Question

A person is observing two trains one coming towards him and other leaving with the same velocity $4\, m / s$. If their whistling frequencies are $240\, Hz$ each, then the number of beats per second heard by the person will be : (if velocity of sound is $320\, m / s$ )

• A. 3
• B. 6
• C. 9
• D. zero

Answer 1

Answer: (B)

6

Answer 2

Observed frequency of first train
$n_{1} =\frac{v}{v-v_{s}} \times n$
$=\frac{320}{320-4} \times 240$
$\frac{320}{316} \times 240$
$=243\, Hz$
Observed frequency of second train
$n_{2}=\frac{v}{v +v_{s}} n$
$=\frac{320}{320+4} \times 240$
$=237\, Hz .$
$\therefore$ Number of beats $=243-237=6$