Answer
Doppler shift in the frequency $\Delta{f}=0$
Work Step by Step
We know that according to the Doppler effect
$f^{-}=f(\frac{v+_-v_D}{v+_-v_S})$
$+,-$ are chosen depending on the direction of motion, if it is directed toward then $+$ is used and if it is directed away then $-$ is used. In our case we will use $-$ sign. $v_D$ represents the detector's speed and $v_S$ stands for source speed. The trooper's speed is the source speed while speeder's speed is the detector's speed.
$v=343\frac{m}{s}$
while $v_D=v_S=1600\frac{km}{h}=\frac{1600\times1000m}{3600s}=44.4\frac{m}{s}$
given that $f=500Hz$
putting the values in the above formula, we get
$f^-=(500)(\frac{343-44.4}{343-44.4})=500Hz$
$f^-=500Hz$
Thus Doppler shift in the frequency $\Delta{f}=0$
