Q96 / 160 · JIPMER 2006 · Physics

A person is observing two trains one coming towards him and other leaving with the same velocity $4 \mathrm{~m} / \mathrm{s}$. 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 $320 \mathrm{~m} / \mathrm{s}$ )

  • A
    3
  • Correct answer
    B
    6
  • C
    9
  • D
    zero

Solution

Observed frequency of first train $\begin{aligned} n_1 & =\frac{v}{v-v_s} \times n \\ & =\frac{320}{320-4} \times 240 \\ & =\frac{320}{316} \times 240 \\ & =243 \mathrm{~Hz} .\end{aligned}$ Observed frequency of second train $\begin{aligned} n_2 & =\frac{v}{v+v_s} n \\ & =\frac{320}{320+4} \times 240 \\ & =237 \mathrm{~Hz} .\end{aligned}$ $\therefore$ Number of beats $=243-237=6$

Hence, the correct answer is option: 'B'.

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