Question

2 trains with 132Hz horns approach one another. The slower train has a speed of 25m/s. what is the speed of the fast train if an observer standing near the tracks between the trains hears a beat frequency of 5.2 Hz?

Answer 1

Fs = ((V + Vr) / (V + Vs))*Vo,
Fs = ((343 + O) / (343 + 25))132,
Fs = (343 / 368)132 = 123Hz = Frequency of slower train's horn,

Ff = 123 - 5.2 = 117.8 = Freq. of fast
train's horn.

Ff=((343 + 0) / (343 + Vs))132=117.8Hz.
45276 / (343 + Vs) = 117.8,
117.8(343 + Vs) = 45276,
343 + Vs = 45276 / 117.8 = 384.35,
Vs = 384.35 - 343 = 41.4m/s = Velocity
of suurce(fast train).











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