An alert physics student stands beside the tracks as a train rolls slowly past.

Question

An alert physics student stands beside the tracks as a train rolls slowly past. He notes that the frequency of the train whistle is 459 Hz when the train is approaching him and 434 Hz when the train is receding from him. Using these frequencies, he calculates the speed of the train. What value does he find? The speed of sound in air is 343m/s. An alert physics student stands beside the tracks as a train rolls slowly past. He notes that the frequency of the train whistle is 459 Hz when the train is approaching him and 434 Hz when the train is receding from him. Using these frequencies, he calculates the speed of the train. What value does he find? The speed of sound in air is 343m/s. An alert physics student stands beside the tracks as a train rolls slowly past. He notes that the frequency of the train whistle is 459 Hz when the train is approaching him and 434 Hz when the train is receding from him. Using these frequencies, he calculates the speed of the train. What value does he find? The speed of sound in air is 343m/s.

Explanation / Answer

apparant freq n' = n[v-vo]/[v-vs]
vo = speed of student =0
v = 343 m/s
vs = train speed
n = real freq of train whistle

here v, vo, vs should be in same direction

approaching train>> v=o (at origin), vs = +vs (moving in +x)
v = +343 (sound reaching student at origin)
n' = 459 = v/(v-vs) = 343*n/(343-vs)

train is receding away from origin
vo =0 (origin), vs = + vs (+x) still, v = - 343 (now sound is coming back to origin so made -ve so that all in one direction
n' = 434 = (-v) /[(-v) - vs] = 343*n/(343+vs)
divide
(343+vs) / (343 - vs) = 459/434
343*434 + 434 vs = 459*343 - 459vs
893 vs = 343[459-434] = 8575
vs = train = 9.6 m/s

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