Explanation:
It is given that,
At a of 14 m from the source the intensity level is 67 db. We need to find the total acoustic power output of the source. The intensity level is given by :
[tex]B=10\ dBlog(\dfrac{I}{I_o})[/tex]
[tex]I_o=10^{-12}\ W/m^2[/tex] (reference intensity)
[tex]67\ dB=10\ dBlog(\dfrac{I}{I_o})[/tex]
[tex]6.7=\ dBlog(\dfrac{I}{10^{-12}})[/tex]
[tex]I=0.00000501\ W/m^2[/tex]
or
[tex]I=5.01\times 10^{-6}\ W/m^2[/tex]
Output power is given by :
[tex]I=\dfrac{P}{A}[/tex]
[tex]P=I\times 4\pi r^2[/tex]
[tex]P=5.01\times 10^{-6}\times 4\pi (14)^2[/tex]
P = 0.0123 watts
So, the output power of the source is 0.0123 watts. Hence, this is the required solution.
