Doppler effect problem

  • Thread starter nickb145
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  • #1
nickb145
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Homework Statement



A parachutist leaps from a hovering helicopter, after 4.0s of freefall shouts back at 425Hz. What frequeny is heard at the helicopter?

Homework Equations



f'=f/(1+Vs/V)


Vsource=?
f'=?
f=425Hz
V=343m/s

The Attempt at a Solution




I'm stuck on finding the speed for the jumper. I'm sure it's simple but i just can't think of it for somereason.
 
Last edited:

Answers and Replies

  • #2
ehild
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Homework Statement



A parachutist leaps from a hovering helicopter, after 4.0s of freefall shouts back at 425Hz. What frequeny is heard at the helicopter?

Homework Equations



f'=f/(1-Vs/V)


Vsource=?
f'=?
f=425Hz
V=343m/s

The Attempt at a Solution




I'm stuck on finding the speed for the jumper. I'm sure it's simple but i just can't think of it for somereason.


Read the problem carefully. It is free-fall, the parachute is not yet opened . What is the speed of a falling body after 4 s?

ehild
 
  • #3
nickb145
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thinking
 
Last edited:
  • #4
nickb145
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Read the problem carefully. It is free-fall, the parachute is not yet opened . What is the speed of a falling body after 4 s?

ehild


vy=Vyi-gt right?

initial speed is 0 so its just -9.8*4=-39.2
 
  • #5
ehild
Homework Helper
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Yes, it is the velocity if you consider upward positive.
The parachuter shouts back. How is his sound heard on the helicopter? (Hm. The helicopter is very noisy --think it is a super helicopter making only low noise so the parachuter's sound can be heard :wink:)


ehild
 
  • #6
nickb145
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But it is strange that i get a higher frequency of what it should be. The answer is 381.4 and i keep getting 479. I'm getting my signs wrong somewhere...

I used f'=f/(1+Vs/V) for a source moving away from reciever.
 
  • #7
ehild
Homework Helper
15,543
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You need to calculate with the speed of moving away. The speed is positive, magnitude of velocity. Substitute Vs=39.2 m/s.

ehild
 

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