Question

If one sound had a decibel reading of 3.2, what would the reading of one that was 24 times as intense?

D=10log(I/10^-12).

I can't figure this homework question out please provide a solution. Grade 12 question.

Answer 1

The equation which relates the intensity of a sound wave to its decibel level is:

$$D=10\log(I/1.0\times10^{-12}W/m^2)$$

Then

$$D_1=10\log(I_1/1.0\times10^{-12}W/m^2)$$

$$D_2=10\log(I_2/1.0\times10^{-12}W/m^2)$$

$$=10\log((\dfrac{I_2}{I_1}\cdot I_1)/1.0\times10^{-12}W/m^2)$$

$$=10\log(\dfrac{I_2}{I_1})+D_1$$

Given $D_1=3.2\ dB,\dfrac{I_2}{I_1}=24.$

Then $D_2=3.2+10\log(24)=17(dB)$

17 dB.