Physics 374
Fall 2005
Prof. E. F. Redish

Solution,  Problem MP09

a.   Taking the origin of the "x" coordinate system to be the point of suspension , with down as positive, one can write Newton's Second Law for mass one (the top) and mass two:
				m(((d^2) (Subscript[x, 1]))/(dt^2)) = -k((Subscript[x, 1]) - (Subscript[L, o])) + k((Subscript[x, 2]) - (Subscript[x, 1]) - (Subscript[L, o])) + mg
				m(((d^2) (Subscript[x, 2]))/(dt^2)) =  -k((Subscript[x, 2]) - (Subscript[x, 1]) - (Subscript[L, o])) + mg
You should be able to think carefully about why each spring force term has the sign that it does.  For example, looking at the first spring force term in the top equation, if the top mass is very near the ceiling, then (Subscript[x, 1]) will be small, so ((Subscript[x, 1]) - (Subscript[L, o])) will be negative, so -k((Subscript[x, 1]) - (Subscript[L, o])) will be positive (i.e. the top spring is pushing the top mass downwards).  Since that spring is obviously compressed in this thought-experiment, that makes sense.
	Most of the constants disappear from the above equations if we choose to switch coordinates to ones measured from the equilibrium positions ((Subscript[x, 10]) and (Subscript[x, 20])) of the two masses:
					(Subscript[y, 1]) = (Subscript[x, 1]) - (Subscript[x, 10])
					(Subscript[y, 2]) = (Subscript[x, 2]) - (Subscript[x, 20])
Setting the force sums in the first two equations above to zero will give a system of two equations to solve for these equilibrium values.  Using these values as you make the coordinate transformation above in the full N2 equations will lead, after some algebra, to
					m(((d^2) (Subscript[y, 1]))/(dt^2)) = -k(Subscript[y, 1]) + k((Subscript[y, 2]) - (Subscript[y, 1]))
					m(((d^2) (Subscript[y, 2]))/(dt^2)) = -k((Subscript[y, 2]) - (Subscript[y, 1]))

b.     First, I'll rearrange the equations from part "a", making use of the substitution ((Subscript[ω, o])^2) = (k/m)
					(((d^2) (Subscript[y, 1]))/(dt^2)) = -2((Subscript[ω, o])^2) (Subscript[y, 1]) + ((Subscript[ω, o])^2) (Subscript[y, 2])
					(((d^2) (Subscript[y, 2]))/(dt^2)) = ((Subscript[ω, o])^2) (Subscript[y, 1]) - ((Subscript[ω, o])^2) (Subscript[y, 2])
Which can be written as
				((d^2)/(dt^2))(({{(Subscript[y, 1])}, {(Subscript[y, 2])}})) = (({{-2 ((Subscript[ω, o])^2), ((Subscript[ω, o])^2)}, {((Subscript[ω, o])^2), -((Subscript[ω, o])^2)}}))(({{(Subscript[y, 1])}, {(Subscript[y, 2])}}))
From our experience in class, we have a hunch that the solutions should be of the form
					(Subscript[y, 1])(t) = α (e^i)ωt
					(Subscript[y, 2])(t) = β (e^i)ωt
Substituting these in gives (noting that the double time derivative brings down two powers of "iω"
					-(ω^2)(({{(Subscript[y, 1])}, {(Subscript[y, 2])}})) = (({{-2 ((Subscript[ω, o])^2), ((Subscript[ω, o])^2)}, {((Subscript[ω, o])^2), -((Subscript[ω, o])^2)}}))(({{(Subscript[y, 1])}, {(Subscript[y, 2])}}))
					
					0 = (({{-2 ((Subscript[ω, o])^2) + (ω^2), ((Subscript[ω, o])^2)}, {((Subscript[ω, o])^2), -((Subscript[ω, o])^2) + (ω^2)}}))(({{(Subscript[y, 1])}, {(Subscript[y, 2])}}))
					
					0 = ((Subscript[ω, o])^2)(({{-2 + ((ω^2)/((Subscript[ω, o])^2)), 1}, {1, -1 + ((ω^2)/((Subscript[ω, o])^2))}}))(({{(Subscript[y, 1])}, {(Subscript[y, 2])}}))
					
					0 = (({{-2 + ((ω^2)/((Subscript[ω, o])^2)), 1}, {1, -1 + ((ω^2)/((Subscript[ω, o])^2))}}))(({{(Subscript[y, 1])}, {(Subscript[y, 2])}}))
And letting λ = ((ω^2)/((Subscript[ω, o])^2)) gives:

0 = (({{-2 + λ, 1}, {1, -1 + λ}}))(({{(Subscript[y, 1])}, {(Subscript[y, 2])}}))
Don't lose track of what we've just done.  We've made an inspired guess that the displacements of the masses from their equilibrium values (i.e. (Subscript[y, 1])(t) and (Subscript[y, 2])(t) ) will be oscillatory in nature and will oscillate at "natural frequencies" ω.  This is just the generalization of the exponential method we learned for single linear ODEs to the coupled case.  Information about just what these natural frequencies are is carried in λ, since λ = ((ω^2)/((Subscript[ω, o])^2)).
	The last matrix equation is just a system of two algebraic equations.  For there to be a non-trivial solution, the determinant of the matrix must vanish.  Thus:
					0 = (-2+λ)(-1+λ) -1
Which is quadratic in λ.  Its solutions are
					λ = ((3   + /-  (Sqrt[5]))/2) = 2.62 and .38
Note that the last matrix equation given above is actually the two algebraic equations
					(-2+λ)(Subscript[y, 1]) +(Subscript[y, 2]) = 0
					(Subscript[y, 1]) +(-1+λ)(Subscript[y, 2]) = 0
You can substitute in each of the lambda values found, in turn, into these two equations (which will end up being redundant) to find the ratio between (Subscript[y, 1]) and (Subscript[y, 2]) that is necessary for the motion that particular λ implies.  Picking some arbitrary value (I chose (Subscript[y, 1])=1) allows you to use this ratio to write down explicit eigenvectors.
	For λ = 2.62 (i.e. ω = 1.62 (Subscript[ω, o])):   	 (({{(Subscript[y, 1])}, {(Subscript[y, 2])}}))  = (({{1}, {- .62}}))
	
	For λ = .38 (i.e. ω = .62 (Subscript[ω, o])):   	 (({{(Subscript[y, 1])}, {(Subscript[y, 2])}}))  = (({{1}, {1.62}}))
	
All together, we have found that the system has two normal modes.  One normal mode has mass one and mass two oscillating together at frequency .62 (Subscript[ω, o]), but with mass two's amplitude being 1.62 times larger than mass one's.  The other normal mode has mass one and mass two oscillating at frequency 1.62 (Subscript[ω, o]), but here they're oscillating opposite each other (as one is going up the other is going down, and vice-versa), with the amplitude of mass two's motion being only .62 times the amplitude of mass one's motion.

c.     I already gave a physical description of each mode's motion in the previous paragraph.  Why are the natural frequencies the size that they are?  
        Let's consider the ω = .62 (Subscript[ω, o])case first.  If the eigenvector had turned out to be (({{1}, {1}}))...a very unphysical possibility in this set-up if you try to think it through, but never mind..., then both masses would be moving perfectly in unison.  The bottom spring would never stretch or compress, so it'd never exert any force on either mass one or mass two.  It would be just like m hanging from one spring of strength k.  The frequency of such an oscillator would be (Sqrt[(k/m)]) = (Subscript[ω, o]).  The eigenvector, however, is (({{1}, {1.62}})).  If the top mass dips down 1cm, the bottom mass simultaneously dips down 1.62cm.  At such a point, the top spring is pulling mass one up but the bottom spring is now extended as is pulling mass one down, partially counteracting the first spring.  This partial counteraction makes for an effectively "softer" spring controlling the oscillation, and we see that .62 (Subscript[ω, o])is indeed less than (Subscript[ω, o]).
        Now for the ω = 1.62 (Subscript[ω, o])case.  Since the eigenvector is (({{1}, {- .62}})), we see that the two masses are moving oppositely.  When one goes down, two is going up.  If mass one goes down, the top spring is extended and pulling upwards on mass one, just like before.  Now, however, mass two is coming up towards mass one, which compresses the bottom spring, which then ALSO exerts an upward force on mass one.  The spring forces add here, not subtract.  That makes for a "stiffer" effective spring controlling the oscillation, and indeed we see that 1.62 (Subscript[ω, o])is greater than (Subscript[ω, o]).

(Solution by T .  Bing)