Question

A sound wave radiates from a source in all directions. If the power of the sound source is 200 watts, then the intensity level of the sound wave 1000 m from the source is (in dB)

Answer 1

I=1000/(4*phi*r^2)

I=1000/12560

I=0.0796

I=0.08 watt /m^2

S Ldb =10*log (I/Io)

S Ldb =10*log (0.08/10^-12)=109 dB

Answer 2

We power   P = Intensity *Area

Area =4?R²

R= 100 m

P = 200W

Then Intensity I= P/Area

I = 200W/4?(100m)² = 1.592*10^-3W/m²

We knowthat intensity level ? = (10dB) log(I/Io)

Here Io= 10^-12W/m²

Then   ? =  (10dB) log(I/Io)

?= (10dB)log( 1.592*10^-3W/m²/10^-12W/m²) =92dB

Answer 3

There is a better way to understand it :

Power = intensity multiply by the area of the sound source, so if the sound wave radiates in all direction, it will form an sphere, with an area of : 4pi*r^2

so then :

Power = 200 = I*4pi*r^2

I = intensity

The intensity is not in Db, is in W / m^2

200 = 4pi*10000*I

I = 1 / 200pi W / m^2

Note, the definition of dB, decibel is this one :

dB = 10log( I / Io )

then, replacing : I on the equation :

dB = 10log(1 /200pi / Io)

Io = 10-^12 W / m^2

dB = 10*[ log ( 1/200pi) + 12 ]

dB = 92.3 aprox

Hope that helps