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

In summary, the problem involves a parachutist shouting back at a frequency of 425Hz while in freefall after 4.0s. The question asks for the frequency heard at the hovering helicopter. Using the equation f'=f/(1+Vs/V), where Vs is the speed of the source (parachutist) moving away from the receiver (helicopter), the speed of the parachutist after 4 seconds is calculated to be 39.2 m/s. Therefore, the frequency heard at the helicopter is 381.4Hz.
  • #1
nickb145
68
0

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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  • #2
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
 
  • #3
thinking
 
Last edited:
  • #4
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
 
  • #5
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
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
You need to calculate with the speed of moving away. The speed is positive, magnitude of velocity. Substitute Vs=39.2 m/s.

ehild
 

1. What is the Doppler effect?

The Doppler effect is a phenomenon observed when there is a change in frequency of a wave (such as sound or light) due to relative motion between the source and the observer.

2. How does the Doppler effect work?

The Doppler effect works by changing the perceived frequency of a wave as the source or observer moves towards or away from each other.

3. What causes the Doppler effect?

The Doppler effect is caused by the relative motion between the source of a wave and the observer. This could be due to the source or observer moving towards or away from each other, or a combination of both.

4. What are some real-life examples of the Doppler effect?

Some real-life examples of the Doppler effect include the change in pitch of an ambulance siren as it passes by, the change in frequency of a train horn as it approaches and then passes by, and the redshift and blueshift of light from stars and galaxies due to their relative motion towards or away from us.

5. How is the Doppler effect used in science and technology?

The Doppler effect is used in various scientific and technological applications, such as in weather radar to detect the movement of storms, in medical ultrasound to measure blood flow, in astronomy to determine the speed and direction of objects in space, and in police radar to measure the speed of moving vehicles.

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