A loudspeaker on a pole is radiating 100 W of sound energy in all directions. You are walking directly toward the speaker at 0.80 m/s. When you are 20 m away, what are (a) the sound intensity level and (b) the rate (dB/s) at which the sound intensity level is increasing? Hint: Use the chain rule and the relationship log10 x = ln x/ln 10.

The power of the sound source P= 100W

The speed of the listener v = 0.80m/s

The distance of the listener from the source is d = 20m

The area covered by the sound source is 4πr²

Substitute the values

$$\begin{gathered} Area=4(3.14){\left(20\right)}^2 \\ =5.24m^2 \end{gathered}$$

The intensity of the sound is

The sound intensity level is given by

$$\begin{gathered} \beta =\left(10dB\right)\log \left(\frac{I}{I_o}\right) \\ =\left(10dB\right)\mathrm{lo}g\left(\frac{1.9\times {10}^{-2}}{1\times {10}^{-12}}\right) \\ =\left(10dB\right)\log \left(1.9\times {10}^{10}\right) \\ =\left(10dB\right)\left[\log 1.9+\log {10}^{10}\right] \\ =\left(10dB\right)\left[0.27+10\right] \\ =\left(10dB\right)(10.27) \\ =102.7dB \end{gathered}$$

Thus, the intensity level of the sound source is 102.7 dB

The time taken by the sound to reach the listener is given by

$$\begin{gathered} Time=\frac{dis\tan ce}{speed} \\ T=\frac{20m}{0.80m/s} \\ T=25sec \end{gathered}$$

In 25 seconds, the sound reaches the listener with an intensity level of 102.7 dB

The intensity level of sound in 1 sec is given by

$\frac{102.7}{25}=4.108dB/s$

Thus, the rate of increase in intensity level is 4.108 dB/s