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Doppler homework problem

  • Thread starter pkossak
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  • #1
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I have been working on this problem for over a half hour. I'm not sure where I'm going wrong with it. If anyone could help me out I would really appreciate it.

You stand by the railroad tracks as a train passes by. You hear a 1000 Hz frequency when the train approaches, which changes to 800 HZ as it goes away. How fast is the train moving? The speed of sound in air is 340 m/s.

I have been trying fo = f*Vsound/Vsound +- Vs
and then trying to set them equal, but it isn't working for some reason :confused: Any ideas? Thanks a lot!
 

Answers and Replies

  • #2
Fermat
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  • #3
ehild
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pkossak said:
You stand by the railroad tracks as a train passes by. You hear a 1000 Hz frequency when the train approaches, which changes to 800 HZ as it goes away. How fast is the train moving? The speed of sound in air is 340 m/s.

I have been trying fo = f*Vsound/Vsound +- Vs
and then trying to set them equal, but it isn't working for some reason
The speed of sound is the same 340 m/s independently of the motion of the train.

When I was a child, my father explained Doppler's effect with an example:
Imagine a ship approaching towards the shore slowly with velocity v, and a cage full of doves on it. Somehow the birds find a small opening on the cage and they can get out one by one, one bird in To seconds. After getting out, each dove flies towards the shore with the same speed c.
There is a man on the shore watching. How often does he see a bird arriving to the shore?

We have to find out the time elapsed till the arrival of the next bird after the previous one. For that we need their distance D.
When a bird gets free, the previous one travelled the distance cTo from the original position of the ship, but the ship travelled vTo. So the distance between those birds is D=(c-v)To When the first bird arrives at the shore, the next one is at a distance of D meters, and arrives after T = (1-v/c)To seconds.

The frequency of the doves getting out is fo=1/To. The frequency the man observes them is f=1/T=fo/(1-v/c), higher than the original frequency.

If the ships moves away from the shore, the velocity is negative, so the frequency decreases.

You have a train instead of a ship, and the doves represent the crests of the sound wave of emitted by the train The distance between the doves corresponds to the wavelength of the sound. The speed of sound is constant, but both the wavelength and frequency changes when the source of sounds moves.

ehild
 
  • #4
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Thank you very much for your help, both of you. The website and the analogy let me visualize this a little better.
 

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