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JIPMER 2006
Exam
JIPMER
Year
2006
Subject
Physics
Chapter
Waves and Sound

JIPMER 2006 · Physics · Waves and Sound

A person is observing two trains one coming towards him and other...

A person is observing two trains one coming towards him and other leaving with the same velocity 4 m/s4 \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 m/s320 \mathrm{~m} / \mathrm{s} )

Options

  1. A3
  2. B6
  3. C9
  4. Dzero

Correct answer

B. 6

Step-by-step solution

Observed frequency of first train n1=vv−vs×n=320320−4×240=320316×240=243 Hz.\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 n2=vv+vsn=320320+4×240=237 Hz.\begin{aligned} n_2 & =\frac{v}{v+v_s} n \\ & =\frac{320}{320+4} \times 240 \\ & =237 \mathrm{~Hz} .\end{aligned}\thereforeNumberofbeatsNumber of beats=243-237=6$
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