When both the source of sound and the observer approach each other with a speed equal to $10 \%$ of the speed of sound,then the percentage change in frequency heard by the observer is nearly (in $\%$)

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).

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 the frequency as the source recedes from the observer with the same speed $\left(V_s < V\right)$ (in $.5$)?

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 $V/V_1 =$

An observer moves towards a stationary source of sound with a speed $\frac{1}{5}$th that of sound. The frequency of the sound emitted by the source is $f$. The apparent frequency recorded by the observer is

$A$ train,standing in a station-yard,blows a whistle of frequency $400\; Hz$ in still air. The wind starts blowing in the direction from the yard to the station with a speed of $10\; m s^{-1}$. What are the frequency,wavelength,and speed of sound for an observer standing on the station's platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of $10\; m s^{-1}$? The speed of sound in still air can be taken as $340\; m s^{-1}$.