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

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Detailed Solution

Apartment frequency heard by the observer due to some S2
$f_{1} = f_{o} (\frac{V + V_{o}}{V}) f_{1} = 120 (\frac{330 + 10}{330}) f_{1} = 123.636 H_{2}$

Apartment frequency heard by the observer due to some S1
$f_{2} = f_{o} (\frac{V + V_{O} \cos 53}{V - V_{S} \cos 37}) f_{2} = f_{o} (\frac{330 + 10 X \frac{3}{5}}{330 - 30 X \frac{4}{5}}) f_{2} = \frac{120 X 336}{306} = 134.4 H 2$

Number of beats
$\begin{array}{l} = f_{2} - f_{1} \\ = 134.4 - 123.636 \\ = 8.13 H 2 \\ = 8 H 2 \end{array}$