Q 11-14-115JEE MainJEE Main 2019 (10 Apr, Shift 2)Easy
A source of sound $S$ is moving with a velocity of $50\ \text{m s}^{-1}$ towards a stationary observer. The observer measures the frequency of the source as $1000\ \text{Hz}$. What will be the apparent frequency of the source when it is moving away from the observer after crossing him? (Take velocity of sound in air is $350\ \text{m s}^{-1}$)
Answer: (A) $750\ \text{Hz}$
Approaching: $1000 = f\,\dfrac{350}{350-50}$, so $f = \dfrac{6000}{7} \approx 857\ \text{Hz}$.
Receding:
$$f' = f\,\frac{350}{350+50} = \frac{6000}{7}\times\frac{7}{8} = 750\ \text{Hz}$$
Solution by Sreeraj P, M.Sc Physics