Determining the Frequency Heard by a Helicopter from a Parachutist's Shout

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SUMMARY

The discussion centers on calculating the frequency of sound heard by a helicopter from a parachutist's shout during free fall. The original frequency of the shout is 425Hz, and the speed of sound is 343m/s. After 4 seconds of free fall, the parachutist reaches a velocity of 39.2 m/s downward. The correct formula to use is f' = f / (1 + Vs/V), leading to a calculated frequency of approximately 381.4Hz at the helicopter, contrasting with an incorrect calculation of 479Hz due to sign errors in velocity.

PREREQUISITES
  • Understanding of the Doppler effect in sound waves
  • Basic kinematics, specifically free fall equations
  • Familiarity with sound speed in air (343 m/s)
  • Ability to manipulate algebraic equations for frequency calculations
NEXT STEPS
  • Study the Doppler effect and its applications in different scenarios
  • Learn about free fall motion and its equations in physics
  • Explore sound wave properties and how they interact with moving sources
  • Practice solving problems involving frequency shifts due to relative motion
USEFUL FOR

Physics students, educators, and anyone interested in understanding sound frequency changes due to motion, particularly in contexts involving free fall and the Doppler effect.

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:
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nickb145 said:

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
 
thinking
 
Last edited:
ehild said:
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
 
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
 
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.
 
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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