Re: resonance tube

[Bernard Cleyet <anngeorg@PACBELL.NET>]

"10:1 voltage-divider, in which the lower
             leg is the speaker itself."

Yes, is a constant current source, as the current is limited by a
constant resistance* (only, therefore, constant impedance, also).
However, the point is to unload the speaker, so it is not damped by the
source.  Unfortunately, one must also measure the impedance of the
speaker away from the the water tube, so its (the speaker's)
resonance(s) may be found and ignored.

* As JD states,  >10 X the impedance of the speaker.  More if one wants
a quantitative result.

BTW a much less expensive method of detecting phase (than a PAR,
Stanford Research, etc.) is to use a center tapped secondary
transformer.  In this case the primary is driven by the speaker,
unloaded (isolated) by a cathode follower (emitter follower for the
modern, op. amp. for the "really" modern).  The output is compared to
the  source  (oscillator output)  using another transformer,  Its
secondary is in series with other's center tap and the center of a
balanced rectifier (the other legs of the rectifier go the the secondary
of the first transformer).  The output of the rectifier is proportional
to the input signal and  the polarity indicates the phase.
Unfortunately, unless the signal (and the reference) are constant, this
device does not give the phase angle, only whether it leads or lags the
reference.  This device was invented for use with strain gauge bridges.

If your library hasn't discarded Hill's Electronics in Engineering,
(because it's so old, 1949)  look on page 247 for the circuit and
description.

bc

John S. Denker wrote:

> On 09/16/2003 02:24 AM, Mark Sylvester wrote:
> > The setup: 1.5m transparent tube about 10 cm diameter, set vertically
> > so that water can be introduced into or drained from the bottom.
> > Loudspeaker mounted over the open upper end; small microphone just
> > below the loudspeaker, connected to oscilloscope. The usual exercise
> > consists of changing the water level to find the resonant lengths of
> > the air column tones of different frequency, thus finding the speed
> > of sound and the end correction. The oscilloscope is used to observe
> > when the sound from the tube is loudest.
> >
> > One notices that with the water just a centimetre or two below the
> > resonant position, the position apparently not affected by the
> > frequency, the trace goes flat, suggesting destructive interference.
> > I have a student looking for a detailed explanation, but he has run
> > out of ideas.
>
> Some ideas:
>
> 1) Does it depend on how far the microphone is below
> the loudspeaker?  Can you reposition the microphone and
> see what happens?  Multiple microphones simultaneously?
>
> 2) You can get additional information by using the
> *speaker* as a point of measurement.  You could drive
> the speaker with a constant-voltage source and measure
> the current, or (probably better) drive the speaker with
> a constant-current source and measure the voltage.
> Hint:  easy way to make a moderately high impedance
> constant-current source:
>   audio source -->
>      audio amplifier -->
>        1:10 step-up transformer -->
>           10:1 voltage-divider, in which the lower
>              leg is the speaker itself.
>
> If you want to get fancy, better circuits are available.
> Emitter follower.....
>
> 3a) Plot out the whole resonant response curve.  Locating
> the peak of this curve is interesting, but it's not the
> only interesting thing.
>
> 3b) Pay attention to the *phase* as well as magnitude of
> the response.  If you can possibly lay hands on a synchronous
> detector ("lockin amplifier") use that to record the in-phase
> and quadrature phasor components.  Otherwise just eyeball it.