Question

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

Answer 1

Answer: 520.00

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