$A$ source with a frequency of $90 \, \text{vibrations/sec}$ moves towards a stationary observer with a velocity of $1/10 \, V$, where $V$ is the speed of sound. What frequency will the observer hear in $\text{vibrations/sec}$?

$A$ source producing sound of frequency $170 \, Hz$ is approaching a stationary observer with a velocity $17 \, ms^{-1}$. The apparent change in the wavelength of sound heard by the observer is (speed of sound in air $= 340 \, ms^{-1}$) ..... $m$

When a source of sound crosses a stationary observer,the change in apparent frequency of sound observed by the observer,when $V_s << V$,will be -

An observer standing at a station observes a frequency of $219 \, Hz$ when a train approaches and $184 \, Hz$ when the train goes away from him. If the velocity of sound in air is $340 \, m/s$,then the velocity of the train and the actual frequency of the whistle will be:

$A$ source of sound of frequency $640 \,Hz$ is moving at a velocity of $\frac{100}{3} \,m/s$ along a road, and is at an instant $30 \,m$ away from a point $A$ on the road. $A$ person standing at $O$, $40 \,m$ away from the road, hears sound of apparent frequency $v^{\prime}$. The value of $v^{\prime}$ is (velocity of sound $= 340 \,m/s$): (in $\,Hz$)

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).