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 83.9 dB at a distance of 6.91 m from the door. At what distance is the music just barely audible to a person with a normal threshold of hearing? Disregard absorption. Answer in units of m.

Given Information:

Intensity level = 83.9 dB

distance = d1 = 6.91 m

Required Information:

distance where sound is barely audible = d2 = ?

d2 = 108262.5 m

Explanation:

As we know the intensity level of sound is

Intensity level = 10*log(I/I₀)

I₀ = 10⁻¹²W/m² is the reference intensity level or threshold

83.9 = 10*log(I/I₀)  eq. 1

The intensity of sound and distance are inversely related as

I ≈ 1/d²

At d1 intensity of sound is 83.9 dB

We want to find the distance d2 where intensity of sound is 0 dB so

I₀/I = (d1/d2)²

or  I/I₀ = (d2/d1)²

Substitute the ratio of I/I₀ into eq. 1

83.9 = 10*log(d2/d1)²

83.9 = 20*log(d2/d1)

83.9/20 = log(d2/d1)

4.195 = log(d2/d1)

10⁴°¹⁹⁵  = d2/d1

d1*15667.51  = d2

d2 = 15667.51*6.91

d2 = 108262.5 m

Therefore, sound of music will barely be heard at a distance of 108262.5 m from the source of music.