Two sound waves having wavelengths 5.0 m and 5.5 m propagate in a gas with velocity 300 m/s. The number of beats produced per second is (a) six (b) two (c) three (d) one

Answer: (a). f = (v)/(λ): f₁ = (300)/(5.0) = 60 Hz, f₂ = (300)/(5.5) = 54.5 Hz Beats = 60 - 54.5 ≈ 5.5 per second, nearest option is 6 So the answer is (a) six.

Two sound waves having wavelengths 5.0 m and 5.5 m propagate in a gas… | MHT CET 2021 (23 Sep, Shift 2)

Q46MHT CET 2021 · 23 SEP · SHIFT 2Easy

Beats

Two sound waves having wavelengths $5.0\text{ m}$ and $5.5\text{ m}$ propagate in a gas with velocity 300 $\text{m/s}$ . The number of beats produced per second is

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Answer: (a)

$f = \frac{v}{\lambda}$ : $f_1 = \frac{300}{5.0} = 60\text{ Hz}$ , $f_2 = \frac{300}{5.5} = 54.5\text{ Hz}$
Beats $= 60 - 54.5 \approx 5.5$ per second, nearest option is 6
So the answer is (a) six.

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