$A$ 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 $36 \ km/h$ towards $S_2$,as shown in the figure. Both the trains are blowing whistles of the same frequency $120 \ Hz$. When $O$ is $600 \ m$ away from $S_2$ and the distance between $S_1$ and $S_2$ is $800 \ m$,the number of beats heard by $O$ is: [Speed of sound $= 330 \ m/s$]

When the source of sound moves away from a stationary listener,we obtain $f_{L} < f_{S}$. Why?

The frequency changes by $10\%$ as a sound source approaches a stationary observer with constant speed $v_s$. What would be the percentage change in frequency as the source recedes from the observer with the same speed? (Given that $v_s < v$,where $v$ is the speed of sound in air)

$A$ train approaching a railway platform with a speed of $20 \ ms^{-1}$ starts blowing the whistle. The speed of sound in air is $340 \ ms^{-1}$. If the frequency of the emitted sound from the whistle is $640 \ Hz$, the frequency of sound to a person standing on the platform will appear to be: (in $Hz$)

Two men are walking along a horizontal straight line in the same direction. The man in front walks at a speed of $1.0 \,m \,s^{-1}$ and the man behind walks at a speed of $2.0 \,m \,s^{-1}$. $A$ third man is standing at a height of $12 \,m$ 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:

$A$ boy is walking away from a wall towards an observer at a speed of $1\, m/s$ and blows a whistle whose frequency is $680\, Hz$. The number of beats heard by the observer per second is (Velocity of sound in air $= 340\, m/s$).