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The loudest sustained snoring on record measures 85 decibels. The town in which the...

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andrewmezzo | Student | Valedictorian

Posted May 2, 2013 at 4:43 AM via web

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The loudest sustained snoring on record measures 85 decibels. The town in which the snorer lived had a noise ordinance that set the maximum traffic noise to 80 decibels. By how many times did the intensity of the snoring compare to the maximum intensity of traffic noise? 

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mathsworkmusic | (Level 3) Associate Educator

Posted May 2, 2013 at 10:37 AM (Answer #1)

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A decibel (dB) is a logarithmic measure here representing ten times the logarithm (base 10) of the ratio of a sound intensity (in W/m^2) to some reference intensity (often the TOH, or threshold of hearing).

The maximum ordained traffic noise T (in dB) is 80dB louder than the implied reference level, on a logarthmic (base 10) scale, so that

`T = 10log_(10)(I_T/I_(ref))`

where `I_T` is the intensity of sound (in W/m^2) the traffic should make at its height of noisiness and `I_(ref)` is the implied reference intensity of sound.

Contrastingly, the snoring noise S (in dB) is 85dB louder than the implied reference level, on a logarithmic (base 10) scale, so that

`S = 10log_(10)(I_S/I_(ref))`

where `I_S` is the intensity of sound (in W/m^2) the snoring creates.

We are interested in how many times more intense the the sound of the snoring is compared to the maximum ordained traffic noise. We have that

` ``I_T/I_(ref) = 10^(80/10) = 10^8`

so that the maximum ordained traffic noise is 10^8 times more intense than the implied reference sound intensity, and that

`I_S/I_(ref) = 10^(85/80) = 10^8.5`

The sound of the snoring is 10^8.5 times more intense than the reference intensity.

Therefore

`I_S/I_T = 10^8.5/10^8 = 10^0.5 = 3.16`   (since the `I_(ref)` terms cancel)

The snoring is 3.16 (to 3sf) times more intense than the maximum ordained traffic noise in the town.

 

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