# 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...

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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### 1 Answer

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.**