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 86.9 dB at a distance of 3.15 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

20939

To solve this problem, we can use the inverse square law, which states that the intensity of sound decreases inversely with the square of the distance from the source. The formula for the inverse square law is:

I₁/I₂ = (r₂/r₁)²

Where I₁ is the initial intensity, r₁ is the initial distance, I₂ is the final intensity, and r₂ is the final distance.

Given that the initial intensity (I₁) is 86.9 dB at a distance (r₁) of 3.15 m, and we want to find the distance (r₂) when the sound is just barely audible, we can rewrite the formula as:

1/10^((I₂ - I₁)/10) = (r₁/r₂)²

Since the sound is just barely audible, we can assume the final intensity (I₂) to be the threshold of hearing, which is typically around 0 dB. Now, let's plug in the values and solve for r₂:

1/10^((0 - 86.9)/10) = (3.15/r₂)²

Simplifying the equation:

1/10^(-86.9/10) = (3.15²)/(r₂²)

Taking the reciprocal of both sides:

10^(86.9/10) = (r₂²)/(3.15²)

Taking the square root of both sides, we get:

r₂ = 3.15 * sqrt(10^(86.9/10))

Calculating the value, we have:

r₂ ≈ 3.15 * sqrt(10^8.69) ≈ 3.15 * 10000

r₂ ≈ 31,500 m

Therefore, the distance at which the music is just barely audible to a person with a normal threshold of hearing is approximately 31,500 meters.