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.

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 109 dB at a distance of 6.86 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

You are asked to find the distance at the which the sound intensity drops to a specified level, equal to the normal hearing threshold for humans.

Intensity of sound is INVERSELY proportional to the Square of the Distance from the source, r. In this problem, the door is the 'source' even though the band is producing the sound. So,

I₁ = 109dB intensity of sound measured at 6.86m from the door

r₁ = 6.86m, distance from the door

I₂ = 1dB Intensity of sound (threshold level)

r₂ = to be determined.

Use the proportionality relationship

I₁/I₂ = r₂²/r₁² ------------ 1

dB is a ratio of sound-intensity to the threshold intensity. It is defined as

β = 10*Log (I/I₀) ----------- 2

Note that based on this equation, β = 1 dB when I = I₀, i.e. when the hearing threshold is reached.

To convert dB into Intensity of sound, you can re-arrange the above defining formula as

I = I₀ * 10^(β/10)

Calculate the ratio of intensities in terms of decibels of sound

I₁/I₂ = 10^(β1-β2)/10 ----------- 3

where, β1 and β2 are the decibel levels of sound corresponding to intensities I₁ and I₂.

Using equations 1 and 3, you can eliminate I₁/I₂ and write

r₂²/r₁² = 10^(β1-β2)/10 ---------- 4

Solve this equation for r₂. You are given all the rest of the quantities. I estimate r₂ = 1,723,154 m !!

You can expect it to be a large distance because sound is not being absorbed.