Question

The sound intensity level at a seat that is a distance of 2.00 meters from a rock concert stage is 120 dB. (a) Calculate the Intensity, I1 , of the sound in W/m2 at that position. (b) Calculate the sound level in dB at a seat that is 32 meters from the stage, given that the intensity decreases with the square of the distance from the source.

Answer 1

Answer:

$1\ \text{W/m}^2$

$0.0039\ \text{W/m}^2$

Explanation:

$I_1$ = Intensity of sound at 2 m away

Sound level = 120 dB

$r_1$ = 2 m

$r_2$ = 32 m

$I_0$ = Threshold of sound = $10^{-12}\ \text{W/m}^2$

Sound level is given by

$dB=10\log (\dfrac{I_1}{I_0})\\\Rightarrow I_1=10^{\dfrac{dB}{10}}I_0\\\Rightarrow I_1=10^{\dfrac{120}{10}}\times 10^{-12}\\\Rightarrow I_1=1\ \text{W/m}^2$

The intensity of sound at 2 m away is $1\ \text{W/m}^2$

$I_1=\dfrac{P}{4\pi r^2}$

$I\propto \dfrac{1}{r^2}$

$\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}\\\Rightarrow I_2=\dfrac{I_1r_1^2}{r_2^2}\\\Rightarrow I_2=\dfrac{1\times 2^2}{32^2}\\\Rightarrow I_2=0.0039\ \text{W/m}^2$

The intensity of the sound 32 m away is $0.0039\ \text{W/m}^2$