How Does the Doppler Effect Alter Perceived Train Whistle Frequencies?

[Coushander]

Homework Statement

A woman is riding a bicycle at 18.0 m/s along a straight road that runs parallel to and right next to some railroad tracks. She hears the whistle of a train that is behind. The frequency emitted by the train is 840 Hz, but the frequency the woman hears is 778 Hz. Take the speed of sound to be 340 m/s.
(a) What is the speed of the train?
(b) What frequency is heard by a stationary observer located between the train and the bicycle?
(c) Is the train traveling away from or toward the bicycle?

Homework Equations

Moving Source

frequency prime = initial frequency / 1 +/- source speed/wave speed

+ for receding - for approaching

Moving observer

frequency prime = frequency/ 1+/- observer speed/wave speed

wave speed = speed of sound (343 m/s)

The Attempt at a Solution

I tried setting frequency primes equal, but that didn't work out. I know you have to combne both equations because both the source and the observer are moving. The train is moving away, but I got that right on a guess so I'd like to know how to tell which way it's moving mathematically.

 

[vela]

Homework Equations

Moving Source

frequency prime = initial frequency / 1 +/- source speed/wave speed

+ for receding - for approaching

Moving observer

frequency prime = frequency/ 1+/- observer speed/wave speed

One of those equations is wrong. You should also be using parentheses to group the terms in the denominator, i.e., f' = f/(1±v/c).

 

[Coushander]

Moving Source

frequency prime = (initial frequency)/(1 ± source speed/wave speed)

+ for receding - for approaching

Moving observer

frequency prime = (initial frequency)(1 ± observer speed/wave speed)

+ for approaching - for receding

I fixed them.

I also had the thought that the frequency prime of the moving source would be equal to the the initial frequency of the moving observer. Am I on the right track with this?

 

[ehild]

You are on the right track.

Look at this page.

http://en.wikipedia.org/wiki/Doppler_effect
ehild

 

[Coushander]

778Hz = {(840Hz)/ 1 ± (train speed/343m/s)}{1 - (18m/s)/(343m/s)}

821.09Hz = (840Hz)/ 1 ± (train speed/343m/s)

1 ± train speed/343 m/s = 1.023

343 m/s ± train speed = 350.8997429 m/s

train speed = 7.899742931 m/s

train speed = 7.9 m/s

The answer is apparently meant to be 7.67 m/s.

Did I do something wrong or is that just a rounding error (I used all raw values)?

 

[vela]

Use 340 m/s for the speed of sound.

 

[Coushander]

Ah I feel like a fool. We usually use 343 in the textbook.

 

[Coushander]

I'd guess that I find that the train is receding by solving the moving observer question for the "initial frequency" and then placing that value in the moving source equation as "frequency prime".

And the frequency heard by someone between the two would be that frequency prime value?

 

[vela]

Yes, f' would be the frequency the observer at rest hears.

Without doing any calculations, you should be able to determine the train is receding just by comparing the frequency of the whistle to what she hears.

 

[PeterO]

Homework Statement

A woman is riding a bicycle at 18.0 m/s along a straight road that runs parallel to and right next to some railroad tracks. She hears the whistle of a train that is behind. The frequency emitted by the train is 840 Hz, but the frequency the woman hears is 778 Hz. Take the speed of sound to be 340 m/s.
(a) What is the speed of the train?
(b) What frequency is heard by a stationary observer located between the train and the bicycle?
(c) Is the train traveling away from or toward the bicycle?

Homework Equations

Moving Source

frequency prime = initial frequency / 1 +/- source speed/wave speed

+ for receding - for approaching

Moving observer

frequency prime = frequency/ 1+/- observer speed/wave speed

wave speed = speed of sound (343 m/s)

The Attempt at a Solution

I tried setting frequency primes equal, but that didn't work out. I know you have to combne both equations because both the source and the observer are moving. The train is moving away, but I got that right on a guess so I'd like to know how to tell which way it's moving mathematically.

I notice that you get an answer of 7.9 - which is probably the 7.67 when v = 340.

I am intrigued by the wording of part c.

Certainly the train is receding from the woman on the bicycle [who is riding really fast], but it seems it is still traveling in the same direction as the woman. ?
So would you call that traveling away from the woman or towards the woman.

eg: If I contested a 100m sprint against Usain Bolt, we would both be running towards the finish line, but I would certainly be receding from him.
We see many examples of sporting people running towards another player, but not actually managing to get any closer to them.

 

[ehild]

There is a third party - the air. The woman travels with 18 m/s with respect to the air. The train travels by -7.9 m/s with respect to the air, so in the opposite direction as the woman. The women experiences the wavefronts coming from the train farther away as they would be from a stationary source.

(By the way, the speed are quite weird, it should be on the opposite way)

ehild

 

[PeterO]

There is a third party - the air. The woman travels with 18 m/s with respect to the air. The train travels by -7.9 m/s with respect to the air, so in the opposite direction as the woman. The women experiences the wavefronts coming from the train farther away as they would be from a stationary source.

(By the way, the speed are quite weird, it should be on the opposite way)

ehild

OK - I hadn't been following the calculations completely. If the train is traveling at -7.9 m/s [that is at a speed of 25.9 relative to her] I have no question.

BUT
Had the frequency shift been smaller that that given here, so that the the train was going the same direction as the woman in the bicycle, but slower, would you say the train was going towards or away from her?

 

[ehild]

The velocities are defined with respect to air. When the train goes in the same direction as the woman the wavefronts are denser in front of the train, the frequency of the sound is f(train)=fo/(1-v(train)/c) at that side where the women is.
The women hears lower frequency, as he goes in the direction away from the wavefronts, so they reach her at longer time intervals:
f=f(train)(1-v(women)).

ehild

 

[PeterO]

The velocities are defined with respect to air. When the train goes in the same direction as the woman the wavefronts are denser in front of the train, the frequency of the sound is f(train)=fo/(1-v(train)/c) at that side where the women is.
The women hears lower frequency, as he goes in the direction away from the wavefronts, so they reach her at longer time intervals:
f=f(train)(1-v(women)).

ehild

I am aware of all that, I was just intrigued with the wording of part (c) from the OP.

(c) Is the train traveling away from or toward the bicycle?

I was querying how you may answer that question if the Train and bicycle were traveling in the same direction, but the train was traveling slower than the bicycle.

 

[ehild]

I would say the train travels toward the bicycle.

ehild

 

[PeterO]

I would say the train travels toward the bicycle.

ehild

I would agree with you - it just seems odd to have a train traveling towards you, but with a downward doppler shift yo its siren.
Sort of seems to make the doppler shift less black and white than I always expected.

 

[ehild]

If the train travels toward you it sends a higher sound in your direction. If you did not move you would hear higher sound. If you travel away from the train the sound becomes deeper. When both you and the train move with equal velocity, you hear the real sound.

ehild

 

[Coushander]

The train is traveling away from the woman on the bike, which gives an altered frequency because of the Doppler effect. This altered frequency is the initial frequency that is altered by her movement away from the train (so it gets altered again).

It's probably a bit more complex than that, but I've only got a beginner's logic on it.

 

[ehild]

The train is traveling away from the woman on the bike, which gives an altered frequency because of the Doppler effect. This altered frequency is the initial frequency that is altered by her movement away from the train (so it gets altered again).

It's probably a bit more complex than that, but I've only got a beginner's logic on it.

You are right, it happens that way. Women and train move in opposite directions with respect to air. PeterO is confused about what happens when the train goes in the same direction as the women but with smaller speed.
Here is a nice picture from Wikipedia http://en.wikipedia.org/wiki/Doppler_effect showing how the wavelength gets shorter in front of a moving source and longer behind it.

ehild