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Question

A body is walking away from a wall towards an observer at a speed of $$1\ m/s$$ and blows a whistle whose frequency is $$680\ Hz$$. The number of beats heard by the observer per second is:-(velocity of second in air $$=340\ m/s$$):


A
4
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B
8
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C
2
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D
zero
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Solution

The correct option is A $$4$$
When source and observer moving close to each other, then apparent frequency
$$n=\left(\dfrac{v+v_0}{v+v_s}\right)\times n_0$$
here,
$$v_s=1m/s\quad v_0=1m/s\quad n_0=680Hz\quad v=340m/s$$
$$n=\left(\dfrac{340+1}{340-1}\right)\times 680$$
$$n=684Hz$$
So,
Number of beats per second$$=684-680=4 beats/sec$$


Physics

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