12-26-2020, 03:48 AM

(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)...

i think.

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)