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+--- Thread: A question on Pendulum experiment (/showthread.php?tid=1202)
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RE: A question on Pendulum experiment - jmnk - 12-02-2020
(12-02-2020, 04:40 AM)solid Wrote: Jmnk, you have to open this forum on your smartphone and click the program. Then choose 'open in phyphox' and save the program in your collection. The editor does not support colors.
yap, thx, indeed I found that out too after few hours of trying and comparing with other .phyphox files with multi-color-graphs. All good.
RE: A question on Pendulum experiment - fbouquet - 12-02-2020
(12-02-2020, 04:40 AM)solid Wrote: Jmnk, you have to open this forum on your smartphone and click the program. Then choose 'open in phyphox' and save the program in your collection. The editor does not support colors.
Another quick solution (that I use a lot): on your PC, when you are on the forum and that you click on the attached phyphox program, you have the option to "show QRcode". This QRcode allows you to rapidly add the experiment on your phone.
Yes, going through the editor changes the program...
Cheers,
Fred
RE: A question on Pendulum experiment - jmnk - 12-02-2020
(12-01-2020, 10:26 PM)Jens Noritzsch Wrote: The resolution in the autocorrelation graph corresponds to that of the sensor, so for iDevices it would be 100Hz, i.e. one dot each 10ms. The analysis algorithm looks out for the right most maximum and divides the time shift by the order of the maximum. This improves the resolution a bit to 10ms/order.
I performed a few tests and could definitely see a 10ms “grid” in 1/frequency, i.e. the period. The associated order of the dots in-between keeps me wondering a bit (in the final example one less than expected)…
Jens,
since I have few smartphones collecting dust..... My previous screenshots were from Samsung Galaxy S6 Active phone. I've tried that very experiment with Samsung Galaxy S8:
and with iPhone 6:
and the graph is exactly as you explained it should be! One can very clearly see the frequencies "grid" - although I see them spaced every ~0.005Hz
I suppose the 'extra' data points I'm seeing on my Samsung Galaxy S6 Active phone are due to some inaccuracies of that device sensor?
RE: A question on Pendulum experiment - Jens Noriʇzsɔɥ - 12-03-2020
(12-02-2020, 12:41 PM)jmnk Wrote: and the graph is exactly as you explained it should be! One can very clearly see the frequencies "grid" - although I see them spaced every ~0.005Hz
Please note that the grid “effect” is in the period. The observed frequencies are not strictly equidistant.
jmnk Wrote:I suppose the 'extra' data points I'm seeing on my Samsung Galaxy S6 Active phone are due to some inaccuracies of that device sensor?
They occur at lower oscillation amplitudes, so apparently at the end of the measurement. An unstable pendulum could be a reason, some torsion, for instance. The algorithm utilizes a simple sum of the gyroscope axes (giving flexibility in the smartphone’s orientation)
RE: A question on Pendulum experiment - jmnk - 12-24-2020
I have yet another question for pendulum experiment. I'm using is as a sort of tutorial on how phyphox file/experiment work - so it is entirely possible I do not understand all of it.
Toward the end of that pendulum experiment there's this code:
"
<average>
<input as="buffer" clear="false">anyGyr</input>
<output as="average" clear="false">avg</output>
<output as="stddev">amplitudeSkewed</output>
</average>
[...]
<divide>
<input as="dividend" clear="false">amplitudeSkewed</input>
<input as="divisor" clear="false">frequency</input>
<input as="divisor" clear="false">frequency</input>
<output as="quotient">amplitude</output>
</divide>
"
so it seems it is calculating average of gyroscope readings (angular velocity), and its standard deviation. And then uses stdDev/(frequency*frequency) to get amplitude.
frequency is the calculated frequency of the pendulum. What does 'amplitude' calculated that way represent?
RE: A question on Pendulum experiment - solid - 12-24-2020
x = A sin(omega t) # deviation angle
d sin(omega t) /dt= omega cos(omega t)
v = dx/dt = (A omega) cos(omega t) # rotation speed
s = dA²omega² # square standard deviation
'amplitude' calculated (dA² omega²)/omega/omega = dA² represents the standard deviation of the angle (for small harmonic oscillations)...
i think.This example is too complicated for a tutorial...
RE: A question on Pendulum experiment - jmnk - 12-26-2020
(12-24-2020, 03:56 AM)solid Wrote: x = A sin(omega t) # deviation angle
d sin(omega t) /dt= omega cos(omega t)
v = dx/dt = (A omega) cos(omega t) # rotation speed
s = dA²omega² # square standard deviation
'amplitude' calculated (dA² omega²)/omega/omega = dA² represents the standard deviation of the angle (for small harmonic oscillations)...
This example is too complicated for a tutorial...
hmm, I think I'm not getting it.
I'm perfectly following up to this: v = dx/dt = (A omega) cos(omega t) # rotation speed
but I do not understand this equation (or perhaps I'm misinterpreting the notation): s = dA²omega² # square standard deviation
the standard deviation of: (A omega) cos(omega t) is, I think, A*omega/√2, so the square standard deviation would be A²omega²/2. What does 'd' in this equation represent: s = dA²omega² ?
Either way, the code at no point appears to be using the _square_ of standard deviation, so why is that stddev divided by (omega * omega) to get amplitude?
If the code was ' stddev * √2 / omega ' then I would understand why the result is 'amplitude'. (Although I think it would need to be stddev * √2 / (2*pi*frequency) since in the code frequency is in 1/s rather then in rad/s)
RE: A question on Pendulum experiment - jmnk - 01-04-2021
^^ Happy New Year to everyone!
anyone with an idea of what that 'amplitude' in the pendulum experiment is to represent? to repeat - I'm wondering about this part of the experiment:
"
Toward the end of that pendulum experiment there's this code:
<average>
<input as="buffer" clear="false">anyGyr</input>
<output as="average" clear="false">avg</output>
<output as="stddev">amplitudeSkewed</output>
</average>
[...]
<divide>
<input as="dividend" clear="false">amplitudeSkewed</input>
<input as="divisor" clear="false">frequency</input>
<input as="divisor" clear="false">frequency</input>
<output as="quotient">amplitude</output>
</divide>
so it seems it is calculating average of gyroscope readings (angular velocity), and its standard deviation. And then uses stdDev/(frequency*frequency) to get amplitude.
frequency is the calculated frequency of the pendulum. What does 'amplitude' calculated that way represent?
"
RE: A question on Pendulum experiment - Jens Noriʇzsɔɥ - 01-06-2021
The standard deviation is a quite convenient measure of the oscillation amplitude of anyGyr. Frequency squared is a (copy) bug (from spring experiment) that gets fixed in the next release. It needs to be just the frequency, of course. Any proportionality factors could be left out for the arbitrary units in the resonance graph.
Thanks for spotting this.
RE: A question on Pendulum experiment - jmnk - 01-07-2021
Makes sense now. All good. Thx for following up.