? (b) What will be the pressure amplitude at this point ? (c ) What will be the displacement amplitude at this point ?

Answer

`44.2 mW//m^(2)`

Explanation

Intensity of sound due to a point source is
`I = (p)/(4pir^(2)) = (20)/(4xx3.14xx(6)^(2))`
= `0.0442 W//m^(2)`
Pressure amplitude is given as
`DeltaP_(0) = sqrt(2pvI)`
=`sqrt(2xx1.2xx340xx0.0442)` = 6Pa
Displacement amplitude is related to intensity as
` I = 2pi^(2)n^(2)A^(2)pv`
`A = sqrt((I)/(2pi^(2)n^(2)pv))`
= `sqrt((0.0442)/(2xx10xx(2000)^(2) xx1.2 xx340))`
= `1.16 xx 10^(-6)`m