1
Theoretical Mechanics
2
3
Vibrations are common phenomena in everyday life and in
engineering practice,
Such as the swinging of a clock pendulum,the bumping of a
traveling car,the vibrations of working machine parts or motors,
and the quake of buildings caused by an earthquake,
3,The aim of the study of vibrations is the elimination or
reduction of harmful oscillations,or the helpful use of
them for special purposes,
2,Advantages and disadvantages of vibrations,Vibrating
feeder,vibrating screen,vibrating pile driver extractor,for
example,are advantageous applications,But vibrations
induce friction loss,affect intensity,make noise,have bad
effects on working conditions,consume unnecessary energy
and reduce the precision of a machine etc,
1,Vibrations are the back and forth movements of a system
around its equilibrium position,
4
振动是日常生活和工程实际中常见的现象。
例如：钟摆的往复摆动,汽车行驶时的颠簸，电动机、机
床等工作时的振动，以及地震时引起的建筑物的振动等。
利,振动给料机 弊,磨损，减少寿命，影响强度
振动筛 引起噪声，影响劳动条件
振动沉拔桩机等 消耗能量，降低精度等。
3,研究振动的目的,消除或减小有害的振动，充分利用振动
为人类服务。
2,振动的利弊,
1,所谓振动就是系统在平衡位置附近作往复运动。
5
The present chapter discusses mainly free vibrations and forced
vibrations of systems with one degree of freedom,
4,Classification of vibrations,Vibrations of systems with one degree of freedom
Vibrations of systems with
more than one degree of
freedom and
Vibrations of elastic bodies
According to the number of
degrees of freedom of a vibrating
system they can be classified into
According to the cause of the vibrations they can be classified into
free vibrations including undamped free vibrations and damped free vibrations
forced vibrations including undamped forced vibrations and
damped forced vibrations
self-excited forced vibrations
6
4,振动的分类, 单自由度系统的振动
按振动系统的自由度分类 多自由度系统的振动
弹性体的振动
按振动产生的原因分类,
自由振动,无阻尼的自由振动
有阻尼的自由振动，衰减振动
强迫振动,无阻尼的强迫振动
有阻尼的强迫振动
自激振动
本章重点讨论单自由度系统的自由振动和强迫振动。
7
§ 18–1 Undamped free vibration of a system
with one degree of freedom
§ 18–2 Methods of determination of the
natural frequency of a system
§ 18–3 Damped free vibration of a system
with one degree of freedom
§ 18–4 Undamped forced vibration of a
system with one degree of freedom
§ 18–5 Damped forced vibration of a
system with one degree of freedom
§ 18–6 The concepts of critical speed of
rotation,vibration reduction and
vibration isolation
Chapter 18,Mechanical vibrations
8
§ 18–1 单自由度系统无阻尼自由振动
§ 18–2 求系统固有频率的方法
§ 18–3 单自由度系统的有阻尼自由振动
§ 18–4 单自由度系统的无阻尼强迫振动
§ 18–5 单自由度系统的有阻尼强迫振动
§ 18–6 临界转速 · 减振与隔振的概念
第十八章 机械振动基础
9
§ 18-1 Undamped free vibration
of a system with one degree of freedom
1.Concept of free vibration
motor
Elastic foundation
10
§ 18-1 单自由度系统无阻尼自由振动
一、自由振动的概念,
11
Equilibrium
position x=0
Equ
ilibr
ium
pos
ition
j=0
Equ
ilib
riu
m
pos
itio
n j
=0
12
13
A force acting on a vibrating body which is always
directed towards its equilibrium position is called a restoring force,
After the action of an initial disturbance,the vibration of a system
around its equilibrium position under the action of only a restoring
force is called an undamped free vibration,
)5,/( 0,
)5,/( 0,
)/( 0,
22
222
22
?
?
????
????
????
??????
??????
?????
j?j?jjj
j?j?jjj
??
Im g am g aI
lgm g lml
mkxxkxxm
nn
nn
nn
For a mass-spring system,
For a simple pendulum,
For a compound pendulum,
14
运动过程中，总指向物体平衡位置的力称为 恢复力 。
物体受到初干扰后，仅在系统的恢复力作用下在其平衡位
置附近的振动称为 无阻尼自由振动 。
)/( 0,22 mkxxkxxm nn ????? ??????质量 —弹簧系统,
单摆,
复摆,)5,/( 0,)5,/( 0,22
222
?
?
????
????
??????
??????
j?j?jjj
j?j?jjj
Im g am g aI
lgm g lml
nn
nn
15
2,Differential equation and its solution for an undamped
free vibration of a system with one degree of freedom
For any system with one degree of freedom,q being the generalized
coordinate measured from the equilibrium position,the differential
equation of a free vibration for a small displacement is
0?? cqqa ??
where a and c are constants related to the physical parameters of
the system,Writing
acn /2 ??
the standard form of the differential equation of free
vibration is
.02 ?? qq n???
Its solution is,)s i n ( ?? ?? tAq
n
16
二、单自由度系统无阻尼自由振动微分方程及其解
对于任何一个单自由度系统，以 q 为广义坐标（从平衡位
置开始量取 ），则自由振动的运动微分方程必将是,
0?? cqqa ??
a,c是与系统的物理参数有关的常数。令 acn /2 ??
则自由振动的微分方程的标准形式,
02 ?? qq n???
解 为,)s i n ( ?? ?? tAq
n
17
Assuming that at t = 0, we get
00 an d qqqq ?? ??
,a r c t g,
0
0
2
2
02
0 q
qqqA n
n ?
? ??
?
???
or,
tCtCq nn ?? s inc o s 21 ??
./,w i t h 02 01 nq CqC ????
.s i nc o s T h e r e f o r e,00 tqtqq n
n
n ???
???
18
0
0
2
2
02
0 a r c t g,q
qqqA n
n ?
? ??
?
???
设 t = 0 时,则可求得,
00,qqqq ?? ??
或,
tCtCq nn ?? s inc o s 21 ??
C1,C2由初始条件决定为
nq CqC ?/,02 01 ???
tqtqq n
n
n ??? s inc o s
0
0
????
19
n?
3.Properties of a free vibration without damping,
A—— The quantity,which is the maximum distance of a vibrating
body from the equilibrium position,is called the amplitude,
?n t + ? —— The quantity,which defines the position of the vibrating
body at any given time,is called the phase of the vibration,
? —— The quantity,which defines the initial phase,at which the
motion starts,
T —— The time T during which the vibrating body makes one
complete vibration is called the period of vibration,
f —— The quantity,which specifies the number of oscillations per
second and is the inverse of the period,is called the frequency
of the vibration,f = 1 / T,
—— The quantity,which specifies the number of oscillations
in 2 seconds,is called natural frequency of the vibration,
It characterizes the dynamics of a given vibrating system and
depends on the inherent parameters describing the system,
.2
n
T ???
?
20
三、自由振动的特点,
A—— 物块离开平衡位置的最大位移，称为振幅。
?n t + ? —— 相位，决定振体在某瞬时 t 的位置
? —— 初相位，决定振体运动的起始位置。
T —— 周期，每振动一次所经历的时间。
f —— 频率，每秒钟振动的次数,f = 1 / T 。
—— 固有频率，振体在 2?秒内振动的次数。
反映振动系统的动力学特性，只与系统本身的固有
参数有关。
n
T ??2?
n?
21
Properties of undamped free vibrations,
4,Other results
1,If a constant force is applied to the system in the direction of its
vibration then this constant force does not affect the law of the
vibration of the system,It only displaces the equilibrium position O
into the direction of this force,The amplitude,frequency and phase
of the vibration do not change,
(2) The amplitude A and the initial phase depend on the initial
conditions (initial displacement and initial velocity),
(1) It is a simple harmonic vibration,
(3)The period T and the natural frequency depend only on the
natural parameters of the system itself (m,k,I,etc.),
n?
?
22
无阻尼自由振动的特点是,
(2) 振幅 A和初相位 ? 取决于运动的初始条件 (初位移和初速度 )；
(1) 振动规律为简谐振动；
(3)周期 T 和固有频率 仅决定于系统本身的固有参数 (m,k,I )。 n?
四、其它
1,如果系统在振动方向上受到某个常力的作用，该常力
只影响静平衡点 O的位置，而不影响系统的振动规律，如振动
频率、振幅和相位等。
23
2,Equivalent
stiffness of series and
parallel spring systems
,t h e r e f o r e
,)( So
,
21
21
21
21
2
2
1
1
kkk
kk
mg
kkmg
FFmg
k
F
k
F
eq
stst
st
??
?
???
????
??
?
For a parallel spring
system 21
21
eq
21
2121
21
k T h e r e f o r e
)
11
(
)
11
(
kk
kk
kk
mg
k
mg
kk
mg
k
mg
k
mg
eq
st
ststst
?
?
???
????
??
?
???
For a series spring system,
pa
rall
el
ser
ies
24
2,弹簧并联系
统和弹簧串联系
统的等效刚度
21
21
21
21
2
2
1
1
,)(
,
kkk
kk
mg
kkmg
FFmg
k
F
k
F
eq
stst
st
???
?
????
????
??
?
并联
21
21
eq
21
2121
21
k
)
11
(
)
11
(
kk
kk
kk
mg
k
mg
kk
mg
k
mg
k
mg
eq
st
ststst
?
??
???
????
??
?
???
串联
并
联
串
联
25
§ 18-2 Methods of determination
of the natural frequency of a system
1,The differential equation of the vibration of a system in
the standard form is
02 ?? qq n???
3,Energy Method,
n?From the equation Tmax=Umax we can determine,
st
n
g
?? ?
st?
where =const,is the static
displacement of the system
under the action of a constant
force,
2,Applying the method of static deformation,we obtain,
26
1,由系统的振动微分方程的标准形式
2,静变形法,
3,能量法,
§ 18-2 求系统固有频率的方法
02 ?? qq n???
st
n
g
?? ?
st?
：集中质量在全部重力
作用下的静变形
n?由 Tmax=Umax,求出
27
A system with undamped free vibration is a conservative system,
The mechanical energy of system does not change during its
motion,
When the vibrating body moves to the maximum distance from
the equilibrium position,its velocity becomes zero,Hence,the
kinetic energy of the system is zero when its potential energy has
its maximum value (choose the static equilibrium position of the
system as the zero point of potential energy),
When the vibrating body moves to the equilibrium position,the
potential energy of the system is zero,while its kinetic energy is
maximum,
m g AAkU stst ???? ])[(21 22m a x ??
2
m a x 2
1 So kAUmgk
st ???
222m a x 2121 nmAxmT ??? ?
Example,
28
无阻尼自由振动系统为保守系统，机械能守恒。
当振体运动到距静平衡位置最远时，速度为零，即系统
动能等于零，势能达到最大值（取系统的静平衡位置为零势
能点）。
当振体运动到静平衡位置时，系统的势能为零，动能达
到最大值。
m g AAkU stst ???? ])[(21 22m a x ??
2m a x 21 kAUmgk st ????
222m a x 2121 nmAxmT ??? ?
如,
29
m
k kA mA
n n ? ? ? ?,,2 1 2 1
2 2 2
U T ? max max From
The energy method is based on the conservation law of the
mechanical energy,It is a simple and convenient method in
order to compute the natural frequency of a more complicated
vibrational system,
[Example 1] The system is shown in the
figure.Assume that the system does not
swing horizontally,The wheel,which is
homogeneous of radius R and mass M,and
the string do not slide one relative to the
other,The load has mass M,Neglecting the
mass of the string and of the spring,write
down the differential equation of vibration
of the system and determine its natural
frequency,
we get
30
m
kkAmA
UT
nn ??
?
??
2
1
2
1
222
m a xm a x由
能量法是从机械能守恒定律出发，对于计算较复杂的振
动系统的固有频率来得更为简便的一种方法。
例 1 图示系统。设轮子无侧向摆动，
且轮子与绳子间无滑动，不计绳子和弹
簧的质量，轮子是均质的，半径为 R，质
量为 M，重物质量 m,试列出系统微幅
振动微分方程，求出其固有频率。
31
Solution 1,Let x be the generalized
coordinate,The origin of the coordinate is at
the position of static equilibrium of the
system,
RkgRmM st 2)( ??? ?
gk mMst ??? 2?
In static equilibrium we have
At an arbitrary position x we get
kxgmMxkF st 22)2( ????? ?
,
32
解,以 x 为广义坐标（静平衡位置为
坐标原点）
RkgRmM st 2)( ??? ?
gk mMst ??? 2?
则任意位置 x 时,
kxgmMxkF st 22)2( ????? ?
静平衡时,
33
Applying the theorem of kinetic energy we have,
k x RRFgRmMFm
xRmM
R
xMRRxMRxmL
A
A
42)()(
)
2
3(
2
1 2
???????
??
???
?
??
From,we get
?? )( FmdtdL AA k x RxRmM 4)
23( ??? ??
The differential equation of vibration is,
Therefore,the natural frequency is
mM
k
x
mM
kx
n 23
8
0
23
8
?
?
?
?
?
?
??
34
应用动量矩定理,
k x RRFgRmMFm
xRmM
R
xMRRxMRxmL
A
A
42)()(
)
2
3(
2
1 2
???????
??
???
?
??
由, 有 ?? )( Fm
dt
dL
AA k x RxRmM 4)23( ??? ??
振动微分方程,
固有频率,
mM
k
x
mM
kx
n 23
8
0
23
8
?
?
?
?
?
?
??
35
Solution 2,Let x be the generalized coordinate,the origin is at the position of static equilibrium
of the system,
Applying the conservation law of mechanical
energy we have
22
2
2
2
1)(
22
1
2
1 xm
R
xMRxMT ??? ???
Choose the position of equilibrium as the zero point of potential
energy,When the displacement of the center of the wheel is x,the
elongation of the spring is 2x,
gxmMxkU stst )(])2[(2 22 ????? ??
In equilibrium,we have
gxmMxk st )(2 ??? 22 k xU ? So
2)
2
3(
2
1 xmM ???
gxmMxkkx st )(22 2 ???? ?
36
解 2, 用机械能守恒定律
以 x为广义坐标（取静平衡位置为原点）
2
22
2
2
)
2
3(
2
1
2
1)(
22
1
2
1
xmM
xm
R
xMRxMT
?
???
??
???
以平衡位置为计算势能的零位置，
并注意轮心位移 x时，弹簧伸长 2x
gxmMxkkx
gxmMxkU
st
stst
)(22
)(])2[(2
2
22
????
?????
?
??
因平衡时
gxmMxk st )(2 ???
22 kxU ??
37
From T+U= we get
const
.2)23(21 22 c o n s tkxxmM ??? ?
04)23( ??? kxxmM ??
mM
k
x
mM
k
x
n 23
8
0
23
8
?
?
?
?
?
?
??
Differentiation with respect to time t and elimination of the
common multiplier results in x?
38
由 T+U= 有,
const
c o n s tkxxmM ??? 22 2)23(21 ?
04)23( ??? kxxmM ??
mM
k
x
mM
k
x
n 23
8
0
23
8
?
?
?
?
?
?
??
对时间 t 求导，再消去公因子,得 x?
39
Solution,choose the
distance x of C from the
equilibrium position as the
generalized coordinate,The
maximum kinetic energy of
the system is
Example 2 A tab wheel,whose mass is M and whose gyroradius with
respect to the center of the wheel is,rotates without sliding on a
horizontal plane,The radii of the big and the small wheels are R and r,
the stiffness of the two springs are and,the load is m,Neglecting
the masses of the wheel and the springs,and assuming the string is not
extensible,determine the natural frequency of the system in small
oscillation,
?
1k 2k
40
例 2 鼓轮：质量 M，对轮心回转半径 ?，在水平面上只滚不滑，
大轮半径 R，小轮半径 r,弹簧刚度,重物质量为 m,不计
轮 D和弹簧质量，且绳索不可伸长。求系统微振动的固有频率。
21,kk
解,取静平衡位置 O为坐标原
点，取 C偏离平衡位置 x为广义
坐标。系统的最大动能为,
41
)
)(
)(
( )(
2
1
])) [ ((
2
1
21
2
m a x21
m a x
22
m a x21m a x
Rkk
rRmg
xkk
x
R
rRmgxkkU
st
stst
?
?
???
??????
?
??
.] [
2
1
)(
2
1
)(
2
1
)(
2
1
2
m a x
222
2
2
m a x
2m a x22
m a xm a x
xr)m ( R)RM(
R
x
R
rR
m
R
x
MxMT
?
?
?
?
????
?
?
??
?
?
The maximum potential energy of the system is,
42
)
)(
)(
( )(
2
1
])) [ ((
2
1
21
2
m a x21
m a x
22
m a x21m a x
Rkk
rRmg
xkk
x
R
rRmgxkkU
st
stst
?
?
???
??????
?
??
2
m a x
222
2
2
m a x
2m a x22
m a xm a x
] [
2
1
)(
2
1
)(
2
1
)(
2
1
xr)m ( R)RM(
R
x
R
rR
m
R
x
MxMT
?
?
?
?
????
?
?
??
?
?
系统的最大势能为,
43
Writing we have
)s i n ( ?? ?? tAx n,,m a xm a x nAxAx ??? ?
,)(21,2 )()( 221m a x222
222
m a x AkkUAR
rRmRMT
n ??
???? ??
From Tmax=Umax we then get
222
2
21
)()(
)(
rRmRM
Rkk
n ???
??
??
44
设 则有
)s i n ( ?? ?? nAx nAxAx ??? m a xm a x,?
)(21 2 )()( 221m a x222 222m a x AkkUAR rRmRMT n ?????? ??
根据 Tmax=Umax,解得
222
2
21
)()(
)(
rRmRM
Rkk
n ???
??
??
45
§ 18-3 Damped free vibration
of a system with one degree of freedom
1,Concept of Resistance,
Resistance,Resistance is the force acting on an oscillation system
restoring its equilibrium,
Viscous Damping,In many cases,the resistance caused by the
surrounding medium is proportional to the first power of
the velocity,Such a resistance is called viscous damping,
vcR ??
Its projection form is xcR
x ???
where C is called the viscous resistance coefficient,or coefficient
of damping,
46
§ 18-3 单自由度系统的有阻尼自由振动
一、阻尼的概念,
阻尼,振动过程中，系统所受的阻力。
粘性阻尼,在很多情况下，振体速度不大时，由于介质粘性
引起的阻尼认为阻力与速度的一次方成正比，这种阻尼称为粘
性阻尼。
vcR ??
投影式,xcR x ???
c —— 粘性阻尼系数，简称阻尼系数。
47
2.Differential equation of a damped free vibration and its solution,
For a spring system with viscous damping we get,xckxxm ??? ???
2 and
2
n ? ? m
c n
m
k ? we write
Using the abbreviations
0 2 2 ? ? ? x x n x n ? ?? ?
This is the differential equation of the damped free vibration in
standard form,
48
二、有阻尼自由振动微分方程及其解,
质量 — 弹簧系统存在粘性阻尼,xckxxm ??? ???
02 2,22n ????? xxnx mcnmk n?? ???则令
有阻尼自由振动微分方程的标准形式。
49
The general solution is discussed for three cases,
1,In the case of small resistance( or ) the
general solution is
mk c 2 < n n < ?
)s i n ( ?? ?? ? tAex dnt
00
22
01
22
2
002
0 tg ;
)(
nxx
nx
n
nxxxA n
n ?
??
?
??? ?
?
? ??
?
22 nnd ?? ??
0 x x ??? 0 x x ? 0 t ?
Where is the circular frequency of the damped free
vibration,Assuming at and,we get
50
其通解分三种情况讨论,
1、小阻尼情形 mkcn
n 2 )( << ?
)s i n ( ?? ?? ? tAex dnt
22 nnd ?? ?? — 有阻尼自由振动的圆频率
则时设,,,0 00 xxxxt ?? ???
00
22
01
22
2
002
0 tg ;
)(
nxx
nx
n
nxxxA n
n ?
??
?
??? ?
?
? ??
?
51
Properties of a damped vibration,
(1) The period of the vibration
increases,the frequency decreases,
mk
cn
n
T
n
n
d
d
2
1
2
22
22
22
??
?
?
?
??
?
?
??
?
?
?
?
?
—— ratio of resistance
Hence
If then and
we get approximately
nn ?<< 1<<?
.,TT dnd ?? ??2
2
2
1
1
1
???
?
?
??
??
?
?
nd
d
d
ff
TT
52
衰减振动的特点,
(1) 振动周期变大,
频率减小 。
mk
cn
n
T
n
n
d
d
2
1
2
22
22
22
??
?
?
?
??
?
?
??
?
?
?
?
?
—— 阻尼比
有阻尼自由振动,
当 时,
可以认为
nn ?<< 1<<?
TT dnd ?? ??2
2
2
1
1
1
???
?
?
??
??
?
?
nd
d
d
ff
TT
53
(2) The amplitude of the vibration decreases in geometric progression,
The logarithmic decrement is defined as
??
?
??
?
?
2
1
2
lnln
2
1
?
?
?
???
?
d
nT
i
i nTe
A
A
d
d
di
i
nT
Ttn
nt
i
i e
Ae
eA
A
A ??
??
?
?
)(
1
The ratio of two adjacent amplitude is
2,In the case of critical damping
or 1,( ?? ?? nn )2 mkc c ?
])([ 000 tnxxxex nt ??? ? ? ).,,0A t ( 00 xxxxt ?? ???
critical viscous resistance coefficient
the general solution is
54
(2) 振幅按几何级数衰减
对数减缩率
??
?
??
?
?
2
1
2
lnln
2
1
?
?
?
???
?
d
nT
i
i nTe
A
A
d
2、临界阻尼情形
临界阻尼系数
) 1,( ?? ?? nn
mkc c 2?
])([ 000 tnxxxex nt ??? ? ? ),,
(at 0 0 x x x x ??? ? t 0 ?
d
di
i
nT
Ttn
nt
i
i e
Ae
eA
A
A ??
??
?
?
)(
1
相邻两次振幅之比
55
We can see that the body tends to move to the position of
equilibrium with increasing time exponentially slowly,
Substituting the initial conditions (at ) we get and 0 0 x x x x ??? ? 0t ?
22
00
22
222
0
22
0
1 2
)(,
2
)(
n
n
n
n
n
xxnnC
n
xnnxC
?
?
?
?
?
?????
?
???? ??
)( 2222 21 tn tnnt nn eCeCex ?? ???? ??
,1,( ?? ?? nn )ccc ?3,In the case of overdamping
the general solution is
The motion is not a periodic one,With increasing time the
quantity x goes exponentially to zero,the body does not vibrate,
56
可见，物体的运动随时间的增长而无限地趋向平衡位置，
不再具备振动的特性。
)( 2222 21 tn tnnt nn eCeCex ?? ???? ??
代入初始条件 ),,0(
00 xxxxt ?? ??? 时
22
00
22
222
0
22
0
1 2
)( ;
2
)(
n
n
n
n
n
xxnnC
n
xnnxC
?
?
?
?
?
?????
?
???? ??
) 1,( ?? ?? nn )( ccc ?3、过阻尼（大阻尼）情形
所示规律已不是周期性的了，随时间的增长,x 0，
不具备振动特性。
57
Example 3 In a spring system W=150N,?st=1cm,A1=0.8cm,
A21=0.16cm,Determine the viscous resistance coefficient c,
20
21
20
3
2
2
1
21
1 )( dnTe
A
A
A
A
A
A
A
A ???? ?
Solution,
20)(16.0 8.0 dn Te ???
21
220205ln
??
?????
??? n
n
dn T
Because is very small we can write approximately
and get ? ??405ln ?
).s / c mN(1 2 2.0
9 8 01
1 5 0
2
40
5ln
2
40
5ln
2
2
??
?
????????
???
?
st
W
g
W
mkc
58
例 3 质量弹簧系统,W=150N,?st=1cm,A1=0.8cm,
A21=0.16cm。 求阻尼系数 c 。
20
21
20
3
2
2
1
21
1 )( dnTe
A
A
A
A
A
A
A
A ???? ?
解,
20)(16.0 8.0 dn Te ???
21
220205ln
??
?????
??? n
n
dn T
由于 很小,? ??405ln ?
)s /c mN(122.0
9801
1502
40
5ln2
40
5ln2 2
??
?
????????
???
?
st
W
g
Wmkc
59
§ 18-4 Undamped forced vibration
of a system with one degree of freedom
1,Concept of forced vibration,
A forced vibration is the oscillation of a body under the action of an
external disturbing force (except the restoring force),
A simple harmonic disturbing force is given by,
where H is the amplitude of the force,? is its circular frequency
and ? is its initial phase,
)s i n ( ?? ?? tHS
)s i n (2 ??? ??? thxx n??
This is the differential equation of an undamped forced vibration in
standard form,It is a linear inhomogeneous differential equation of
second order,its solution is
)s i n ( ?? ???? tHkxxm ??
with we can write,2 m H h m k n ? ? ?
2,Differential equation and its solution,
60
§ 18-4 单自由度系统的无阻尼强迫振动
一、强迫振动的概念
强迫振动：在外加激振力作用下的振动。
简谐激振力,
H— 力幅； ?— 激振力的圆频率 ； ? — 激振力的初相位。
)s i n ( ?? ?? tHS
)s i n ( ?? ???? tHkxxm ??
则令,2 mHhmkn ???
)s i n (2 ??? ??? thxx n??
无阻尼强迫振动微分方程的标准形式，
二阶常系数非齐次线性微分方程。
二、无阻尼强迫振动微分方程及其解
61
)s i n ()s i n ( 22 ?????? ????? thtAx
n
n
The total solution is,
x2 is a stable state force vibration,
3,Main properties of stable state force vibration,
1)It is the additional vibration of a system in one dimension,caused
by a simple harmonic disturbing fore,
2)The frequency of this part is equal to the frequency of the
simple harmonic disturbing force,it does not depend on the mass and
the stiffness of the system,
3)The amplitude of it depends on the natural frequency
of the system and of the frequency and the amplitude of the disturbing
force,but not the initial conditions,
)s i n (,w it h 22222 ?????? ????? thxhb
nn
w i t h21 xxx ??
) sin( 2 ? ? ? ? t b x
) sin( 1 ? ? ? ? t A x n the general solution of the corresponding
homogeneous differential equation and
a particular solution of the complete equation,
62
21 xxx ??
)s i n (
)s i n (
2
1
??
??
??
??
tbx
tAx n
为对应齐次方程的通解
为特解
)s in (,22222 ?????? ????? thxhb
nn
)s i n ()s i n ( 22 ?????? ????? thtAx
n
n
全解为,
稳态强迫振动
3、强迫振动的振幅大小与运动初始条件无关，而与振动系统
的固有频率、激振力的频率及激振力的力幅有关。
三、稳态强迫振动的主要特性,
1、在简谐激振力下，单自由度系统强迫振动亦为简谐振动。
2、强迫振动的频率等于简谐激振力的频率，与振动系统的
质量及刚度系数无关。
63
? is the amplitude ratio (or the
dynamic coefficient),? is the
frequency ratio,This is the ? - ?
curve,the dependence of the
amplitude on the frequency,
1
(1) At ? =0
k
Hhb
n
?? 20 ?
(3) At, the phase of the forced vibrations is inverse to
the phase of the disturbing force,i.e.,the phase shift is rad ?n?? ?
22 ?? ??
n
hb
(2)At, the amplitude b of this vibration increase with ?,
and goes to infinity with ? ??n,
n??<
where b decreases with increasing ?,
If ? goes to,b goes to b0,If ? goes to,b goes to zero,n?2 ?
64
(1) ? =0时
k
Hhb
n
?? 20 ?
(2) 时，振幅 b随 ? 增大而增大；当 时,
n?? ? ??bn??<
(3) 时，振动相位与激振力相位反相，相差 。 rad ?
n?? ?
22 ?? ??
n
hb
b 随 ? 增大而减小； 0 ;,2
0 ????? bbbn 时时 ???
? — 振幅比或称动力系数
? — 频率比
?— ? 曲线 幅频响应曲线
（幅频特性曲线） 1
65
At resonance we get
)co s (2 ?? ?? tBtx n
)c o s (
2
So
2
2
??
?
?
???
??
tt
h
x
h
B
n
n
n
4,Resonance
then n ? ? ? b is the infinite( ),this phenomenon is called
resonance,
If ?
66
4、共振现象
,时n?? ? ??b,这种现象称为共振。
此时,)co s (2 ?? ?? tBtx n
)c o s (
2
2
2
??
?
?
????
??
tt
h
x
h
B
n
n
n
67
§ 18-5 Damped forced vibration
of a system with one degree of freedom
1,Differential equation of a damped forced
vibration and its solution,
,s i n,,tHQxcRkxF xxx ?????? ?
.s in tHxckxxm ????? ???
thxxnx n ?? s i n2 2 ??? ???
This equation is the differential equation of a damped forced
vibration in standard form,It is a linear inhomogeneous
differential equation of second order,Its solution is
21 xxx ??
Dividing both sides of the last equation by m
and introducing the notations
m
Hh
m
cn
m
k
n ???,2,
2?
we obtain
68
§ 18-5 单自由度系统的有阻尼强迫振动
一、有阻尼强迫振动微分方程及其解
tHQxcRkxF xxx ?s i n,,????? ?
tHxckxxm ?s in???? ???
将上式两端除以 m,并令
mHhmcnmkn ??? ; 2 ; 2?
thxxnx n ?? s i n2 2 ??? ???
有阻尼强迫振动微分方程的标准形式，二阶常系数非齐次微
分方程。
21 xxx ??
69
22
22222
2
tg
4)(
??
?
?
???
?
?
??
?
n
n
n
n
h
b — the amplitude of the forced vibration,
? is the phase shift of the forced vibration,the
difference to the phase of the disturbing force,
)s in ()s in ( 22 ????? ????? ? tbtAex nnt
The total solution of the differential equation of the vibration is
damped vibration forced vibration
in the case of small resistance ),s in ( 22
1 ??? ??? ? tAex nnt
and x2 is a particular solution of the complete equation
)s i n (2 ?? ?? tbx
where x1 is the general solution of the corresponding linear
homogeneous equation
),02( 2 ??? xxnx n????
Substituting this expression into the differential equation in
standard form we obtain
70
x
1是齐次方程的通解 )02( 2 ??? xxnx n????
小阻尼,
)s in ( 221 ??? ??? ? tAex nnt
（ A,? 积分常数，取决于初始条件）
x2 是特解,)s i n (
2 ?? ?? tbx
代入标准形式方程并整理
22
22222
2
tg
4)(
??
?
?
???
?
?
??
?
n
n
n
n
h
b — 强迫振动的振幅
— 强迫振动相位滞后激振力相位角
振动微分方程的全解为
)s in ()s in ( 22 ????? ????? ? tbtAex nnt
衰减振动 强迫振动
71
At the beginning of the vibration,the process with both the natural and the forced vibration simultaneously,is called the transient
process,
The other process of forced vibration after the transient process,is called the stable state process,It is the subject to be discussed in
detail,
2,Effect of resistance on forced vibration,
frequency ratio,the amplitude ratio and the resistance ratio n n
n
b
b
? ? ? ?
? ? ? ? ? and are the,
0
We get
.1 2 t g,
4)1(
1
22222 ?
???
???
? ??
??
?
1,The vibration given by is a simple harmonic
motion,。
2,Frequency,The frequency of the forced vibration is equal to the
frequency of the disturbing force,
3,Amplitude,
)s i n (2 ?? ?? tbx
72
振动开始时，二者同时存在的过程 —— 瞬态过程。
仅剩下强迫振动部分的过程 —— 稳态过程。需着重讨论部分。
nn
nbb ?????? ??? ;,
0
令
频率比 振幅比 阻尼比
因此,
22222 1
2 t g;
4)1(
1
?
???
???? ?????
二、阻尼对强迫振动的影响
1、振动规律 简谐振动。
2、频率,有阻尼强迫振动的频率，等于激振力的频率。
3、振幅
)s i n (2 ?? ?? tbx
73
(3)
(1) ? 1 << ? ) << ? ( ?
n
0 b (b
At tends to 1
tends to )and the resistance can be neglected,
(2) ? 1 ?? ? ) ?? ? ( ? n At tends to 0
and the resistance can also be neglected,
If tends to 1 and the
resistance has a remarkable effect on
the amplitude,
? 70.0<?
When ? does not change,the amplitude
decreases strongly with increasing
resistance,
2 2 2 2 1 2 ? ? ? ? ? ? ? ? n n n we obtain 0 ? ? d db From
At this frequency
2
0
m a x22m a x 12o r 2 ??? ????
bb
nn
hb
n
— the resonance frequency
74
(1)
,1,)(1 ?<<<< ???? 时n
可不计阻尼。,0bb ?
(2)
,0,)(1 ????? ???? 时n
阻尼也可忽略。
时时 0, 7 0,)(1 <?? ???? n
(3)
阻尼对振幅影响显著。 ?一
定时，阻尼增大，振幅显著
下降。
222 212,0 ????? ????? nn nddb 得由
— 共振频率
此时,
2
0
m a x22m a x 12 2 ??? ????
bb
nn
hb
n
或
75
4,Phase difference,
Damped forced vibrations always have a phase behind that of the
disturbing force,This phase difference is given by the angle ? with
21
2tg
?
???
??
(1) The value of ? is always between 0 and ? 。
(2) The phase-frequency curve (the ? - ? - curve) is a monotonic
increasing curve,? increases with ?,
(3) At resonance,? =1 and, The curve increases near
resonance rapidly,all curves of different resistance pass this point,
(4) For ?>1,? increases with ?, If ?, 1,,(inverse
phase),
???
2???
b
? 2
0
max b,? ? n ? ? 1<< At and ?
76
???? 2,,1
0m a x bbn ??<< 时当
4、相位差
有阻尼强迫振动相位总比激振力滞后一相位角 ?,? 称为 相位差 。
21
2tg
?
???
??
(1) ?总在 0至 ? 区间内变化。
(2) 相频曲线（ ? - ?曲线）是一条单调上升的曲线。 ? 随 ? 增
大而增大。
(3) 共振时 ? =1，,曲线上升最快，阻尼值不同的曲线,
均交于这一点。
(4) ?>1时,? 随 ? 增大而增大。当 ?, 1时,反相。 ???
2???
77
Example 1 Knowing P=3500N,k=20000N/m,
H=100N,f=2.5Hz and c=1600N·s/m,determine b,?
and the equation of the forced vibration,
Solution,
r a d / s 58.103 5 0 0 8.92 0 0 0 022 ?????? Pkgmk eqn?
m 105.22 0 0 0 02 1 0 02 30 ??????? kHkHb
eq
4 8 5.1
58.10
5.222; 2 1 2.0
58.10
24.2
r a d /s 24.2
8.9/3 5 0 02
1 6 0 0
2
?
?
??????
?
?
??
?
?
?
?
?
?
?
?
nnn
fn
m
c
n
,
,
,
,
78
例 1 已知 P=3500N,k=20000N/m,
H=100N,f=2.5Hz,c=1600N·s/m,求 b,?,强迫
振动方程。
解,
r a d / s 58.103 5 0 0 8.92 0 0 0 022 ?????? Pkgmk eqn?
m 105.22 0 0 0 02 1 0 02 30 ??????? kHkHb
eq
4 8 5.1
58.10
5.222; 2 1 2.0
58.10
24.2
r a d /s 24.2
8.9/3 5 0 02
1 6 0 0
2
?
?
??????
?
?
??
?
?
?
?
?
?
?
?
nnn
fn
m
c
n
79
m m,84.15.27 3 6.0
,7 3 6.0
4 8 5.12 1 2.04)4 8 5.11(
1
4)1(
1
0
2222222
????
?
????
?
??
?
bb ?
???
?
).8 4 7.05s i n (84.1
),r a d( 8 4 7.0)5 2 2.0(a r c t g)]1/(2[a r c t g
2
22
??
?????
??
?????
tx
80
mm 84.15.2736.0
736.0
485.1212.04)485.11(
1
4)1(
1
0
2222222
????
?
????
?
??
?
bb ?
???
?
)8 4 7.05s in (84.1
)r a d( 8 4 7.0)5 2 2.0(a r c t g)]1/(2[a r c t g
2
22
??
?????
??
?????
tx
81
§ 18-6 The concepts of critical speed of
rotation,vibration reduction and vibration isolation
1,Critical speed of rotation of a rotator
The special speed of rotation at which the rotator will vibrate
drastically is called critical speed of rotation, This phenomenon is
caused by resonance,In course of designing of a higher speed shaft,
such checking computations should be taken into account,
Rotator of single circle disc,
Point C is the center of mass of a circular disc whose
mass is m,The geometric center A of the disc is on the
axis of rotation,AC=e,the disc is rotating around its
axis with a constant angular velocity ?,When ?< ?n
(?n is the natural frequency of the system consisting of
the disc and the axis of rotation),OC= x+e (x is the
flexural deformation of the mid-point of the axis),
82
§ 18-6 临界转速 ? 减振与隔振的概念
一、转子的临界转速
引起转子剧烈振动的特定转速称为 临界转速 。这种现象是
由共振引起的，在轴的设计中对高速轴应进行该项验算。
单圆盘转子,
圆盘：质量 m,质心 C点；转轴过盘的几
何中心 A点,AC= e,盘和轴共同以匀角速
度 ? 转动。 当 ?< ?n（ ?n为圆盘转轴所组
成的系统横向振动的固有频率）时,
83
kxexm ?? 2)( ?
11 2
2
2 ?
?
?
?
?
?
?
n
e
m
k
ex
Critical angular velocity is
Critical speed of rotation is
cc
nc
n
m
k
?
?
??
30?
??
From,where k is the equivalent
stiffness of the axis of rotation,we obtain,
n ? ? When goes to,x goes to infinity,
84
kxexm ?? 2)( ? （ k为转轴相当刚度系数）
11 2
2
2 ?
?
?
?
?
?
?
n
e
m
k
ex
??? xn,时当 ??
临界角速度,
临界转速,
cc
nc
n
m
k
?
?
??
30?
??
85
n ? ? ? C of mass is between the point O and the
2
2 )(11 ?
?
?
n
e
m
k
ex
?
?
?
?
If the center
point A,OC= x- e,
When the speed of rotation is very high,the center C of mass of
the circular disc approaches the line through two bracing point,
and the circular disc almost rotates a round the center C of mass,
Therefore,it rotates steadily,
n ?? ? ? n ? ? ? If,x decreases with increasing ?,When,x goes to e,
In order to ensure safety,the working speed of the axis must evade
its critical speed of rotation,
86
,运转时当 n?? ? 质心 C位于 O,A之间 OC= x- e
2
2 )(11 ?
?
?
n
e
m
k
ex
?
?
?
?
exx nn ??????,;,,时当时当 ?????
当转速 ? 非常高时，圆盘质心 C与两支点的连线相接近，
圆盘接近于绕质心 C旋转，于是转动平稳。
为确保安全，轴的工作转速一定要避开它的临界转速。
87
2,Concept of reduction and isolation of vibration
Drastic vibrations affect not only the normal work of a machine,but
also the normal work of the instrumental equipments around it,The
basic measure to reduce the danger and harm of vibrations is a
reasonable design,Try to reduce the vibrations and prevent them
from acting in the region of resonance,
Many factors causing vibrations are impossible to be reduced
effectively or difficult to be avoided,In these cases we can take
measures of vibration reduction and isolation,
Reduction of vibration,Various dampeners can be equipped on
vibrating bodies to reduce vibrations,
For example,we can use various resistance dampeners to consume
energy,then the goal is attained,
88
二、减振与隔振的概念
剧烈的振动不但影响机器本身的正常工作，还会影响周围
的仪器设备的正常工作。减小振动的危害的根本措施是合理设
计，尽量减小振动，避免在共振区内工作。
许多引发振动的因素防不胜防，或难以避免，这时，可以
采用减振或隔振的措施。
减振, 在振体上安装各种减振器，使振体的振动减弱。例如,
利用各种阻尼减振器消耗能量达到减振目的。
89
Isolation of vibration,the equipments which need to be isolated are equipped on proper vibration isolators (elastic devices) in order
to make most vibrations to be absorbed by the vibration isolators,
Isolation of vibration is
classified into
Positive isolation of vibration,which refers to isolating
the source of vibration from its foundation,
Passive isolation of vibration,which refers to isolating
the equipment,which needs to be prevented from
vibration,from the source of vibration,
90
隔振,将需要隔离的仪器、设备安装在适当的隔振器（弹性
装置）上，使大部分振动被隔振器所吸收。
隔振 主动隔振：将振源与基础隔离开。
被动隔振：将需防振动的仪器、设备单独与振源隔离开。
91
92
Theoretical Mechanics
2
3
Vibrations are common phenomena in everyday life and in
engineering practice,
Such as the swinging of a clock pendulum,the bumping of a
traveling car,the vibrations of working machine parts or motors,
and the quake of buildings caused by an earthquake,
3,The aim of the study of vibrations is the elimination or
reduction of harmful oscillations,or the helpful use of
them for special purposes,
2,Advantages and disadvantages of vibrations,Vibrating
feeder,vibrating screen,vibrating pile driver extractor,for
example,are advantageous applications,But vibrations
induce friction loss,affect intensity,make noise,have bad
effects on working conditions,consume unnecessary energy
and reduce the precision of a machine etc,
1,Vibrations are the back and forth movements of a system
around its equilibrium position,
4
振动是日常生活和工程实际中常见的现象。
例如：钟摆的往复摆动,汽车行驶时的颠簸，电动机、机
床等工作时的振动，以及地震时引起的建筑物的振动等。
利,振动给料机 弊,磨损，减少寿命，影响强度
振动筛 引起噪声，影响劳动条件
振动沉拔桩机等 消耗能量，降低精度等。
3,研究振动的目的,消除或减小有害的振动，充分利用振动
为人类服务。
2,振动的利弊,
1,所谓振动就是系统在平衡位置附近作往复运动。
5
The present chapter discusses mainly free vibrations and forced
vibrations of systems with one degree of freedom,
4,Classification of vibrations,Vibrations of systems with one degree of freedom
Vibrations of systems with
more than one degree of
freedom and
Vibrations of elastic bodies
According to the number of
degrees of freedom of a vibrating
system they can be classified into
According to the cause of the vibrations they can be classified into
free vibrations including undamped free vibrations and damped free vibrations
forced vibrations including undamped forced vibrations and
damped forced vibrations
self-excited forced vibrations
6
4,振动的分类, 单自由度系统的振动
按振动系统的自由度分类 多自由度系统的振动
弹性体的振动
按振动产生的原因分类,
自由振动,无阻尼的自由振动
有阻尼的自由振动，衰减振动
强迫振动,无阻尼的强迫振动
有阻尼的强迫振动
自激振动
本章重点讨论单自由度系统的自由振动和强迫振动。
7
§ 18–1 Undamped free vibration of a system
with one degree of freedom
§ 18–2 Methods of determination of the
natural frequency of a system
§ 18–3 Damped free vibration of a system
with one degree of freedom
§ 18–4 Undamped forced vibration of a
system with one degree of freedom
§ 18–5 Damped forced vibration of a
system with one degree of freedom
§ 18–6 The concepts of critical speed of
rotation,vibration reduction and
vibration isolation
Chapter 18,Mechanical vibrations
8
§ 18–1 单自由度系统无阻尼自由振动
§ 18–2 求系统固有频率的方法
§ 18–3 单自由度系统的有阻尼自由振动
§ 18–4 单自由度系统的无阻尼强迫振动
§ 18–5 单自由度系统的有阻尼强迫振动
§ 18–6 临界转速 · 减振与隔振的概念
第十八章 机械振动基础
9
§ 18-1 Undamped free vibration
of a system with one degree of freedom
1.Concept of free vibration
motor
Elastic foundation
10
§ 18-1 单自由度系统无阻尼自由振动
一、自由振动的概念,
11
Equilibrium
position x=0
Equ
ilibr
ium
pos
ition
j=0
Equ
ilib
riu
m
pos
itio
n j
=0
12
13
A force acting on a vibrating body which is always
directed towards its equilibrium position is called a restoring force,
After the action of an initial disturbance,the vibration of a system
around its equilibrium position under the action of only a restoring
force is called an undamped free vibration,
)5,/( 0,
)5,/( 0,
)/( 0,
22
222
22
?
?
????
????
????
??????
??????
?????
j?j?jjj
j?j?jjj
??
Im g am g aI
lgm g lml
mkxxkxxm
nn
nn
nn
For a mass-spring system,
For a simple pendulum,
For a compound pendulum,
14
运动过程中，总指向物体平衡位置的力称为 恢复力 。
物体受到初干扰后，仅在系统的恢复力作用下在其平衡位
置附近的振动称为 无阻尼自由振动 。
)/( 0,22 mkxxkxxm nn ????? ??????质量 —弹簧系统,
单摆,
复摆,)5,/( 0,)5,/( 0,22
222
?
?
????
????
??????
??????
j?j?jjj
j?j?jjj
Im g am g aI
lgm g lml
nn
nn
15
2,Differential equation and its solution for an undamped
free vibration of a system with one degree of freedom
For any system with one degree of freedom,q being the generalized
coordinate measured from the equilibrium position,the differential
equation of a free vibration for a small displacement is
0?? cqqa ??
where a and c are constants related to the physical parameters of
the system,Writing
acn /2 ??
the standard form of the differential equation of free
vibration is
.02 ?? qq n???
Its solution is,)s i n ( ?? ?? tAq
n
16
二、单自由度系统无阻尼自由振动微分方程及其解
对于任何一个单自由度系统，以 q 为广义坐标（从平衡位
置开始量取 ），则自由振动的运动微分方程必将是,
0?? cqqa ??
a,c是与系统的物理参数有关的常数。令 acn /2 ??
则自由振动的微分方程的标准形式,
02 ?? qq n???
解 为,)s i n ( ?? ?? tAq
n
17
Assuming that at t = 0, we get
00 an d qqqq ?? ??
,a r c t g,
0
0
2
2
02
0 q
qqqA n
n ?
? ??
?
???
or,
tCtCq nn ?? s inc o s 21 ??
./,w i t h 02 01 nq CqC ????
.s i nc o s T h e r e f o r e,00 tqtqq n
n
n ???
???
18
0
0
2
2
02
0 a r c t g,q
qqqA n
n ?
? ??
?
???
设 t = 0 时,则可求得,
00,qqqq ?? ??
或,
tCtCq nn ?? s inc o s 21 ??
C1,C2由初始条件决定为
nq CqC ?/,02 01 ???
tqtqq n
n
n ??? s inc o s
0
0
????
19
n?
3.Properties of a free vibration without damping,
A—— The quantity,which is the maximum distance of a vibrating
body from the equilibrium position,is called the amplitude,
?n t + ? —— The quantity,which defines the position of the vibrating
body at any given time,is called the phase of the vibration,
? —— The quantity,which defines the initial phase,at which the
motion starts,
T —— The time T during which the vibrating body makes one
complete vibration is called the period of vibration,
f —— The quantity,which specifies the number of oscillations per
second and is the inverse of the period,is called the frequency
of the vibration,f = 1 / T,
—— The quantity,which specifies the number of oscillations
in 2 seconds,is called natural frequency of the vibration,
It characterizes the dynamics of a given vibrating system and
depends on the inherent parameters describing the system,
.2
n
T ???
?
20
三、自由振动的特点,
A—— 物块离开平衡位置的最大位移，称为振幅。
?n t + ? —— 相位，决定振体在某瞬时 t 的位置
? —— 初相位，决定振体运动的起始位置。
T —— 周期，每振动一次所经历的时间。
f —— 频率，每秒钟振动的次数,f = 1 / T 。
—— 固有频率，振体在 2?秒内振动的次数。
反映振动系统的动力学特性，只与系统本身的固有
参数有关。
n
T ??2?
n?
21
Properties of undamped free vibrations,
4,Other results
1,If a constant force is applied to the system in the direction of its
vibration then this constant force does not affect the law of the
vibration of the system,It only displaces the equilibrium position O
into the direction of this force,The amplitude,frequency and phase
of the vibration do not change,
(2) The amplitude A and the initial phase depend on the initial
conditions (initial displacement and initial velocity),
(1) It is a simple harmonic vibration,
(3)The period T and the natural frequency depend only on the
natural parameters of the system itself (m,k,I,etc.),
n?
?
22
无阻尼自由振动的特点是,
(2) 振幅 A和初相位 ? 取决于运动的初始条件 (初位移和初速度 )；
(1) 振动规律为简谐振动；
(3)周期 T 和固有频率 仅决定于系统本身的固有参数 (m,k,I )。 n?
四、其它
1,如果系统在振动方向上受到某个常力的作用，该常力
只影响静平衡点 O的位置，而不影响系统的振动规律，如振动
频率、振幅和相位等。
23
2,Equivalent
stiffness of series and
parallel spring systems
,t h e r e f o r e
,)( So
,
21
21
21
21
2
2
1
1
kkk
kk
mg
kkmg
FFmg
k
F
k
F
eq
stst
st
??
?
???
????
??
?
For a parallel spring
system 21
21
eq
21
2121
21
k T h e r e f o r e
)
11
(
)
11
(
kk
kk
kk
mg
k
mg
kk
mg
k
mg
k
mg
eq
st
ststst
?
?
???
????
??
?
???
For a series spring system,
pa
rall
el
ser
ies
24
2,弹簧并联系
统和弹簧串联系
统的等效刚度
21
21
21
21
2
2
1
1
,)(
,
kkk
kk
mg
kkmg
FFmg
k
F
k
F
eq
stst
st
???
?
????
????
??
?
并联
21
21
eq
21
2121
21
k
)
11
(
)
11
(
kk
kk
kk
mg
k
mg
kk
mg
k
mg
k
mg
eq
st
ststst
?
??
???
????
??
?
???
串联
并
联
串
联
25
§ 18-2 Methods of determination
of the natural frequency of a system
1,The differential equation of the vibration of a system in
the standard form is
02 ?? qq n???
3,Energy Method,
n?From the equation Tmax=Umax we can determine,
st
n
g
?? ?
st?
where =const,is the static
displacement of the system
under the action of a constant
force,
2,Applying the method of static deformation,we obtain,
26
1,由系统的振动微分方程的标准形式
2,静变形法,
3,能量法,
§ 18-2 求系统固有频率的方法
02 ?? qq n???
st
n
g
?? ?
st?
：集中质量在全部重力
作用下的静变形
n?由 Tmax=Umax,求出
27
A system with undamped free vibration is a conservative system,
The mechanical energy of system does not change during its
motion,
When the vibrating body moves to the maximum distance from
the equilibrium position,its velocity becomes zero,Hence,the
kinetic energy of the system is zero when its potential energy has
its maximum value (choose the static equilibrium position of the
system as the zero point of potential energy),
When the vibrating body moves to the equilibrium position,the
potential energy of the system is zero,while its kinetic energy is
maximum,
m g AAkU stst ???? ])[(21 22m a x ??
2
m a x 2
1 So kAUmgk
st ???
222m a x 2121 nmAxmT ??? ?
Example,
28
无阻尼自由振动系统为保守系统，机械能守恒。
当振体运动到距静平衡位置最远时，速度为零，即系统
动能等于零，势能达到最大值（取系统的静平衡位置为零势
能点）。
当振体运动到静平衡位置时，系统的势能为零，动能达
到最大值。
m g AAkU stst ???? ])[(21 22m a x ??
2m a x 21 kAUmgk st ????
222m a x 2121 nmAxmT ??? ?
如,
29
m
k kA mA
n n ? ? ? ?,,2 1 2 1
2 2 2
U T ? max max From
The energy method is based on the conservation law of the
mechanical energy,It is a simple and convenient method in
order to compute the natural frequency of a more complicated
vibrational system,
[Example 1] The system is shown in the
figure.Assume that the system does not
swing horizontally,The wheel,which is
homogeneous of radius R and mass M,and
the string do not slide one relative to the
other,The load has mass M,Neglecting the
mass of the string and of the spring,write
down the differential equation of vibration
of the system and determine its natural
frequency,
we get
30
m
kkAmA
UT
nn ??
?
??
2
1
2
1
222
m a xm a x由
能量法是从机械能守恒定律出发，对于计算较复杂的振
动系统的固有频率来得更为简便的一种方法。
例 1 图示系统。设轮子无侧向摆动，
且轮子与绳子间无滑动，不计绳子和弹
簧的质量，轮子是均质的，半径为 R，质
量为 M，重物质量 m,试列出系统微幅
振动微分方程，求出其固有频率。
31
Solution 1,Let x be the generalized
coordinate,The origin of the coordinate is at
the position of static equilibrium of the
system,
RkgRmM st 2)( ??? ?
gk mMst ??? 2?
In static equilibrium we have
At an arbitrary position x we get
kxgmMxkF st 22)2( ????? ?
,
32
解,以 x 为广义坐标（静平衡位置为
坐标原点）
RkgRmM st 2)( ??? ?
gk mMst ??? 2?
则任意位置 x 时,
kxgmMxkF st 22)2( ????? ?
静平衡时,
33
Applying the theorem of kinetic energy we have,
k x RRFgRmMFm
xRmM
R
xMRRxMRxmL
A
A
42)()(
)
2
3(
2
1 2
???????
??
???
?
??
From,we get
?? )( FmdtdL AA k x RxRmM 4)
23( ??? ??
The differential equation of vibration is,
Therefore,the natural frequency is
mM
k
x
mM
kx
n 23
8
0
23
8
?
?
?
?
?
?
??
34
应用动量矩定理,
k x RRFgRmMFm
xRmM
R
xMRRxMRxmL
A
A
42)()(
)
2
3(
2
1 2
???????
??
???
?
??
由, 有 ?? )( Fm
dt
dL
AA k x RxRmM 4)23( ??? ??
振动微分方程,
固有频率,
mM
k
x
mM
kx
n 23
8
0
23
8
?
?
?
?
?
?
??
35
Solution 2,Let x be the generalized coordinate,the origin is at the position of static equilibrium
of the system,
Applying the conservation law of mechanical
energy we have
22
2
2
2
1)(
22
1
2
1 xm
R
xMRxMT ??? ???
Choose the position of equilibrium as the zero point of potential
energy,When the displacement of the center of the wheel is x,the
elongation of the spring is 2x,
gxmMxkU stst )(])2[(2 22 ????? ??
In equilibrium,we have
gxmMxk st )(2 ??? 22 k xU ? So
2)
2
3(
2
1 xmM ???
gxmMxkkx st )(22 2 ???? ?
36
解 2, 用机械能守恒定律
以 x为广义坐标（取静平衡位置为原点）
2
22
2
2
)
2
3(
2
1
2
1)(
22
1
2
1
xmM
xm
R
xMRxMT
?
???
??
???
以平衡位置为计算势能的零位置，
并注意轮心位移 x时，弹簧伸长 2x
gxmMxkkx
gxmMxkU
st
stst
)(22
)(])2[(2
2
22
????
?????
?
??
因平衡时
gxmMxk st )(2 ???
22 kxU ??
37
From T+U= we get
const
.2)23(21 22 c o n s tkxxmM ??? ?
04)23( ??? kxxmM ??
mM
k
x
mM
k
x
n 23
8
0
23
8
?
?
?
?
?
?
??
Differentiation with respect to time t and elimination of the
common multiplier results in x?
38
由 T+U= 有,
const
c o n s tkxxmM ??? 22 2)23(21 ?
04)23( ??? kxxmM ??
mM
k
x
mM
k
x
n 23
8
0
23
8
?
?
?
?
?
?
??
对时间 t 求导，再消去公因子,得 x?
39
Solution,choose the
distance x of C from the
equilibrium position as the
generalized coordinate,The
maximum kinetic energy of
the system is
Example 2 A tab wheel,whose mass is M and whose gyroradius with
respect to the center of the wheel is,rotates without sliding on a
horizontal plane,The radii of the big and the small wheels are R and r,
the stiffness of the two springs are and,the load is m,Neglecting
the masses of the wheel and the springs,and assuming the string is not
extensible,determine the natural frequency of the system in small
oscillation,
?
1k 2k
40
例 2 鼓轮：质量 M，对轮心回转半径 ?，在水平面上只滚不滑，
大轮半径 R，小轮半径 r,弹簧刚度,重物质量为 m,不计
轮 D和弹簧质量，且绳索不可伸长。求系统微振动的固有频率。
21,kk
解,取静平衡位置 O为坐标原
点，取 C偏离平衡位置 x为广义
坐标。系统的最大动能为,
41
)
)(
)(
( )(
2
1
])) [ ((
2
1
21
2
m a x21
m a x
22
m a x21m a x
Rkk
rRmg
xkk
x
R
rRmgxkkU
st
stst
?
?
???
??????
?
??
.] [
2
1
)(
2
1
)(
2
1
)(
2
1
2
m a x
222
2
2
m a x
2m a x22
m a xm a x
xr)m ( R)RM(
R
x
R
rR
m
R
x
MxMT
?
?
?
?
????
?
?
??
?
?
The maximum potential energy of the system is,
42
)
)(
)(
( )(
2
1
])) [ ((
2
1
21
2
m a x21
m a x
22
m a x21m a x
Rkk
rRmg
xkk
x
R
rRmgxkkU
st
stst
?
?
???
??????
?
??
2
m a x
222
2
2
m a x
2m a x22
m a xm a x
] [
2
1
)(
2
1
)(
2
1
)(
2
1
xr)m ( R)RM(
R
x
R
rR
m
R
x
MxMT
?
?
?
?
????
?
?
??
?
?
系统的最大势能为,
43
Writing we have
)s i n ( ?? ?? tAx n,,m a xm a x nAxAx ??? ?
,)(21,2 )()( 221m a x222
222
m a x AkkUAR
rRmRMT
n ??
???? ??
From Tmax=Umax we then get
222
2
21
)()(
)(
rRmRM
Rkk
n ???
??
??
44
设 则有
)s i n ( ?? ?? nAx nAxAx ??? m a xm a x,?
)(21 2 )()( 221m a x222 222m a x AkkUAR rRmRMT n ?????? ??
根据 Tmax=Umax,解得
222
2
21
)()(
)(
rRmRM
Rkk
n ???
??
??
45
§ 18-3 Damped free vibration
of a system with one degree of freedom
1,Concept of Resistance,
Resistance,Resistance is the force acting on an oscillation system
restoring its equilibrium,
Viscous Damping,In many cases,the resistance caused by the
surrounding medium is proportional to the first power of
the velocity,Such a resistance is called viscous damping,
vcR ??
Its projection form is xcR
x ???
where C is called the viscous resistance coefficient,or coefficient
of damping,
46
§ 18-3 单自由度系统的有阻尼自由振动
一、阻尼的概念,
阻尼,振动过程中，系统所受的阻力。
粘性阻尼,在很多情况下，振体速度不大时，由于介质粘性
引起的阻尼认为阻力与速度的一次方成正比，这种阻尼称为粘
性阻尼。
vcR ??
投影式,xcR x ???
c —— 粘性阻尼系数，简称阻尼系数。
47
2.Differential equation of a damped free vibration and its solution,
For a spring system with viscous damping we get,xckxxm ??? ???
2 and
2
n ? ? m
c n
m
k ? we write
Using the abbreviations
0 2 2 ? ? ? x x n x n ? ?? ?
This is the differential equation of the damped free vibration in
standard form,
48
二、有阻尼自由振动微分方程及其解,
质量 — 弹簧系统存在粘性阻尼,xckxxm ??? ???
02 2,22n ????? xxnx mcnmk n?? ???则令
有阻尼自由振动微分方程的标准形式。
49
The general solution is discussed for three cases,
1,In the case of small resistance( or ) the
general solution is
mk c 2 < n n < ?
)s i n ( ?? ?? ? tAex dnt
00
22
01
22
2
002
0 tg ;
)(
nxx
nx
n
nxxxA n
n ?
??
?
??? ?
?
? ??
?
22 nnd ?? ??
0 x x ??? 0 x x ? 0 t ?
Where is the circular frequency of the damped free
vibration,Assuming at and,we get
50
其通解分三种情况讨论,
1、小阻尼情形 mkcn
n 2 )( << ?
)s i n ( ?? ?? ? tAex dnt
22 nnd ?? ?? — 有阻尼自由振动的圆频率
则时设,,,0 00 xxxxt ?? ???
00
22
01
22
2
002
0 tg ;
)(
nxx
nx
n
nxxxA n
n ?
??
?
??? ?
?
? ??
?
51
Properties of a damped vibration,
(1) The period of the vibration
increases,the frequency decreases,
mk
cn
n
T
n
n
d
d
2
1
2
22
22
22
??
?
?
?
??
?
?
??
?
?
?
?
?
—— ratio of resistance
Hence
If then and
we get approximately
nn ?<< 1<<?
.,TT dnd ?? ??2
2
2
1
1
1
???
?
?
??
??
?
?
nd
d
d
ff
TT
52
衰减振动的特点,
(1) 振动周期变大,
频率减小 。
mk
cn
n
T
n
n
d
d
2
1
2
22
22
22
??
?
?
?
??
?
?
??
?
?
?
?
?
—— 阻尼比
有阻尼自由振动,
当 时,
可以认为
nn ?<< 1<<?
TT dnd ?? ??2
2
2
1
1
1
???
?
?
??
??
?
?
nd
d
d
ff
TT
53
(2) The amplitude of the vibration decreases in geometric progression,
The logarithmic decrement is defined as
??
?
??
?
?
2
1
2
lnln
2
1
?
?
?
???
?
d
nT
i
i nTe
A
A
d
d
di
i
nT
Ttn
nt
i
i e
Ae
eA
A
A ??
??
?
?
)(
1
The ratio of two adjacent amplitude is
2,In the case of critical damping
or 1,( ?? ?? nn )2 mkc c ?
])([ 000 tnxxxex nt ??? ? ? ).,,0A t ( 00 xxxxt ?? ???
critical viscous resistance coefficient
the general solution is
54
(2) 振幅按几何级数衰减
对数减缩率
??
?
??
?
?
2
1
2
lnln
2
1
?
?
?
???
?
d
nT
i
i nTe
A
A
d
2、临界阻尼情形
临界阻尼系数
) 1,( ?? ?? nn
mkc c 2?
])([ 000 tnxxxex nt ??? ? ? ),,
(at 0 0 x x x x ??? ? t 0 ?
d
di
i
nT
Ttn
nt
i
i e
Ae
eA
A
A ??
??
?
?
)(
1
相邻两次振幅之比
55
We can see that the body tends to move to the position of
equilibrium with increasing time exponentially slowly,
Substituting the initial conditions (at ) we get and 0 0 x x x x ??? ? 0t ?
22
00
22
222
0
22
0
1 2
)(,
2
)(
n
n
n
n
n
xxnnC
n
xnnxC
?
?
?
?
?
?????
?
???? ??
)( 2222 21 tn tnnt nn eCeCex ?? ???? ??
,1,( ?? ?? nn )ccc ?3,In the case of overdamping
the general solution is
The motion is not a periodic one,With increasing time the
quantity x goes exponentially to zero,the body does not vibrate,
56
可见，物体的运动随时间的增长而无限地趋向平衡位置，
不再具备振动的特性。
)( 2222 21 tn tnnt nn eCeCex ?? ???? ??
代入初始条件 ),,0(
00 xxxxt ?? ??? 时
22
00
22
222
0
22
0
1 2
)( ;
2
)(
n
n
n
n
n
xxnnC
n
xnnxC
?
?
?
?
?
?????
?
???? ??
) 1,( ?? ?? nn )( ccc ?3、过阻尼（大阻尼）情形
所示规律已不是周期性的了，随时间的增长,x 0，
不具备振动特性。
57
Example 3 In a spring system W=150N,?st=1cm,A1=0.8cm,
A21=0.16cm,Determine the viscous resistance coefficient c,
20
21
20
3
2
2
1
21
1 )( dnTe
A
A
A
A
A
A
A
A ???? ?
Solution,
20)(16.0 8.0 dn Te ???
21
220205ln
??
?????
??? n
n
dn T
Because is very small we can write approximately
and get ? ??405ln ?
).s / c mN(1 2 2.0
9 8 01
1 5 0
2
40
5ln
2
40
5ln
2
2
??
?
????????
???
?
st
W
g
W
mkc
58
例 3 质量弹簧系统,W=150N,?st=1cm,A1=0.8cm,
A21=0.16cm。 求阻尼系数 c 。
20
21
20
3
2
2
1
21
1 )( dnTe
A
A
A
A
A
A
A
A ???? ?
解,
20)(16.0 8.0 dn Te ???
21
220205ln
??
?????
??? n
n
dn T
由于 很小,? ??405ln ?
)s /c mN(122.0
9801
1502
40
5ln2
40
5ln2 2
??
?
????????
???
?
st
W
g
Wmkc
59
§ 18-4 Undamped forced vibration
of a system with one degree of freedom
1,Concept of forced vibration,
A forced vibration is the oscillation of a body under the action of an
external disturbing force (except the restoring force),
A simple harmonic disturbing force is given by,
where H is the amplitude of the force,? is its circular frequency
and ? is its initial phase,
)s i n ( ?? ?? tHS
)s i n (2 ??? ??? thxx n??
This is the differential equation of an undamped forced vibration in
standard form,It is a linear inhomogeneous differential equation of
second order,its solution is
)s i n ( ?? ???? tHkxxm ??
with we can write,2 m H h m k n ? ? ?
2,Differential equation and its solution,
60
§ 18-4 单自由度系统的无阻尼强迫振动
一、强迫振动的概念
强迫振动：在外加激振力作用下的振动。
简谐激振力,
H— 力幅； ?— 激振力的圆频率 ； ? — 激振力的初相位。
)s i n ( ?? ?? tHS
)s i n ( ?? ???? tHkxxm ??
则令,2 mHhmkn ???
)s i n (2 ??? ??? thxx n??
无阻尼强迫振动微分方程的标准形式，
二阶常系数非齐次线性微分方程。
二、无阻尼强迫振动微分方程及其解
61
)s i n ()s i n ( 22 ?????? ????? thtAx
n
n
The total solution is,
x2 is a stable state force vibration,
3,Main properties of stable state force vibration,
1)It is the additional vibration of a system in one dimension,caused
by a simple harmonic disturbing fore,
2)The frequency of this part is equal to the frequency of the
simple harmonic disturbing force,it does not depend on the mass and
the stiffness of the system,
3)The amplitude of it depends on the natural frequency
of the system and of the frequency and the amplitude of the disturbing
force,but not the initial conditions,
)s i n (,w it h 22222 ?????? ????? thxhb
nn
w i t h21 xxx ??
) sin( 2 ? ? ? ? t b x
) sin( 1 ? ? ? ? t A x n the general solution of the corresponding
homogeneous differential equation and
a particular solution of the complete equation,
62
21 xxx ??
)s i n (
)s i n (
2
1
??
??
??
??
tbx
tAx n
为对应齐次方程的通解
为特解
)s in (,22222 ?????? ????? thxhb
nn
)s i n ()s i n ( 22 ?????? ????? thtAx
n
n
全解为,
稳态强迫振动
3、强迫振动的振幅大小与运动初始条件无关，而与振动系统
的固有频率、激振力的频率及激振力的力幅有关。
三、稳态强迫振动的主要特性,
1、在简谐激振力下，单自由度系统强迫振动亦为简谐振动。
2、强迫振动的频率等于简谐激振力的频率，与振动系统的
质量及刚度系数无关。
63
? is the amplitude ratio (or the
dynamic coefficient),? is the
frequency ratio,This is the ? - ?
curve,the dependence of the
amplitude on the frequency,
1
(1) At ? =0
k
Hhb
n
?? 20 ?
(3) At, the phase of the forced vibrations is inverse to
the phase of the disturbing force,i.e.,the phase shift is rad ?n?? ?
22 ?? ??
n
hb
(2)At, the amplitude b of this vibration increase with ?,
and goes to infinity with ? ??n,
n??<
where b decreases with increasing ?,
If ? goes to,b goes to b0,If ? goes to,b goes to zero,n?2 ?
64
(1) ? =0时
k
Hhb
n
?? 20 ?
(2) 时，振幅 b随 ? 增大而增大；当 时,
n?? ? ??bn??<
(3) 时，振动相位与激振力相位反相，相差 。 rad ?
n?? ?
22 ?? ??
n
hb
b 随 ? 增大而减小； 0 ;,2
0 ????? bbbn 时时 ???
? — 振幅比或称动力系数
? — 频率比
?— ? 曲线 幅频响应曲线
（幅频特性曲线） 1
65
At resonance we get
)co s (2 ?? ?? tBtx n
)c o s (
2
So
2
2
??
?
?
???
??
tt
h
x
h
B
n
n
n
4,Resonance
then n ? ? ? b is the infinite( ),this phenomenon is called
resonance,
If ?
66
4、共振现象
,时n?? ? ??b,这种现象称为共振。
此时,)co s (2 ?? ?? tBtx n
)c o s (
2
2
2
??
?
?
????
??
tt
h
x
h
B
n
n
n
67
§ 18-5 Damped forced vibration
of a system with one degree of freedom
1,Differential equation of a damped forced
vibration and its solution,
,s i n,,tHQxcRkxF xxx ?????? ?
.s in tHxckxxm ????? ???
thxxnx n ?? s i n2 2 ??? ???
This equation is the differential equation of a damped forced
vibration in standard form,It is a linear inhomogeneous
differential equation of second order,Its solution is
21 xxx ??
Dividing both sides of the last equation by m
and introducing the notations
m
Hh
m
cn
m
k
n ???,2,
2?
we obtain
68
§ 18-5 单自由度系统的有阻尼强迫振动
一、有阻尼强迫振动微分方程及其解
tHQxcRkxF xxx ?s i n,,????? ?
tHxckxxm ?s in???? ???
将上式两端除以 m,并令
mHhmcnmkn ??? ; 2 ; 2?
thxxnx n ?? s i n2 2 ??? ???
有阻尼强迫振动微分方程的标准形式，二阶常系数非齐次微
分方程。
21 xxx ??
69
22
22222
2
tg
4)(
??
?
?
???
?
?
??
?
n
n
n
n
h
b — the amplitude of the forced vibration,
? is the phase shift of the forced vibration,the
difference to the phase of the disturbing force,
)s in ()s in ( 22 ????? ????? ? tbtAex nnt
The total solution of the differential equation of the vibration is
damped vibration forced vibration
in the case of small resistance ),s in ( 22
1 ??? ??? ? tAex nnt
and x2 is a particular solution of the complete equation
)s i n (2 ?? ?? tbx
where x1 is the general solution of the corresponding linear
homogeneous equation
),02( 2 ??? xxnx n????
Substituting this expression into the differential equation in
standard form we obtain
70
x
1是齐次方程的通解 )02( 2 ??? xxnx n????
小阻尼,
)s in ( 221 ??? ??? ? tAex nnt
（ A,? 积分常数，取决于初始条件）
x2 是特解,)s i n (
2 ?? ?? tbx
代入标准形式方程并整理
22
22222
2
tg
4)(
??
?
?
???
?
?
??
?
n
n
n
n
h
b — 强迫振动的振幅
— 强迫振动相位滞后激振力相位角
振动微分方程的全解为
)s in ()s in ( 22 ????? ????? ? tbtAex nnt
衰减振动 强迫振动
71
At the beginning of the vibration,the process with both the natural and the forced vibration simultaneously,is called the transient
process,
The other process of forced vibration after the transient process,is called the stable state process,It is the subject to be discussed in
detail,
2,Effect of resistance on forced vibration,
frequency ratio,the amplitude ratio and the resistance ratio n n
n
b
b
? ? ? ?
? ? ? ? ? and are the,
0
We get
.1 2 t g,
4)1(
1
22222 ?
???
???
? ??
??
?
1,The vibration given by is a simple harmonic
motion,。
2,Frequency,The frequency of the forced vibration is equal to the
frequency of the disturbing force,
3,Amplitude,
)s i n (2 ?? ?? tbx
72
振动开始时，二者同时存在的过程 —— 瞬态过程。
仅剩下强迫振动部分的过程 —— 稳态过程。需着重讨论部分。
nn
nbb ?????? ??? ;,
0
令
频率比 振幅比 阻尼比
因此,
22222 1
2 t g;
4)1(
1
?
???
???? ?????
二、阻尼对强迫振动的影响
1、振动规律 简谐振动。
2、频率,有阻尼强迫振动的频率，等于激振力的频率。
3、振幅
)s i n (2 ?? ?? tbx
73
(3)
(1) ? 1 << ? ) << ? ( ?
n
0 b (b
At tends to 1
tends to )and the resistance can be neglected,
(2) ? 1 ?? ? ) ?? ? ( ? n At tends to 0
and the resistance can also be neglected,
If tends to 1 and the
resistance has a remarkable effect on
the amplitude,
? 70.0<?
When ? does not change,the amplitude
decreases strongly with increasing
resistance,
2 2 2 2 1 2 ? ? ? ? ? ? ? ? n n n we obtain 0 ? ? d db From
At this frequency
2
0
m a x22m a x 12o r 2 ??? ????
bb
nn
hb
n
— the resonance frequency
74
(1)
,1,)(1 ?<<<< ???? 时n
可不计阻尼。,0bb ?
(2)
,0,)(1 ????? ???? 时n
阻尼也可忽略。
时时 0, 7 0,)(1 <?? ???? n
(3)
阻尼对振幅影响显著。 ?一
定时，阻尼增大，振幅显著
下降。
222 212,0 ????? ????? nn nddb 得由
— 共振频率
此时,
2
0
m a x22m a x 12 2 ??? ????
bb
nn
hb
n
或
75
4,Phase difference,
Damped forced vibrations always have a phase behind that of the
disturbing force,This phase difference is given by the angle ? with
21
2tg
?
???
??
(1) The value of ? is always between 0 and ? 。
(2) The phase-frequency curve (the ? - ? - curve) is a monotonic
increasing curve,? increases with ?,
(3) At resonance,? =1 and, The curve increases near
resonance rapidly,all curves of different resistance pass this point,
(4) For ?>1,? increases with ?, If ?, 1,,(inverse
phase),
???
2???
b
? 2
0
max b,? ? n ? ? 1<< At and ?
76
???? 2,,1
0m a x bbn ??<< 时当
4、相位差
有阻尼强迫振动相位总比激振力滞后一相位角 ?,? 称为 相位差 。
21
2tg
?
???
??
(1) ?总在 0至 ? 区间内变化。
(2) 相频曲线（ ? - ?曲线）是一条单调上升的曲线。 ? 随 ? 增
大而增大。
(3) 共振时 ? =1，,曲线上升最快，阻尼值不同的曲线,
均交于这一点。
(4) ?>1时,? 随 ? 增大而增大。当 ?, 1时,反相。 ???
2???
77
Example 1 Knowing P=3500N,k=20000N/m,
H=100N,f=2.5Hz and c=1600N·s/m,determine b,?
and the equation of the forced vibration,
Solution,
r a d / s 58.103 5 0 0 8.92 0 0 0 022 ?????? Pkgmk eqn?
m 105.22 0 0 0 02 1 0 02 30 ??????? kHkHb
eq
4 8 5.1
58.10
5.222; 2 1 2.0
58.10
24.2
r a d /s 24.2
8.9/3 5 0 02
1 6 0 0
2
?
?
??????
?
?
??
?
?
?
?
?
?
?
?
nnn
fn
m
c
n
,
,
,
,
78
例 1 已知 P=3500N,k=20000N/m,
H=100N,f=2.5Hz,c=1600N·s/m,求 b,?,强迫
振动方程。
解,
r a d / s 58.103 5 0 0 8.92 0 0 0 022 ?????? Pkgmk eqn?
m 105.22 0 0 0 02 1 0 02 30 ??????? kHkHb
eq
4 8 5.1
58.10
5.222; 2 1 2.0
58.10
24.2
r a d /s 24.2
8.9/3 5 0 02
1 6 0 0
2
?
?
??????
?
?
??
?
?
?
?
?
?
?
?
nnn
fn
m
c
n
79
m m,84.15.27 3 6.0
,7 3 6.0
4 8 5.12 1 2.04)4 8 5.11(
1
4)1(
1
0
2222222
????
?
????
?
??
?
bb ?
???
?
).8 4 7.05s i n (84.1
),r a d( 8 4 7.0)5 2 2.0(a r c t g)]1/(2[a r c t g
2
22
??
?????
??
?????
tx
80
mm 84.15.2736.0
736.0
485.1212.04)485.11(
1
4)1(
1
0
2222222
????
?
????
?
??
?
bb ?
???
?
)8 4 7.05s in (84.1
)r a d( 8 4 7.0)5 2 2.0(a r c t g)]1/(2[a r c t g
2
22
??
?????
??
?????
tx
81
§ 18-6 The concepts of critical speed of
rotation,vibration reduction and vibration isolation
1,Critical speed of rotation of a rotator
The special speed of rotation at which the rotator will vibrate
drastically is called critical speed of rotation, This phenomenon is
caused by resonance,In course of designing of a higher speed shaft,
such checking computations should be taken into account,
Rotator of single circle disc,
Point C is the center of mass of a circular disc whose
mass is m,The geometric center A of the disc is on the
axis of rotation,AC=e,the disc is rotating around its
axis with a constant angular velocity ?,When ?< ?n
(?n is the natural frequency of the system consisting of
the disc and the axis of rotation),OC= x+e (x is the
flexural deformation of the mid-point of the axis),
82
§ 18-6 临界转速 ? 减振与隔振的概念
一、转子的临界转速
引起转子剧烈振动的特定转速称为 临界转速 。这种现象是
由共振引起的，在轴的设计中对高速轴应进行该项验算。
单圆盘转子,
圆盘：质量 m,质心 C点；转轴过盘的几
何中心 A点,AC= e,盘和轴共同以匀角速
度 ? 转动。 当 ?< ?n（ ?n为圆盘转轴所组
成的系统横向振动的固有频率）时,
83
kxexm ?? 2)( ?
11 2
2
2 ?
?
?
?
?
?
?
n
e
m
k
ex
Critical angular velocity is
Critical speed of rotation is
cc
nc
n
m
k
?
?
??
30?
??
From,where k is the equivalent
stiffness of the axis of rotation,we obtain,
n ? ? When goes to,x goes to infinity,
84
kxexm ?? 2)( ? （ k为转轴相当刚度系数）
11 2
2
2 ?
?
?
?
?
?
?
n
e
m
k
ex
??? xn,时当 ??
临界角速度,
临界转速,
cc
nc
n
m
k
?
?
??
30?
??
85
n ? ? ? C of mass is between the point O and the
2
2 )(11 ?
?
?
n
e
m
k
ex
?
?
?
?
If the center
point A,OC= x- e,
When the speed of rotation is very high,the center C of mass of
the circular disc approaches the line through two bracing point,
and the circular disc almost rotates a round the center C of mass,
Therefore,it rotates steadily,
n ?? ? ? n ? ? ? If,x decreases with increasing ?,When,x goes to e,
In order to ensure safety,the working speed of the axis must evade
its critical speed of rotation,
86
,运转时当 n?? ? 质心 C位于 O,A之间 OC= x- e
2
2 )(11 ?
?
?
n
e
m
k
ex
?
?
?
?
exx nn ??????,;,,时当时当 ?????
当转速 ? 非常高时，圆盘质心 C与两支点的连线相接近，
圆盘接近于绕质心 C旋转，于是转动平稳。
为确保安全，轴的工作转速一定要避开它的临界转速。
87
2,Concept of reduction and isolation of vibration
Drastic vibrations affect not only the normal work of a machine,but
also the normal work of the instrumental equipments around it,The
basic measure to reduce the danger and harm of vibrations is a
reasonable design,Try to reduce the vibrations and prevent them
from acting in the region of resonance,
Many factors causing vibrations are impossible to be reduced
effectively or difficult to be avoided,In these cases we can take
measures of vibration reduction and isolation,
Reduction of vibration,Various dampeners can be equipped on
vibrating bodies to reduce vibrations,
For example,we can use various resistance dampeners to consume
energy,then the goal is attained,
88
二、减振与隔振的概念
剧烈的振动不但影响机器本身的正常工作，还会影响周围
的仪器设备的正常工作。减小振动的危害的根本措施是合理设
计，尽量减小振动，避免在共振区内工作。
许多引发振动的因素防不胜防，或难以避免，这时，可以
采用减振或隔振的措施。
减振, 在振体上安装各种减振器，使振体的振动减弱。例如,
利用各种阻尼减振器消耗能量达到减振目的。
89
Isolation of vibration,the equipments which need to be isolated are equipped on proper vibration isolators (elastic devices) in order
to make most vibrations to be absorbed by the vibration isolators,
Isolation of vibration is
classified into
Positive isolation of vibration,which refers to isolating
the source of vibration from its foundation,
Passive isolation of vibration,which refers to isolating
the equipment,which needs to be prevented from
vibration,from the source of vibration,
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隔振,将需要隔离的仪器、设备安装在适当的隔振器（弹性
装置）上，使大部分振动被隔振器所吸收。
隔振 主动隔振：将振源与基础隔离开。
被动隔振：将需防振动的仪器、设备单独与振源隔离开。
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