Answer:
A. 468 Hz while approaching, 440 Hz through the crossing, 415 Hz while receding
Explanation:
When bus is approaching the stand where boy is standing then the frequency is given as
now we know that
[tex]f_0 = 440 Hz[/tex]
[tex]v = 330 m/s[/tex]
[tex]v_b = 20 m/s[/tex]
frequency is given by
[tex]f_1 = f_0 \frac{v}{v - v_b}[/tex]
now plug in the data in the equation
[tex]f_1 = 440 \times \frac{330}{330 - 20}[/tex]
[tex]f_1 = 468 Hz[/tex]
Now when bus is moving away from the bus stop then frequency is given as
[tex]f_2 = f_0 \frac{v}{v + v_b}[/tex]
[tex]f_2 = 440 \times \frac{330}{330 + 20}[/tex]
[tex]f_2 = 415 Hz[/tex]
so here correct answer would be
A. 468 Hz while approaching, 440 Hz through the crossing, 415 Hz while receding