Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed $1.0 m s ^{-1}$ and the man behind walks at a speed $2.0 ms ^{-1}$. A third man is standing at a height $12 m$ above the same horizontal line such that all three men are in a vertical plane. The two walking men are blowing identical whistles which emit a sound of frequency $1430 Hz$. The speed of sound in air is $330 m s ^{-1}$. At the instant, when the moving men are $10 m$ apart, the stationary man is equidistant from them. The frequency of beats in $Hz$, heard by the stationary man at this instant, is. . . . .
IIT 2018, Advanced
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$f_A=f\left(\frac{v}{v-2 \cos \theta}\right)=1430\left[\frac{330}{330-2 \cos \theta}\right]=1430\left[\frac{1}{1-\frac{2 \cos \theta}{330}}\right]=1430\left[1+\frac{2 \cos \theta}{330}\right]$
$f_B=f\left(\frac{v}{v+\cos \theta}\right)=1430\left[\frac{330}{330+\cos \theta}\right]=1430\left[1-\frac{\cos \theta}{330}\right]$
$\Delta f =1430\left[\frac{3 \cos \theta}{330}\right]=13 \cos \theta$
$=13\left(\frac{5}{13}\right)=5 Hz \quad\left[\text { From } \triangle CAO , \cos \theta=\frac{5}{13}\right]$
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