A rock group is playing in a bar. Sound emerging from the door spreads uniformly in all directions. The intensity level of the music is 64 dB at a distance of 4.08 m from the door. At what distance is the music just barely audible to a person with a normal threshold of hearing

d2 = 4080 m

Explanation:

The intensity level of sound is given by

Intensity level = 10*log(I/I₀)

where I₀ = 10⁻¹²W/m² is the threshold of hearing

60 = 10*log(I/I₀)  eq. 1

The intensity of sound decreases with the increase in distance squared

I ≈ 1/d²

Let d1 is the distance where intensity of sound is 64 dB and d2 is the distance where intensity of sound is 0 dB (barely audible) so

I₀/I = (d1/d2)²

or  I/I₀ = (d2/d1)² put it in eq. 1

60 = 10*log(d2/d1)²

60 = 20*log(d2/d1)

60/20 = log(d2/d1)

3 = log(d2/d1)

10³  = d2/d1

d1*10³  = d2

d2 = 4.08*10³

d2 = 4080 m

Therefore, at a distance of 4080 m the sound of music will barely be audible to a person with normal hearing.

answer; acquired mainly during early childhood;