A point source of sound emits a constant power with intensity inversely proportional to the square of the distance from the source. By how many decibels does the sound intensity level drops when you move from point $P_{1}$ to $P_{2} ?$ Distance of $P_{2}$ from the source is two times the distance of source from $P_{1}$ .

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-6.0 dB

4.0 dB

2.0 dB

5.0 dB

answer is A.

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Detailed Solution

We label the two points 1 and 2 , and we use the equation $β = 10 \log \frac{I}{I_{0}}$ (dB) twice.

The difference in sound intensity level $β_{2} - β_{1}$ is given by

$β_{2} - β_{1} = (10 d B) (\log \frac{I_{2}}{I_{0}} - \log \frac{I_{1}}{I_{0}})$

$= (10 d B) [(\log I_{2} - \log I_{0}) - (\log I_{1} - \log I_{0})]$

$= (10 d B) \log \frac{I_{2}}{I_{1}}$

Now,

$I \propto \frac{1}{r^{2}} \Rightarrow ∴ \frac{I_{2}}{I_{1}} = {(\frac{r_{1}}{r_{2}})}^{2} = \frac{1}{4} as r_{2} = 2 r_{1}$

$β_{2} - β_{1} = (10 d B) \log (\frac{1}{4}) = - 6.0 d B$

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