Question

of having intensity 1,000,0001), where l, is the initial intensity (a) Find the decibel rating of a sound, d = 10 log (b) If the intensity of a sound is doubled, by how much is the decibel rating increased? (a) Find the decibel rating of a sound having intensity 1,000,0001, where lo is the initial intensity The decibel rating of this sound is nts

Answer 1

Given that the decibel rating of a sound is, d = 10 log(I/I₀) where I₀ is the initial intensity.

(a) Now when I = 1,000,000I₀ i.e. (I/I₀) = 1000000 then we have,

d = 10 * log (1000000)

i.e. d = 10 * log(10⁶)

i.e. d = 10 * 6

i.e. d = 60

So the decibel rating of this sound is 60​​​​​​

.

(b) Now if the intensity of a sound is doubled, then we have, I = 2I₀ which implies, (I/I₀) = 2

So, the decibel rating becomes,

d = 10 * log(I/I₀)

i.e. d = 10 * log(2)

i.e. d ≈ 3.01 (rounded to two decimals)

So if the intensity of a sound is doubled then the decibel rating is increased by 3.01 dB.