At what frequency does a stationary listener hear the sound 1) as the plane approaches and 2) after it passes?I'm stumped as to how to solve this....

The speed at which the sound travels remains same
irrespective of the speed of an object, like an aeroplane, making the sound. Therefore,
because of a phenomenon called Doppler effect the frequency of sound made by an object
approaching a listener increases, while that be a receding object
decreases.

The relation between the observed frequency of
sound (f) and the emitted frequency (fo) is given by:

f =
[v/(v + vs)]*fo

Where v is velocity of sound and vs is the
velocity of source respective to the medium.

Also value of
fo is given as 4.66 kHz.

It is given that speed of plane is
half that of speed.

Therefore when the plane approaches the
observer, it is like the intervening medium between the observer and the plane moving
towards the observer making vs= -v/2

Therefore frequency of
sound when the plane approaching the observer is given
by:

f = [v/(v - v/2)]*fo = 2*fo = 2*4.66 = 9.32
kHz.

Similarly, after the plane passes the observer, it is
like the intervening medium between the observer and the plane moving towards the
observer making vs= v/2

Therefore frequency of sound after
the plane passes the observer is given by:

f = [v/(v +
v/2)]*fo = 2/3*fo = 2/3*4.66 = 3.10666667   kHz.