Answer:
Distance from the jackhammer is given as
[tex]r_2 = 4(I^{\frac{7}{16}})[/tex]
where I is intensity at x = 4 m
Explanation:
As we know that the loudness of the sound is given as
[tex]L = 10 Log(\frac{I}{I_o})[/tex]
So we have
we have to change the intensity becomes
[tex]I' = I^{1/8}[/tex]
so the loudness becomes 1/8 times of initial
now we know that intensity inversely varies with the square of the distance
so we have
[tex]I = \frac{P}{4\pi(4^2)}[/tex]
[tex]I' = \frac{P}{4\pi r^2}[/tex]
so we have
[tex]\frac{I}{I'} = \frac{r^2}{16}[/tex]
[tex]\frac{I}{I^{1/8}} = \frac{r^2}{16}[/tex]
[tex]I^{7/8} = \frac{r^2}{16}[/tex]
[tex]r = 4(I^{7/16})[/tex]
