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Doppler Effect on a Train

  • Thread starter merlos
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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Well, yeah! The frequency of the whistle of a train approaching you is higher, not lower. You have a sign error.
 
  • #3
Andrew Mason
Science Advisor
Homework Helper
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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
 
  • #4
23
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i dont get it///
 

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