Wave Motion 5 Question 1
1. A small speaker delivers $2 W$ of audio output. At what distance from the speaker will one detect $120 dB$ intensity sound? [Take, reference intensity of sound as $10^{-12} W / m^{2}$
(2019 Main, 12 April II)]
(a) $40 cm$
(b) $20 cm$
(c) $10 cm$
(d) $30 cm$
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Answer:
Correct Answer: 1. (a)
Solution:
- Loudness of sound in decible is given by
$$ \beta=10 \log _{10}\left(\frac{I}{I _0}\right) $$
where, $I$ = intensity of sound in $W / m^{2}$,
$I _0=$ reference intensity $\left(=10^{-12} W / m^{2}\right)$, chosen because it is near the lower limit of the human hearing range.
Here, $\beta=120 dB$
So, we have $120=10 \log _{10}\left(\frac{I}{10^{-12}}\right)$
$$ \Rightarrow \quad 12=\log _{10}\left(\frac{I}{10^{-12}}\right) $$
Taking antilog, we have
$$ \begin{aligned} \Rightarrow & 10^{12} & =\frac{I}{10^{-12}} \\ \Rightarrow & I & =1 W / m^{2} \end{aligned} $$
This is the intensity of sound reaching the observer.
Now, intensity,
$$ I=\frac{P}{4 \pi r^{2}} $$
where, $r=$ distance from source,
$$ P=\text { power of output source. } $$
Here, $P=2 W$, we have
$1=\frac{2}{4 \pi r^{2}} \Rightarrow r^{2}=\frac{1}{2 \pi} \Rightarrow r=\sqrt{\frac{1}{2 \pi}} m=0.398 m \approx 40 cm$
