Remember that sound intensity decreases in inverse proportion to the distance squared. So, to solve this we are going to use the inverse square formula: [tex] \frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2 [/tex] where [tex]I_{2}[/tex] is the intensity at distance 2 [tex]I_{1}[/tex] is the intensity at distance 1 [tex]d_{2}[/tex] is distance 2 [tex]d_{1}[/tex] is distance 1
We can infer for our problem that [tex]I_{1}=270[/tex], [tex]d_{1}=1[/tex], and [tex]d_{2}=3[/tex]. Lets replace those values in our formula to find [tex]I_{2}[/tex]: [tex] \frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2 [/tex] [tex] \frac{I_{2} }{270} =( \frac{1}{3} )^2[/tex] [tex]\frac{I_{2} }{270} = \frac{1}{9} [/tex] [tex]I_{2}= \frac{270}{9} [/tex] [tex]I_{2}=30[/tex] dB
We can conclude that the intensity of the sound when is 3 m from the source is 30 dB.
