$A$ source of sound of frequency $90 \text{ vibrations/sec}$ is approaching a stationary observer with a speed equal to $1/10$ the speed of sound. What will be the frequency heard by the observer in $\text{vibrations/sec}$?

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 speed of sound in air and $F_1/F_2 = 2$, then $V/V_1 =$

$A$ train blowing the whistle moves with a constant velocity $V$ away from an observer standing on the platform. The ratio of the natural frequency of the whistle $n$ to the apparent frequency is $1.2: 1$. If the train is at rest and the observer moves away from it at the same velocity $V$,the ratio of $n$ to the apparent frequency is (in $: 1$)

$A$ source of sound emits a sound wave of frequency '$f$' and moves towards an observer with a velocity $\frac{V}{3}$,where $V$ is the velocity of sound. If the observer moves away from the source with a velocity $\frac{V}{5}$,the apparent frequency heard by him will be:

An observer and a source emitting sound of frequency $120 \,Hz$ are on the $X$-axis. The observer is stationary while the source of sound is in motion given by the equation $x=3 \sin \omega t$ (where $x$ is in metres and $t$ is in seconds). If the difference between the maximum and minimum frequencies of the sound observed by the observer is $22 \,Hz$,then the value of $\omega$ is (speed of sound in air $=330 \,ms^{-1}$):

$A$ train moves towards a stationary observer with a speed of $34 \ m/s$. The train sounds a whistle and its frequency registered by the observer is $f_1$. If the train's speed is reduced to $17 \ m/s$,the frequency registered is $f_2$. If the speed of sound is $340 \ m/s$,then the ratio $f_1/f_2$ is: