How Does the Doppler Effect Alter the Frequency Heard by the Engineer?

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Homework Help Overview

The discussion revolves around a problem involving the Doppler Effect, specifically how the frequency of a sound emitted by a moving train is perceived by another moving train's engineer. The scenario includes two trains moving in the same direction, with one emitting a whistle at a known frequency while the other is attempting to determine the frequency heard.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the Doppler Effect formula, questioning the correct signs to use based on the relative motion of the source and observer. There are attempts to clarify the conditions under which the frequency changes, including the direction of movement and the speed of sound.

Discussion Status

The discussion is active, with participants providing insights into the correct application of the Doppler Effect equations. There is a focus on clarifying misunderstandings regarding the signs in the formulas and the relative motion of the trains. Some participants acknowledge errors in their previous statements, indicating a collaborative effort to refine their understanding.

Contextual Notes

Participants are navigating through the complexities of the Doppler Effect, particularly in the context of moving sources and observers. There is an emphasis on ensuring that the velocity of sound is always considered in relation to the observer's position, which is a crucial aspect of the problem.

phEight
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Frequency Problem, need urgent help please

Homework Statement



A train moving west at a speed of 30 m/s emits a whistle at a frequency of 348 Hz. On another train behind the first train and moving west at a speed of 20 m/s, the engineer hears the whistle from the first train. If the speed of sound in air is 343 m/s what is the frequency of the sound heard by the second train engineer?

Homework Equations



(I believe this is it)

f' = (1/1+- u/v)f

The Attempt at a Solution



To be honest I've tried my best to figure this out but I'm simply lost on it. Thanks.
 
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f’ = f(v+vo)/(v+vs), where v is the speed of sound.

The observer is moving toward the source in the air frame, and the source is moving away from the observer. Hence, this particular choice of signs.
 
Ah that makes sense, so if they were moving away from each other it would f' = f(v-vo)/(v+vo) ?
 
You are right, excepting your typo. The bottom one should be vs.
 
Ah that makes sense, so if they were moving away from each other it would f' = f(v-vo)/(v+vs) ? This formula is wrong.
The direction of the velocity of sound must be always towards the observer because is the person who listens the apperent change in the frequency. In this question source and observer are moving towards west and velocity of sound is towards east. Therefore
f' = f(v+vo)/(v+vs)
 
These small techinal errors I make! In this example, they are actually moving away from each other, and I have given the correct formula.

For the 2nd case, by "moving away from each other", I meant that when they are moving in opp directions wrt the air.
 
In all such cases we have to consider the relative velocities of sound with respect to source and the observer. If they are moving in the same direction take -ve sign in the general expression which I have written. Other wise put +ve sign. And the direction of the velocity of sound must be always towards the observer.
 
I have not said anything to the contrary. If the observer moves wrt air toward the source, the sign in the numerator is +ve, else –ve. If the source moves wrt air toward the observer, then the sign in the denominator is –ve, else +ve.
 
I am sorry. It is my fault. You have written it correctly.
 

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