Physics 374
Fall 2004
Prof. E. F. Redish

HW3, Problem 4

a.  Given:
					m(((d^2) x)/(dt^2)) = -kx - bv
	So it's seen that m has dimensions M, k has dimensions M/T\(TraditionalForm\`\^2\), and b has dimensions M/T.  Using these, we can construct two natural times since the time dimension can come either from the k or b:
	(Subscript[t, b]) = (m/b)  and  (Subscript[t, k]) = (Sqrt[(m/k)]) 
It's impossible to construct a natural length from m, k, and b since none of them carries any length dimension at all.  There's no way to get [L] into the mix. The oscillator is "length-scale free".  This is because the model we are using for the spring does not distinguish lengths.  This is clearly nonsense.  Stretch any spring too much and it will break.  Compress any spring too much and it will collapse fully and not be able to compress any more (without much more force, now squeezing the solid metal).  However, these lengths are not expressed anywhere in our force model -- "kx".  The scale of the oscillation comes from the starting value.  If we start with just stretching the spring to a distance A and releasing it, that distance is the maximum of the oscillation and sets the scale.  If we start by hitting the mass to give it a velocity at its equilibrium position, the scale is set by the inital velocity times the period.

b.   In a qualitative argument, if (Subscript[t, b]) >> (Subscript[t, k]), that means the damping occurs on a much longer time scale than the spring force's effects take to change significantly.  In other words, any change caused by the restoring force (say, one full period of oscillation) should be much quicker than any significant change caused by the damping force (say, the time to reduce the maximum amplitude by a factor of 2).  Thus, the restoring term should dominate for small times.
      In a more quantitative argument, (Subscript[t, b]) >> (Subscript[t, k]) implies that (m/b) >> (Sqrt[(m/k)]) , which is to say that 
      (Sqrt[mk]) >> b...which implies that the damping is relatively weak, compared to the restoring force (assuming no extreme masses are involved).  It should take a comparatively long time to see significant effects of the damping force, so again we'd expect the restoring term to dominate for small times.  A reasonable guess at the oscillator's position as a function of time would be:

c.   Let t* = t / (Subscript[t, k]).  Then:
					m(((d^2) x)/(dt^2)) = -kx - bv
					(m/((Subscript[t, k])^2))\(TraditionalForm\`\(\(d\^2\) x\)\/\(\(dt\)\(*\^2\)\)\) = -kx - bv
					k\(TraditionalForm\`\(\(d\^2\) x\)\/\(\(dt\)\(*\^2\)\)\) = -kx - bv
					\(TraditionalForm\`\(\(d\^2\) x\)\/\(\(dt\)\(*\^2\)\)\) = -x - (b/k)v
Since the v term carries with it a time, we will want to take out that time dimension as well, which turns v (with dimensions L/T ) into v* (with dimensions L ).
	
		v  = (dx/dt) =  ((1/(Subscript[t, k])))(dx/(dt*))  =  (Sqrt[(k/m)])(dx/(dt*))  = (Sqrt[(k/m)])v*
Thus:
				\(TraditionalForm\`\(\(d\^2\) x\)\/\(\(dt\)\(*\^2\)\)\) = -x - (b/k)	(Sqrt[(k/m)])v*
				\(TraditionalForm\`\(\(d\^2\) x\)\/\(\(dt\)\(*\^2\)\)\)  = - (x + (b/(Sqrt[km]))v*)
And we can now see the convenient dimensionless constant we're seeking: (b/(Sqrt[km])).

		If (b/(Sqrt[km])) << 1, (note this is the same thing arrived at earlier in part b) then  the x term (i.e. restoring force) dominates over times, and we expect a position graph like the one above.
		If (b/(Sqrt[km])) >>  1, then the v* (damping term) dominates over small times, and you'd expect a position graph like:

(Solution by T .  Bing)