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 sound in air = 340 m/s )

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 sound in air = 340 m/s ) (A) 4 (B) 8 (C) 2 (D) zero

Tardigrade Admin · Accepted Answer

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

n = (\frac{v + v _{o}}{v - v _{s}}) \times n \dots (1)

Here,

v_{s} = 1 m / s

,

v_{0} = 1 m / s

,

n = 680 Hz

So, from equation (1)

n = (\frac{340 + 1}{340 - 1}) 680 = 684

Number of beats per second

= 684 - 680 = 4

beats / sec

Q. 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 sound in air =340m/s )

4

8

2

zero

When source and observer are moving close to each other, then apparent frequency n=(v−vs​v+vo​​)×n…(1) Here, vs​=1m/s, v0​=1m/s, n=680Hz So, from equation (1) n=(340−1340+1​)680=684 Number of beats per second =684−680=4 beats / sec

Q. 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 sound in air $= 340 m / s$ )