Question

From a distance of 4.0 m, a bystander listens to a jackhammer breaking concrete. How far would he need to move from the jackhammer so that its perceived loudness decreases by a factor of 8?

Answer 1

Answer:

Distance from the jackhammer is given as

$r_2 = 4(I^{\frac{7}{16}})$

where I is intensity at x = 4 m

Explanation:

As we know that the loudness of the sound is given as

$L = 10 Log(\frac{I}{I_o})$

So we have

we have to change the intensity becomes

$I' = I^{1/8}$

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

$I = \frac{P}{4\pi(4^2)}$

$I' = \frac{P}{4\pi r^2}$

so we have

$\frac{I}{I'} = \frac{r^2}{16}$

$\frac{I}{I^{1/8}} = \frac{r^2}{16}$

$I^{7/8} = \frac{r^2}{16}$

$r = 4(I^{7/16})$