If a source emitting waves of frequency $F$ moves towards an observer with a velocity $V/3$ and the observer moves away from the source with a velocity $V/4$,the apparent frequency as heard by the observer will be ($V=$ velocity of sound).

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:

$A$ source of sound is moving with a constant velocity of $20\, m/s$ emitting a note of frequency $1000\, Hz.$ The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him will be (Speed of sound $v = 340\, m/s$).

Two factories are sounding their sirens at $800 \, Hz$. $A$ man walks from one factory to the other at a speed of $2 \, m/s$. The velocity of sound is $320 \, m/s$. The number of beats heard by the person in one second will be:

$A$ person is observing two trains,one coming towards him and the other leaving with the same speed $4\, m/s$. If their whistling frequencies are $240\, Hz$ each,then the number of beats per second heard by the person will be: (if the velocity of sound is $320\, m/s$)

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)