Travelling along the Y-axis with a speed of 22 m s^ -1 , a source...

Let the source cross the origin at $t = 0$. The sound detected by the listener at $t = 0$ was emitted by the source at an earlier time $t = -t_0$.

During the time interval $t_0$, the sound travels a distance $d = v t_0$, where $v$ is the speed of sound. At $t = -t_0$, the source was at a distance $y = v_s t_0$ from the origin along the Y-axis.

The distance from the source at the time of emission to the listener at $(660, 0)$ is $d = \sqrt{660^2 + y^2}$. Thus, we have $d = v t_0$.

The component of the source's velocity along the line of sight towards the listener is $v_s \cos \theta$, where $\theta$ is the angle between the source's velocity and the direction to the listener. From the geometry of the setup, we have:

$\cos \theta = \dfrac{y}{d} = \dfrac{v_s t_0}{v t_0} = \dfrac{v_s}{v}$

The apparent frequency $f$ detected by the listener is given by the Doppler effect formula:

$f = f_0 \left( \dfrac{v}{v - v_s \cos \theta} \right)<br><br>Substituting$\cos $\theta = \dfrac{v_s}{v} into the formula, we get:<br><br>$f = f_0 \left( \dfrac{v}{v - \dfrac{v_s^2}{v}} \right) = f_0 \left( \dfrac{v^2}{v^2 - v_s^2} \right)$<br><br>Given$f_0 = 4000 \text{ Hz}$,$v = 330 \text{ m s}^{-1}$, and$v_s = 22 \text{ m s}^{-1}$, we can substitute these values:<br><br>$f = 4000 \left( \dfrac{330^2}{330^2 - 22^2} \right)

Dividing the numerator and the denominator by $22^2$, we get:

$f = 4000 \left( \dfrac{15^2}{15^2 - 1^2} \right) = 4000 \left( \dfrac{225}{224} \right)$

$f \approx 4017.86 \text{ Hz}$

Rounding to the nearest integer, the frequency detected by the listener is $4018 \text{ Hz}$.

Answer: $4018 \text{ Hz}$