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25 April, 13:39

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 64 dB at a distance of 4.08 m from the door. At what distance is the music just barely audible to a person with a normal threshold of hearing

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  1. 25 April, 13:42
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    d2 = 4080 m

    Explanation:

    The intensity level of sound is given by

    Intensity level = 10*log (I/I₀)

    where I₀ = 10⁻¹²W/m² is the threshold of hearing

    60 = 10*log (I/I₀) eq. 1

    The intensity of sound decreases with the increase in distance squared

    I ≈ 1/d²

    Let d1 is the distance where intensity of sound is 64 dB and d2 is the distance where intensity of sound is 0 dB (barely audible) so

    I₀/I = (d1/d2) ²

    or I/I₀ = (d2/d1) ² put it in eq. 1

    60 = 10*log (d2/d1) ²

    60 = 20*log (d2/d1)

    60/20 = log (d2/d1)

    3 = log (d2/d1)

    10³ = d2/d1

    d1*10³ = d2

    d2 = 4.08*10³

    d2 = 4080 m

    Therefore, at a distance of 4080 m the sound of music will barely be audible to a person with normal hearing.
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