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 $V/V_1 =$

What should be the velocity of a sound source moving towards a stationary observer so that the apparent frequency is double the actual frequency? (Velocity of sound is $v$)

An engine whistling at a constant frequency $n_0$ and moving with a constant velocity goes past a stationary observer. As the engine crosses him,the frequency of the sound heard by him changes by a factor $f$. The actual difference in the frequencies of the sound heard by him before and after the engine crosses him is

$A$ whistle of frequency $500 \, Hz$ tied to the end of a string of length $1.2 \, m$ revolves at $400 \, rev/min$. $A$ listener standing some distance away in the plane of rotation of the whistle hears frequencies in the range (speed of sound $= 340 \, m/s$):

$A$ small source of sound vibrating at a frequency $500 Hz$ is rotated along a circle of radius $\frac{100}{\pi} cm$ at a constant angular speed of $5$ revolutions per second. The minimum and maximum frequency of the sound observed by a listener situated in the plane of the circle is (Speed of sound is $332 ms^{-1}$)

$A$ source of sound $S$ is moving with a velocity $50 \ m/s$ towards a stationary observer. The observer measures the frequency of the source as $1000 \ Hz$. What will be the apparent frequency of the source when it is moving away from the observer after crossing him? The velocity of sound in the medium is $350 \ m/s$.