Question

A jet fighter flies in a straight line over an airfield line over an airfield with a speed $264\, m / s$ level with the ground, at a height of $600\, m$. If it emits a continuous note of frequency $1 \,k H z$, what is the change in apparent frequency (in $Hz$ ) as experienced by one of the ground crew over a $1.5\, s$ interval as the jet flies overhead? (Assume speed of sound is $340 \,m s^{-1}$ )

Answer 1

$\cos \theta=\frac{198}{\sqrt{399204}}=0.31337$
The frequency heard by crew in initial position
$v' =v\left(\frac{v}{v-v_{ S } \cos \theta}\right) $
$=(1)\left(\frac{340}{340-264 \times 0.31337}\right) $
$=1.321\, KHz$
Frequency heard by crew in final position
$v'' =v\left(\frac{v}{v+ v_{ S } \cos \theta}\right) $
$=(1)\left(\frac{340}{340+264 \times 0.31337}\right) $
$=0.804 \,KHz$
Change in apparent frequency
$=v''-v' $
$=1.321-0.804=0.517\, KHz $
$\approx 520 \,Hz$