Question

The end of the runway at Princess Juliana International Airport on the Caribbean island of St Maarten is right next to a beach, and watching the planes land just overhead is a popular tourist attraction. In this video posted on Youtube, www.youtube.com/watch?v=zAfQwDizpRo a KLM Boeing 747, prepares to land. It approaches tourists on the beach who hear a high pitched sound from the engines of frequency 1900 Hz. After the jet has passed overhead, the frequency drops to 1020 Hz (I used a frequency analyzer to determine these values directly from the video). Assuming the air temperature that day was 86O F (which is 30.0O C), find the speed of the jet as it prepared to land.

Answer 1

let f is the actual frequency of the engine produced.

f' = 1900 Hz

f'' = 1020 Hz

speed of sound, v_sound = 331 + 0.6*30

= 349 m/s

v_jet = ?

when jet approches,

f' = f*v_sound/(v_sound - v_jet)

1900 = f*349/(349 - v_jet) --------(1)

when jet moves away,

f'' = f*v_sound/(v_sound + v_jet)

1020 = f*349/(349 + v_jet) --------(2)

take equation(1)/equation(2)

1900/1020 = (349 + v_jet)/(349 - v_jet)

1900*(349 - v_jet) = 1020*(349 + v_jet)

1900*349 - 1900*v_jet = 1020*349 + 1020*v_jet

(1900*349 - 1020*349) = v_jet*(1020 + 1900)

v_jet = (1900*349 - 1020*349)/(1020 + 1900)

= 105 m/s