Question

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 1

Correct Answer - A::B::C::D
(a) Here given, `V_(air) = 340 m//s`
power ` =(e )/(t) = 20 W`
`f =2,000 Hz, p = 1.2 kg//m^3`
So, intensity,
`I = (e )/(t.A) = (20)/(4pir^2) = (20)/(4 xx pi xx 6^2)`
` = 44 mw//m^2 [because r = 8m]`
(b) We know that,
`I = (p_0^2)/(2p V_(air))`
`rArr P_0 = sqrt(1 xx 2p V_(air))`
`= sqrt(2xx 1.2 xx 340 xx 44 xx 10^(-3))`
`=6.0 N//m^2`
(c ) We know that, `I = 2pi^2 S_0^2 upsilon^2 P V`
where `S_0` = displacement amplitude
`rArr S_0 = sqrt((I)/(2pi^2 P^2 p V_(air))`
Putting the value we get,
`S_0 = 1.2 xx 10^(-6) m`