Question
What is the change in frequency when a source goes past a stationary observer with velocity $V_s$? (Given: velocity of sound is c and the frequency is v).
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Answer
When a source approaches a stationary observer,
$\text{v}'=\text{v}\Big(\frac{\text{c}}{\text{c}-\text{v}_\text{s}}\Big)$
When the source goes away,
$\text{v}"=\text{v}\Big(\frac{\text{c}}{\text{c}+\text{v}_\text{s}}\Big)$
$\therefore$ Change when the source goes past stationary observer is,
$\text{v}'-\text{v}"=\text{vc}\Big[\frac{\text{c}}{\text{c}-\text{v}_\text{s}}-\frac{\text{c}}{\text{c}+\text{v}_\text{s}}\Big]$
$=\frac{2\text{vcv}_\text{s}}{(\text{c}^2-\text{v}^2_\text{s})}$
$\text{v}'=\text{v}\Big(\frac{\text{c}}{\text{c}-\text{v}_\text{s}}\Big)$
When the source goes away,
$\text{v}"=\text{v}\Big(\frac{\text{c}}{\text{c}+\text{v}_\text{s}}\Big)$
$\therefore$ Change when the source goes past stationary observer is,
$\text{v}'-\text{v}"=\text{vc}\Big[\frac{\text{c}}{\text{c}-\text{v}_\text{s}}-\frac{\text{c}}{\text{c}+\text{v}_\text{s}}\Big]$
$=\frac{2\text{vcv}_\text{s}}{(\text{c}^2-\text{v}^2_\text{s})}$
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