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Help with log problem

  1. Jul 20, 2004 #1
    Here is the question as it follows:

    The decibel scale, defined as

    [tex] L = 10 log (\frac{I}{I_0}) [/tex]

    where L is the Loudness (or comparative intensity) in decibels (dB),
    [tex] I_0 [/tex] is the reference level (10^-12 watts per m^2)
    I is the actual intensity of the sound measured (watts per m^2).

    now heres the question part:
    An owner of an industrial plant next to a residential suburb agreed to limit noise to 75dB with a variation of up to 15% at a distance of 100m from the factory gates. Residents subsequently complained that sounds often reached 85dB. The manager replied, "It's not far over the limit - it's under the 15% variation agreed." Discuss whether or not the mangers statement is justified.

    Dont quite know what there getting at, any help would be greatly appreciated.
  2. jcsd
  3. Jul 20, 2004 #2


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    85- 75= 10 so the variation is 10. 10/75= 0.133333.. which is less than 0.15. Yes, the variation is less than 15% of the noise limit. I am interpreting the statement "75dB with a variation of up to 15%" literally: 15% of 75dB.

    Of course, 75 decibels means 10 log(I/I0)= 75 or I= I0107.5= 10-1210[7.5]= 10-4.5 watts/m2 and
    85 decibels means 10 log(I/I0)= 85 or I= I0108.5= 10-1210[8.5]= 10-3.5 watts/m2 so the actual difference in intensity is 0.000285 so the percentage variation is 0.000285/10-4.5= 9 (900%) which is far more than 15%.

    But, the original agreement was about decibels, not intensity. The manager's statement is justified. (and that second paragraph is irrelevant.)
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