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Question

A boy blowing a whistle, is running away from a wall towards an observer with a speed of  $$1\ ms^{-1}$$. The frequency of whistle is $$680\ Hz$$. The number of beats heard per second by the observer will be $$($$given $$ v = 340\ ms^{-1})$$


A
zero
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B
2
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C
4
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D
8
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Solution

The correct option is A 2
Here a boy blowing a whistle, is running away from a wall towards an observer hence, the apparent frequency (of approaching) detected by observer is
$$n' = \dfrac{v}{v - u} n $$
where$$,\ u$$ is the speed of speed of boy running towards observer$$,\ n$$ is frequency of whistle sound$$,\ v$$ is velocity of sound in air.
$$n' = \dfrac{340}{340 - 1} (680)$$ .......................(given)
$$n' = \dfrac{340}{339} (680)$$
$$n' \approx 682  Hz$$
Hence, the number of beats per second heard by the observer is  $$682 - 680 = 2$$

Physics

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