Question

you are at a rock concert sitting 10m from the stage at your position the sound is 120dB. To protect your ears, you want to reduce the sound intensity level by 20dB. How far away from the stage should you move?

Answer 1

Sound Intensity from a Point Source obeys the inverse square law i.e the intensity of sound is inversely proportional to the distance from the point source.

i.e

I   (1/ r² ) ----------------(1) here I in W / m²

the conversion from the W / m² to dB is

I (dB) = 10 log₁₀^{[I / I₀]}   ---------------(2)

here I₀ is Threshold of Hearing (TOH)

if I₁ is intensity of the sound when the person at r₁ distance from the sound source and I₂ is final intensity of the sound when the person at r₂ distance from the sound source then

from (1) we can write

I₁ r₁² = I₂ r₂²

I₁ / I₂ = [ r₂ / r₁]²  ------------(3)

and from (2) we can write as

I₁ (dB) = 10 log₁₀^{[I₁ / I₀]}   --------------(4)

I₂ (dB) = 10 log₁₀^{[I₂ / I₀]}   --------------(5)

from (4) - (5)

I₁ (dB) - I₂ (dB) =10 log₁₀^{[I₁ / I₀]} - 10 log₁₀^{[I₂ / I₀]}  

I₁ (dB) - I₂ (dB) = 10 [  log₁₀^{[I₁ / I₂]} ]   

from (3) replace I₁ / I₂ with  [ r₂ / r₁]² then

I₁ (dB) - I₂ (dB) = 20 [  log₁₀^{[ r₂ / r₁]} ]

the diffrence in sound intensity level can be written as

I₁ - I₂ = 20 log₁₀^{[ r₂ / r₁]} here I₁ , I₂ are in dB

here r₁ = 10 m , I₁ = 120 dB , I₂ =120 dB - 20 dB =100 dB

and I₁ - I₂ = 20 dB on substituting all values we get

20 = 20 log log₁₀^{[ r₂ / 10m]}

  log₁₀^{[ r₂ / 10m]} = 1

r₂ / 10 m = 10 ¹

r₂ = 100 m ( from the sound source )

Distance moved by the person away from the stage = r₂ - r₁ = 100 m - 10m = 90 m .