A train travelling at 108 km h^ -1 towards east whistles with a...

Velocity of the train, $v_s = 108 \text{ km h}^{-1} = 108 \times \dfrac{5}{18} = 30 \text{ m/s}$

Velocity of the wind, $v_w = 36 \text{ km h}^{-1} = 36 \times \dfrac{5}{18} = 10 \text{ m/s}$

For the passenger sitting in the train, the relative velocity between the source (train whistle) and the observer (passenger) is zero. Therefore, there is no Doppler shift, and the frequency heard by the passenger is equal to the actual frequency, which is $500 \text{ Hz}$.

For the person standing near the track whom the train has just passed, the train is moving away from the person. The sound travels from the train to the person, which is opposite to the direction of the wind (since the train is moving east and the wind is blowing east, the person is to the west of the train).

Effective speed of sound, $v_{eff} = v - v_w = 340 - 10 = 330 \text{ m/s}$

Since the source is moving away from the stationary observer, the apparent frequency $f'$ is given by:

$f' = f_0 \left( \dfrac{v_{eff}}{v_{eff} + v_s} \right)<br><br>Substituting the values:<br><br>$f' = 500 \left( \dfrac{330}{330 + 30} \right)$<br><br>$f' = 500 \left( \dfrac{330}{360} \right) = 500 \times \dfrac{11}{12} \approx 458.33 \text{ Hz}$<br><br>Thus, the frequency heard by the person near the track is approximately$458 \text{ Hz}$.$