An engine moving towards a wall with a velocity $50 \ m/s$ emits a note of $1.2 \ kHz$. The speed of sound in air is $350 \ m/s$. The frequency of the note after reflection from the wall as heard by the driver of the engine is: (in $kHz$)

$A$ bus driving along at $39.6 \,km/h$ is approaching a person who is standing at the bus stop,while honking repeatedly at an interval of $30 \,s$. If the speed of sound is $330 \,ms^{-1}$,at what interval will the person hear the horn?

With what velocity should an observer approach a stationary sound source,so that the apparent frequency of sound appears double the actual frequency? (Let $v$ be the velocity of sound).

When the observer moves towards a stationary source with velocity $V_1$,the apparent frequency of the emitted note is $F_1$. When the observer moves away from the source with velocity $V_1$,the apparent frequency is $F_2$. If $V$ is the velocity of sound in air and $F_1 / F_2 = 2$,then $V / V_1$ is equal to:

$A$ motorcyclist is approaching a large wall with a velocity of $90\,km/hr$. $A$ car is chasing him with a velocity of $108\,km/hr$. If the car sounds a horn at $30\,Hz,$ the beat frequency heard by the motorcyclist is . . . . . . $Hz$. Take the velocity of sound $= 330\,m/s$.

$A$ train blowing its whistle moves with constant speed on a straight track towards an observer and then crosses him. If the ratio of the difference between the actual and apparent frequencies is $3:2$ in the two cases (approaching and receding),then the speed of the train is (where $v$ is the speed of sound).