A train $S_{1}$ , moving with a uniform velocity of 108 km/h approaches another train $S_{2}$ standing on a platform. An observer O moves with a uniform velocity of 36km/h towards $S_{2}$ as shown in figure. Both the trains are blowing whistles of same frequency 120Hz. When O is 600m away from S₂ and distance between $S_{1}$ and $S_{2}$ is 800m, the number of beats heard by O is_____
[ Speed of the sound =330m/s]

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answer is 8.13.

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Detailed Solution

$\tan θ_{1} = \frac{800}{600} = \frac{4}{3}$

$\Rightarrow θ_{1} = 53^{\circ} a n d θ_{2} = 37^{\circ}$

$Now, Let say Freq. Heard fromm train at S_{2} = f_{2}$

$f_{2} = (\frac{V + V_{0}}{V}) f_{s} = (\frac{330 + 10}{330}) \times f_{s} = \frac{340}{330} f_{s}$

$Now, Let say freq. Heard from train at S_{1} = f_{1}$

$\Rightarrow f_{1} = (\frac{V + V_{0} \cos θ_{1}}{V - V_{s} \cos θ_{2}}) f_{s} = (\frac{330 + 10 \times \frac{3}{5}}{330 - 30 \times \frac{4}{5}}) f_{s} = (\frac{336}{306}) f_{s}$

$So, Beat freq = (\frac{336}{306}) f_{s} - \frac{340}{330} f_{s} = (\frac{336}{306} - \frac{340}{330}) \times 120 = 8.13 H z$

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