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)((I_A)/(I_o))\rightarrow(1) \\ B=3A=60db=10\log _(10)((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)=(I_A)/(I_o)=(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)((I_B)/(I_o)) \\ 10^(60)=(I_B)/(I_o)=(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}`