H
Hein ten Horn
Jim Kelley wrote
Not a pure sine oscillation (rather than wave), but a near sine
oscillation at an exact period of 1/222 s. The closer the source
frequenties, the better the sine fits a pure sine. Thus if you
wish to get a sufficient near harmonic oscillation, conditions like
"slow changing envelope" are essential.
Hmm, let's examine this.
From the two composing oscillations you get the overall
displacement:
y(t) = sin(2 pi 220 t) + sin(2 pi 224 t)
From the points of intersection of y(t) at the time-axes you
can find the period of the function, so examine when y(t) = 0.
sin(2 pi 220 t) + sin(2 pi 224 t) = 0
(..)
(Assuming you can do the math.)
(..)
The solutions are:
t = k/(220+224) with k = 0, 1, 2, 3, etc.
so the time between two successive intersections is
Dt = 1/(220+224) s.
With two intersections per period, the period is
twice as large, thus
T = 2/(220+224) s,
hence the frequency is
f = (220+224)/2 = 222 Hz,
which is the arithmetic average of both composing
frequencies.
As seen above, the particle oscillates (or vibrates) at 222 Hz.
Since the oscillation is non-harmonic (not a pure sine),
it needs several harmonic oscillations (frequencies,
here 220 Hz and 224 Hz) to compose the oscillation at 222 Hz.
To be precise, this is nonsense, but I suspect you're trying
to state somewhat else, and since I'm not able to read your
mind today, I skip that part.
No peculiarities found.
gr, Hein
Under the stated conditions there is no sine wave oscillating at 222 Hz.
The wave has a complex shape and contains spectral components at two
distinct frequencies (neither of which is 222Hz).
Not a pure sine oscillation (rather than wave), but a near sine
oscillation at an exact period of 1/222 s. The closer the source
frequenties, the better the sine fits a pure sine. Thus if you
wish to get a sufficient near harmonic oscillation, conditions like
"slow changing envelope" are essential.
The particle also does not average the two frequencies.
Hmm, let's examine this.
From the two composing oscillations you get the overall
displacement:
y(t) = sin(2 pi 220 t) + sin(2 pi 224 t)
From the points of intersection of y(t) at the time-axes you
can find the period of the function, so examine when y(t) = 0.
sin(2 pi 220 t) + sin(2 pi 224 t) = 0
(..)
(Assuming you can do the math.)
(..)
The solutions are:
t = k/(220+224) with k = 0, 1, 2, 3, etc.
so the time between two successive intersections is
Dt = 1/(220+224) s.
With two intersections per period, the period is
twice as large, thus
T = 2/(220+224) s,
hence the frequency is
f = (220+224)/2 = 222 Hz,
which is the arithmetic average of both composing
frequencies.
The waveform which results from the sum of two pure sine waves is not a pure
sine wave, and therefore cannot be accurately described at any single
frequency.
As seen above, the particle oscillates (or vibrates) at 222 Hz.
Since the oscillation is non-harmonic (not a pure sine),
it needs several harmonic oscillations (frequencies,
here 220 Hz and 224 Hz) to compose the oscillation at 222 Hz.
Matter would move in the same way the sound pressure wave does,
To be precise, this is nonsense, but I suspect you're trying
to state somewhat else, and since I'm not able to read your
mind today, I skip that part.
the amplitude of which is easily plotted versus time using Mathematica,
Mathcad, Sigma Plot, and even Excel. I think you should still give that a
try.
No peculiarities found.
gr, Hein