Upload
mimi-lui
View
58
Download
1
Embed Size (px)
Citation preview
Concept Test 1
V=0extreme point
V=0extreme point
amplitdue, A
equilibrium positionFnet = 0V = max
• 由 equilibrium 去到 extreme point 需時 02.05T41 ×= , 所以 frequency=2.5 Hz
• max speed 出現於 equilibrium position, ωA= ( )( )f215.0 π= = 2.36 m s-1
Concept Test 2
θO
θA
θ
A2
1
ωAv =
21
A
A21
cos ==θ
θ = 60° u = vsinθ u = vsin 60°
=
23vu
3u2v =
Concept Test 3 • 若果 object 仍與 platform remains in contact 的話,即代表 normal reaction 0≥R 。 • 從自由體圖所知,
R
mg
+ve
[ ]
[ ]
cm
.
gA
AgAgmR
mAmgRAmRmg
maFnet
1
2012
10
0
22
2
2
2
2
=
=≤
≥≥−=
−==−
=
πω
ωωωω
Concept Test 4
θθ6 s
2 sB
x’ 0.3 mx
C
A
θcosA
Period, T = (2 + 6) s = 8 s
From x to x’, T41s2t ==
°↔ 360T
θ2T41 ↔ ∴ 2θ = 90°, θ = 45°
CB = A – Acos 45° = 0.3 m
3.0)cos45-A(1 o =
)45cos1(
3.0A o−= = 1.02 m
Concept Test 6 (1) 在 simple harmonic motion 中,acceleration, a 是會隨著 displacement, x 的大小而改變。
( )xa ∝Q (2) 在 projectile motion 中,物件受到向下的勻加速度(i.e. constant acceleration = g),令其
垂直方向的速度(vertical velocity)改變。
ux=constant
uy resultant velocity
a=g=10 m s-1
(3) 在 uniform circular motion 中,物件的 velocity 方向會隨著時間而改變。而向心加速
度(centripetal acceleration)的數值r
vac
2
= 是維持不變的。
vB
vA
Concept Test 7
2maxmax mv
21KE =
2)r(m21 ω=
)fπ4r(m21 222=
)fπ4025.005.0(21100.3 2223 ×××=× −
Hz2.2f =
Concept Test 8
• 因為 amplitude=1000
5.0,f = 250 Hz,所以 22 XAv −= ωm
( )22
10004.0
10005.02502
−
= π = 0.47 m s-1
Concept Test 9
(1)
O
mg
xo
T = kxo
at equilibrium position, mg = kxo
∴ tension ∝ m (2)
mg
xo
T = k(xo+A)
OA
tension = k(xo + A) tension = kA + kxo
y = mx + c ∴ Tension is directly related to A, but it is not proportional to A.
Concept Test 10
O
X
Y
Z
(1) 因為 point Y 為 equilibrium position, 所以 acceleration 在 point Y 應為零。
(2) 從上圖得知: 物體在 level Y 時彈簧己伸延 xo 長度用於抵銷 mg,所以 strain energy 在 level Y 應
該是 0kx21 2
o ≠
(3) 當物件處於 level Z 時,其 displacement 的數值是最大。(Q xa ∝ ) ∴ displacement = max a = max Fnet = max
Concept Test 11 ∵ a = −ω2x
a1x 2ω−=
y = mx ∴
0 a
x
xokxo
mg
level Y(equilibrium position)
Concept Test 12 (1) v = Aω
∴ v ∝ A (2) Total energy of the system = KEmax of the system
∴ Total energy, 2mv21E =
vmE2 =
∴ Ev ∝
(3) KEmax of the system = PEmax of the system
∴ max2 PEmv
21 =
mPE2
v max=
∴ maxPEv ∝
Concept Test 13
• total energy = 2kA21
( )
( )( )
( ) ( )
( ) ( )
2T
42.005.0
008.0T2
2.005.0008.0
05.0m21004.0
mk(05.0k
21004.0
2
22
22
2
π
π
ω
ω
ω
=
=
=
=
=
== Q
Concept Test 14 同學須知當 mass 在 forced oscillation(強迫振盪)時,該物件會跟從 driving force 的頻率振
動。 Concept Test 15
X
A
0
Period Period
time, t
(1) 在阻尼諧動中週期是維持不變的(i.e. constant)。
(2) 在 damped oscillation 中,該系統的 energy 會隨著時間而減少。
Q 2amplitudeenergy ∝ energy ↓ , amplitude ↓
(3) ∵ 振幅(Amplitude, A)↓, 2kA21E ↓=↓
∴ 機械能(Mechanical energy) ↓.
Concept Test 16 (1) 題目中說明小球經數次反彈後最終靜止於彈簧上端,從而得知此振動並非簡諧運
動(S.H.M.),應為阻尼運動(damped oscillation)。 由於小球靜止在 equilibrium position, 即 FNet = 0
kx = mg x ∝ m.
(2) 正如敘述句(1)以上所解釋,小球經數次反彈後最終靜止於彈簧上。因此彈簧的壓
縮量與高度 h 無關。 (3) 正如敘述句(1)以上所說明,該運動為阻尼運動(damped oscillation),固此小球所捐
失的重力勢能是用於扺銷摩力所作的功。 Concept Test 17
• 1201.0
12mk ===ωQ
• 在 equilibrium position, v = max = 0.5 m s-1, 所以 max acceleration, ωva =
= ( )( )1205.0
= 5.5 m s-2
Concept Test 18
• ( )ωω
=→=
2Av,2
AAif,Av newmaxQ
maxv21=
• 2AAif,kA
21.E.Pmax 2 →=Q
2
new 2Ak
21.E.P
=
4Ak
21 2
=
max.E.P41=
• 因為 Period,Km2T π= ,所以週期不受 amplitude 的大小影響。
Concept Test 19 (1) ∵ LB = LHeavy bob
∴ fB = fdriving (resonance) 當 driving frequency = natural frequency 時,其振幅 amplitude, A = max
(2) 所有紙單擺的振盪頻率皆為相同。 (3) Out of syllabus
Concept Test 20
x
equilibriumposition
restoring force, kx
(1) 因為 restoring force 與 displacement 成正比(i.e. F = k x),所以 max displacement 等於
max restoring force。
(2) ∵ Elastic P.E. = 2kx21
∴ Elastic P.E.↑ , as x↑ (3) 當物件處於 max displacement 時(i.e. extreme point),其所受的速率為零。
Concept Test 21
ωAv* max =Q 1 = (0.5)(ω ) ω = 2 rad s-1
* 當物件距離 equilibrium position 0.3 m 時,速度 22 xAv −= ω
= ( ) ( )22 3.05.02 −
= 0.8 m s-1
Concept Test 22 * 同學可由右圖了解:
tension
mgsinθmgcosθ
mg
(1) (2) (3) 物件只受到兩個力,一是 weight(地心吸力),二是 tension 張力。
同學們切記誤以為有第三個力叫 θsinmg ,其實此力是由 mg 分拆出來而成的。 Concept Test 33 在 equilibrium positions 時,K.E. = max = E = 2kA
21
Let the displacement of the object from the equilibrium position be x 22 kx
21kA
21E
43 −=
222 kx21kA
21kA
21
43 −=
2Ax =
Concept Test 34 A
若變數 y 為位移(displacement)、變數 x 為加速度(acceleration)時, Q a = –ω2x
x = – 21ω
a
應該為直線(straight line) B
aFnet ∝Q 若變數 y 為力(force)、變數 x 為位移(displacement)時,可將變數 y 視作為加速度
(acceleration) Q a = –ω2x
固此亦應為直線(straight line) C
若變數 y 為動能(kinetic energy)、變數 x 為位移(displacement)時,
Q v = 22 xA −ωm
KE =21 mv2
=21 mω2(A2 – x2)
KE = –21 mω2x2 +
21 mω2 + A2
y = –ax2 + bx + c i.e. concave downward
D 若變數 y 為速率(speed)、變數 x 為位移(displacement)時,
v = 22 xA −ωm v2 = ω2A2 – ω2x2
1Ax
)A(v
2
2
2
2=+
ω (ellipse)
Concept Test 36 當方塊離牆壁的位移最大時(i.e. extreme point),在其上輕放一塊膠泥
A 1° Amplitude = constant (∵ Amplitude depends on initial condition)
2° Max. PE =21 kA2 = constant
∵ Max PE = Max KE = constant
∴ 21 mv2 = constant
∴ when mass, m↑, velocity, v↓.
3° T = 2πkm
∴ m↑, T↑
4° amax = vω = v⋅Tπ2
∵ v↓ and T↑ ∴ amax↓. ∴ mass↑ ⇒ v↓, T↑, amax↓.
Concept Test 37
(1) At steady state, driving force = damping force, Energy input = Energy loss The system would oscillate with driving freq.
(2) The amplitude of oscillation would become infinite if there are no damping force and driving freq. = natural freq.
(3) out of syllabus.
Concept Test 38 設每段彈簧的 spring constant 為 k.
k
k
m串聯彈簧(connected in series)
兩條彈簧的等效力常數(effective force constant) = 2k
在 equilibrium position : (2k )(e) = mg 1
kk
xo
2m並聯彈簧(connected in parallel)
兩條彈簧的等效力常數(effective force constant) = 2k 在 equilibrium position: 2kxo = 2mg 2
∴ From 2, xo =k
mg
From 1, k
mg =2e
∴ xo = 2e
Concept Test 39 Force - displacement graph 可視作為 acceleration - displacement graph. a = –ω2x y = mx i.e.
F
x
Concept Test 41
P Q
Q’P’
Q’’
0.2 s
0.1 s
0.3 s
0.2 s
180° - 2θ
θ θ
θO
∵ The object passes points P and Q with the same speed, ∴ ∠P’OP =∠Q’OQ = θ, (∵ v = Aωsinθ)
∠P’OQ’’ = θ + θ + 180° – 2θ = 180° ∴ 180° ↔ 0.3 s
T ↔ 0.6 s Concept Test 42
(1) T = 2πkm , ∴ m↑, T↑.
(2) 由於連接兩方塊 P 及 Q 的兩條彈簧相同(i.e. same tension)與及由靜止釋放(i.e. released from rest),因此兩方塊 P 及 Q 的振幅應該相同。
再者,amplitude(振幅) 受制於初始情況(i.e. initial condition)。 (3) Max KE = Max PE
Max PE = 21 kA 2
∵ same A, ∴ same Max PE and Max KE.
Concept Test 43
xUF∆∆−=Q
x
U
0
當物體相對某固定點移離時,其勢能 U↑
when x > 0, 0xU >∆∆
,F < 0
when x < 0, 0xU <∆∆
,F > 0
因此,物體上的力指向該固定點。 Concept Test 46 同學可參考 Lecture Notes Page 18。 Concept Test 48
P QO XX'
2 s
2 s
3 s
3 s3 s
3 s 3 s
1st
2nd
3rd
Concept Test 49 (1) 簡諧運動的振幅可為任何數值(any magnitude)。 (2) 加速度總是指向一個固定點稱為 equilibrium position。 (3) 其振盪頻率與振幅無關(i.e. isochronous oscillation)。
Concept Test 51
k
k
k k
( )12
k2m2
2km2
TT
parallel
series =
=π
π
Concept Test 52
450 mm 475 mm 500 mm 525 mm 550 mm
A
A/2
60o
30o
Period, T = 1.5 s
°↔ 360s5.1
o30t ↔ ∴ t = s81
Concept Test 53
T = 2πkm
, A 2A
A
由以上方程得知週期與振幅無關(i.e. period is independent of amplitude)。 B
最大動量(maximum momentum) = MVmax = M( ωA )
A 2A M( ωA ) M( ωA2 )
C D
2max kA
21PE =
2max APE ∝
A 2A ↑maxPE 4 maxPE