Question

The formula for finding the decibels of a sound is Decibels= 10log(I/I0) . If the decibel rating for sound A is 20 decibels and sound B’s intensity is 3 times larger, what is the decibel rating for sound B?

Answer 1

Answer: We have to find the sound Intensity of sound B in Decibels, provided the sound Intensity of sound A is 20 Decibels. The Sound Intensity of B is 3 times the sound Intensity of sound A:

$\begin{gathered} A=20db=10\log _{10}(\frac{I_A}{I_o})\rightarrow(1) \\ B=3A=60db=10\log _{10}(\frac{I_B}{I_o})\rightarrow(2) \end{gathered}$

The threshold Intensity in equations (1) and (2) is as follows:

$I_o=10^{-12}watts\/m^2$

Solving the equation (1) gives:

$\begin{gathered} 10^{20}=\frac{I_A}{I_o}=\frac{I_A}{10^{-12}watts\/m^2} \\ I_A=10^{20}\cdot10^{-12}watts\/m^2 \\ I_A=10^8watts\/m^2\text{ } \end{gathered}$

Same is true for the equation (2):

$\begin{gathered} 60db=10\log _{10}(\frac{I_B}{I_o}) \\ 10^{60}=\frac{I_B}{I_o}=\frac{I_B}{10^{-12}watts\/m^2} \\ I_B=10^{60}\cdot10^{-12}watts\/m^2 \\ I_B=10^{48}watts\/m^2 \end{gathered}$

The answer, therefore, is as follows:

$\begin{gathered} I_A=10^8watts\/m^2=20db \\ I_B=10^{48}watts\/m^2=60db \end{gathered}$