Answer:
(a) 9.36 kHz
(b) 3.12 kHz
Explanation:
(a)
V = speed of sound
[tex]v[/tex] = speed of airplane = (0.5) V
f = actual frequency of sound emitted by airplane = 4.68 kHz = 4680 Hz
f' = Frequency heard by the stationary listener
Using Doppler's effect
[tex]f' = \frac{Vf}{V-v}[/tex]
[tex]f' = \frac{V(4680)}{V-(0.5)V)}[/tex]
f' = 9360 Hz
f' = 9.36 kHz
(b)
V = speed of sound
[tex]v[/tex] = speed of airplane = (0.5) V
f = actual frequency of sound emitted by airplane = 4.68 kHz = 4680 Hz
f' = Frequency heard by the stationary listener
Using Doppler's effect
[tex]f' = \frac{Vf}{V+v}[/tex]
[tex]f' = \frac{V(4680)}{V+(0.5)V)}[/tex]
f' = 3120 Hz
f' = 3.12 kHz