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Skylab. C hams: {0(1344 % ya,» ) (C) v \c’L‘ Section 9.1, Problem 17 Problem. The equation of motion of a springmass system with damping (see
Section 3.8) is
d2u du m—Cu—2+czi?+ku=0 where m, c, and k are positive. Write this second order equation as a system of ﬁrst
order equations for :1: = u, y = 53%. Show that x = 0,11; = 0 is a critical point, and
analyze the nature and stability of the critical point as a function of the parameters m, c, and k. A similar analysis can be applied to the electrical circuit equation (see Section 3.8) d2] d1 1 ’
L— R— —I = 0.
dt2 + dt + 0
Solution. The ﬁrst equation in the system is easy; it will be y = %’;—‘, or
in the new variables x and y, y = 5%. This is common to all problems in which one changes a single highorder differential equation into a system of ﬁrstorder
differential equations. To ﬁnd the second equation, write the original differential equation, entirely
in terms of the new variables. Instead of 93%, write 3;. Instead of 3%, write gig. This
makes the second equation: dy mE+cy+kx=Q Generally, a system of differential equations is written as ‘13”; = f (as, y), it =
g(ac, y), as that makes it more convenient to write the system as a matrix multi
plication. So, moving all the ﬁrst derivatives to the left side of the equation, and
everything else to the right, gives one the pair of equations, and so the system of ﬁrstorder differential equations: d_w_
dt_y @_£,_£
dt_ m y This can then be written in the form of a matrix equation, using the vector M rw/ x 0 1 a? = and the matrix A = k c — so that :E" = .A56.
31 ‘5 ‘R
The point a; = 0, y = 0 has to be a critical point; at these values, % = 0 and
43 = 0
dt ' To analyze this system, look at the eigenvalues for this matrix — which, despite
there being symbols in the second row, are calculated the same way as if there were numerals. As a result, the eigenvalues are A___c_i\/02—4mk 2m 2m It’s assumed in this problem that c is always positive, so the only thing that
changes behavior is the plus or minus sign of (:2 — 4mk: If c2 — 4mk > 0, then both
eigenvalues will be distinct, real, negative numbers. This means that the equilibrium
at a; = 0, y = 0 will be an asymptotically stable, proper, node. If 02 — 4mk = 0, there is only a single, real, negative eigenvalue. This will still
be an asymptotically stable node, but it will be an improper one. I If 02 ——4mk < 0, there will be a pair of complexvalued eigenvalues. The system
will be an asymptocially stable (because the real component, —§%, will be less than zero) spiral. ...
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 Spring '04
 Yoon
 Differential Equations, Equations

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