Question

A source of frequency $n$ and an observer are moving on a straight line with velocity $x$ and $y$ respectively. If the source is ahead of the observer and the medium is also moving in the same direction with velocity $Z$, then the apparent frequency of the sound heard by the observer will be:
($V$ = velocity of sound)

• A. $\frac{V-Z+y}{V-Z+x}n$
• B. $\frac{V+Z+y}{V+Z+x}n$
• C. $\frac{V-Z-y}{V-Z-x}n$
• D. $\frac{V+Z-y}{V+Z-x}n$

Answer 1

The correct option is A $\frac{V-Z+y}{V-Z+x}n$

The distance traveled by the waves relative to the source in $1sec$
$=(V+x-Z)$
Wavelength, $\lambda =\frac{V+x-Z}{n}$
Now, the apparent frequency
$f^{{}^{\prime }}=\frac{V+y-Z}{{\lambda }^{{}^{\prime }}}$
$f^{{}^{\prime }}=\frac{V+y-Z}{V+x-Z}n$
$f^{{}^{\prime }}=\frac{V-Z+y}{V-Z+x}n$