A person observes two moving trains, $$A$$ reaching the station and $$B$$ leaving the station with equal speed of $$30$$ m s$$^{-1}$$. If both trains emit sounds with frequency $$300$$ Hz, (Speed of sound: $$330$$ m s$$^{-1}$$) approximate difference of frequencies heard by the person will be:

Train A approaches the station and Train B leaves the station, both at speed $$v_s = 30$$ m/s. Both emit sound at frequency $$f_0 = 300$$ Hz and the speed of sound is $$v = 330$$ m/s. We wish to find the approximate difference in frequencies heard by the person at the station.

The source (Train A) is moving toward the observer. Using the Doppler formula: $$f_A = f_0 \times \frac{v}{v - v_s} = 300 \times \frac{330}{330 - 30} = 300 \times \frac{330}{300} = 330 \text{ Hz}$$

The source (Train B) is moving away from the observer: $$f_B = f_0 \times \frac{v}{v + v_s} = 300 \times \frac{330}{330 + 30} = 300 \times \frac{330}{360} = 275 \text{ Hz}$$

$$\Delta f = f_A - f_B = 330 - 275 = 55 \text{ Hz}$$ Answer: Option B (55 Hz)