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Potato-chan
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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
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