Q 11-14-121JEE MainJEE Main 2019 (9 Apr, Shift 2)Easy
Two cars A and B are moving away from each other in opposite directions. Both the cars are moving with speed of $20\ \text{m s}^{-1}$ with respect to the ground. If an observer in car A detects a frequency $2000\ \text{Hz}$ of the sound coming from car B, what is the natural frequency of the sound source in car B? (speed of sound in air $= 340\ \text{m s}^{-1}$)
Answer: (C) $2250\ \text{Hz}$
Observer moving away from the source and source moving away from the observer:
$$f' = f\,\frac{v - v_o}{v + v_s} = f\,\frac{340 - 20}{340 + 20} = \frac{8}{9}f$$
$$f = \frac98\times2000 = 2250\ \text{Hz}$$
Solution by Sreeraj P, M.Sc Physics