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Standard XII

Physics

Apparent Frequency:Listener Moving

Two sound sou...

Question

Two sound sources are moving in opposite directions with velocities $v_{1}$ and $v_{2} (v_{1} > v_{2})$ . Both are moving away from a stationary observer. The frequency of both the sources is 900 Hz. What is the value of $v_{1} - v_{2}$ so that the beat frequency observed by the observer is 6 Hz? Speed of sound v = 300 m/s. Given that $v_{1}$ and $v_{2}$ ≪ v.

1 m/s

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2 m/s

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3 m/s

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4 m/s

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Solution

The correct option is B 2 m/s
$f_{1} = 900 (\frac{300}{300 + v_{1}}) ≅ 900 {(1 + \frac{r_{1}}{300})}^{- 1} 900 - 3 v_{1}$
Similarly,
$f_{2} = 900 (\frac{300}{300 + v_{2}}) = 900 - 3 v_{2} f_{2} - f_{1} = 6 ∴ 3 (v_{1} - v_{2}) = 6 ∴ 3 (v_{1} - v_{2}) = 6 o r v_{1} - v_{2} = 2 m / s$

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Similar questions

Q. Two sound sources are moving in opposite directions with velocities

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The source producing sound and an observer both are moving along the

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The correct option is B 2 m/sf1=900(300300+v1)≅900(1+r1300)−1900−3v1 Similarly, f2=900(300300+v2)=900−3v2f2−f1=6∴3(v1−v2)=6∴3(v1−v2)=6or v1−v2=2m/s

The correct option is B 2 m/s
$f_{1} = 900 (\frac{300}{300 + v_{1}}) ≅ 900 {(1 + \frac{r_{1}}{300})}^{- 1} 900 - 3 v_{1}$
Similarly,
$f_{2} = 900 (\frac{300}{300 + v_{2}}) = 900 - 3 v_{2} f_{2} - f_{1} = 6 ∴ 3 (v_{1} - v_{2}) = 6 ∴ 3 (v_{1} - v_{2}) = 6 o r v_{1} - v_{2} = 2 m / s$