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.

The distance at which the music is just barely audible is 108262.50 m

Explanation:

Sound intensity is the energy transmitted by a source in a unit area. The unit is in decibels.

Recall L2 - L1 = - 20 log r2/r1 with the log in base 10

L = sound level in decibels, r = distance from source of sound

the sound becomes barely audible at location 0 decibel, applying this concept

0 - 83.9 = - 20 log r2/6.91

20log r2/6.91 = 83.9

log r2/6.91 = 83.9/20 = 4.195, applying the laws of logarithms

r2/6.91 = 10^4.195 = 15667.5107

r2 = 6.91 * 15667.5107 = 108262.50 m

Given Information:

Intensity level = 83.9 dB

distance = d1 = 6.91 m

Required Information:

distance where sound is barely audible = d2 = ?

Answer:

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.