On the same path, the source and observer are moving in such a way that the distance between these two…
On the same path, the source and observer are moving in such a way that the distance between these two increases with the time. The speeds of source and observer are same and equal to $10\text{ ms}^{-1}$ with respect to the ground while no wind is blowing. The apparent frequency received by observer is $1950\text{ Hz}$, then the original frequency must be (Take, the speed of sound in present medium is $340\text{ ms}^{-1}$) [AIIMS 2015]
$2068\text{ Hz}$
$2100\text{ Hz}$
$1903\text{ Hz}$
$602\text{ Hz}$
Solution
Let the original frequency be $f$, then apparent frequency,
$f' = \frac{v - u_o}{v + u_s} \times f$
where, $u_o = \text{speed of observer}$,
$u_s = \text{speed of source}$
and $v = \text{speed of sound wave}$.
$\Rightarrow 1950 = \frac{340 - 10}{340 + 10} f$
$\Rightarrow \text{Frequency}, f = \frac{35}{33} \times 1950 = 2068.18\text{ Hz} \approx 2068\text{ Hz}$