Question

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

• A. $4$
• B. $8$
• C. $2$
• D. $zero$

Answer 1

The correct option is A $4$

When source and observer moving close to each other, then apparent frequency

$n=\left(\frac{v+v_0}{v+v_s}\right)\times n_0$

here,

$v_s=1m/s{v}_0=1m/s{n}_0=680Hzv=340m/s$

$n=\left(\frac{340+1}{340-1}\right)\times 680$

$n=684Hz$

So,

Number of beats per second$=684-680=4beats/sec$