review
Chapter 1,Introduction,
Measurement,Estimating
Page1-16
了解
Chapter 2&3,Kinematics
2-1—2-7,一维运动，了解概念。（ p16-34）
2-8,Use of Calculus,Variable Acceleration
会用积分求速度和加速度。 （ p34-35）
3-1,3-2,3-3,3-4 and 3-5 （ p44-51）； 7-2 （ p151-
152） ; 11-1 （ p275-277） Vectors：知道如何计算。
3-6,Vector Kinematics （ p52-53）
3-7 & 3-8 projectile Motion,了解，（ p54-p61）
3-9 Uniform Circular Motion,5-4 Nonuniform
Circular Motion & Dynamics of Uniform
Circular Motion p62-63; p119-120; p113-115
3-10 Relative Velocity,不要求 (p64-67)
Position Vector(位矢）
kzjyixr
ktzjtyitxtrr )()()()(
For a Cartesian system:
Motion Function ( 运动方程 )：
P (x,y,z,t )
ik
j
x
y
z
(t0)
r(t)
(t?)
z
x
y
Instantaneous velocity (瞬时速度 ),
ktzjtyitx
dt
d
dt
rd
t
r
v
t
)()()(
lim
0
)( tvkvjviv zyx
Instantaneous Acceleration:
2
2
0 d
d
d
d
d
d
d
dl i m
t
r
t
r
tt
v
t
va
t
In a natural coordinate （ 自然坐标系）：
反映速度大小变化的快慢
--Tangential acceleration（切向加速度）
--Radial acceleration（法向加速度）
naτaa nt
2
2
dt
sd
dt
dva
t
R
va
n
2
反映速度方向的变化
Circular Motion
1,Uniform Circular Motion (p62):
2,Nonuniform Circular Motion (p119):
tn anaa
where
dt
dva
t? r
va
n
2
or 22
nt aaaa
)/a r c t a n (
nt aa
1,第一类问题
a,v已知运动学方程，求
2,第二类问题已知加速度和初始条件，求 r, v
运动学的两类问题：
1,第一类问题 a,v已知运动学方程，求
(1) t =1s 到 t =2s 质点的位移
(3) 轨迹方程
(2) t =2s 时 a,v
jir 21 jir 242
jijirrr 321)2(2)(412
已知一质点运动方程 jtitr )2( 2 2
求例解 (1) 由 运动方程得
jtitr
22ddv
ji 4 22v
(2)
当 t =2s 时 ja 2 2
jtt ra?
2dddd 2
2
v
222 tytx 4/2 2xy(3) 轨迹方程为
2,第二类问题 已知加速度和初始条件，求 r, v
ttv d2d?
A particle moves along the direction of x axis,a=2t,At t =0,
x0=0,v0=0,What are its velocity and position at t =2s?
Solution,Acceleration,a=2t,is not constant.
)1(dd;d2d 2
00
ttxvttv tv
tx ttxttx 0 202 dd;dd )2(3
1; 3tx?
Eliminate t from (1) and (2),we can also get:
3/2)3(;)( xvxvv
m67.238;m / s4 xv
So,
txvtva d)62(d;dd 2
vvxa dd?
t
v
t
va
d
d
d
d
x
vv
d
d?
xd
xd
Solution:
vx vvxx 00 2 dd)62( 23 21)(2 vxx
32 xxv
―-” no physical meaning
If a(v) is given,we may also find v(x) by
x
v
t
x
t
vva
d
d
d
d
d
d)(;
d
d
x
vv?
Note:
A particle moves along x direction,,
At t =0,x0=0,v0=0,What is its v(x)?
262 xa
)(
dd
va
vvx?
Example:
Chapter 4&5,Dynamics,
Newton’s Laws
Page77-132
Newton’s Second Law
amF n e t
t
PF
d
d
t vmd )(d
vtm?dd tvm dd
vmP
Where is momentum (动量 ) of body.
So in classical mechanics,it can be expressed as:
zzn e t
yyn e t
xxn e t
maF
maF
maF
,
,
,
In rectangular coordinate:
Applying Newton’s Laws
1,Pick a body based on the feature of problem (研究对象 );
2,Draw a free-body diagram (画出隔离体的示力图 p89);
3,Choose coordinate axes (建立较方便的坐标系 );
4,Resolve vectors along chosen axes (沿轴分解矢量 );
5,Apply Newton’s laws of motion (列方程,统一单位,求解 );
6,Decide whether or not the solution makes sense.
Problem Solving Guide:(p94 –95,problem solving)
重点放在极限问题、积分上。 题目主要看 CHAPTER 5中的例题。
A ring screen of radius R is horizontally fastened on a
frictionless surface,A slippy block of mass m moves along
its inner-wall,The frictional constant between them is?.
Find,(a) when the speed of block is v,Frictional force f=?
and at =? (b) How long will it take as its(滑块 )speed
decreases v? v/3?
N?
v?
tF
R
O
mv?
(a) Using natural coordinate frame
Solution:
Chapter 6 Gravitation
Page132-146
不要求
Chapter 7&8,Work and Energy
p147-160
7-1 Work Done by a constant Force P147-P151
if
4321 FFFFF
321
321
WWW
rdFrdFrdFrdFW
The work done by several forces = algebraic
sum of the work done by the individual force.
Work done by several forces:
a n d, kFjFiFF zyx, kdzjdyidxrd
fz
iz z
fy
iy y
fx
ix x
B
A zyx
B
A
dzFdyFdxF
dzFdyFdxFrdFW )(
Special expression in scalar product:
7-3 work done by a general variable force P152-155
BABA rFrFW dc o sd
EXAMPLE 7-5； 7-6
质量为 10kg 的质点，在外力作用下做平面曲线运动，该质点的速度为 jit 164 2v
解 24dd ttxxv ttx d4d 2?
16dd tyyv ty 16?
ttmF xx 80dd v 0dd?tmF yy v
J 1 20 0d3 2021 3 tt
在质点从 y = 16m 到 y = 32m 的过程中，外力做的功。求例
,开始时质点位于坐标原点。
时16?y 1?t
时32?y 2?t
yFxFW yx dd
F?
L
缓慢拉质量为 m 的小球，
解 0s in θTF
0c o s mgθT
θmgF ta n?
sθθmg dc o st a n
x
y
θ
G?
T?
0c o s1 θm g L
例
=?0 时，求已知用力 F? 保持方向不变F?
作的功。F?
00 c o st a nθ θθθL m g d
sθFrFW dc o sd
质点系 动能定理
0kk
inex EEWW
7-4 Work-Kinetic Energy Theorem (P156-p160)
7-5 Kinetic Energy at Very High Speed
不要求
8-3 & 8-4 Mechanical Energy and its Conservation
P174-181
cUKE m e c
8-1 & 8-2 basic concepts (P168-p174)
m e ciiffince x t EUKUKWW )()(
m e cn o n c o n EW
or
8-6 Work-Energy Principle(P182-p184)
8-5 The law of Conservation of Energy 不要求
8-7,8-8 & 8-9 不要求
Chapter 9,Linear Momentum
and Collisions
p200-233
linear momentum
vmP?
dt
PdF
n e t
Conservation of linear momentum
0?n e tF? P
= constant vector
常量常量常量
ziziz
yiyiy
xixix
PmF
PmF
PmF
v
v
v
0
0
0
9-1& 9-2 Momentum and Its Relation of Force
and Conservation of Momentum p200-p205
Impulse-Momentum Theorem:
)(e x t vmdtddt PdaMF?
,pddtF ppppddtF ifp
p
t
t
f
i
f
i
fitt dtFJ
—–The impulse(冲量 )of force
vmP —– Momentum (动量 )
Impulse-Momentum Theoremif PPPJ
9-3 Impulse and Collisions P205-208
ixfxxx PPPJ
iyfyyy PPPJ
izfzzz PPPJ
9-4,9-5,9-6 p208-212了解
9-7 9-8,9-9 & 9-10 p213-223 不要求
Chapter 10 & 11,Rotation
Motion About a Fixed Axis;
General Rotation
10-1,Angular Quantities
10-2 ; 10-4 &10-5 Torque; (P235-238; P241-243; )
掌握基本概念,
10-3,10-11 & 10-12,11-6 ; 11-8; 11-9 ; 11-10 不要求 ;
10-6,10-7,10-8,10-9,10-10; 11-3; 11-4; 11-5; 11-6;
11-7 P241-256 ; p277- 285
重点重点,复习课件
11-2; 11-3; 11-4 11-5 ; 11-7 P276-282; P284-285
10-6&10-8N-II Law for Rotation (定轴转动定律 ) and Moments of Inertia
(P243-244)
In e t? N-II Law for Rotation(p244)
2ii rmI
dmrI 2
dl
dS
dV
dm
where,. dmrI 2
Page 245,example:10-6
)2,(
12
1 2
0
22
2
2
LL
Lc dIIormLdxxL
mI
2
0
2
3
1 mLdxx
L
mI L
o
Solution(1),dxdm
c
x
x
dmxdI 2?dxx?2 dx
L
mx 2
Try to find out I of a thin rod of mass m,length L to
(1) C and (2) O axes.
O x dm xLSolution(2)：
10-7 Solving Problems in Rotational Dynamics
(P245-248)
As always,draw a clear and complete diagram.
1,Draw a free-body diagram.
2,Identify the axis of rotation and calculate the
torques about it.
3,Apply Newton’s second law for rotation and for
translation.
4,Solve the resulting equations.
例 2：质量为 m1和 m2两个物体，跨在定滑轮上 m2 放在光滑的桌面上，滑轮半径为 R，
质量为 M，求,m1 下落的加速度，和绳子的张力 T1,T2。
1m
2m RM,
T1
T2
解：受力分析
1m以 为研究对象
amTgm 111（ 1）
1m
gm1
1T
2m以 为研究对象
RM,
1T
2T
2T2m
2m以 为研究对象
amT 22? （ 2）
M以 为研究对象
JRTT )( 21 （ 3）
补充方程,?Ra? （ 4）
联立方程（ 1） ---（ 4）求解得
2/21
1
Mmm
gma
2/
)2/(
21
21
1 Mmm
gMmmT
2/21
21
2 Mmm
gmmT
讨论：当 M=0时
21
21
21 mm
gmmTT
Fig,Shows a uniform disk,with mass M=2.5kg and radius
R=20cm,mounted（安放） on a fixed horizontal axle,A block
with mass m=1.2kg hangs from a mass-less cord that is wrapped
（缠绕） around the rim of the disk,Find the acceleration of the
falling block,the angular acceleration of the disk,and the tension
in the cord,The cord does not slip and there is no friction at the
axle,(I=MR2/2)
Axis O
R
m
M
α
T
·R
mg
T = - T′
maM
For M,? = TR = I? (1)
For m,mg – T = ma (2)
a = R? (3)
From above equations,we find the answer:
a=4.8 m/s2
=24 rad/s2
T=6.0 N
Solution:
mg
T = - T′
ma
α
T
·RM
In fig.,two blocks with mass m1 and m2 (m1>m2),two
pulleys (with radius of RA and RB,mass of mA and mB) are
mounted in horizontal frictionless bearings,assuming that
the pulley is a disk,the cord is not slipping,Please find the
blocks’ acceleration.
m1 m2
mA mBR
A
RB
Solution:
Example:
Draw free-diagram:
m2
m2g
T2
m1
m1g
T1
T1-m1g=m1a (1)
m2g-T2=m2a (2)
mB
RB
T
T2
mA R
A
T
T1
(T-T1)RA=IA?A (3)
(T2-T)RB=IB?B (4)
Where,a= RA?A = RB?B
m1 m2
mA mBR
A
RB
Solve equations (1)-(5),yields:
BA mmmm
gmma
2
1
2
1
)(
21
12
2
2
1
BBB RmI?
2
2
1
AAA RmI?
A body rotating about an fixed axis is said to
have rotational kinetic energy,By analogy
with translational kinetic energy,we have
221 ii vmK 222 )(2121 iiii rmrm
Using equation of I, 2
ii rmI
.21 2 IK
1,The Kinetic Energy of Rotation:
10-10 Rotational Kinetic Energy (P254-256)
then,
2,Work-Kinetic Energy Theorem for Fixed-axis Rotation
21 dMW
力矩的功
From the kinetic theorem of a single particle,
,KmvmvW if 22 2121
.2121 22 KIIW if
KIIW if 22 2121 f
i
dWwhere
So,we get the work-kinetic energy theorem for
rigid body rotated about a fixed axis:
Solution:
As the Fig.,For the spring,k=2.0?103 N/m;
For the wheel,I=0.5 kg?m2,R=30cm;
m
R
k
What is the speed of the body (m=60kg),
when it falls 40cm,Initially,it is at rest
and the spring is not stretched.
maTmg
IkxRTR
Ra?
2/ RIm
kxmga
EXAMPLE:
x
RIm
kxmgvv hv d
/
d 0 20
m / s5.1
/
2
2
2
RIm
khmg hv
or,
2
2
1 kh m g h?221?I?221 mv?
,dd tva?,dd txv? vxav dd
2/ RIm
kxmga
Torque about a point:
Fr
vmP
The linear momentum of a particle:
Angular momentum of a Particle
The Angular momentum relative to a fixed point：
)( vrmPrl
11-2 The Torque Vector p276
11-3 Angular momentum of a Particle(p277
Solution:
A particle moves in the xOy-plane,Write down
its angular momentum in terms of its
coordinates and its components of velocity,
vrmprl
)(or xyz yvxvml
, jyixr, jvivv yx
)( kyvkxvm xy
EXAMPLE:
11-4 Relation between Angular Momentum
and Torque:(p278-280)
.dtldn e t
The angular Momentum of a system of Particles
.
1
,
1
n
i
in et
n
i
i
dt
ld
dt
Ld
,0If CLn et
11-7 The Conservation Law of Angular Momentum (p284)
If net external torque acting on particle’s system is zero,
the total angular momentum of the system remains
conserved,合外力矩为零，质点系总角动量守恒 !
v1 r
1 r2
O
v2
F?
A cord makes a ball,mass m,move in a circle (r1
v1) on the surface of frictionless table,(1) When
one pulls the cord downward slowly and makes
the radius equal to r2,v2=? (2) The work done by
the during r1?r2.F?
2211 rmvrmv?
)(
2
1
12 r
rvv?
12 vv?
F is central force,and v
r,L=mvr =constant
Solution,(1)
EXAMPLE:
11-5 Angular Momentum and Torque for a Rigid
Body (Page:280-282)
1,Angular Momentum & Torque of Rigid Body
(P280)
刚体定轴转动的角动量
i
iiii
i
i rmrmL )(
2v IL?
Summing over all the particles to obtain:
dtLdn e t
Using the angular momentum principle of
particle’s system,
For a rigid body,the component along the
rotation axis is:
dtdLa x i s
I
dt
dII
dt
d )(
刚体所受的外力矩等于刚体角动量的变化率。
2 刚体定轴转动的角动量定理
10-9 Conservation of Angular Momentum (p251-
254)
c o n st a n tL fi LL?
or,ffii II
If no net external torque acts on a rigid body,
IdtdLa x i s
o
1
o
2
例 3,人 与 转 盘 的 转 动 惯 量
J0=60kg·m2,伸臂时臂长为 1m，
收臂时臂长为 0.2m。 人站在摩擦可不计的自由转动的圆盘中心上,每只手抓有质量 m=5kg的哑铃 。 伸臂时转动角速度?1 = 3 s-1,
求收臂时的角速度?2,
解：整个过程合外力矩为 0，
角动量守恒，
2211 JJ?
2
101 2 mlJJ
215260
2
202 2 mlJJ
22.05260
2mkg70 2mkg4.60
2211 JJ?
2
11
2 J
J
由转动惯量的减小，角速度增加。
在此过程中机械能不守恒，因为人收臂时做功。
4.60
703
1-s5.3?
10-11,10-12,11-6不要求
Chapter 32,Special Theory of
Relativity
P731-762
32-1 & 32-2 P731-P736 了解
32-3 & 32-4 P738 基本概念
31-5 Time Dilation P739-743
2
0
)/(1 cv
tt
固有 时间,同一 地点发生的 两 事件的时间间隔,
When two events occur at the same location in an
inertial reference frame,the time interval between
them,measured in that frame,is called the proper
time interval or the proper time.
Measurements of the same time interval from
any other inertial reference frame are always
greater,（在其他任何惯性系中观测者看来，这两个事件为异地事件，其之间的时间间隔?t 总是比
t0 要大。）
0' ttt 固有时间
02
0
2
0
1)/(1
tt
cv
tt
1
1
1
2
2
c
v
例 3 设想有一光子火箭以 速率相对地球作直线运动,若火箭上宇航员的计时器记录他观测星云用去 10 min,则地球上的观察者测得此事用去多少时间？
c95.0?v
m i n01.32m i n
95.01
10
1
'
22
tt
运动的钟似乎走慢了,
解,设火箭为 系、地球为 S 系'S
m i n10' t
例介子是一种不稳定的粒子，从它产生到它衰变为介子经历的时间即为它的寿命，已测得静止 介子的平均寿命?0 = 2? 10?8s,某加速器产生的 介子以速率 u = 0.98 c 相对实验室运动。
求介子衰变前在实验室中通过的平均距离。
解 对实验室中的观察者来说，运动的介子的寿命?为
s7?
1000519801 1021 2
8
2
0,
.β
ττ
因此,介子衰变前在实验室中通过的距离 d' 为
m5.2910005.1 98.0 7cu τd'
0?ud?
32-6 Length Contraction P743-744
The length contraction,
02
0 1
LLL
Where L0 is proper length（固有长度,measured
in the rest frame of the object）
The length L0 of an object measured in the rest frame
of the object is its proper length or rest length。
Measurements of the length from any reference frame
that is in relative motion parallel to that length are
always less than the proper length。
例 1 设想有一光子火箭,相对于地球以速率 飞行，若以火箭为参考系测得火箭长度为 15 m,问以地球为参考系，此火箭有多长？
c95.0?v
s'
s
火箭参照系地面参照系解,固有长度 'm150 ll
21' ll m68.4m95.0115 2l
m150?l
v?
x
'x
y 'y
o 'o
32-7 Four-Dimensional Space-Time
32-8 Lorentz Transformations P746-748
了解基本思想
32-9,32-11 Relativistic Momentum,Mass,Energy
Ｐ７４９－７５４
2
2
0
1
c
v
m
m
相对论质量
2
0
1
mm
Where,m0 is rest mass （ 静止质量 ）,m is
relativistic mass (相对论质量,总质量 total mass)
相对论动量：
2
2
0
1
c
v
vm
vmP
相对论动力学基本方程：
2
2
0
1
c
v
vm
dt
d
dt
Pd
F
relativistic kinetic energy（相对论动能）,
202 cmmcE k
relativistic energy （相对论能量）,
20 cMEE k
relation of momentum and energy（相对论动量和能量的关系）
20222 EcpE
32-10,32-12 不要求
Chapter 12 Oscillations
p297-324
12-2 Simple Harmonic Motion (SHM) (简谐振动
P299-304）
tAx c o s
0
d
d 2
2
2
x
t
x?
Equation of motion for the
simple harmonic oscillator.
简谐运动的动力学方程
—— Oscillation Equation
Feature of Oscillatory Motion:
Dynamics,xaorkxf 2
12-1 Oscillations of a Spring 了解
Basic Quantities of SHM,(P301)
m
k
k
mT?
22
m
k
Tf?2
11
Phase or Phase Angle & Initial
t)c os ( tAx在 中,称为振动的相位。
Phase difference of two SHM with same frequency:
12
= 2k?,(k=0,?1,?2,?3…),two oscillators
have same phase (in phase);
当 =?(2k+1)?,( k= 0,1,2,… )，
两振动步调相反,称 反相 。 opposite (out of) phase
,)co s ()()( txdtddt tdxtv m
)2c o s (s i n txtxtv mm
The velocity of oscillating particle is also SHM and varies
between the limits ± vm=±xm,v(t) is ahead of x(t) for?/2.
The Velocity of SHM (p302)
简谐运动的速度函数式
The Acceleration of SHM,(p302)
dt
tdvta?)(.c o s
2 txta m
The acceleration of the oscillating particle is also
SHM varies between the limits ± am = ±?2?xm.
简谐运动的加速度函数式
)c o s (
)c o s (
2
2
2
2
tA
tA
dt
xd
a
tx? 图
t?v 图
ta? 图
T
A
A?
2?A
2?A?
x
v
a
t
t
t
A
A?
o
o
o T
T
)c os ( tAx
)
2
πc o s ( tA
)πc o s (2 tAa
振动曲线
2
2
02
0?
v xA
0
0t an
x?
v
常数 和 的确定 (p301)A?
000 vv xxt
初始条件
co s0 Ax?
s i n0 Av
对于给定的振动系统，周期由系统本身性质决定，
振幅和初相由初始条件决定,
)s i n ( tAv
)c os ( tAx
12-4,Uniform Circular Motion (旋转矢量法 ),(P306)
co s0
)co s (
A
tAx
2
π
0s i n0Av?
2
π 0s i n 取
0,0,0 vxt
已知 求 。
x
v?
o
)
2
π c o s ( tAx?
A
A?
x
T
2
T
t
o
A body (m=10g) is in SHM along x axis,A=20cm,
T=4s,When t = 0,xo= -10cm,and v < 0 (moves
along negative x direction),Find,(1) x(t=1s)=?; (2)
At what time will the body pass the position of
x=10cm first time (何时物体第一次 通过 ） (3) How long
does the body need to pass x=10cm second time(再经多少时间物体第二次运动到 x=10cm处）?
Example:
24
22
T
mx 173.0)322c o s (2.0
or as the Fig.,mx 1 7 3.0c o s2.0
(2)
sttt o 2
2
(1) When t=1s,
st
3
4
2
3
2
(3)
2A?
0
X
O
6
32
2A
mtx )322c o s (2.0
As the Fig,
3
2
0
Solution:
用匀速圆周运动表示简谐运动的 速度变化 (306)
A?mv
做匀速圆周运动的质点的速率是：
在时刻 t,它在 x轴上的投影是
)
2
π
co s (
)s i n (
tA
tmvv
这就是简谐运动的速度公式。
2
πt
mv?
v?
x
y
0
At
)c os ( tAx
Potential Energy:
2
1)( 2kxtU?
§ 12-3 Energy in Simple Harmonic Motion (P304)
)s i n (
)co s (
tA
tAx
v
kxF
)(c o s
2
1 22 tkA
Kinetic Energy:
)(s i n
2
1
2
1 222 tkAmK v
2
2
1 kAUKE
弹簧振子的总的机械能
Total mechanical energy
弹簧振子的总能量不随时间改变，即作 简谐 运动的系统 机械能守恒。
Total energy of oscillatory system is a constant.
2A?
振幅的动力学意义,振幅不仅给出了简谐运动的运动范围，而且还反映了 振动系统总能量的大小,或者说反映了振动的 强度 。
2
2
1 kAUKE
例 质量为 的物体，以振幅作简谐运动，其最大加速度为,求,kg10.0
m100.1 2
2sm0.4
（ 1） 振动的周期；
（ 2） 通过平衡位置的动能；
（ 3） 总能量；
（ 4） 物体在何处其动能和势能相等？
解 （ 1）
2
m a x?Aa?
A
a m a x 1s20
s314.0π2T
（ 2） J100.2 3
222
m a xm a x,k 2
1
2
1 AmmE v
（ 3）
m a x,kEE?
J100.2 3
（ 4）
pk EE?
时，
J100.1 3pE
由
222
p 2
1
2
1 xmkxE
2
p2 2
m
E
x?
24 m105.0
cm7 0 7.0x
小结：简谐运动的描述和特征
xa 24） 加速度与位移成正比而方向相反
xt x 22
2
d
d2） 简谐运动的动力学描述
)s i n ( tAv
)c o s( tAx3） 简谐运动的运动学描述
mk弹簧振子
kxF1） 物体受线性回复力作用 平衡位置 0?x
§ 12-5 The Simple Pendulum单摆 (P307-
308) 了解在角位移很小的情况下，单摆的振动是简谐运动。
Is undergoing SHM
)c o s (m t
l
g?2?
glT π2?
l
g
mgmgF t s i n由
Period does not depend on the
mass of the pendulum bob
§ 12-5,12-6,12-7,12-8不要求
Chapter 13 Wave Motion
p325-353
13-1,13-2,了解基本概念
§ 13-3 Energy,Power and Intensity (波强） of
Traveling Wave P331-332 了解动能 2
2
1 vdmdE
k?
)/(si n)(21 222 uxtAdV
kinetic energy
弹性势能 Elastic Potential Energy
)(s i nd21dd 222pk uxtVAWW
Both K & U of element dm varies periodically with time,同时达到最大值,又同时达到最小值,体积元的机械能不守恒 。
An important feature of wave motion is the transfer of
energy,
13-4Math’s Expression of Traveling waves (P332-
335):
])(c o s [),( vxtDtxD Mp
v沿 x 轴 负 向波函数
v沿 x轴 正 向 ])(c o s [),(
v
xtDtxD
Mp
Neglecting the subscript p,we have equivalent
expressions for SH wave equation:
])(2s i n [),( oTtxAtxy
)s i n (),( otkxAtxy
])(2s i n [),( otfxAtxy
In general,wave function of an arbitrary
shape of traveling wave has a form,
)s i n (),( tkxDtxD M (See P334)
The Physical Meaning of Wave Equation:
波 函数 具有 时间周期性 （ T )o t
A
y
T
(1) Fixed x,corresponding to the oscillating curve
of medium element at position x,i.e,y(t,xo).
(2) Fixed t,corresponding to the y-x curve
(波形图 ) at to.
波 函数 具有 空间周期性 （? )o
y
x
2x1xA
在下列情况下试求波函数：
)]81(π4c o s [ tAy A
(3) 若 u 沿 x 轴负向，以上两种情况又如何？
例 2
(1) 以 A 为原点；
(2) 以 B 为原点； B A
1x
x
已知 A 点的振动方程为：
(1) 在 x 轴上任取一点 P,该点振动方程为：
)]81(π4c o s [ uxtAy p
)]81(π4c o s [),( uxtAtxy波函数为：
解
u?
P
1x
B A x
(2) B 点振动方程为,)]8
1(π4c o s [)( 1
u
xtAty
B
)]81(π4c o s [),( 1 u xxtAtxy
)]81(π4c o s [),( uxtAtxy(3) 以 A 为原点：
以 B 为原点：
波函数为,)]8
1(π4c o s [),( 1
u
xxtAtxy
13-6 The principle of superposition for waves（波的叠加原理 P337,） P338
结合 13-8
13-5,13-7,13-8,P13-10,13-11不要求
13-9 Standing Waves,Resonance
1 the produce of standing waves 驻波的产生
13-6 The principle of superposition for waves（波的叠加原理 P337,） P338
结合 13-8
*13-8 Interference (干涉 ) of Waves (P339-340)
Interference of Waves (波的干涉 ):
)c o s (21 tDDDD M
The resultant oscillation in point P is also SHM:
Based on the superposition of oscillation,
c o s2 212 22 1 MMMMM DDDDD
2211
2211
c o sc o s
s i ns i nt a n
MM
MM
DD
DD
1212 2 rr
The in expression A is the phase difference
arisen by two waves at point P,It does not vary
with time,so does not amplitude at every points.
1s
2s
P
*1r
2r
(1) when (k=0,1,2,…)?
k
rr 22 12
12
21 MMM DDD
The greatest possible DM —– Fully
constructive interference (相干加强 ),
(2) When (k=0,1,2,…), )12( k
MMM DDD 21
The lest possible A —– Fully
destructive interference (相干减弱 ).
(3) When (k=0,1,2,…), )12(2 kk
Then DM is between —–
intermediate interference (neither fully
constructive nor fully destructive).
2121 MMMM DDandDD
Special case,(i)
21
MM
MMM
DDD
DDD
kk
k
rrL
21
21
21,.,,)2,1,0(
2
)12(
两列相干波源同相位时,其相干加强和减弱情况仅取决于波程差。
(ii) When,
021 MMM DDD
)c o s1(2c o s2 0212 22 1 MMMMMM DDDDDD
2c o s2 0
MM DD
x 2
13-9 Standing Waves,Resonance
1 the produce of standing waves 驻波的产生振幅、频率、传播速度都相同的两列相干波，在同一直线上沿 相反 方向传播时叠加而形成的一种特殊的干涉现象,
If two sinusoidal waves of the same amplitude and
wavelength travel in opposite directions along a
stretched string,their interference with each other
produces a standing wave.
tkxDtxD m?co s]s i n2[),('?
驻波的振幅与位置有关各质点都在作同频率的简谐运动
)(,...2,1,0, nnkx or 2 nx –— nodes
2)2
1( nxor –— antinodes
The interval between pairs of two nodes and
antinodes is half wavelength.
13-5,13-7,13-8,P13-10,13-11不要求
Chapter 14 也不做要求
Chapter 30 The Wave
Nature of Light; interference
p680-701
30-1 Huygens’ Principle and Diffraction
P680-681
了解
30-1 P682-683; 30-2,30-7,30-8 P695 不要求
30-3,30-4,30-5 Interference-Young’s Double-Slit
Experiment p683-690
D
xdsi n12 drrr
1 The path length difference波程差
=
m=0,1,2,…
m=0,1,2,…
m?
)21( m
(Bright fringe明 ); constructive interference
(Dark fringe暗 ); destructive interference
sind
2 interference pattern lines
rdDx
=?dDm?
dDm )21(
m=0,1,2,… 明
m=0,1,2,… 暗
3.条纹间距 (spacing between neigboring points
of constructive or destructive interference
a.相邻明纹间距,b.相邻暗纹间距：
可以看出相邻明纹与相邻暗纹的间距都相同，所以条纹 明暗相间平行等距 。
d
Dx
Intensity in the Double-Slit Interference
Pattern (p687-690)
)π2c o s ( 1111
rtAy p
)π2c os ( 2222
rtAy p
点 P 的两个分振动
co s2 212221 AAAAA
12
12 π2
rr 常量
intensity pattern 双缝干涉光强分布
)c o s (2 122010220210 AAAAA
)c o s (2 122121 IIIII合光强干涉项若
021 III
r π2
12其中
)( πc o s4 20
rII?krI,4 0
2)12(,0 kr
则
30-6 Interference in Thin Films p691-694
光程 =折射率?几何路程 = nr定义：
例,在双缝干涉实验中,用波长为 632.8 nm 的激光照射一双缝,将一折射率为 n=1.4 的透明的介质薄片插入一条光路,发现屏幕上中央明纹移动了
3.5个条纹,求介质薄片的厚度 d 。
o’
D
S1
S2
S d
r1
r2 I
oo
d
解,由于中央明纹移动了 3.5 个条纹，则插入的介质薄片所增加的光程差为 3.5 个波长，对应原屏幕中央 o 点两条光线的光程差也为 3.5? 。
D
S1
S2
a
r1
r2 o
d
在原屏幕中央 o点两光线的光程差为：
5.3?
5.3)1( dn
dnd
1
5.3
nd
m105,5 6- 14.1
108.6 3 25.3 9
2.半波损失 phase shift of? p691-692
interference in thin films（薄膜干涉） P691-692
For near-normal incidence
（垂直入射）
dn 22
未考虑半波损失时考虑半波损失：
321 nnn
321 nnn
321 nnn
321 nnn
光程差不附加 2
光程差附加 2
2
2 2 dnΔ
1n
1n
2n
Example 30-7 p693
22 2
dnΔ
12 nn?
当 时
2)12(
m
m )2,1(m
)2,1,0(m
加强减弱 22 2?dnΔ
Minimum thickness
22
12
nd
1?m
24 n
d
Wedge-shape（劈尖） with a thickness that varies
uniformly (example 30-6,p692)
（ Perpendicularly incident）
2)12(
m
m )2,1(m
)2,1,0(m
加强 bright
band
减弱 dark
band
22 knd
劈尖 (the point of contact)
2
Δ
0?d
为暗纹,
e?
b
'b
由于同一条纹下的空气薄膜厚度相同，当待测平面上出现沟槽时条纹弯曲。
待测平面
The bright and dark
bands will be straight
only if the glass
plates are extremely
flat,If they are not,
the pattern is uneven,
Newton’s rings（ 牛顿环 ） (p691) 不要求
Chapter 31 Diffraction and
Polarization
P702-730
31-1 diffraction by a single slit（单缝衍射） p703-704
half wave zone method （半波带法）
s ina
2)12(
m
m dark fringes
0 Central bright fringe
Bright fringes
a
31-7 diffraction Grating (p713-715)
Grating spacing （光栅常数） d =a+b
The feature of fringes（条纹特征）
1 明条纹细而亮，有较宽的暗区,N越大明纹越细。
2 各级明纹亮度不同，有缺级现象。
I
1 2 4 5-1-2-4-5
缺级缺级-2 2
N=5
光栅方程
(maximum-bright 主极大 )k=0,1,2,3…
k )2,1,0(k s i n)( ba 加强例：分光计作光栅实验，用波长? = 632.8 nm的激光照射光栅常数 d = 1/300 mm的光栅上，问最多能看到几条谱线。
解：在分光计上观察谱线，最大衍射角为 90°,
k s i n)( ba
90si n)(
m a x
bak
d?
f
o
x
P
取 5m a xk
能观察到的谱线为 11条：
。，，，，，，，，，,5 4 3 2 1 0 12345
9
3
108.632300
101
3.5
90si n)(
m a x
bak
单缝衍射 减弱 条件, 'si n ka
光栅衍射 加强 条件, kba si n)(
两式相比 'k
k
a
ba m? )321(?，，?m
缺级条件 miss order
m为整数时，光栅谱线中 m,2m,3m等处缺级。
例：一光栅,b=2a，则缺级的明纹：
'32'' k
a
aak
a
bakk,,,3.2.1'k
故,..9.6.3'3 kk
31-3 diffraction by a double slit (p708-709)
Example 31-4 Diffraction plus interference,Given d=6a,
Show why the central diffraction peak contain 11
interference fringes,(p708)
Solution,The first minimum in the diffraction pattern
occurs where
a
s in
Since d=6a 6)(6s i n
aad
Interference peaks (bright fringes) occurs for
Where m can be 1,2,3,or any integer,
6s in?d
31-11 polarization (p720-725)
自然光
Polarized light（偏振光、线偏振光）
符号表示
3,partially polarized light 部分偏振光符号表示
Malus’ law:(cosine-squared rule)
212 c o sII20 c o s2
1 I?
马吕斯定律
1.当 或0 时,12 II?
2.当 2
3
2
或时,02?I
例 1 有两个偏振片,一个用作起偏器,一个用作检偏器,当它们偏振化方向间的夹角为 时,一束单色自然光穿过它们,出射光强为 ; 当它们偏振化方向间的夹角为 时,另一束单色自然光穿过它们,出射光强为,且,求两束单色自然光的强度之比,
30
60
1I
2I 21 II?
10I 20I解 设两束单色自然光的强度分别为 和,
经过起偏器后光强分别为 和,
2
20I
2
10I
经过检偏器后?30co s
2
210
1
II60c o s
2
220
2
II?
3
1
60co s
30co s
2
2
20
10
21
I
III
212 co sII?
3p1p0I
0I 1I 3p2
p1p
2I 3I
2p?
3p
1p
01 2
1 II20 c o s
2
I?
在两块正交偏振片 之间插入另一块偏振片,光强为 的自然光垂直入射于偏振片,
讨论转动 透过 的光强 与转角的关系,I
31 p,p
2p
2p 3p
0I 1p
)2π(c o s 223 II?202 co s2II?
220223 s i nc o s21s i n III
2s i n81 203 II?
若 在 间变化,如何变化？ π2~0
3I
0,2π3,π,2π,0 3 I? 8,4π7,4π5,4π3,4π 03 II
2p?
3p
1p
Polarization by reflection (反射光的偏振 P723-724)
当自然光入射到媒质表面时，
光的传播方向会发生变化，
形成折射光和反射光，同时折射光和反射光的偏振状态也发生变化。反射光和折射光都是部分偏振光。
i
1n
2n
The reflected light is
polarized light（反射光是线偏振光，振动方向垂直于入射面）,
and the reflected and
refracted rays are
perpendicular to each
other（反射光和折射光互相垂直）,
When light incident at the Brewster angleB=i0
1
21t a n
n
n
B
Brewster law（布儒斯特定律）,p724
0i
1n
2n
0i
0r
当 入射角满足
1
2
0tg n
ni?
时，反射光为 完全偏振光，且振动面垂直入射面，折射光为部分偏振光。
这个特定的入射角称为起偏振角或者布儒斯特角。
当光线以 布儒斯特角入射时,反射线与折射线垂直。
即 200
ri
31-2,31-4,31-5,31-6,31-8,31-9,31-10,31-12不做要求
Chapter 1,Introduction,
Measurement,Estimating
Page1-16
了解
Chapter 2&3,Kinematics
2-1—2-7,一维运动，了解概念。（ p16-34）
2-8,Use of Calculus,Variable Acceleration
会用积分求速度和加速度。 （ p34-35）
3-1,3-2,3-3,3-4 and 3-5 （ p44-51）； 7-2 （ p151-
152） ; 11-1 （ p275-277） Vectors：知道如何计算。
3-6,Vector Kinematics （ p52-53）
3-7 & 3-8 projectile Motion,了解，（ p54-p61）
3-9 Uniform Circular Motion,5-4 Nonuniform
Circular Motion & Dynamics of Uniform
Circular Motion p62-63; p119-120; p113-115
3-10 Relative Velocity,不要求 (p64-67)
Position Vector(位矢）
kzjyixr
ktzjtyitxtrr )()()()(
For a Cartesian system:
Motion Function ( 运动方程 )：
P (x,y,z,t )
ik
j
x
y
z
(t0)
r(t)
(t?)
z
x
y
Instantaneous velocity (瞬时速度 ),
ktzjtyitx
dt
d
dt
rd
t
r
v
t
)()()(
lim
0
)( tvkvjviv zyx
Instantaneous Acceleration:
2
2
0 d
d
d
d
d
d
d
dl i m
t
r
t
r
tt
v
t
va
t
In a natural coordinate （ 自然坐标系）：
反映速度大小变化的快慢
--Tangential acceleration（切向加速度）
--Radial acceleration（法向加速度）
naτaa nt
2
2
dt
sd
dt
dva
t
R
va
n
2
反映速度方向的变化
Circular Motion
1,Uniform Circular Motion (p62):
2,Nonuniform Circular Motion (p119):
tn anaa
where
dt
dva
t? r
va
n
2
or 22
nt aaaa
)/a r c t a n (
nt aa
1,第一类问题
a,v已知运动学方程，求
2,第二类问题已知加速度和初始条件，求 r, v
运动学的两类问题：
1,第一类问题 a,v已知运动学方程，求
(1) t =1s 到 t =2s 质点的位移
(3) 轨迹方程
(2) t =2s 时 a,v
jir 21 jir 242
jijirrr 321)2(2)(412
已知一质点运动方程 jtitr )2( 2 2
求例解 (1) 由 运动方程得
jtitr
22ddv
ji 4 22v
(2)
当 t =2s 时 ja 2 2
jtt ra?
2dddd 2
2
v
222 tytx 4/2 2xy(3) 轨迹方程为
2,第二类问题 已知加速度和初始条件，求 r, v
ttv d2d?
A particle moves along the direction of x axis,a=2t,At t =0,
x0=0,v0=0,What are its velocity and position at t =2s?
Solution,Acceleration,a=2t,is not constant.
)1(dd;d2d 2
00
ttxvttv tv
tx ttxttx 0 202 dd;dd )2(3
1; 3tx?
Eliminate t from (1) and (2),we can also get:
3/2)3(;)( xvxvv
m67.238;m / s4 xv
So,
txvtva d)62(d;dd 2
vvxa dd?
t
v
t
va
d
d
d
d
x
vv
d
d?
xd
xd
Solution:
vx vvxx 00 2 dd)62( 23 21)(2 vxx
32 xxv
―-” no physical meaning
If a(v) is given,we may also find v(x) by
x
v
t
x
t
vva
d
d
d
d
d
d)(;
d
d
x
vv?
Note:
A particle moves along x direction,,
At t =0,x0=0,v0=0,What is its v(x)?
262 xa
)(
dd
va
vvx?
Example:
Chapter 4&5,Dynamics,
Newton’s Laws
Page77-132
Newton’s Second Law
amF n e t
t
PF
d
d
t vmd )(d
vtm?dd tvm dd
vmP
Where is momentum (动量 ) of body.
So in classical mechanics,it can be expressed as:
zzn e t
yyn e t
xxn e t
maF
maF
maF
,
,
,
In rectangular coordinate:
Applying Newton’s Laws
1,Pick a body based on the feature of problem (研究对象 );
2,Draw a free-body diagram (画出隔离体的示力图 p89);
3,Choose coordinate axes (建立较方便的坐标系 );
4,Resolve vectors along chosen axes (沿轴分解矢量 );
5,Apply Newton’s laws of motion (列方程,统一单位,求解 );
6,Decide whether or not the solution makes sense.
Problem Solving Guide:(p94 –95,problem solving)
重点放在极限问题、积分上。 题目主要看 CHAPTER 5中的例题。
A ring screen of radius R is horizontally fastened on a
frictionless surface,A slippy block of mass m moves along
its inner-wall,The frictional constant between them is?.
Find,(a) when the speed of block is v,Frictional force f=?
and at =? (b) How long will it take as its(滑块 )speed
decreases v? v/3?
N?
v?
tF
R
O
mv?
(a) Using natural coordinate frame
Solution:
Chapter 6 Gravitation
Page132-146
不要求
Chapter 7&8,Work and Energy
p147-160
7-1 Work Done by a constant Force P147-P151
if
4321 FFFFF
321
321
WWW
rdFrdFrdFrdFW
The work done by several forces = algebraic
sum of the work done by the individual force.
Work done by several forces:
a n d, kFjFiFF zyx, kdzjdyidxrd
fz
iz z
fy
iy y
fx
ix x
B
A zyx
B
A
dzFdyFdxF
dzFdyFdxFrdFW )(
Special expression in scalar product:
7-3 work done by a general variable force P152-155
BABA rFrFW dc o sd
EXAMPLE 7-5； 7-6
质量为 10kg 的质点，在外力作用下做平面曲线运动，该质点的速度为 jit 164 2v
解 24dd ttxxv ttx d4d 2?
16dd tyyv ty 16?
ttmF xx 80dd v 0dd?tmF yy v
J 1 20 0d3 2021 3 tt
在质点从 y = 16m 到 y = 32m 的过程中，外力做的功。求例
,开始时质点位于坐标原点。
时16?y 1?t
时32?y 2?t
yFxFW yx dd
F?
L
缓慢拉质量为 m 的小球，
解 0s in θTF
0c o s mgθT
θmgF ta n?
sθθmg dc o st a n
x
y
θ
G?
T?
0c o s1 θm g L
例
=?0 时，求已知用力 F? 保持方向不变F?
作的功。F?
00 c o st a nθ θθθL m g d
sθFrFW dc o sd
质点系 动能定理
0kk
inex EEWW
7-4 Work-Kinetic Energy Theorem (P156-p160)
7-5 Kinetic Energy at Very High Speed
不要求
8-3 & 8-4 Mechanical Energy and its Conservation
P174-181
cUKE m e c
8-1 & 8-2 basic concepts (P168-p174)
m e ciiffince x t EUKUKWW )()(
m e cn o n c o n EW
or
8-6 Work-Energy Principle(P182-p184)
8-5 The law of Conservation of Energy 不要求
8-7,8-8 & 8-9 不要求
Chapter 9,Linear Momentum
and Collisions
p200-233
linear momentum
vmP?
dt
n e t
Conservation of linear momentum
0?n e tF? P
= constant vector
常量常量常量
ziziz
yiyiy
xixix
PmF
PmF
PmF
v
v
v
0
0
0
9-1& 9-2 Momentum and Its Relation of Force
and Conservation of Momentum p200-p205
Impulse-Momentum Theorem:
)(e x t vmdtddt PdaMF?
,pddtF ppppddtF ifp
p
t
t
f
i
f
i
fitt dtFJ
—–The impulse(冲量 )of force
vmP —– Momentum (动量 )
Impulse-Momentum Theoremif PPPJ
9-3 Impulse and Collisions P205-208
ixfxxx PPPJ
iyfyyy PPPJ
izfzzz PPPJ
9-4,9-5,9-6 p208-212了解
9-7 9-8,9-9 & 9-10 p213-223 不要求
Chapter 10 & 11,Rotation
Motion About a Fixed Axis;
General Rotation
10-1,Angular Quantities
10-2 ; 10-4 &10-5 Torque; (P235-238; P241-243; )
掌握基本概念,
10-3,10-11 & 10-12,11-6 ; 11-8; 11-9 ; 11-10 不要求 ;
10-6,10-7,10-8,10-9,10-10; 11-3; 11-4; 11-5; 11-6;
11-7 P241-256 ; p277- 285
重点重点,复习课件
11-2; 11-3; 11-4 11-5 ; 11-7 P276-282; P284-285
10-6&10-8N-II Law for Rotation (定轴转动定律 ) and Moments of Inertia
(P243-244)
In e t? N-II Law for Rotation(p244)
2ii rmI
dmrI 2
dl
dS
dV
dm
where,. dmrI 2
Page 245,example:10-6
)2,(
12
1 2
0
22
2
2
LL
Lc dIIormLdxxL
mI
2
0
2
3
1 mLdxx
L
mI L
o
Solution(1),dxdm
c
x
x
dmxdI 2?dxx?2 dx
L
mx 2
Try to find out I of a thin rod of mass m,length L to
(1) C and (2) O axes.
O x dm xLSolution(2)：
10-7 Solving Problems in Rotational Dynamics
(P245-248)
As always,draw a clear and complete diagram.
1,Draw a free-body diagram.
2,Identify the axis of rotation and calculate the
torques about it.
3,Apply Newton’s second law for rotation and for
translation.
4,Solve the resulting equations.
例 2：质量为 m1和 m2两个物体，跨在定滑轮上 m2 放在光滑的桌面上，滑轮半径为 R，
质量为 M，求,m1 下落的加速度，和绳子的张力 T1,T2。
1m
2m RM,
T1
T2
解：受力分析
1m以 为研究对象
amTgm 111（ 1）
1m
gm1
1T
2m以 为研究对象
RM,
1T
2T
2T2m
2m以 为研究对象
amT 22? （ 2）
M以 为研究对象
JRTT )( 21 （ 3）
补充方程,?Ra? （ 4）
联立方程（ 1） ---（ 4）求解得
2/21
1
Mmm
gma
2/
)2/(
21
21
1 Mmm
gMmmT
2/21
21
2 Mmm
gmmT
讨论：当 M=0时
21
21
21 mm
gmmTT
Fig,Shows a uniform disk,with mass M=2.5kg and radius
R=20cm,mounted（安放） on a fixed horizontal axle,A block
with mass m=1.2kg hangs from a mass-less cord that is wrapped
（缠绕） around the rim of the disk,Find the acceleration of the
falling block,the angular acceleration of the disk,and the tension
in the cord,The cord does not slip and there is no friction at the
axle,(I=MR2/2)
Axis O
R
m
M
α
T
·R
mg
T = - T′
maM
For M,? = TR = I? (1)
For m,mg – T = ma (2)
a = R? (3)
From above equations,we find the answer:
a=4.8 m/s2
=24 rad/s2
T=6.0 N
Solution:
mg
T = - T′
ma
α
T
·RM
In fig.,two blocks with mass m1 and m2 (m1>m2),two
pulleys (with radius of RA and RB,mass of mA and mB) are
mounted in horizontal frictionless bearings,assuming that
the pulley is a disk,the cord is not slipping,Please find the
blocks’ acceleration.
m1 m2
mA mBR
A
RB
Solution:
Example:
Draw free-diagram:
m2
m2g
T2
m1
m1g
T1
T1-m1g=m1a (1)
m2g-T2=m2a (2)
mB
RB
T
T2
mA R
A
T
T1
(T-T1)RA=IA?A (3)
(T2-T)RB=IB?B (4)
Where,a= RA?A = RB?B
m1 m2
mA mBR
A
RB
Solve equations (1)-(5),yields:
BA mmmm
gmma
2
1
2
1
)(
21
12
2
2
1
BBB RmI?
2
2
1
AAA RmI?
A body rotating about an fixed axis is said to
have rotational kinetic energy,By analogy
with translational kinetic energy,we have
221 ii vmK 222 )(2121 iiii rmrm
Using equation of I, 2
ii rmI
.21 2 IK
1,The Kinetic Energy of Rotation:
10-10 Rotational Kinetic Energy (P254-256)
then,
2,Work-Kinetic Energy Theorem for Fixed-axis Rotation
21 dMW
力矩的功
From the kinetic theorem of a single particle,
,KmvmvW if 22 2121
.2121 22 KIIW if
KIIW if 22 2121 f
i
dWwhere
So,we get the work-kinetic energy theorem for
rigid body rotated about a fixed axis:
Solution:
As the Fig.,For the spring,k=2.0?103 N/m;
For the wheel,I=0.5 kg?m2,R=30cm;
m
R
k
What is the speed of the body (m=60kg),
when it falls 40cm,Initially,it is at rest
and the spring is not stretched.
maTmg
IkxRTR
Ra?
2/ RIm
kxmga
EXAMPLE:
x
RIm
kxmgvv hv d
/
d 0 20
m / s5.1
/
2
2
2
RIm
khmg hv
or,
2
2
1 kh m g h?221?I?221 mv?
,dd tva?,dd txv? vxav dd
2/ RIm
kxmga
Torque about a point:
Fr
vmP
The linear momentum of a particle:
Angular momentum of a Particle
The Angular momentum relative to a fixed point：
)( vrmPrl
11-2 The Torque Vector p276
11-3 Angular momentum of a Particle(p277
Solution:
A particle moves in the xOy-plane,Write down
its angular momentum in terms of its
coordinates and its components of velocity,
vrmprl
)(or xyz yvxvml
, jyixr, jvivv yx
)( kyvkxvm xy
EXAMPLE:
11-4 Relation between Angular Momentum
and Torque:(p278-280)
.dtldn e t
The angular Momentum of a system of Particles
.
1
,
1
n
i
in et
n
i
i
dt
ld
dt
Ld
,0If CLn et
11-7 The Conservation Law of Angular Momentum (p284)
If net external torque acting on particle’s system is zero,
the total angular momentum of the system remains
conserved,合外力矩为零，质点系总角动量守恒 !
v1 r
1 r2
O
v2
F?
A cord makes a ball,mass m,move in a circle (r1
v1) on the surface of frictionless table,(1) When
one pulls the cord downward slowly and makes
the radius equal to r2,v2=? (2) The work done by
the during r1?r2.F?
2211 rmvrmv?
)(
2
1
12 r
rvv?
12 vv?
F is central force,and v
r,L=mvr =constant
Solution,(1)
EXAMPLE:
11-5 Angular Momentum and Torque for a Rigid
Body (Page:280-282)
1,Angular Momentum & Torque of Rigid Body
(P280)
刚体定轴转动的角动量
i
iiii
i
i rmrmL )(
2v IL?
Summing over all the particles to obtain:
dtLdn e t
Using the angular momentum principle of
particle’s system,
For a rigid body,the component along the
rotation axis is:
dtdLa x i s
I
dt
dII
dt
d )(
刚体所受的外力矩等于刚体角动量的变化率。
2 刚体定轴转动的角动量定理
10-9 Conservation of Angular Momentum (p251-
254)
c o n st a n tL fi LL?
or,ffii II
If no net external torque acts on a rigid body,
IdtdLa x i s
o
1
o
2
例 3,人 与 转 盘 的 转 动 惯 量
J0=60kg·m2,伸臂时臂长为 1m，
收臂时臂长为 0.2m。 人站在摩擦可不计的自由转动的圆盘中心上,每只手抓有质量 m=5kg的哑铃 。 伸臂时转动角速度?1 = 3 s-1,
求收臂时的角速度?2,
解：整个过程合外力矩为 0，
角动量守恒，
2211 JJ?
2
101 2 mlJJ
215260
2
202 2 mlJJ
22.05260
2mkg70 2mkg4.60
2211 JJ?
2
11
2 J
J
由转动惯量的减小，角速度增加。
在此过程中机械能不守恒，因为人收臂时做功。
4.60
703
1-s5.3?
10-11,10-12,11-6不要求
Chapter 32,Special Theory of
Relativity
P731-762
32-1 & 32-2 P731-P736 了解
32-3 & 32-4 P738 基本概念
31-5 Time Dilation P739-743
2
0
)/(1 cv
tt
固有 时间,同一 地点发生的 两 事件的时间间隔,
When two events occur at the same location in an
inertial reference frame,the time interval between
them,measured in that frame,is called the proper
time interval or the proper time.
Measurements of the same time interval from
any other inertial reference frame are always
greater,（在其他任何惯性系中观测者看来，这两个事件为异地事件，其之间的时间间隔?t 总是比
t0 要大。）
0' ttt 固有时间
02
0
2
0
1)/(1
tt
cv
tt
1
1
1
2
2
c
v
例 3 设想有一光子火箭以 速率相对地球作直线运动,若火箭上宇航员的计时器记录他观测星云用去 10 min,则地球上的观察者测得此事用去多少时间？
c95.0?v
m i n01.32m i n
95.01
10
1
'
22
tt
运动的钟似乎走慢了,
解,设火箭为 系、地球为 S 系'S
m i n10' t
例介子是一种不稳定的粒子，从它产生到它衰变为介子经历的时间即为它的寿命，已测得静止 介子的平均寿命?0 = 2? 10?8s,某加速器产生的 介子以速率 u = 0.98 c 相对实验室运动。
求介子衰变前在实验室中通过的平均距离。
解 对实验室中的观察者来说，运动的介子的寿命?为
s7?
1000519801 1021 2
8
2
0,
.β
ττ
因此,介子衰变前在实验室中通过的距离 d' 为
m5.2910005.1 98.0 7cu τd'
0?ud?
32-6 Length Contraction P743-744
The length contraction,
02
0 1
LLL
Where L0 is proper length（固有长度,measured
in the rest frame of the object）
The length L0 of an object measured in the rest frame
of the object is its proper length or rest length。
Measurements of the length from any reference frame
that is in relative motion parallel to that length are
always less than the proper length。
例 1 设想有一光子火箭,相对于地球以速率 飞行，若以火箭为参考系测得火箭长度为 15 m,问以地球为参考系，此火箭有多长？
c95.0?v
s'
s
火箭参照系地面参照系解,固有长度 'm150 ll
21' ll m68.4m95.0115 2l
m150?l
v?
x
'x
y 'y
o 'o
32-7 Four-Dimensional Space-Time
32-8 Lorentz Transformations P746-748
了解基本思想
32-9,32-11 Relativistic Momentum,Mass,Energy
Ｐ７４９－７５４
2
2
0
1
c
v
m
m
相对论质量
2
0
1
mm
Where,m0 is rest mass （ 静止质量 ）,m is
relativistic mass (相对论质量,总质量 total mass)
相对论动量：
2
2
0
1
c
v
vm
vmP
相对论动力学基本方程：
2
2
0
1
c
v
vm
dt
d
dt
Pd
F
relativistic kinetic energy（相对论动能）,
202 cmmcE k
relativistic energy （相对论能量）,
20 cMEE k
relation of momentum and energy（相对论动量和能量的关系）
20222 EcpE
32-10,32-12 不要求
Chapter 12 Oscillations
p297-324
12-2 Simple Harmonic Motion (SHM) (简谐振动
P299-304）
tAx c o s
0
d
d 2
2
2
x
t
x?
Equation of motion for the
simple harmonic oscillator.
简谐运动的动力学方程
—— Oscillation Equation
Feature of Oscillatory Motion:
Dynamics,xaorkxf 2
12-1 Oscillations of a Spring 了解
Basic Quantities of SHM,(P301)
m
k
k
mT?
22
m
k
Tf?2
11
Phase or Phase Angle & Initial
t)c os ( tAx在 中,称为振动的相位。
Phase difference of two SHM with same frequency:
12
= 2k?,(k=0,?1,?2,?3…),two oscillators
have same phase (in phase);
当 =?(2k+1)?,( k= 0,1,2,… )，
两振动步调相反,称 反相 。 opposite (out of) phase
,)co s ()()( txdtddt tdxtv m
)2c o s (s i n txtxtv mm
The velocity of oscillating particle is also SHM and varies
between the limits ± vm=±xm,v(t) is ahead of x(t) for?/2.
The Velocity of SHM (p302)
简谐运动的速度函数式
The Acceleration of SHM,(p302)
dt
tdvta?)(.c o s
2 txta m
The acceleration of the oscillating particle is also
SHM varies between the limits ± am = ±?2?xm.
简谐运动的加速度函数式
)c o s (
)c o s (
2
2
2
2
tA
tA
dt
xd
a
tx? 图
t?v 图
ta? 图
T
A
A?
2?A
2?A?
x
v
a
t
t
t
A
A?
o
o
o T
T
)c os ( tAx
)
2
πc o s ( tA
)πc o s (2 tAa
振动曲线
2
2
02
0?
v xA
0
0t an
x?
v
常数 和 的确定 (p301)A?
000 vv xxt
初始条件
co s0 Ax?
s i n0 Av
对于给定的振动系统，周期由系统本身性质决定，
振幅和初相由初始条件决定,
)s i n ( tAv
)c os ( tAx
12-4,Uniform Circular Motion (旋转矢量法 ),(P306)
co s0
)co s (
A
tAx
2
π
0s i n0Av?
2
π 0s i n 取
0,0,0 vxt
已知 求 。
x
v?
o
)
2
π c o s ( tAx?
A
A?
x
T
2
T
t
o
A body (m=10g) is in SHM along x axis,A=20cm,
T=4s,When t = 0,xo= -10cm,and v < 0 (moves
along negative x direction),Find,(1) x(t=1s)=?; (2)
At what time will the body pass the position of
x=10cm first time (何时物体第一次 通过 ） (3) How long
does the body need to pass x=10cm second time(再经多少时间物体第二次运动到 x=10cm处）?
Example:
24
22
T
mx 173.0)322c o s (2.0
or as the Fig.,mx 1 7 3.0c o s2.0
(2)
sttt o 2
2
(1) When t=1s,
st
3
4
2
3
2
(3)
2A?
0
X
O
6
32
2A
mtx )322c o s (2.0
As the Fig,
3
2
0
Solution:
用匀速圆周运动表示简谐运动的 速度变化 (306)
A?mv
做匀速圆周运动的质点的速率是：
在时刻 t,它在 x轴上的投影是
)
2
π
co s (
)s i n (
tA
tmvv
这就是简谐运动的速度公式。
2
πt
mv?
v?
x
y
0
At
)c os ( tAx
Potential Energy:
2
1)( 2kxtU?
§ 12-3 Energy in Simple Harmonic Motion (P304)
)s i n (
)co s (
tA
tAx
v
kxF
)(c o s
2
1 22 tkA
Kinetic Energy:
)(s i n
2
1
2
1 222 tkAmK v
2
2
1 kAUKE
弹簧振子的总的机械能
Total mechanical energy
弹簧振子的总能量不随时间改变，即作 简谐 运动的系统 机械能守恒。
Total energy of oscillatory system is a constant.
2A?
振幅的动力学意义,振幅不仅给出了简谐运动的运动范围，而且还反映了 振动系统总能量的大小,或者说反映了振动的 强度 。
2
2
1 kAUKE
例 质量为 的物体，以振幅作简谐运动，其最大加速度为,求,kg10.0
m100.1 2
2sm0.4
（ 1） 振动的周期；
（ 2） 通过平衡位置的动能；
（ 3） 总能量；
（ 4） 物体在何处其动能和势能相等？
解 （ 1）
2
m a x?Aa?
A
a m a x 1s20
s314.0π2T
（ 2） J100.2 3
222
m a xm a x,k 2
1
2
1 AmmE v
（ 3）
m a x,kEE?
J100.2 3
（ 4）
pk EE?
时，
J100.1 3pE
由
222
p 2
1
2
1 xmkxE
2
p2 2
m
E
x?
24 m105.0
cm7 0 7.0x
小结：简谐运动的描述和特征
xa 24） 加速度与位移成正比而方向相反
xt x 22
2
d
d2） 简谐运动的动力学描述
)s i n ( tAv
)c o s( tAx3） 简谐运动的运动学描述
mk弹簧振子
kxF1） 物体受线性回复力作用 平衡位置 0?x
§ 12-5 The Simple Pendulum单摆 (P307-
308) 了解在角位移很小的情况下，单摆的振动是简谐运动。
Is undergoing SHM
)c o s (m t
l
g?2?
glT π2?
l
g
mgmgF t s i n由
Period does not depend on the
mass of the pendulum bob
§ 12-5,12-6,12-7,12-8不要求
Chapter 13 Wave Motion
p325-353
13-1,13-2,了解基本概念
§ 13-3 Energy,Power and Intensity (波强） of
Traveling Wave P331-332 了解动能 2
2
1 vdmdE
k?
)/(si n)(21 222 uxtAdV
kinetic energy
弹性势能 Elastic Potential Energy
)(s i nd21dd 222pk uxtVAWW
Both K & U of element dm varies periodically with time,同时达到最大值,又同时达到最小值,体积元的机械能不守恒 。
An important feature of wave motion is the transfer of
energy,
13-4Math’s Expression of Traveling waves (P332-
335):
])(c o s [),( vxtDtxD Mp
v沿 x 轴 负 向波函数
v沿 x轴 正 向 ])(c o s [),(
v
xtDtxD
Mp
Neglecting the subscript p,we have equivalent
expressions for SH wave equation:
])(2s i n [),( oTtxAtxy
)s i n (),( otkxAtxy
])(2s i n [),( otfxAtxy
In general,wave function of an arbitrary
shape of traveling wave has a form,
)s i n (),( tkxDtxD M (See P334)
The Physical Meaning of Wave Equation:
波 函数 具有 时间周期性 （ T )o t
A
y
T
(1) Fixed x,corresponding to the oscillating curve
of medium element at position x,i.e,y(t,xo).
(2) Fixed t,corresponding to the y-x curve
(波形图 ) at to.
波 函数 具有 空间周期性 （? )o
y
x
2x1xA
在下列情况下试求波函数：
)]81(π4c o s [ tAy A
(3) 若 u 沿 x 轴负向，以上两种情况又如何？
例 2
(1) 以 A 为原点；
(2) 以 B 为原点； B A
1x
x
已知 A 点的振动方程为：
(1) 在 x 轴上任取一点 P,该点振动方程为：
)]81(π4c o s [ uxtAy p
)]81(π4c o s [),( uxtAtxy波函数为：
解
u?
P
1x
B A x
(2) B 点振动方程为,)]8
1(π4c o s [)( 1
u
xtAty
B
)]81(π4c o s [),( 1 u xxtAtxy
)]81(π4c o s [),( uxtAtxy(3) 以 A 为原点：
以 B 为原点：
波函数为,)]8
1(π4c o s [),( 1
u
xxtAtxy
13-6 The principle of superposition for waves（波的叠加原理 P337,） P338
结合 13-8
13-5,13-7,13-8,P13-10,13-11不要求
13-9 Standing Waves,Resonance
1 the produce of standing waves 驻波的产生
13-6 The principle of superposition for waves（波的叠加原理 P337,） P338
结合 13-8
*13-8 Interference (干涉 ) of Waves (P339-340)
Interference of Waves (波的干涉 ):
)c o s (21 tDDDD M
The resultant oscillation in point P is also SHM:
Based on the superposition of oscillation,
c o s2 212 22 1 MMMMM DDDDD
2211
2211
c o sc o s
s i ns i nt a n
MM
MM
DD
DD
1212 2 rr
The in expression A is the phase difference
arisen by two waves at point P,It does not vary
with time,so does not amplitude at every points.
1s
2s
P
*1r
2r
(1) when (k=0,1,2,…)?
k
rr 22 12
12
21 MMM DDD
The greatest possible DM —– Fully
constructive interference (相干加强 ),
(2) When (k=0,1,2,…), )12( k
MMM DDD 21
The lest possible A —– Fully
destructive interference (相干减弱 ).
(3) When (k=0,1,2,…), )12(2 kk
Then DM is between —–
intermediate interference (neither fully
constructive nor fully destructive).
2121 MMMM DDandDD
Special case,(i)
21
MM
MMM
DDD
DDD
kk
k
rrL
21
21
21,.,,)2,1,0(
2
)12(
两列相干波源同相位时,其相干加强和减弱情况仅取决于波程差。
(ii) When,
021 MMM DDD
)c o s1(2c o s2 0212 22 1 MMMMMM DDDDDD
2c o s2 0
MM DD
x 2
13-9 Standing Waves,Resonance
1 the produce of standing waves 驻波的产生振幅、频率、传播速度都相同的两列相干波，在同一直线上沿 相反 方向传播时叠加而形成的一种特殊的干涉现象,
If two sinusoidal waves of the same amplitude and
wavelength travel in opposite directions along a
stretched string,their interference with each other
produces a standing wave.
tkxDtxD m?co s]s i n2[),('?
驻波的振幅与位置有关各质点都在作同频率的简谐运动
)(,...2,1,0, nnkx or 2 nx –— nodes
2)2
1( nxor –— antinodes
The interval between pairs of two nodes and
antinodes is half wavelength.
13-5,13-7,13-8,P13-10,13-11不要求
Chapter 14 也不做要求
Chapter 30 The Wave
Nature of Light; interference
p680-701
30-1 Huygens’ Principle and Diffraction
P680-681
了解
30-1 P682-683; 30-2,30-7,30-8 P695 不要求
30-3,30-4,30-5 Interference-Young’s Double-Slit
Experiment p683-690
D
xdsi n12 drrr
1 The path length difference波程差
=
m=0,1,2,…
m=0,1,2,…
m?
)21( m
(Bright fringe明 ); constructive interference
(Dark fringe暗 ); destructive interference
sind
2 interference pattern lines
rdDx
=?dDm?
dDm )21(
m=0,1,2,… 明
m=0,1,2,… 暗
3.条纹间距 (spacing between neigboring points
of constructive or destructive interference
a.相邻明纹间距,b.相邻暗纹间距：
可以看出相邻明纹与相邻暗纹的间距都相同，所以条纹 明暗相间平行等距 。
d
Dx
Intensity in the Double-Slit Interference
Pattern (p687-690)
)π2c o s ( 1111
rtAy p
)π2c os ( 2222
rtAy p
点 P 的两个分振动
co s2 212221 AAAAA
12
12 π2
rr 常量
intensity pattern 双缝干涉光强分布
)c o s (2 122010220210 AAAAA
)c o s (2 122121 IIIII合光强干涉项若
021 III
r π2
12其中
)( πc o s4 20
rII?krI,4 0
2)12(,0 kr
则
30-6 Interference in Thin Films p691-694
光程 =折射率?几何路程 = nr定义：
例,在双缝干涉实验中,用波长为 632.8 nm 的激光照射一双缝,将一折射率为 n=1.4 的透明的介质薄片插入一条光路,发现屏幕上中央明纹移动了
3.5个条纹,求介质薄片的厚度 d 。
o’
D
S1
S2
S d
r1
r2 I
oo
d
解,由于中央明纹移动了 3.5 个条纹，则插入的介质薄片所增加的光程差为 3.5 个波长，对应原屏幕中央 o 点两条光线的光程差也为 3.5? 。
D
S1
S2
a
r1
r2 o
d
在原屏幕中央 o点两光线的光程差为：
5.3?
5.3)1( dn
dnd
1
5.3
nd
m105,5 6- 14.1
108.6 3 25.3 9
2.半波损失 phase shift of? p691-692
interference in thin films（薄膜干涉） P691-692
For near-normal incidence
（垂直入射）
dn 22
未考虑半波损失时考虑半波损失：
321 nnn
321 nnn
321 nnn
321 nnn
光程差不附加 2
光程差附加 2
2
2 2 dnΔ
1n
1n
2n
Example 30-7 p693
22 2
dnΔ
12 nn?
当 时
2)12(
m
m )2,1(m
)2,1,0(m
加强减弱 22 2?dnΔ
Minimum thickness
22
12
nd
1?m
24 n
d
Wedge-shape（劈尖） with a thickness that varies
uniformly (example 30-6,p692)
（ Perpendicularly incident）
2)12(
m
m )2,1(m
)2,1,0(m
加强 bright
band
减弱 dark
band
22 knd
劈尖 (the point of contact)
2
Δ
0?d
为暗纹,
e?
b
'b
由于同一条纹下的空气薄膜厚度相同，当待测平面上出现沟槽时条纹弯曲。
待测平面
The bright and dark
bands will be straight
only if the glass
plates are extremely
flat,If they are not,
the pattern is uneven,
Newton’s rings（ 牛顿环 ） (p691) 不要求
Chapter 31 Diffraction and
Polarization
P702-730
31-1 diffraction by a single slit（单缝衍射） p703-704
half wave zone method （半波带法）
s ina
2)12(
m
m dark fringes
0 Central bright fringe
Bright fringes
a
31-7 diffraction Grating (p713-715)
Grating spacing （光栅常数） d =a+b
The feature of fringes（条纹特征）
1 明条纹细而亮，有较宽的暗区,N越大明纹越细。
2 各级明纹亮度不同，有缺级现象。
I
1 2 4 5-1-2-4-5
缺级缺级-2 2
N=5
光栅方程
(maximum-bright 主极大 )k=0,1,2,3…
k )2,1,0(k s i n)( ba 加强例：分光计作光栅实验，用波长? = 632.8 nm的激光照射光栅常数 d = 1/300 mm的光栅上，问最多能看到几条谱线。
解：在分光计上观察谱线，最大衍射角为 90°,
k s i n)( ba
90si n)(
m a x
bak
d?
f
o
x
P
取 5m a xk
能观察到的谱线为 11条：
。，，，，，，，，，,5 4 3 2 1 0 12345
9
3
108.632300
101
3.5
90si n)(
m a x
bak
单缝衍射 减弱 条件, 'si n ka
光栅衍射 加强 条件, kba si n)(
两式相比 'k
k
a
ba m? )321(?，，?m
缺级条件 miss order
m为整数时，光栅谱线中 m,2m,3m等处缺级。
例：一光栅,b=2a，则缺级的明纹：
'32'' k
a
aak
a
bakk,,,3.2.1'k
故,..9.6.3'3 kk
31-3 diffraction by a double slit (p708-709)
Example 31-4 Diffraction plus interference,Given d=6a,
Show why the central diffraction peak contain 11
interference fringes,(p708)
Solution,The first minimum in the diffraction pattern
occurs where
a
s in
Since d=6a 6)(6s i n
aad
Interference peaks (bright fringes) occurs for
Where m can be 1,2,3,or any integer,
6s in?d
31-11 polarization (p720-725)
自然光
Polarized light（偏振光、线偏振光）
符号表示
3,partially polarized light 部分偏振光符号表示
Malus’ law:(cosine-squared rule)
212 c o sII20 c o s2
1 I?
马吕斯定律
1.当 或0 时,12 II?
2.当 2
3
2
或时,02?I
例 1 有两个偏振片,一个用作起偏器,一个用作检偏器,当它们偏振化方向间的夹角为 时,一束单色自然光穿过它们,出射光强为 ; 当它们偏振化方向间的夹角为 时,另一束单色自然光穿过它们,出射光强为,且,求两束单色自然光的强度之比,
30
60
1I
2I 21 II?
10I 20I解 设两束单色自然光的强度分别为 和,
经过起偏器后光强分别为 和,
2
20I
2
10I
经过检偏器后?30co s
2
210
1
II60c o s
2
220
2
II?
3
1
60co s
30co s
2
2
20
10
21
I
III
212 co sII?
3p1p0I
0I 1I 3p2
p1p
2I 3I
2p?
3p
1p
01 2
1 II20 c o s
2
I?
在两块正交偏振片 之间插入另一块偏振片,光强为 的自然光垂直入射于偏振片,
讨论转动 透过 的光强 与转角的关系,I
31 p,p
2p
2p 3p
0I 1p
)2π(c o s 223 II?202 co s2II?
220223 s i nc o s21s i n III
2s i n81 203 II?
若 在 间变化,如何变化？ π2~0
3I
0,2π3,π,2π,0 3 I? 8,4π7,4π5,4π3,4π 03 II
2p?
3p
1p
Polarization by reflection (反射光的偏振 P723-724)
当自然光入射到媒质表面时，
光的传播方向会发生变化，
形成折射光和反射光，同时折射光和反射光的偏振状态也发生变化。反射光和折射光都是部分偏振光。
i
1n
2n
The reflected light is
polarized light（反射光是线偏振光，振动方向垂直于入射面）,
and the reflected and
refracted rays are
perpendicular to each
other（反射光和折射光互相垂直）,
When light incident at the Brewster angleB=i0
1
21t a n
n
n
B
Brewster law（布儒斯特定律）,p724
0i
1n
2n
0i
0r
当 入射角满足
1
2
0tg n
ni?
时，反射光为 完全偏振光，且振动面垂直入射面，折射光为部分偏振光。
这个特定的入射角称为起偏振角或者布儒斯特角。
当光线以 布儒斯特角入射时,反射线与折射线垂直。
即 200
ri
31-2,31-4,31-5,31-6,31-8,31-9,31-10,31-12不做要求
