Basics of SHM (undamped, under-driven)

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An object undergoing undamped, under-driven harmonic motion does not have its maxima at the points where the displacement curve touches the exponential envelope. Instead, the maxima correspond to points where the velocity (dx(t)/dt) is zero, which aligns with the peaks of the cosine curve. The discussion highlights that while the ratios of successive maxima are constant, the relationship between the displacement and its velocity is crucial for understanding this phenomenon. The exponential envelope does not have a tangent with zero slope, reinforcing that maxima occur at different points than visually interpreted. Clarifications on terminology, such as "undriven" versus "undamped," were also noted in the conversation.
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My textbook says the an object undergoing undamped, under-driven
harmonic motion (http://romano.physics.wisc.edu/lab/manual/img279.gif)
does NOT have its maxima at the points where the displacement
curve makes contact with the exponential envelope curve.

How can this be the case?? Doesn't the graph clearly imply that
the maxima are indeed the peaks of the decaying cosine curve (that
do make contact with the exponential wrapper)??

The text goes on to say that the maxima actually correspond not
to the x(t) vs. t plot -- but to the dx(t)/dt (the velocity) plot,
specifically where dx(t)/dt = 0. I can partially understand this since
at the maxima -- velocity does equal 0!


It then states that the displacement ratios between successive
maxima are constant.

I can see the constancy of the maxima ratios, but not the
basis on dx(t)/dt over the visual interpretation -- let alone
the assertion that successive maxima ratios are constant.


Comments? Thanks!
 
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That exponential curve is an envelope for that graph if the tangents of the curve and the graph agree at all contact points. No point on the exponential curve has a tangent line with zero slope.
 
Is the displacement maxima (which occurs at the time
where v(t) = dx(t)/dt = 0) the same point as the displacement
at t=0, T, 2T, etc., where T = 2*pi/omega (the under-damped
version of omega)?

robphy said:
That exponential curve is an envelope for that graph if the tangents of the curve and the graph agree at all contact points. No point on the exponential curve has a tangent line with zero slope.
 
graph

Just try drawing an exponentially damped sine wave and then the smooth exponential -- you will see that's correct.
ymax is dy/dt =0 for the sine , byt dy/dt is never =0 for the exponential.
 
My textbook says the an object undergoing undamped, under-driven

Sorry to be nit picky but this is bugging me. Could you possibly have meant to say.

Undriven, under damped?

That is what your graph looks like.
 
Apparently, per rigorous research, the correct form is "undriven, underdamped."

Integral said:
Sorry to be nit picky but this is bugging me. Could you possibly have meant to say.

Undriven, under damped?

That is what your graph looks like.
 
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