# College Physics problem

1. Jan 26, 2008

### Potato-chan

1. The problem statement, all variables and given/known data
A tuning forking vibrates at 512Hz falls from rest and accelerates at 9.80m/s^2. How far below the point of release is the tuning fork when waves of the frequency 485Hz reach the release point? Take the sound of sound in air to be 340m/s.

2. Relevant equations
f'=((V+Vo)/(V-Vs))f [doppler effect equ]

Xf=Xi+vt+(1/2)at^2

Vf=Vi+at

3. The attempt at a solution

-Vs = ((V+Vo)f/f')-V

((340+0)512Hz/485)-340

Vs = -18.93 m/s

Last edited: Jan 26, 2008
2. Jan 26, 2008

### hage567

The question is asking for the distance the tuning fork has fallen. Have you tried to calculate that yet?

3. Jan 26, 2008

### Potato-chan

Yeah I have but it give me an outragious answer.

4. Jan 26, 2008

### hage567

Like what?

5. Jan 26, 2008

### Potato-chan

So I plug in my found Velocity. into the Vf=Vi+at equation. I know Vi has to be 0 or at least assumed to be. Acceleration is given to me;9.8 m/s^2. and I have the velocity.

Disregarding the sign. 18.93=0+(9.8)t therefore t = 1.93 seconds

Plug that into the distance equation Yf = Yi + vt + .5at^2

Assume Yi = 0

Yf = -18.75

The real answer is suppose to be 19.3

6. Jan 26, 2008

### hage567

Well, I don't think there's an error in your approach. I did get a slightly different answer than you (18.2 m) and I found the distance by a bit more direct method. Maybe rounding error is involved? I don't know.