Travelling along a straight line at a speed of 170 m s^ -1 , a...

Let the source move along the x-axis and the detector be at $(0, d)$.

The source reaches the closest point to the detector at $t = 0$. The sound received by the detector at this instant was emitted by the source at an earlier time $-t_0$.

At time $-t_0$, the position of the source was $(-v_s t_0, 0)$. The distance between the source at the time of emission and the detector is $D = \sqrt{(-v_s t_0)^2 + d^2}$.

The time taken for the sound to travel this distance is $t_0$, so $D = v t_0$.

From the geometry, the angle $\theta$ between the velocity of the source and the line of sight to the detector satisfies $\cos \theta = \dfrac{v_s t_0}{D} = \dfrac{v_s t_0}{v t_0} = \dfrac{v_s}{v}$.

The apparent frequency received by the detector is given by the Doppler effect formula:

$f = f_0 \dfrac{v}{v - v_s \cos \theta}$

Substituting $\cos \theta = \dfrac{v_s}{v}$:

$f = f_0 \dfrac{v}{v - v_s \left(\dfrac{v_s}{v}\right)} = f_0 \dfrac{v^2}{v^2 - v_s^2}$

Given $v = 340 \text{ m s}^{-1}$, $v_s = 170 \text{ m s}^{-1} = \dfrac{v}{2}$, and $f_0 = 1200 \text{ Hz}$:

$f = 1200 \times \dfrac{v^2}{v^2 - \left(\dfrac{v}{2}\right)^2} = 1200 \times \dfrac{1}{1 - \dfrac{1}{4}} = 1200 \times \dfrac{4}{3} = 1600 \text{ Hz}$