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17 December, 00:22

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.

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  1. 17 December, 03:53
    0
    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
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