Calculating Distance of a Falling Tuning Fork using the Doppler Effect

[Potato-chan]

Homework Statement

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.

Homework Equations

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

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

Vf=Vi+at

The Attempt at a Solution

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

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

Vs = -18.93 m/s

 

[hage567]

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

 

[Potato-chan]

Yeah I have but it give me an outragious answer.

 

[hage567]

Like what?

 

[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

 

[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.