Physics AP - Victor's Car Horn Frequency (#21)

Background:
The distance and how fast a sound is coming towards you or away from
you can change the frequency that you hear.  The frequency also
changes if the object is moving or if you the listener are moving.
The frequency also can depend on the temperature.  For instant the
equation u = vo + .6 × (°C).  This equation goes into the equation
f = fo × (u"vs)/(u"vs).  This equation tells you the either the
frequency of the sound coming toward you or the sound going away
from you or you moving toward a stationary sound or you moving away
from a stationary sound.  The fo in the equation stands for the
original frequency of the sound nothing or no one is moving. U is the
velocity of sound through air.  The vs is the velocity of the
listener or of the sound moving.

Problem:
What frequency did we hear coming towards and away from us?

Procedure:
1. Video taping the car coming toward us then away from us at a
   constant unknown speed.
2. Video tape the car's horn while the car is not moving.
3. From that video use the oscilloscope to find the Fo of the car.
4. Knowing that use the equation u = vo + .6 × (°C) to find the
   velocity of the sound through air then use the equation
   fn = fo × (u"vs)/( u"vs) to find the unknown speed.  Manipulate
   the equation to look like -1 × ((fo × u)/fn) - u = vs.
5. Now use the equation f = fo × (u"vs)/( u"vs) to find the frequency
   coming toward you, then frequency away from us.

Data:
fo = 350Hz T = 4.4°C fn = 366Hz

Frequency Car Moving Toward Ear   Frequency Car Moving Away From Ear
ft=fo×(u/(u-vs))                  ft=fo×(u/(u+vs))
----------------------------------------------------------------------
ft=350×(333.64/(333.64-14.6))     ft=350×(333.64/(333.64+14.6))
ft=350×(333.64/(319.04))          ft=350×(333.64/(348.24))
ft=350×1.05                       ft=350×.958
----------------------------------------------------------------------
ft=255Hz                          ft=335Hz