Answer:
f'=5.58kHz
Explanation:
This is an example of the Doppler effect, the formula is:
[tex]f'=\frac{(v+v_o)}{(v+v_s)}f[/tex]
Where f is the actual frequency, [tex]f'[/tex] is the observed frequency, [tex]v[/tex] is the velocity of the sound waves, [tex]v_o[/tex] the velocity of the observer (which is negative if the observer is moving away from the source) and [tex]v_s[/tex] the velocity of the source (which is negative if is moving towards the observer). For this problem:
[tex]f=3.72kHz\\v=342m/s\\v_o=0m/s\\v_s=-114m/s[/tex]
[tex]f'=\frac{(342+0)}{(342-114)}3.72\times10^3\\f'=\frac{342}{228}3.72\times10^3\\f'=(1.5)3.72\times10^3\\f'=5580Hz=5.58kHz[/tex]
