第 四 篇 振动和波动
Part Four Oscillation and Waves
第 16章电磁振荡和电磁波
Chapter 16 Electromagnetic Oscillation &Wave
Chapter 17 Wave Optics
第 17章 波动光学
第 14章 机械振动
第 15章 机械波
Chapter 14 Mechanical Oscillation
Chapter 15 Mechanical Wave
第 14章 简谐振动
Chapter 14 Simple Harmonic Motion
§ 14-4 Damped Vibration& Forced Vibration
Resonance 阻尼振动 受迫振动 共振
§ 14-5 Superposition of two SHM with same
Frequency in same Direction
同方向同頻率的简谐振动的合成
§ 14-6 Superposition of two Perpendicular SHM
相互垂直的简谐振动的合成
§ 14-3 The Energy of SHM
简谐振动的能量
§ 14-2 Amplitude Period Frequency & Phase
简谐振动的振幅,周期,頻率和位相
§ 14-1 Simple Harmonic Motion (SHM)
简谐振动
教学要求
1、确切理解描述谐振动的特征量的物理意义,并能熟练
地确定振动系统的特征量,从而建立谐振动方程;
2、掌握描述谐振动的旋转矢量法和图线表示法;
3、掌握谐振动的特征和规律;
4、了解阻尼振动、强迫振动和共振的发生条件和规律;
5、掌握同方向、同频率谐振动的合成的特点和规律,了
解互相垂直谐振动的合成的特点,
Introduction
Varying force,)(tF? An infinite number ofways in which a force
may vary.
Periodic motion Vibration,a restoring force
particularly
四季的变化 共振现象
振动
In this chapter,we will study the most important
periodic motion-------Simple Harmonic Motion(简谐
振动 ),which is base to investigate( 研究 ) the
complex periodic motion.
简谐 振动
周期运动
复杂运动
波动、电磁波、电压电流等等
§ 14-1 Simple Harmonic Motion
简谐振动
1,The simple harmonic motion( abbreviate:SHM)
The system ( black-
spring) in Fig.15-1 is
called as a simple
harmonic oscillator( 谐
振子 ),
k m x
115,?F igSpring,不计质量。
body:
k
m
Equilibrium position,point o(和外力等于零) ;
物体:在平衡位置附近作周期性往复运动;
坐标原点:通常取为平衡位置 o;
Spring,deformation x;
Body:
displacement x;
force,(frictionless);
acceleration:
kxF ??
xmFa ???? ?
which solution is
?
???
?????
???
???? )ts i n (Axor)tco s (Ax
This equation is called the harmonic equation.
k m x
115,?F ig
xdt xda ??
?
???? ? mk??
Summary:
If the body is subject to:
A restoring force that is proportional to the
displacement but opposite in sign( ) ;kxF ??
or
The acceleration is proportional to its displacement
and of opposite sign( ) ;xa
2???
or
The motion equation is ; )c os ( ?? ?? tAx
k m x
115,?F ig
Example 14-1:prove the system is a simple harmonic
oscillator as shown in Fig.15-2.
l
0
x
0l
x
k
m
k
mgl ?
0
kxF ??
l
0
x
0l
k
m
215,?F i g
Example 14-2,In Fig.15-3,prove the
simple pendulum(单摆 ) is a simple
harmonic oscillator when ? is small.
Prove,(1) Displacement,xor?
(2)Restoring torque(恢复力矩),
?s i nm glM ??
(3) Angular(角量) acceleration:
???? ?????? lgIM ??? mlIlg?
Note,?lx ? x
dt
xda ?
?
? ??? ? (linear线量 )
x
315,?F ig
?? s m a l lf o rm gl??
2,The velocity and acceleration of SHM:
)tc o s (Ax ?? ??
Its velocity is given by:
)ts i n (Adtdxv ??? ????
and its acceleration is equal to:
)xor()tc o s (Adt xda ???
?
?????? ????
k m x
115,?F ig
Note,(1) Maximal values:
Aa
Av
Ax
m a x
m a x
m a x
??
?
?
?
?
(2) The curves of, )()()( taan dtvtx,
o T t
x,??,a
x
? 2A
?? > 0 < 0 < 0 > 0
a < 0 < 0 > 0 > 0
减速 加速 减速 加速
?A
A
-A
-?A
-? 2A
??
a
)( ???
(3)看图,.aan dvx,
3,The rotating vector representation(旋转矢量
表示法 ) of SHM
A vector with a length A is rotating about point
o at an angular velocity ? in Fig.15-4.
A?
0?t
????.F ig
?
?
?? ?t
The projection P of this
rotating vector in x-axis is
given by
)tc o s (Ax ?? ??
which is as same as the equation of SHM.
A rotating vector one to one
一一对应
SHM
0?t
????.F ig
?
?
?? ?t
x
4,Remarks:
A simple harmonic oscillator has to have two parts,
(1)‘Spring’ which
provides a restoring
force( that is it store
the potential energy);
(2) ‘Body’ which
has inertia( that is
it store the kinetic
energy).
k
PE
m
KE
§ 14-2 Amplitude (振幅 ) Period (周期 ) Frequency
(頻率 ) and Phase (位相 ) of SHM
1.Introduction
From
)tc o s (Ax ?? ??
We can see that:
(1) The displacement is determined by three
quantities,A,?,? ;
(2) is a periodic function of time t.)c os ( ?? ?tA
k m x
115,?F ig
2,描述简谐振动的特征量
振幅 A:简谐振动的物体离开平衡位置的最大位移
的绝对值(由初始条件决定) (代表系统总
能量的多少)。
115,?F ig
A2
周期 T:完成一次全振动所需时间
?
?2?T
由系统的力学参数决定 — 固有周期
k
mT ???
For black-spring system:
g
lT ???
and for simple pendulum system:
k m x
115,?F ig
))(2(1 s?
角频率,2?秒内 完成全振动的次数?
T
???? ????
频率,单位时间完成全振动的次数?
m
k
T ?? ?
????
固有频率
相位,决定任意时刻 t 运动状态的物理量?? ?t
初相,决定初始时刻 t=0 运动状态的物理量?
0?t
????.F ig
?
?
?? ?t
x
??
?
?
?
?
?
???
???
vv
xx
t
t
2,振幅和初相的确定
Initial conditions:
初始时刻的位置和速度
?
?
A
vs i n ???
A
xc o s ???
?
?
?
??
?
?
?
??
?
s i nAv
c o sAx
115,?F ig
?x
?v?
???
? ?? )
v(xA
?
可解出:
or
关键问题:决定 ?
( 1)由 决定 的两个可能值,;
A
xc o s ??? ?
?? ??
( 2)由 的正负选出正确的 。?
?? A
vs i n ???
关键:正确判断初始时刻
的位置和速度的 正负 !!0?t
????.F ig
?
?
?? ?t
x
Example 14-3:一个质量为 10g的物体作简谐振动, 周
期为 4s,t=0时坐标为 24cm速度为零 。 计算,( 1)
t=0.5s时物体的位置; (2) t=0.5s时物体受到的力的大
小和方向; (3)从初始位置运动到 x=-12cm所需的最少
时间; (4)x=12cm时物体速度的大小 。
解:本题已知物体作简谐振动。周期为 4s,振幅为 24cm,则:
??
?? ???
T )tc o s (Ax ?? ???
由初始条件:
?
?
?
?
??
0
24.024
0
0
v
mcmx
??
?
?
?
0
24.0
?
mA
2c o s24.0
tx ??
)(1018.4 3 NkxF ??????
( 1)当 t=0.5s时,)m(.c o s.x ?????
?????
?
( 2)当 t=0.5s时,物体受力,)
MN(.mk ? ? ?????????
??? ???
与 x轴正向相反。
( 3)从初始位置运动到 x=-12cm的最少时间:
o 24cm-12cm
?????
??? m i nt
)s(t m i n ????
( 4) x=-12cm物体的速度大小:
?????????
tc o s.,? 232s in ??t?
)(33.02s in smtAv ?? ??
注意:单位换算。
?
???
??
tc o s ?
Example 14-4:如图 14-5,为质点作谐振动的 x随时间
的变化曲线 。 求质点的振动方程和初速度 。
X(cm)
t(s)
2
4
2 5o
解:( 1)由 x-t曲线知:
sT 6? mcmA 04.04 ??
??
?
?
?
0
02.0
0
0
v
mx 32?T
( 2)设质点的振动方程为
)3c o s (04.0 ?? ?? tx
T
?? 2?
t=0时:
332
1c o s ???? ???? or 3?? ??
0s in 0 ??? Av??
所以:
)33c o s (04.0 ?? ?? tx
(IS)
( 2)当 t=0时,速度等于:
)sm()s i n (.v ? ? ?????????????? ???
注意:( 1)看图识‘量’;
( 2)正确写出初始条件;
( 3) ?的选择。
1k 2k
1k 2
k
Example 14-5:给出下列系统的等效弹性系数。
1k
2k
k
21 kkk ??
21 kkk ??
21
21
kk
kkk
??
Example 14-6,在一轻弹簧下端悬挂 m0=100g砝码时, 弹
簧伸长 8cm,现在这弹簧下悬挂 m=250g的物体, 将物体
从平衡位置向下拉动 4cm,并给予向上的 21cm/s的初速
度 ( 此时 t=0).选 x轴向下, 如图 15-6,求振动方程的数值
表达式 。
解:( 1)由题可得
)/(25.12
0
0 mN
l
gmk ??
l
0
x
0l
k
0m
615,?F i g
l
0
x
k
m
0l
0x
0v
( 2)园频率:
)/.(725.0 25.12 sr a dmk ????
( 3)设运动方程为,)7c o s ( ??? tAx
初始条件:
??
?
????
??
smscmv
mcmx
/21.0/21
04.04
0
0
可给出:
?
?
?
?
?
.64.0
05.0
r a d
mA
?
则:
)64.07c o s (05.0 ?? tx
( IS)
Example 14-7,一轻弹簧在 60N的拉力下伸长 30cm,现在
这弹簧下悬挂 m=4kg的物体, 使其静止,再将物体从平衡
位置向下拉动 10cm由静止释放 ( 此时 t=0),如图 15-7,
求,(1)振动方程的数值表达式 ;(2)物体在平衡位置上方
5cm时 弹簧对物体的拉力 ;(3)物体从第一次越过平衡位置
时刻起到它运动到上房 5cm处所需要的最短时间 。
715,?F i g
l
0
x
k
m
0l
0x
解:( 1)由题可得, )m/N(
l
Fk ?????
?
)s/.r a d(.mk ???????????
(2)选 x轴向下为正, )t.c o s (Ax ??????
??
?
?
??
0
1.010
0
0
v
mcmx
??
?
?
?
0
1.0
?
mA )07.7c o s (1.0 tx ?
(3)设物体在平衡位置上方 5cm时 弹簧对物体的拉力 f:
mg
fmafmg ??
22 /5.2 smxa ??? ?
Nmgmaf 2.29????
负号‘ -’表方向向上,为拉力,
(4)分两步,第一次到达平衡位置的时间 t1
0?x 21 ?? ?t st 225.021 ?? ??
第一次到达平衡位置上方 5cm的 的时间 t2
2
1)c o s (
2 ??t? st 296.03
2
2 ?? ?
?
)(0 7 4.012 sttt ???
o
§ 14-3 The Energy of SHM 简谐振动的能量
115,?F ig
x vPE
KE
The potential energy of
the system is:
)(c o s2121 222 ?? ??? tkAkxE P
and its kinetic energy is equal to
)(s in2121 2222 ??? ??? tAmmvE K
It can be proved that its total energy is constant:
222
2
1
2
1
2
1 kAmvkxEEE
KP ?????
You might now understand why an oscillating(振动 )
system normally contains an element of springiness and
an element inertia.
① 谐振动的总能量与振幅的平方成正比;
②如图 15-8,谐振动的动能和势能相互转化,总能
量守恒,
Conclusions:
x
tT
E
E p
E k(1 / 2) kA 2
o
kp EE ?PE
KE
AA?
x
815,?F ig
PE KE
③ 关系,xv ~
222
2
1
2
1
2
1 kAmvkx ??
22 xAv ??? ?
Example 14-8:x为何值时谐振子系统的动能等于势能?
解, 设 x=xP处
2
2
1
PPK kxEE ??
则,
?
??
?
?
?
?
?
?
?
?
?
??
kA
kxkx
EEE
PP
PK
Ax P 22??
x
tT
E
E p
E k(1 / 2) kA 2
o
kp EE ?PE
KE
AA?
x
Example 14-9:如果谐振子的振动频率为 ?.问其动能和
势能的变化频率为多少?
解, 因为
)t(c o skAkxE P ?? ??????? ???
)t(s i nAmmvE K ??? ??????? ????

)tc os (Ax ?? ??
所以 动能和势能的变化频率为,2?.
§ 14-4 Damped Vibration Forced Vibration and
Resonance(阻尼振动 受迫振动 共振 )
1,Damped Vibration 阻尼振动
x
In the air
Drag force
The motion dies
out eventually!!
x
In the water
When the motion of an oscillator is reduced by an
external force,the oscillator and its motion are said
to be damped,
Drag force
x
In the water
In the following,take the harmonic oscillator as an
example and consider only the frictions of
surrounding media,For the case of low speed,the
friction of media can be expressed as
dt
dxvf ?? ????
dt
dxkxvkxF ?? ??????

02
2
??? dtdxkxdt xdm ?
which has three solutions,underdamped vibration,
overdamped vibration and critical damping.
Underdamped vibration(欠阻尼),
x
0
t
特征,
??小,振动很多次 ;
?振幅逐渐减小 ;
?不是周期运动,但可引入周期,
)c o s ()c o s (2 ???? ?
?
???? ?? tAetAex tm
t
—— 阻尼系数,由阻力系数决定。
m2
?? ?
x
0 t
临界阻尼
过阻尼
弱阻尼
Overdamped vibration(过阻尼) and critical damping:
特征,
?Critical damping:能回到平
衡位置 ;
?Overdamped vibration:不能
回到平衡位置 (需无限长时
间 );
x
In the oil
f
2,Forced Vibration and Resonance(受迫振动 共振 )
x
In the air
Drag force
The motion dies
out eventually!!
Provide energy
(force it vibrate)
Add an additional force called as a driving force to a
harmonic oscillator,This motion is a forced vibration.
Example,a girl(child) sitting on a swing(秋千 ).
By giving the girl a little push once each cycle,you
can maintain a nearly constant amplitude.
gently
OK!
阻尼部分稳态部分 ??x
In general,a driving force is chosen as
tFF ?co s0?
In the following,the natural ( or intrinsic 固有的 )
angular frequency of the system is denoted by ?0 such
as, Therefore,we have
m
k?
0?
tFdtdxkxdt xdm ?? c o s02
2
???
Its solution consists of two parts:
)co s ( 0 ??? ?? tAe t
)c o s (),,,( 00 ????? ?tFA
?
A
o
0?
)c o s (),,,( 00 ????? ?? tFAx
特征,
?频率与驱动力的频率相同 ;
?振幅 A与 ?,? 0,?和 F0有关,特别是 ?的函数。
large ?
Small ?
??0
Resonance(共振 ):
From the following figure,the amplitude is
greatest when
0?? ?
A condition called Resonance.
depending on ?.
?
A
o
0?
Applications:
(1)All mechanical structures have one or more
natural angular frequencies;
(2) strengthen vibration(增强 ):
(3) reduced vibration(减弱 ):
Aircraft designer
Earthquake
§ 14-5 The composition of two harmonic vibrations
with the same direction and same frequency
两个 同方向同频率简谐振动的合成
1.Introduction
? A particle may takes part in two harmonic
oscillations such as the earthquake of 1988,Yunnan;
? For the next chapter and many fields;
)tc o s (Ax ??? ?? ?? )c o s ( 222 ?? ?? tAx
2.The composition of two harmonic vibrations
with the same direction and same frequency
The displacements resulting from two harmonic
vibrations in the same straight line (x-axis) are
)c o s (21 ?? ???? tAxxx
It can be proved that the displacement of composition
vibration is given by
with the same angular frequency ?.
1A
?
2A
?
A?
1?
2?
?
?
o
????
????
?
??
??
???
c o sAc o sA
s i nAs i nAa r c t g
)c o s (AAAAA ???????? ????? ??
It is easy to prove using the rotating vectors:
1A
?
2A
?
A??
o 2x 1x x
)c o s (AAAAA ???????? ????? ??
depend on A1,A2 and ?1- ?2.
There are two special situations:
(1) when
,.....),,k(k ????????? ?? ???
2121
2
2
2
1 2 AAAAAAA ?????
Enhancing(加强)
(2) when
,.,,,, ),,k()k( ??????????? ?? ???
2121
2
2
2
1 2 AAAAAAA ?????
Weakening (减弱)
In general
???? ???? AAAAA
)c o s (AAAAA ???????? ????? ??
Example 14-9:The two harmonic vibrations are
)tco s (.x
)tco s (.x
?????
?
??????
?
?
?
Find their composition vibration.
Solution,)5c o s (
21 ????? tAxxx
? ? ??????
????
????,
c o sAc o sA
s i nAs i nAa r c t g
??
???
????????? ????????,)c o s (AAAAA ??
Hence:
)643.05c o s (05.0 ?? tx