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
B
8
C
2
D
zero

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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