$A$ source emits a sound of frequency $400 \ Hz$,but the listener hears it to be $390 \ Hz$. Then:

When an engine passes near a stationary observer,its apparent frequencies occur in the ratio $5/3$. If the velocity of sound is $340 \ m/s$,then the velocity of the engine is .... $m/s$.

When the observer moves towards the 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 $\frac{F_1}{F_2}=2$,then $\frac{V}{V_1}=?$

Two sources $A$ and $B$ are producing notes of frequency $680 \,Hz$. $A$ listener moves from $A$ to $B$ with a constant velocity $v$. If the speed of sound in air is $340 \,ms^{-1}$, the value of $v$ so that he hears $10$ beats per second is: (in $\,ms^{-1}$)

With what velocity an observer should move relative to a stationary source so that a sound of double the frequency of the source is heard by the observer?

$A$ stationary sound source $S$ of frequency $334 \, Hz$ and a stationary observer $O$ are placed near a reflecting surface moving away from the source with velocity $u = 2 \, m/s$ as shown in the figure. If the velocity of the sound waves in air is $V = 330 \, m/s$,the apparent frequency of the echo heard by the observer is ... $Hz$.