ページ1:

實體擺(小角度,即日<5°⇒ simd = tand≈日) (Physical Pendulum)
【力矩与轉動慣量】:
ㄈㄛ=ㄩˋ ㄨ:
=>
U-
d
=CM.
-mg.d(sim
8
【能量規定】:
說明:ZE=const.
mg
=
lö
+8=
mgd
0=0($HM)
Io
x Io = Icn + Md² =w=
mgd
Icx + Md²
ICM+MJ²
∴T=
2T
w
T=2TU
imgd
ZE = ㄜˊ Io.co² + mg d (1-cose) = const.
200
E=Iow
dt
st
+ mg·d smo
Sm00
= 0
dt
"
9.
B
0
mgd
1.
0+ 0=〇同理可得上述結果
註:(1)實体擺又稱「複擺」(Compound pendulum)
(2) 複擺用途之一:測量「剛体繞定軸轉動之轉動慣量(Io)
Io
T = 2 TV
↑
= °I =
mgd.
4 TV2
.. I CM = mgd 412
·md⋅
已知
扭擺 (Torsional Pendulum) (無小角度之限制)
【力矩与轉動慣量】:
施力
說明:此回復力遵守虎克定律
(振動幅度仍在細弦線之線性範圍內)
0
施力
5 回復力
==
Σ7 = 11.α = - k = 10
70
0 = 0
dt z
11
e
(SH.M)
2TV
I
T
"
T=2Tv.

ページ2:

§第二節 S.H.M-彈簧振盪(Spring Oscillicator)
无
mmmmm
e F
x
【牛頓力学】:ΣF=mā ⇒ -Ex=mx+
= 0
dix
dt2
: - RW² cos (wt + 4 ) + for Roos (wt + $):
0
m
"
W
|x=0>即a = -
x
⇒R
= ( — - w²)] cos (wt + 4) = 1
解有:①R=0
無物理意義解
@cos(wt+中)=0
☑
2πV
W =
m
。
= 3 = 0 = 3-
2元
m
m
无
natural
11: x (t) = R cos (wt + $ ) = R [cos (wt) cost - sin (wt) sind ]
= (R cost) cos(wt) - (Rsnø) sin(wt)
同理v(t),a(t)亦可整理成此型式
。
R
【能量觀】:
U
³ C M ↳ 7 Å ¥ + ¿ ³ ¯ = = k [ R cos (wt + 4 ) ]]
E
+ Ex = 2 m -Rwsin (wt + )]
x
E = 1/2 kk² = = mw³R² = const.
Uma
|Ex(max) Unmax
·次微分:
It
IL
+12
E
m
dt
V
H
a.
= v(ma + kt)
18
ㄧˋ
6
a
→
dt.
RE
U(x) =
dx dt
x
the same result

ページ3:

問題二
sr, where A, B are positive
I consider a one-dimensional potential U(r) = Br-9-
(a) Find the equilibrium point to in terms of A and B.
constants. There is a static equilibrium point y = ro for this potential.
(b) a partide with mass m and velocity v - dr = r moves in this potential.
Write down the total energy Exot (Assume the total energy is conserved), and find
the equation of motion for this system.
(c) Near the equilibrium point, the system can be approximated
x = r-ro, rewrite the equation of
as a S.H.M. Let
motion in part (b) as function of x by using the
formula (ro+ x)=^.
nx
= Yo
B
], if x «ro.
(d) Find the period of this particle in terms of ro, A and/or B.
(a) U(r)
-B A
=
-ri
r
==
U(r) =
dr
--9B
ro
A
+
=
9B + Ar²
r²
10
Static equilibrium point:
dr
U (r) = 0 -9B + Ar² = 0
(b) Ex₁ = 12 m r² + U (r) = const.
0/12
Et mŕŕ
+
d
dt
- U (r) = 0 => mr + √r
UJ (Y) = 0
dr
0
Ucr
(c) r = r+x
⇒ (r.+x)
"
2:
+
=
m
(r₁+x)
-2/1-22]
x.ro³ (r. -2x)
..
☑
&:
84
+
(d) w =
mr³
W
BA
1 m²r.³
x = 0
AB
(1+x)
=0
9 B
-r
m
r." (r.-10x) =
m
1
=TV
mr
-2A-
一介
B
r =
A
r
32
-9B
= 0
110
A
9B
-
+
0
my 2
m
-1°
"
[A(r. -2x) +91 (r.-10%)
-8-A-
mp3.x
(r-r.) +
Tro
JU
-21 dr²
(r-ro) +
JU
dr
Taylor expansion
U (r) = U (r.) +
For) = - F (r.)
-
PE
dr
JF
dr (r-r.)+
"
Ar
0
JU
dr
1
F=-x

ページ4:

₤2 134 377 17
【例題-【HW11-17
【HW 11-1)
A 0.650 kg mass oscillates according to the equation X = 0.25 sin (5.50 t) where X is n
meters and t is in seconds. Determine.
(a) the amplitude (b) the frequency (c) the period (d) the total energy
(e) the kinetic and potential energy when x = 15 cm (0.15m).
x(t)=0.25. sin (5.5t)
11
W
(a) The amplitude A = 0.25m
v(t)
21
w
5.5
(b) f =
=
≈0.8751/5
d
2 TV
2 TV
=
dt.
-x(t) 1.375 cos (5.5t)
11
21
27
(c) T =
=
≈1.14 S
Umax
w
5.5
ㄨ
0.15
0.25
(d) E Ek (max)
0.15
11
11/27 -0.65. (1.375) ² ≈ 0.614J
(e) sing = 0.15 = 33 ⇒ ca 50 =
5.
5
v = 1.375
+4
= 1.17/s
x = 15 m = Ek = 12.0.65. (1.1)² ≈ 0.393J
U ≈0.614 -0.393 = 0. 221 J
or mat
20
ma+x
-82x
= P < 0 =
=
mw²x dx
²
2
U=
|= mw².
☑
2
= 0.65.5.5². (0.15)
≈ 0.221 J

ページ5:

第八章 振盪Oscillations
§第一節簡諧運動(Simple Harmonic Motion
-130 + ( Angular frequency)
Equation of motion for S.H.M:
J²x +w w²³. x =
=0 或ā=-
dt 2
(x + w2 x = 0)
即「a∝x 但方向相反」。(證明詳見第二節)
註:(1)上式中,以 x=0為振盪中心(equilibrium position)。
(for SHM)
(for circular motion).
(2)上式中,w為Angular frequency,物理意義上与Angular velocity不同。
(3)上式為- differential equation,需依其運動軌跡(大图)猜測
其解 x(x)可能為 sin/cos出权。
【S.H.M与圓周運動】:依等效性將圓周運動之投影視為S.H.M.。
Vt
。
Rˊ
x(t)
~t
aci
t = 0
wt.
→x
-x-
。
x(x)
可得: x(t) = R cos (wt+4)
Umak = Ut→平衡
d
v(t) =
-RW sin(wt+)
x(t)
It
鉛直面投影作法相同!
amax = ac→端点
d
a(t) = = - Rw³ cos(wt +$) =
(七)
dt
註:(1)0点→ equilibrium position
(2)wt+中⇒ 相位(phase):即以三角函叔描述波之角度。
(對於一個波,其循环中任一時刻之位置)
(3) 4 ⇒ 相角(phase angle)/相位常取(phase constant):初始位置之一種描述。
(4) SH.M之週期(Period) 圓周運動之週期:T= v
2TUR
=
2πV
W

ページ6:

§第三節單擺、实体擺、扭擺
=
單擺(小角度,即日<5° → sme≈tand≈日) (Simple Pendulum)
【牛頓力学】:
LcosB
-U =0
【力矩与轉動慣量】
F = ma
* - * mgsme =
max = m.c.d°0
= mié
dt²
Lsing
恢復力
R&
× Sine = 0 ⇒ - mg 0 = mLo
g
mg
w, α
0 + 6=0>符合SH.M運動方程式
L
w
說明:∑ㄥ=ㄩˇㄡ: -ng.bsme = ml*@
(mgsmB-L)
..
e+
ō
•
【能量】:
說明:ZE = const
ZE = · mo² + mg. L (1 - cose) = const.
2
Ⅲ
(RW)
dt
E = mL²ol
dw
+
It
dt
masing.
20
sm00
= 0
dt
0
◎切線速度ㄩ之分析:
x = L smo⇒x= L cose e
⇒ v²= (L.è)
y = Lose>y=-Lsme.é
W
11: Expansion →此項用於誤差之計算)
0
sino 0-
+31
+]
-051
日陞小⇒0
T = 2元.
9.
L
-w²
| = 0
同理可得上述結果女
10=0. + cos(wt+8)
代入推出(0)
+8←
L
8-7100
18 = 0
同理可得上述結果,

ページ7:

2.09KB/s ♫ O
←
19:13
407220048-1-...
DA14%
QA:
國立中正大學物理系普通物理答案卷
學年度 第
學期
班別: 01班週一、三10:00-12:00
□ 02班週二、四10:00-12:00
□ 03班週二、四14:00-16:00
□04班週二、四16:00-18:00
□ 05班週一、三10:00-12:00
□ 06班週二、四10:00-12:00
系別:
物理一
姓名: 鐘紘
學號: 4-7220048
注意事項:
1. 請遵守考場規則。
2.計算題只寫答案而無計算過程者不予計分。
3. 答案必須填寫於指定的答案空格內,才予計分。
4. 考試四十五分鐘後才可交卷離場。
5. 手機一律關機。
6.其他未盡事宜由試場人員處置。
A
B
選擇題
35
14
49
1
2
3
4
計算題
選擇題作答區(請依題號位置依序作答)
52 51
XD.
總分
100
1
2
3
4
5
6
7
8
9
10
E
C
C
F
B
BG
D
11
12
13
14
15
16
17
18
19
20
F
D
C
EC

ページ8:

【二階O.DE之解】:(僅討論常係)
Linear/Nonlinear:
線性O.DE. 之判斷: ①未知函权間(y)or其導函間(y)。
③未知函权与導函权之次方权皆為一次。
③不包含非線性函(siny, cosy, ery".....
齐次(Homogeneous)、非齐次(Inhomogeneous)
:
(n)
(n-1)
n階O.DE:any + an-, Y
(常係or变係权)
•+ay'+ axy = R(x)
⇒ If R(x)=0,則為 H. O. D. E.
·無外力項(S.H.M. Damped oscillation)
R(X) ≠ 0,則為I.O.D. E. (Forced oscillation)
y
二階線性O.DE.可寫成: +a. •+by=k(x)或i+ay+by = R(x)
dx2
dy
dx
(y") (y')
討論 R(x) =0之解一次解:(Homogenious Solution, n)
若y:(x),Vi=ln為其L.I.解⇒符合封閉性:
即:yh = diyi(x) ⇒ 向量加法与量乘法之封閉性。
2=1
快速判断解是否為L.I. :
即:fyi,vi=1...nare L.I. Wronskian 行列式值≠0
Y₁ Y2
yn
W (y......, Yn)
(n-1)
·積分几次
· 結論:n必含Ci, vi=in等n個積分常权。
討論R(x)≠0之解一特解:(Particular solution, yp)
特解:即非齐次解,几何上代表通过某些特定奌之曲線。
注意:hyp必互為L.I
結論:p不含未定係权。
“通”
[y(x)=yn(x)+f(x)
we are an
adventure Story

ページ9:

Euler 恆等式与其几何意義
Euler恆等式:e
ix = cos x + i sinx
VXER
證明如下:由定義法古展開式
⇒ eix = 1 + ix + (ix) (x)(ix) (x)+
21
1+--
2!
+
+
+
3!
4!
x
i
+
✗
? +
✗
3! 4! 5'
+ix-
=[-*+**]+[*-+++]
21 4!
31 5!
= cos x + i sinx
[幾何意義:複和平面上,由ex(rx∈R)集合形成一半徑=1之圓・(Fig)
I
Heint
=
= cos nx+isin nx (cosx tisinx)
說明:以n=3為例
ix
e = cos x + i sinx
⇒ (cosx+ismx) ³ = cos ³x + 3 i co s ²x smx - 3 cosx smx -
=
=
=
i sin x
(cos 3x-3 cos x sin²x) + i (3 cos ³x smx - sin³x)
= cosx (cosx-35m²x) + i sinx (3 cos²x - sin 2x)
= cosx (40sx−3) + i sinx (3-4 sinx)
(4 cos³x-3006x) + i (3 sinx - 4 sin²x)
cos 3x+i sinex Q. E.V.
@ Z, Z₂ = reid. 2. e202 = (ritz) e² (0₁+02)
I
Z.Z2
2. nicio - [n].
' 22
I
=
re
rzei
=
(0,-02)
e
£2
01+02
£2
82
E₁
£2
k
0.-02
you mustn't be freid to
dream a little
bigger.
darling

ページ10:

第四節 旅制振盪与阻尼振盪
【秋字应用補充】:
虚权与授权:(Complex number)
工(虛軸)
Z₁ = x + i
23
Z₁
+) Ze = xe+iye
Z3 = (x+x2)+i(y+ye)
E2
→ R(实軸)
> 坐標表示法
由此衍伸共軛葭叔 (Complex conjugate)之定義与性質:
對稱窨軸之矣。
Z
定義: Letz=x+y.
即区 = x - iy (Fig.)
性質: 0 Z. Z = | 8 |² © Z.Zz
=
證明如下:①zz=(x+y)(x-y)
=
x²-xy+dxx-3y²
2
-1
=
x² + y ² = [z][2
ZZ2 = (x+iy)(x2+iyz) = (x,x2-yy2)-i(x,y2+xzy.)
Zzz = (x,y)(x2-iye)=(x,xz-y, ye) -i (X, Y2 + X2 Y₁) Q.E.D.
i与各展開式之关係:(Complex function)
五種常見之展開式:
① e* = 1 + x +
+ +
2! 3!
@ln(x+1)=x-号+号-ㄊㄨㄩㄝˊ
⑤sm x = x
-
31
+
2
7!
4
+
④cosx = 1
+
2! 41
6!
ste) E
微分
CAD F
李+x+375x
In general (即泰勒展開):
|X|<
f(x) = f(x)+ fixe).(x-xe) + f(x)(x-x)²+
fe)(x) (x-xok.
=
k=0
k!
2!
+ Rn(x)
Lagrange Remainder
ni
f(x)(x)(x-xx)+1
n
f(n+1)(c)
(n+1)!
-(x-x)
·n+1