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Dopler without a wave?



Suppose one end of a yardstick is clamped.
The other end is weighted and can oscillate
at a low frequency, such as 2 or 3 Hz. The
amplitude is 3 cm. To measure the frequency,
f=1/T, we use a Vernier motion detector (*see
the footnote) and record several periods of
the harmonic trace.

Then the experiment is repeated several
times. Each time the motion detector box
(observer) is moved rapidly (by hand) either
toward the oscillating yardstick ("source") or
away from it. And each time the harmonic
trace is used to determine both the apparent
frequency, f', and the velocity of the
observer, v_o.

I think that the results would be consistent
with the Dopler effect formula:

f' = f * (v + v_o) / v

where v is the speed of ultrasonic pulses of the
motion detector (about 340 m/s). Note that the
ultrasonic traces are not observable (because
the amplitude is too small and frequency is too
high). Therefore I think that such demonstration
could be labeled "Dopler effect without a wave."
Do you agree?

Yes, I know that each pulse is a wave packet.
But we are not talking about frequencies of
these waves, we are talking about apparent
frequencies of a single harmonic oscillator
observed from different frames of reference.
Note that v_o can be either positive or negative.
I am assuming that the absolute value of v_o is
smaller than that of v (to avoid speculations
about the meaning of negative f').

Footnote:
(*) Vernier motion detector, connected to a
computer, sends short ultrasonic pulses toward
an object, for example, at a rate of 20 kHz, and
detects the returning "echo" from each pulse.
The time delays (between pulses and echos)
are used to calculate distances, one after another.
Ludwik Kowalski