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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 112 dB at a distance of 5.1 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.

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akande212

Answer:

A distance of 7m.

Explanation:

I1 = 112dB = intensity of sound produced.

r1 = 5.1m = distance of hearer from sound source.

I2 = normal hearing threshold for humans 60dB

r2 = unknown

From inverse square law

I1/I2 = r2²/r1²

r2 = √(I1/I2×r1²) = √(112/60×5.1²) = 6.97m

Approximately 7m

The distance where the music is just barely audible to a person with a normal threshold of hearing should be 7 m.

Calculation of the distance:

Since

I1 = 112dB = intensity of sound produced.

r1 = 5.1m = distance of hearer from sound source.

I2 = normal hearing threshold for humans 60dB

r2 = unknown

Now we apply inverse square law

I1/I2 = r2²/r1²

r2 = √(I1/I2×r1²) = √(112/60×5.1²)

= 6.97m

= 7m

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