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:

  1. 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

$1=\frac{2}{4 \pi r^{2}} \Rightarrow r^{2}=\frac{1}{2 \pi} \Rightarrow r=\sqrt{\frac{1}{2 \pi}} m=0.398 m \approx 40 cm$



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