# Basics of SHM (undamped, under-driven)

1. Jul 23, 2004

### cj

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.

Last edited: Jul 23, 2004
2. Jul 23, 2004

### robphy

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.

3. Jul 24, 2004

### cj

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)?

4. Jul 24, 2004

### rayjohn01

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.

5. Jul 24, 2004

### Integral

Staff Emeritus
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.

6. Jul 25, 2004

### cj

Apparently, per rigorous research, the correct form is "undriven, underdamped."