Question

Sound with intensity larger than $120dB$appears painful to a person. A small speaker delivers $2·0W$ of audio output. How close can the person get to the speaker without hurting his ears?

Answer 1

Step 1: Given data:

The level of the sound is =$120dB$

The audio output = $2·0W$

Step 2: Formula used:

It is understood that,

$\mathrm{Sound}\mathrm{intensity}\mathrm{level}\left(I_L\right)=10\cdot \log \left(\frac{I}{I_{\circ }}\right)$

Where, $I$ is sound intensity

$I_o$ is reference intensity

$Intensity\left(I\right)=\frac{power\left(P\right)}{Area\left(A\right)}$

Step 3: Compute the intensity of sound:

On putting the given values in the formula of sound intensity level-

$$\begin{gathered} 120=10·{\log }_{10}\left(\frac{I}{I_{\circ }}\right) \\ {10}^{12}=\frac{I}{I_{\circ }} \end{gathered}$$

Since, $I_{\circ }={10}^{-12}\raisebox{1ex}{$w$}\!\left/ \!\raisebox{-1ex}{$m^2$}\right.$. So,

The value of sound intensity will be-

$I=1\raisebox{1ex}{$w$}\!\left/ \!\raisebox{-1ex}{$m^2$}\right.$

Step 4: Compute the closest distance:

We know that,

$Intensity=\frac{power}{Area}$

Suppose the nearest distance is $x$.

The area will be- $A=4\pi x^2$

So, using the formula we get-

$$\begin{gathered} I=\frac{P}{4\pi x^2} \\ 1=\frac{2}{4\pi x^2} \\ {x}^2=\frac{1}{2\pi } \\ x=0.4m \end{gathered}$$

Thus, the nearest distance that a person can go is $0·4\mathrm{m}$.