# Homework Help: Frequency of a falling persons scream

1. Jun 9, 2012

### IanIanIan

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
f'=f(v+-v_d)/(v-+v_s)

3. 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 thats 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?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 9, 2012

### harts

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

3. Jun 9, 2012

### IanIanIan

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

4. Jun 9, 2012

### 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

5. Jun 10, 2012

### 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 travelled 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 eachother 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.

6. Jun 10, 2012

### 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

7. Jun 10, 2012

### 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

8. Jun 10, 2012

### ehild

It is not that simple...

ehild