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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 $320 \mathrm{~m} / \mathrm{s}$ )
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The correct answer is:
6
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 $=243-237=6$
$\begin{aligned} n_1 & =\frac{v}{v-v_s} \times n \\ & =\frac{320}{320-4} \times 240 \\ & =\frac{320}{316} \times 240 \\ & =243 \mathrm{~Hz} .\end{aligned}$
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 $=243-237=6$
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