Chapter 12, Problem 9P

At a rock concert, a dB meter registered 130 dB when placed 2.5 m in front of a loudspeaker on stage.

(a) What was the power output of the speaker, assuming uniform spherical spreading of the sound and neglecting absorption in the air?

(b) How far away would the sound level be 85dB?

To determine

(a)The power output of the speaker

(b)The intensity level of^$85dB$

Answer to Problem 9P

Solution:

(a)The power output of the speaker will be $785W$

(b)The intensity level of^$85dB$ will be $445m$

Explanation of Solution

Given:

$\begin{array}{l}I=130dB\\ I_0=85dB\end{array}$

(a)To determine the power output of the speaker

$\beta =10\log \frac{I}{I_0}$

Where,

$\beta$=sound level

$I_1$=intensity

$I_0$=power per unit

$\begin{array}{l}130=10\log \frac{I}{I_0}\\ 13=\log I-\log I_0\\ 13=\log I+12\\ 1=\log I\\ 10\frac{W}{m{}^2}=I\\ I=\frac{p}{4\pi r^2}\\ P=I\left(4\pi r^2\right)=\left(10\frac{W}{m^2}\right)4\pi {\left(2.5m\right)}^2=785W\end{array}$

(b)The determine the intensity level of $85dB$

$\beta =10\log \frac{I}{I_0}$

put

$\beta =85$

$\begin{array}{l}85\log =10\log \frac{I}{1\times {10}^{12}}\\ \frac{I_1}{I_2}=\frac{r_2^2}{r_1^2}\\ I_1=10\\ I_2=3.2\\ r_1=2.5\\ \end{array}$

Therefore

$r_2=445m$