When the observer moves towards a stationary source with velocity $V_1$, the apparent frequency of emitted note is $F_1$. When 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 $F_1 / F_2=2$ then $V / V_1$ is equal to

Apparent frequency shifts occur due to the Doppler effect when an observer moves relative to a stationary sound source. When the observer approaches the source with velocity $v_1$, the apparent frequency is $F_1 = F \left( \frac{V + v_1}{V} \right)$.

When moving away, the apparent frequency becomes $F_2 = F \left( \frac{V - v_1}{V} \right)$.

Given that $\frac{F_1}{F_2} = 2$, the ratio simplifies to $\frac{F_1}{F_2} = \frac{V + v_1}{V - v_1} = 2$.

Solving for $\frac{V}{v_1}$:
$2(V - v_1) = V + v_1$
$2V - 2v_1 = V + v_1$
$V = 3v_1$
$\frac{V}{v_1} = 3$.

The ratio of the speed of sound to the observer's speed is $\boxed{3}$.