Joel Kammet
Posted on Wednesday, 28 May, 2003 - 10:28 pm:
Everybody (well, maybe not everybody :) ) likes to model damped harmonic motion with a damping term that's a function of velocity, so we start with something like
m d2x
dt2
 + b dx
dt
 + k x = 0
and end up with
x(t) = Ae-(b/2m)t cos(wt + f)
But what if the damping is not a function of velocity? All the textbooks like to teach SIMPLE harmonic motion with blocks and springs on a frictionless table, but as soon as they want to deal with damping, the venue conveniently changes to things flying through the air or slogging through some viscous fluid. What happened to all those tables?

Suppose there's friction between the block and the table, so the damping force is just a multiple of m, and hence constant (that is, its magnitude is constant, but its sign must always be opposite to the sign of dx/dt). So ideally, it needs a damping term that is some constant always multiplied by +or- 1 depending on where we are in the cycle. Is there a way to set up and solve a differential equation to model that?

So far, the best idea that I've thought of to deal with this (and I don't like it) is to start with
x = Ae-(b/2m)tcos(wt + f)
and tinker around with the value of b, and maybe play with different powers of t in the exponential term (i.e. Ae-(b/2m)t2, or something like that) by trial and error; testing by multiplying the distance covered each cycle times the frictional force, subtracting that result from the initial potential energy to arrive at theoretical potential energy at the end of the cycle, and comparing that result to potential energy computed by (1/2)kx2 at the end of the cycle to see if the equation is a reasonable approximation of the facts.

Yeccchhh. There must be a better way. Please!!

David Loeffler
Posted on Thursday, 29 May, 2003 - 12:16 am:
Isn't it rather more likely that the solution would be defined piecewise, with separate formulae for each of the intervals [0,t1 ], [t1 ,t2 ], [t2 ,t3 ] etc, where the ti 's are the times at which the direction of motion changes?

For example, suppose y'' = -y + K(y') where K(y') = -1 if y' > 0, 1 if y' < 0.

Then (for argument's sake) let's start with t = 0, y = 5, y' = 0.

On this stretch y'' = 1 - y, so y'' + y = -1. That is (given the initial conditions) y = 1 + 4 cos t.

Now this will be fine until y' reaches zero, which is when t = pi . Then y(t) = -3.

We have to check that the block is going to move at all: this is clear as the attractive force on it is 3 and the resisting force is 1.

So it continues to move, and now we have y'' + y = -1, y = -1 + 2 cos t, and this continues until time t = pi .

Now we have y = 1, and the block simply stops, as the frictional force is sufficient to prevent it returning.

(You might like to see what happens if you choose different starting positions. Can you see why the block will always stop moving after a finite time?)

David
Joel Kammet
Posted on Thursday, 29 May, 2003 - 04:55 am:
Yes, of course. A piecewise solution does make more sense. Actually, now that you mentioned it, it's clear that yet another function should be used to determine when the motion stops completely, since the coefficient of static friction is generally larger than m k .

But, still, it would be convenient to be able to approximate it with some sort of continuous function, even though that wouldn't be precisely correct...

Anyway, thanks David.