Mathematical Methods for Physicists: A concise introduction - Site Map

ORDINARY DIFFERENTIAL EQUATIONS
Obviously p 2 < 0 also. Thus, y…t† !0ast !1. This means that the oscillation
dies out with time and eventually the mass will assume the static equilibrium
position.
Case 2 b 2 4km ˆ 0 (critical damping)
The solution is of the form
y…t† ˆe bt=2m …c 1 ‡ c 2 t†:
As both b and m are positive, y…t† !0ast !1as in case 1. But c 1 and c 2 play a
signi®cant role here. Since e bt=2m 6ˆ 0 for ®nite t, y…t† can be zero only when
c 1 ‡ c 2 t ˆ 0, and this happens when
t ˆc 1 =c 2 :
If the number on the right is positive, the mass passes through the equilibrium
position y ˆ 0 at that time. If the number on the right is negative, the mass never
passes through the equilibrium position.
It is interesting to note that c 1 ˆ y…0†, that is, c 1 measures the initial position.
Next, we note that
y 0 …0† ˆc 2 bc 1 =2m; or c 2 ˆ y 0 …0†‡by…0†=2m:
Case 3 b 2 4km < 0 (underdamping)
The auxiliary equation now has complex roots
p 1 ˆ b
2m ‡ i p
4km b 2 ; p
2m
2 ˆ b
2m i p
4km b 2
2m
and the solution is of the form
h p t
p
y…t† ˆe bt=2m c 1 cos 4km b 2
2m ‡ c 2 sin 4km b 2 t
i
;
2m
which can be rewritten as
y…t† ˆce bt=2m cos…!t †;
where
q
c ˆ c 2 1 ‡ c2 2;ˆ tan 1 c 2
; and ! ˆ
c 1
p
4km b 2 =2m:
As in case 2, e bt=2m ! 0ast !1, and the oscillation gradually dies down to
zero with increasing time. As the oscillator dies down, it oscillates with a frequency
!=2. But the oscillation is not periodic.
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