### Basic Theory of Vibrations

```2007.7.23
Basic Theory of Vibrations
Assinment 4-1 Confirm the fact with the following procedure.
Fact : Even in a case of the damping ratio ζ > 1, the free vibration of a one DOF
vibration system shows an overshoot with a range of initial conditions.
(1)
Consider a case of the damping ratio ζ > 1. Both of two eigenvalues λ1,2 are negative
and λ1 < λ2 . Derive a free vibration response of the displacement x(t) with a set of
initial conditions of x(0) = x0 and x(0)
˙
= v0 .

x(t) =
(2)
λ2 x0 − v0 λ1 t −λ1 x0 + v0 λ2 t
e +
e
λ2 − λ1
λ2 − λ1
Derive a free vibration response of the velocity x(t)
˙
with a set of initial conditions of
x(0) = x0 and x(0)
˙
= v0 .

x(t)
˙
=
(λ2 x0 − v0 )λ1 λ1 t (−λ1 x0 + v0 )λ2 λ2 t
e +
e
λ2 − λ1
λ2 − λ1
(3)
Consider a time t1 (s) when x(t
˙ 1 ) = 0, and derive a relation of x0 , v0 and t1 .

λ1 λ2 (eλ1 t1 − eλ2 t1 )x0 + (λ2 eλ2 t1 − λ1 eλ1 t1 )v0 = 0
(4)
The initial displacement is positive; x0 > 0. Show a range of the initial velocity v0 with
no overshoot.

1
2 x0 −v0 )λ1
ln (λ
(式 A) と書ける．x が t > 0 において極値を持たないためには，
λ2 −λ1
(λ1 x0 −v0 )λ2
(λ2 x0 −v0 )λ1
(λ1 x0 −v0 )λ2
< 1 (式 B) である必要がある．分母の符号で場合を分けて，
(a) (λ1 x0 − v0 )λ2 > 0 の場合，v0 > λ1 x0 と仮定したことになり，式 B より (λ2 −
λ1 )v0 < 0 すなわち v0 < 0 を得る．このことから極値を持たない v0 の範囲のうちの 1
つは λ1 x0 < v0 < 0 となる．
(b) (λ1 x0 − v0 )λ2 < 0 の場合，v0 < λ1 x0 と仮定したことになり，式 B より (λ2 −
λ1 )v0 > 0 すなわち v0 > 0 を得る．仮定と式 B を同時に満たす初速度 v0 は存在しな
い．
v0 > 0 のとき x(t1 ) は正であり，極大であることが証明できるので，極小値を持た
ない初速度 v0 の範囲としては v0 > λ1 x0 となる．
Assinment 4-2 Consider a system is modeled as a one DOF vibration system with a
viscous damping, where, the mass is m(kg), the
√ stiffness of the spring is k(N/m), and the
damping coefficient is c(Ns/m) { c : 0 < c < 2 mk }. The equation of motion is
represented by m¨
x + cx˙ + kx = 0. As you studied at the class, its logarithmic decrement δ
and damping ratio ζ are identified from the measured displacement of its free vibration
response. Suppose a case that you have a velocity sensor instead of a displacement sensor.
(1)
Show an equation of the velocity v(t) of the free vibration response for an initial
displacement x(0) = x0 and an initial velocity v(0) = x(0)
˙
= 0.

x(t) = e−ζωn t x0 cos ωd t +
x0 ζωn + v0
sin ωd t
ωd
(1)
と書けるので，v(0) = x(0)
˙
= 0 を考慮すると，速度の応答は
−x0 ωn −ζωn t
v(t) = √
e
sin ωd t
1 − ζ2
(2)
となる．
(2)
Derive the period of the local maximum of the velocity theoretically.

Td =
2π
ωd
(3)
である．これは隣り合う変位の極大の間隔と同じである．
(3)
Derive the ratio of the local maximum value of the velocity to the next one in closed
form.

x0 ωn −ζωn tn
vn = v(tn ) = √
e
(4)
1 − ζ2
x0 ωn −ζωn tn+1
vn+1 = v(tn+1 ) = √
e
1 − ζ2
(5)
tn+1 = tn + Td
(6)
であるから，隣り合う速度の極大の比は
vn
= e−ζωn tn eζωn tna+1
vn+1
= eζωn Td
= e
√2πζ
1−ζ 2
である．これは隣り合う変位の極大の比と同じである．
(7)
(4)
Figure 4-2 shows an example of the velocity of a free vibration response. Compute
its logarithmic decrement δ and damping ratio ζ.
√
√
√
√

√
δ = ln 2 = 0.35（有効桁 2 桁)
(8)
ζ=
δ
2π 1 + {δ/(2π)}2
= 0.055（有効桁 2 桁)
(9)
となる．
(5)
Identify the spring constant k and the damping coefficient c for the mass is m = 0.1
kg.
√

ωn2 =
k =
=
=
c =
(2π)2
Td2 (1 − ζ 2 )
mωn2
(2π)2
m 2
Td (1 − ζ 2 )
1.6 × 103 N/m （有効桁 2 桁）
√
ζ2 mk
v (cm/s)
= 1.4 × 100 Ns/m （有効桁 2 桁）
3
2
1
0
−1
−2
−3
0
0.05
0.1
0.15
0.2
t(s)
Fig.4-2 Free vibration velocity response of a one DOF vibration system
(10)
(11)
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