Re: [Phys-L] overdamping

[Carl Mungan <mungan@usna.edu>]

Okay yes, after I hit SEND I started thinking along those lines.

That means the stopping time goes like 2m/b. Big viscosity or little mass makes it faster to stop.

So the story for big m is that although it oscillates with more difficulty, that is more than made up for by the extra difficulty viscosity has in stopping the object.

Just to really nail it, without going through the math, is there an intuitive way to see what I just said: viscosity increases in difficulty (to stop a mass) FASTER than the spring increases in difficulty (to oscillate the mass) as the MASS of the object increases? I know both increase in difficulty. Without doing math, could you have intuitively predicted that the spring wins out over the viscosity as the mass increases such that the oscillations become more (rather than less) underdamped? -Carl

ps: Is ajm@cpp.edu <mailto:ajm@cpp.edu> a new email address for you? I think you used to have a different one… or am I remembering incorrectly?

> On Oct 21, 2015, at 4:35 PM, A. John Mallinckrodt <ajm@cpp.edu> wrote:
>
> How about ... b/2m is something like an inverse viscous stopping time?
>
> John Mallinckrodt
> Cal Poly Pomona
>
> On Oct 21, 2015, at 1:20 PM, Carl Mungan wrote:
>
> > I’m probably just not thinking about things clearly, but the following seems unintuitive to me.
> >
> > Imagine a mass m on a spring k in a viscous fluid so that we have Stokes’ drag -bv where v is the velocity of the object.
> >
> > Okay we know how this all goes. If we start with values of m, k, and b all near 1 (in usual units of m, kg, s) we are near the critical damping limit. Now let’s imagine changing one parameter and seeing what happens:
> >
> > * if we increase b, we get overdamping - that makes sense because the fluid is more viscous and so it’s harder for the mass to oscillate through it
> >
> > * if we decrease k, we get overdamping - that makes sense because the spring is too weak to overcome the drag
> >
> > * but the one that puzzles me is m - I know the answer is you have to decrease m to get overdamping, but my intuition says it should be opposite because it should be harder for a bigger mass to oscillate
> >
> > Which of course it is. It’s just that the drag properly normalized decreases less rapidly than the natural frequency decreases, as one decreases m, so the drag always eventually wins.
> >
> > To put it mathematically, in the limit of small enough m, sqrt(k/m) < b/2m is guaranteed to hold (if k and b are held constant).
> >
> > So I guess my real question is: Why intuitively does the effective “drag frequency” (if I can call b/2m that) depend on the object’s mass?
> >
> > To put it another way: What is an appropriate intuitive way of understanding b/2m? I know mathematically it has units of frequency. But what does that mean physically?
> >
> > -----
> > Carl E Mungan, Assoc Prof of Physics  410-293-6680 (O) -3729 (F)
> > Naval Academy Stop 9c, 572C Holloway Rd, Annapolis MD 21402-1363 
> > mailto:mungan@usna.edu     http://usna.edu/Users/physics/mungan/
> >
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-----
Carl E Mungan, Assoc Prof of Physics  410-293-6680 (O) -3729 (F)
Naval Academy Stop 9c, 572C Holloway Rd, Annapolis MD 21402-1363 
mailto:mungan@usna.edu     http://usna.edu/Users/physics/mungan/