Dynamics of Machines - Part II - IFS.pdf
Dynamics of Machines - Part II - IFS.pdf
Dynamics of Machines - Part II - IFS.pdf
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1 yini − A<br />
det<br />
λ1 vini − A<br />
C2 = =<br />
1 1<br />
det<br />
−λ1(yini − A) + (vini − iωA)<br />
λ2 − λ1<br />
λ1 λ2<br />
Summarizing, below is the analytical solution <strong>of</strong> second order differential equation, which is<br />
responsible for describing the movements <strong>of</strong> the mass-damping-spring system in time domain,<br />
as a function <strong>of</strong> the excitation force and initial condition <strong>of</strong> displacement and velocity:<br />
y(t) = C1e λ1t + C2e λ2t + Ae iωt<br />
where<br />
<br />
λ1 = −ξωn − ωn 1 − ξ2 · i<br />
<br />
λ2 = −ξωn + ωn 1 − ξ2 · i<br />
A =<br />
f/m<br />
−ω 2 + ω 2 n + i2ξωnω<br />
C1 = λ1(yini − A) − (vini − iωA)<br />
λ2 − λ1<br />
C2 = −λ1(yini − A) + (vini − iωA)<br />
λ2 − λ1<br />
Numerical Solution According to Taylor’s expansion, an equation can be approximated by:<br />
f(t) ⋍ f(t0) + df<br />
<br />
<br />
<br />
<br />
dt<br />
t=t0<br />
(t − t0) + d2 f<br />
dt 2<br />
<br />
<br />
<br />
<br />
t=t0<br />
(t − t0)... + dn f<br />
dt n<br />
<br />
<br />
<br />
<br />
t=t0<br />
(40)<br />
(t − t0) (41)<br />
Assuming a very small time step t −t0 ≪ 1, the higher order terms <strong>of</strong> eq.(42) can be neglected.<br />
It turns:<br />
f(t) ⋍ f(t0) + df<br />
<br />
<br />
<br />
<br />
dt<br />
t=t0<br />
(t − t0) (42)<br />
Knowing the initial conditions <strong>of</strong> the movement when t = 0,<br />
y(0) = y0 = yini<br />
˙y(0) = ˙y0 = vini<br />
and the equation <strong>of</strong> motion, which has to be solved,<br />
¨y(t) = − 2ξωn ˙y(t) − ω 2 ny(t) + f<br />
m eiωt<br />
13<br />
(43)