Two electric trains travel in the same direction along the same...

Speed of the trains, $v_s = 72 \text{ km/h} = 72 \times \dfrac{5}{18} = 20 \text{ m/s}$.

The separation between the two trains is $2.4 \text{ km} = 2400 \text{ m}$. Since the person is equidistant from both trains at the instant of whistling, the distance along the track from the person's projection to each train is $\dfrac{2400}{2} = 1200 \text{ m}$.

The perpendicular distance of the person from the track is $500 \text{ m}$.

The straight-line distance from each train to the person is:
$d = \sqrt{1200^2 + 500^2} = 1300 \text{ m}$

Let $\theta$ be the angle between the track and the line joining the train to the person. Then:
$\cos \theta = \dfrac{1200}{1300} = \dfrac{12}{13}$

One train is approaching the person, and the other is moving away. The component of the velocity of the trains along the line of sight to the person is:
$v_{s \parallel} = v_s \cos \theta = 20 \times \dfrac{12}{13} = \dfrac{240}{13} \text{ m/s}$

Using the Doppler effect formula, the apparent frequency of the approaching train is:
$f_1 = f$\left( \dfrac{v}{v - v_{s \parallel}} \right) =$500 \left( \dfrac{340}{340 - \dfrac{240}{13}} \right)<br><br>$f_1 = 500 \left( \dfrac{340 \times 13}{340 \times 13 - 240} \right) =$500 \left( \dfrac{4420}{4180} \right) =$500 \left( \dfrac{221}{209} \right) \approx$528.7$\text{ Hz} \approx$529$\text{ Hz}

The apparent frequency of the receding train is:
$f_2 = f \left( \dfrac{v}{v + v_{s \parallel}} \right) = 500 \left( \dfrac{340}{340 + \dfrac{240}{13}} \right)$

$f_2 = 500 \left( \dfrac{340 \times 13}{340 \times 13 + 240} \right) = 500 \left( \dfrac{4420}{4660} \right) = 500 \left( \dfrac{221}{233} \right) \approx 474.2 \text{ Hz} \approx 474 \text{ Hz}$

Thus, the frequencies heard by the person are $529 \text{ Hz}$ and $474 \text{ Hz}$.