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Re: speed of sound



On Sat, 6 Apr 1996, Darren Watsky (813)751-7004 wrote:

A source of sound is moving 343 m/s in air (speed of sound = 345 m/s)
emitting a frequency of 500 Hz. An observer passes the source at a speed of
344 m/s. What will be heard by the observer?
My initial thought was to simply consider the 1 m/s separation of the
source/observer causing a doppler shift reduction in the observed
frequency (498.6 Hz). But, if I consider both of their motions with respect
to the speed of sound I get an observed frequency of 250 Hz. Somebody
straighten me out please!

Darren,

A common mistake in sonic Doppler shift problems is not to appreciate the
importance of the fact that sound waves move with respect to the air.
Because of this all source and observer "velocities" must be expressed
with respect to the preferred reference frame in which the air is
motionless; a "stationary" source in a 30 m/s wind is moving at 30 m/s as
far as the Doppler formula goes.

Then there's the nontrivial matter of remembering how the formula goes.
We all remember that it involves v_sound +/- v_source and v_sound +/-
v_observer with one on the top and one on the bottom, but which is on the
top and which is on the bottom and when is it plus and when is it minus?
Here's how I "remember" it:

Only motion of the source can cause the observed frequency to become
infinite since it can cause the waves to pile up on top of each other as
it approaches "the sound barrier." So its velocity has to be on the
bottom (where it can do the most damage!) and the forumla is

/v_sound +/- v_observer\
f_observed = f_source | ---------------------- |
\ v_sound +/- v_source /

The individual motions can now be analyzed in order to choose the signs:
If the motion of either object is "toward" the other, then we expect the
observed frequency to be greater than that of the source and vice versa.
Now we simply choose the signs accordingly.

In your problem, *before* the observer passes the source, the observer is
moving toward the source (=> + on top) and the source is moving away from
the observer (=> + on the bottom.) Thus,

f_observed = 500 Hz(345 + 344)/(345 + 343) = 500.7 Hz

*After* the observer passes the source, the observer is moving away from
the source (=> - on top) and the source is moving toward the observer (=>
- on the bottom.) Thus,

f_observed = 500 Hz(345 - 344)/(345 - 343) = 250 Hz

It is less conventional, but equally correct, to work the problem in the
frame of the source using the same formula as follows: Before passing the
observer is effectively moving at 1 m/s toward a stationary source in a
world where the speed of sound is (345 + 343) m/s = 688 m/s. Afterward
the observer is moving away from a stationary source in a world in which
the speed of sound is (345 - 343) m/s = 2 m/s.

Sorry for the long-winded reply; guess I'm feeling particularly pedantic
this morning.

John
----------------------------------------------------------------
A. John Mallinckrodt email: mallinckrodt@csupomona.edu
Professor of Physics voice: 909-869-4054
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web: http://www.sci.csupomona.edu/~mallinckrodt/