- #1
FranzDiCoccio
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Homework Statement
A Doppler flow meter is used to measure the speed of red blood cells.
The frequency of the apparatus is f = 12 MHz. The sensor in the apparatus measure 1.8 kHz beats between the emitted frequency and the frequency of the ultrasound reflected back by the blood cells.
The speed of sound in blood is 1520 m/s.
Homework Equations
General formula for the Doppler effect:
[tex]f_r = f_e \frac{1\pm \frac{v_r}{v_w}}{1\pm \frac{v_e}{v_w}}[/tex]
where [itex]e[/itex] is for "emitter", [itex]r[/itex] is for "receiver" and [itex]w[/itex] is for "wave", and the sign must be chosen appropriately.
EDIT: the problem assumes that everything happens in 1D, that's why I'm using this formula.Frequency of the beats between two different frequencies.
[tex]f_b = |f_1-f_2|[/tex]
The Attempt at a Solution
I think this problem is a bit tricky. In fact, I suspect that the solution I was provided is correct, but its derivation may be wrong.
I think that the blood cells should be considered at rest, and the emitter/receiver should be considered in motion with the same speed as the blood. This is because I think the Doppler effect formula makes sense in the frame of reference of the medium (blood).
Then, if my assumption is correct, the frequency "received" by the blood cells is
[tex]f_r = f_e \frac{1}{1\pm\frac{v}{v_w}}[/tex]
where the sign is - if the blood cells are approaching the flow meter and + otherwise. This because the emitter (flow meter) is in motion.
The blood cells reflect back the waves with the same frequency as they were received. Since the flow meters which receives the reflected waves is in motion, one gets
[tex]f_r' = f_e \frac{1\mp\frac{v}{v_w}}{1\pm\frac{v}{v_w}}[/tex]
where the minus in the denominator corresponds to the plus in the numerator, and the other way round.
The equation for the beats can be inverted for [itex]v[/itex] with a little algebra, which gives
[tex]v= v_w \frac{f_b }{2f_e \pm f_b}[/tex]
The solution I was given is basically the same
[tex]v = \frac{f_b}{2 f_e} v_w [/tex]
because the [itex]f_b[/itex] term in my denominator is negligible.
However the reason provided for this solution does not convince me, because it is based on the equations
[tex]f_r = f_e (1\pm\frac{v}{v_w})[/tex]
[tex]f_r' = f_e (1\mp\frac{v}{v_w})[/tex]
[tex]f_b = |f_r-f_r'|[/tex]
While this approach does result in the proposed solution, it does not make any sense, in my opinion.
The second Doppler shift should depend on the first, right? Not simply on the emitted frequency...
Also, why calculating the beats between the Doppler shifted frequencies?
The exercise text says that the beats is between the emitted frequency (unshifted) and the received frequency ("twice" shifted).
Thanks a lot for your help.
Franz