Re: [Phys-L] overdamping

[Paul Nord <paul.nord@valpo.edu>]

Intuitively, an object's response to drag forces is about the relationship
between its mass and its geometry.  A racket ball and a billiard ball both
have the same diameter (approximately) and at a given velocity will have an
identical drag.  However, the terminal velocity of the billiard ball will
be much higher.  This is because the amount of drag force required to match
the gravitational force will be greater.

Your intuition tells you that applying your critical damping formula should
be different for a grain of sand than for a billiard ball.  But you've
changed two things: both the mass and the geometry.  If you imagine a
billiard ball with the mass of a sand grain, you should see how little
force would be required to damp its motion.

Paul


On Wed, Oct 21, 2015 at 3:20 PM, Carl Mungan <mungan@usna.edu> 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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