Determining Distance from Release Point Using Doppler Effect

[lackos]

Homework Statement

A tuning fork vibrating at 506 Hz falls from rest and accelerates at 9.80 m/s2. How far below the point of release is the tuning fork when waves of frequency of 488 Hz reach the release point? (Take the speed of sound in air to be 343 m/s).

Homework Equations

f(prime)=(v/(v+v(source)))*f(initial)
x=v(source)^2/19.6

The Attempt at a Solution

im actually fairly happy with my answer (8.17), because i worked it backwards from a textbook question, after it said i was wrong. but the system says i was wrong ( but within 10 percent).

any insight?

btw i subbed in my value for v(source) from question 1 into question 2 to get my answer.

 

[Doc Al]

Did you distinguish between:
(1) Where the tuning fork was when the 488 Hz waves were emitted
(2) Where the tuning fork was when the 488 Hz waves reached the release point

 

[lackos]

Did you distinguish between:
(1) Where the tuning fork was when the 488 Hz waves were emitted
(2) Where the tuning fork was when the 488 Hz waves reached the release point

i thought that was already factored into the the doppler equation?

 

[Doc Al]

i thought that was already factored into the the doppler equation

No. The Doppler equation will give you the speed at which the 488 Hz waves are emitted. That's just the first step in finding the answer. You need to figure out how long it takes for the sound to reach the top, then figure out where the tuning fork is at that time.

 

[lackos]

No. The Doppler equation will give you the speed at which the 488 Hz waves are emitted. That's just the first step in finding the answer. You need to figure out how long it takes for the sound to reach the top, then figure out where the tuning fork is at that time.

okay thanks for that info (note to self).

so with my value (8.17), i would divide that by the speed of sound to get the extra time taken for travel. i would then form another kinematic equation factoring this in.

doing this i get 8.47m does this look correct

 

[Doc Al]

okay thanks for that info (note to self).

so with my value (8.17), i would divide that by the speed of sound to get the extra time taken for travel. i would then form another kinematic equation factoring this in.

doing this i get 8.47m does this look correct

Looks good to me.

 

[lackos]

Looks good to me.

thanks for the help and time