A person observes two moving trains. First reaching the station and another leaves the station with equal speed of 30 ~m / s . If both trains emit sounds of frequency 300 Hz , difference of frequencies heard by the person will be (speed of sound in air =330 ~m / s )

Step-by-step solution

The Doppler effect governs the frequency shift for a moving source relative to a stationary observer. For a source approaching the observer, the observed frequency is f₁ = f₀ v v - v_s , while for a receding source, it is f₂ = f₀ v v + v_s , where f₀ = 300 , Hz , v = 330 , m/s , and v_s = 30 , m/s . For the approaching train: f₁ = 300 330 330 - 30 = 300 330 300 = 330 , Hz . For the receding train: f₂ = 300 330 330 + 30 = 300 330 360 = 275 , Hz . The difference in frequencies is |f₁ - f₂| = |330 - 275| = 55 , Hz . F