Sound level: At a distance of 2.00 m from a point source of sound, the intensity level is 80.0 dB. What will be the intensity level at a distance of 4.00 m from this source? The lowest detectable intensity is 1.0 × 10^-12 W/m². A) 77.0 dB B) 74.0 dB C) 60.0 dB D) 40.0 dB E) 20.0 dB

Question

Quizplus · Accepted Answer

The sound intensity obeys the inverse square law, so at a distance of 4.00 m the intensity will be 1/4 times the intensity at a distance of 2.00 m.

$I_2 = I_1\left(\frac{r_1}{r_2}ight)^2$
where, 
$I_1$ is the initial sound intensity
$I_2$ is the new sound intensity
$r_1$ is the initial distance
$r_2$ is the new distance

Substituting the given values, we get:

$I_2 = I_1\left(\frac{r_1}{r_2}ight)^2 = 10^{8.0} 	imes \left(\frac{2.00}{4.00}ight)^2 = 10^{6.0} 	ext{ W/m}^2$

Converting to decibels using the formula:

$	ext{Intensity level} = 10\log_{10}\left(\frac{I}{I_0}ight)$

where $I_0$ is the lowest detectable sound intensity.

$	ext{Intensity level} = 10\log_{10}\left(\frac{10^{6.0}}{1.0 	imes 10^{-12}}ight) = 74.0 	ext{ dB}$

Therefore, the answer is B.

Sound Level: at a Distance of 2