Question

0 A sound point source, situated 2.0 m above a brick floor, is emitting sound at 4x10 watts. (a) Find the sound intensity level due to that source at a listening noint 3.5 m above the floor and 15 m away. (b) Find the sound intensity level at that point when the reflection from the floor is taken into account.

Answer 1

a] intensity = power / area where area is 4pi*r^2

= 4e-5/[4pi*(15^2+(3.5-2)^2)]

= 1.4*10^-8 W/m^2

intensity level = 10 log (I/i0) = 10*log(1.4e-8/1e-12) = 41.46 dB answer

b] intensity = power / area where area is 4pi*r^2

= 4e-5/[4pi*(15^2+(3.5-2)^2)] + 4e-5/[4pi*(15^2+(3.5+2)^2)] (the other term is because of reflection)

= c2.6477e-8 W/m^2

intensity level = 10 log (I/i0) = 10*log(2.6477e-8/1e-12)

= 44.23 dB answer