Frequency of a falling persons scream

[IanIanIan]

Homework Statement

Someone is free falling off a cliff. 20 m below the clifftop they scream with a frequency f. a) What does a person standing at the top of the cliff hear? b) what will they hear when the person is 40 m below the cliff?

Homework Equations

f'=f(v+-v_d)/(v-+v_s)

The Attempt at a Solution

I have no problem whatsoever with part a, but our professor gave us part b to think about over the weekend and said it would likely be on our midterm that's coming up. I looked at it and at first thought it easy as well, but he pointed out that they started screaming at 20 m below, not at 40 m. So they have already been screaming for the last 20 m. I can't fathom what this does to the frequency though. Does anyone have any idea where to start on something like this?

 

[harts]

The person is falling at a faster speed at 40 meters than at 20 meters, because of gravity.

 

[IanIanIan]

Yeah but my professor said it was a very tricky problem, and i feel like that is too obvious :S

 

[ehild]

The speed of the sound is about 340 m/s. The sound emitted at distance h needs t=h/340 m time to reach the observer. When the falling person is 40 m deep, the observer on the cliff top hears his earlier scream.

ehild

 

[IanIanIan]

I thought of that too, but then wouldn't the answer to part A be the observer hears no sound since it technically hasn't traveled to him yet? So I feel like that isn't it. We took part A up in class so I know that's it.

I know the whole nature of the doppler equation comes from stuff moving away/towards each other at constant velocities, could the fact that the source is now accelerating make this different? But then what makes it different from Part A? This is why I posted here I am at a total loss of where to start on this.

 

[ehild]

Question a) does not refer to the time. It asks the frequency heard from the first scream - at any time.
Question b) asks the frequency of the scream heard at that time instant when the falling man is at 40 m depth. It is at t=sqrt(2h/g) s after he dropped from the cliff. But the sound heard that time instant came from an earlier stage of fall, when the man was at depth h'.

ehild

 

[IanIanIan]

Ah actually you are right..I still feel like that's too simple though, I just wrote the question out by memory though. I'll double check and get back to you :)

Thanks for the quick replies

 

[ehild]

It is not that simple...

ehild