Re: [Phys-L] overdamping

[Jeffrey Schnick <JSchnick@Anselm.Edu>]

I think the idea is that with increasing mass the contribution to the acceleration made by the spring decreases in magnitude for one reason and the contribution to the acceleration made by the fluid decreases in magnitude for that same reason and another reason.  The first reason is that from F=ma, the greater the mass for the same force, the smaller the acceleration.  The additional reason for the case of the fluid is that, if we release the object from rest at maximum stretch, from the same position with different masses, even without dampening, at every position of the object on its way to the equilibrium position, the greater the mass, the smaller the velocity at that position.  With the fluid in place, the smaller velocity results in a smaller retarding force.  Thus, greater mass leads to less dampening.

> -----Original Message-----
> From: Phys-l [mailto:phys-l-bounces@www.phys-l.org] On Behalf Of Carl
> Mungan
> Sent: Wednesday, October 21, 2015 4:20 PM
> To: PHYS-L
> Subject: [Phys-L] overdamping
>
> 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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