Question

An audio system 1 produces a sound level of 90.0 dB and system 2 produces a sound level of 93.0 dB. (a) Determine the ratio of the intensities (in W/m 2 ) . (b) Suppose a recording device was placed 2 m away from s ystem 2 . H ow far away from system 1 must that recording device be placed in order for the intensity to be the same as it was near system 2?

Answer 1

a) Relation between the sound level (S) and the intensity (I) is

S =(10 dB)log (I /I_ref)

here I_ref is the intensity of sound at 0 dB

for sound level (S₁) of 90.0 dB having intensity (I₁)

S₁ =(10 dB)log (I₁ /I_ref)

for sound level (S₂) of 93.0 dB having intensity (I₂)

  S₂ =(10 dB)log (I₂ /I_ref)

from S₁ -S₂ =(10 dB)log (I₁ /I_ref) - (10 dB)log (I₂ /I_ref)  

  S₁ -S_{2 =}(10 dB)log (I₁/ I₂)

90.0 dB - 93.0 dB =(10 dB)log (I₁/ I₂)

-3.0 dB =(10 dB)log (I₁/ I₂)

  I₁/ I₂ =10^-0.30

   I₁/ I₂ =1 /10^0.30

and I₂ / I₁ =10^0.30

(b) sound intensity decreases with (1/r²) from the source so

I   (1/r²)

new sound intensity at a distance r₂  is I₂^| = I₂ / r₂²  

new sound intensity at a distance r₁ is I₁^| =I₁ / r₁²  

I₂^| /  I₁^| =( I₂ / r₂² ) / (I₁ / r₁² ) here    I₂^| =  I₁^| (because same intensity in both the cases)

1 = ( I₂ / r₂² ) / (I₁ / r₁² )

(I₁ / r₁² ) =  ( I₂ / r₂² )

I₂ / I₁ = r₂² /  r₁² -----------------------(1)

here    I₂ / I₁ = 10^0.30 =1.99526 (from part (a))

and r₂ = 2 m

from (1) 1.99526 =   (2 m)² / r₁²

r₁² =2.00475 m and

r₁ =1.42 m

recording device should be placed at 1.42 m from the system 1