# Doppler Effect on a Train

A train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz.
a. What frequency (fapproach) is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction opposite to the first train and approaching it?

a. fapproach = (v-vL)/(v+vs) * fs

f= [(345m/s-18m/s)/(345m/s+30m/s)]*(262 Hz)
= 228.46 Hz

Wrong

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HallsofIvy
Homework Helper
Well, yeah! The frequency of the whistle of a train approaching you is higher, not lower. You have a sign error.

Andrew Mason
Homework Helper
A train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz.
a. What frequency (fapproach) is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction opposite to the first train and approaching it?

a. fapproach = (v-vL)/(v+vs) * fs

f= [(345m/s-18m/s)/(345m/s+30m/s)]*(262 Hz)
= 228.46 Hz

Wrong

Careful. There are two sets of relative motion here. The source relative to the air and the observer relative to the air. You need to do a doppler calculation for each. As Halls pointed out, the observed frequency should be higher.

AM

i dont get it///