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

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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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