Answer:
f'= 607.8 Hz
Step-by-step explanation:
This is a Doppler effect exercise due to the relative velocity of the sound source and the observer.
By the time the source and the observer are getting closer the expression is
f ’=
where vs is the speed of the source, vo is the speed of the observer, if the bodies move away the signs are exchanged
in this case, train 1 emits sound, so its speed is v_s = 30.3 m / s and train 2 is the receiver of the sound v₀ = 22.5 m / s
let's calculate
f ’=
520 (343+ 22.5 / 343 - 30.3)
f'= 607.8 Hz