When you buy earplugs at the store, they have a \"dB rating\". What it tells you

Question

When you buy earplugs at the store, they have a "dB rating". What it tells you is the reduction of SIL that the plugs will provide, if properly worn. Suppose you're helping some friends, working the stage at a rock concert wearing earplugs that provide a reduction of 40 dB. What is the ratio of the sound intensity just outside the plugs to the intensity reaching your eardrums? (40 dB may not seem huge, the music IS still pretty loud, but look what an impact this has on the energy being deposited in your ears!)

Explanation / Answer

as we know, the intensity of sound in dB is given as

10*log10(I/I0)

where log10 is logarithm with base 10

where I=intensity of the sound

I0=base intensity=10^(-12) W/m^2

so if intensity outside the earplug is I1 and intensity inside earplug is I2

==>10*log10(I1/I0)-10*log10(I2/I0)=40

using the formula, log(a)-log(b)=log(a/b)

we get

10*log10(I1/I2)=40

==>log10(I1/I2)=40/10=4

==>I1/I2=10^4=10000

hence the intensity is being reduced by 10000 times as compared to outside.

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