Intensity of sound and pain level

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SUMMARY

The intensity of sound at a pain level of 120 dB is calculated to be 1012 W/m2, while a whisper at 20 dB has an intensity of 102 W/m2. This results in the loud noise being 1010 times more intense than the whisper. The formula used for these calculations is I(dB) = 10 log10(I/I0), where I0 is the reference intensity level, typically set to 1 W/m2. Understanding the logarithmic nature of decibels is crucial for comparing sound intensities.

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  • Understanding of logarithmic functions
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needhelp83
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What is the intensity of sound at the pain level of 120 dB? Compare this to that of a whisper at 20 dB?


I know this should be so simple, but it has me confused.

I have the formula but I am just not sure how to apply it.

I(db)=10 log _ 10 (I/I _ 0)
 
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I would probably just assume that the base level (I_0) is 1... then solve each sound level in dB for the sound level in a linear intensity scale... then compare the two sound levels by saying something like... the loud noise is x times louder than the whisper.
 
Wiki has some fascinating information about decibels.

1 bel is a 10-fold increase in sound level. It is dimensionless (like percent, it is only for comparison). 1 decibel is 1/10th of a bel*.


*(So, 1 decibel is a 1-fold increase in sound? i.e. no increase at all? :biggrin: )
 
120 dB=10 log (I/1) vs. 20 dB=10 log (I/ 1)
 
b=10 log (I/ I0)
120 dB=10 log (I/1)
10^12=(I/ 1)
I= 10^12 W/m^2


b=10 log (I/I0)
20 dB=10 log (I/1)
10^2=(I/1)
I=10^2 W/m^2


The intensity of 120 dB would be 10^12 W/m^2 compared to the whisper with the intensity at 10^2 W/m^2 which equals 10^10 more intense.

Sound good?
 

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