Question

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 116 dB at a distance of 2.82 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.

Answer 1

Answer:

1779299.7m

Explanation:

From formulas in acoustic sound, we know that sound intensity is inversely proportional to the square of the distance away.

Thus;

I2/I1 = r2²/r 1²

So,

∆L = 10 log (I2/I1)

Where ∆L is the intensity of music and r1 and r2 are distances away.

∆L=10log 10(r1²/r2²)

∆L=10log 10(r1/r2)²

∆L= - 20log 10(r1/r2)

r2 = r1•10^(-∆L/20)

​From the question,

∆L = 116 Db

r1 = 2.82m

Thus,

r2 = 2.82 x 10^(116/20)

r2 = 2.82 x 630957.34 = 1779299.7m

Answer 2

Answer / Explanation:

To properly answer this question, let us first define what sound intensity is:

Sound intensity which can also be refereed to as also known as acoustic intensity can be described as the calculated power or energy needed to transmit sound waves per unit area in a direction at 90 degree to that area. We should also note that the SI unit of sound intensity is the watt per square meter.

Referring back to the question asked,

To find where the music is barely audible, we need to use the equation

L₂ = L₁ - 20 log r₂/r₁

where  

L₁ = Sound level in decibels, as well as the

r₁ =  distance from the source that sound level is heard, and  

L₂ = Sound level, while

r₂ =  distance at a different point.

Moving forward,

To be able to find where the sound is barely audible, we need to find the location at which the sound level is zero db (o db).

Therefore,

L₂ = L₁ - 20 log r₂ / r₁

0dB  =  116 dB  −  20log  r₂ / 2.82 m 20  log  r₂ / 2.82 m

= 116 dB r₂ / 2. 82m = 10 ∧ 116 / 20

r₂ = (2.82m) 10 ∧ 116 / 20

r₂ =  178.7m