C:/KENAR/0p1.dvi

Similar documents
1.1 ft t 2 ft = t 2 ft+ t = t+ t d t 2 t + t 2 t 2 = lim t 0 t = lim t 0 = lim t 0 t 2 + 2t t + t 2 t 2 t + t 2 t 2t t + t 2 t 2t + t = lim t 0

ma22-9 u ( v w) = u v w sin θê = v w sin θ u cos φ = = 2.3 ( a b) ( c d) = ( a c)( b d) ( a d)( b c) ( a b) ( c d) = (a 2 b 3 a 3 b 2 )(c 2 d 3 c 3 d

Gmech08.dvi

d (K + U) = v [ma F(r)] = (2.4.4) t = t r(t ) = r t 1 r(t 1 ) = r 1 U(r 1 ) U(r ) = t1 t du t1 = t F(r(t)) dr(t) r1 = F dr (2.4.5) r F 2 F ( F) r A r

t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ

Gmech08.dvi

< 1 > (1) f 0 (a) =6a ; g 0 (a) =6a 2 (2) y = f(x) x = 1 f( 1) = 3 ( 1) 2 =3 ; f 0 ( 1) = 6 ( 1) = 6 ; ( 1; 3) 6 x =1 f(1) = 3 ; f 0 (1) = 6 ; (1; 3)

4 4 4 a b c d a b A c d A a da ad bce O E O n A n O ad bc a d n A n O 5 {a n } S n a k n a n + k S n a a n+ S n n S n n log x x {xy } x, y x + y 7 fx

1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0

I 1

sec13.dvi

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)

24.15章.微分方程式

1 [ 1] (1) MKS? (2) MKS? [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0 10 ( 1 velocity [/s] 8 4 O

1 180m g 10m/s v 0 (t=0) z max t max t z = z max 1 2 g(t t max) 2 (6) r = (x, y, z) e x, e y, e z r = xe x + ye y + ze z. (7) v =

i

1 1.1 [ 1] velocity [/s] 8 4 (1) MKS? (2) MKS? 1.2 [ 2] (1) (42.195k) k 2 (2) (3) k/hr [ 3] t = 0

TCSE4~5

Note.tex 2008/09/19( )

() n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (5) (6 ) n C + nc + 3 nc n nc n (7 ) n C + nc + 3 nc n nc n (

応力とひずみ.ppt

pdf

f(x) = x (1) f (1) (2) f (2) f(x) x = a y y = f(x) f (a) y = f(x) A(a, f(a)) f(a + h) f(x) = A f(a) A x (3, 3) O a a + h x 1 f(x) x = a

Acrobat Distiller, Job 128


B. 41 II: 2 ;; 4 B [ ] S 1 S 2 S 1 S O S 1 S P 2 3 P P : 2.13:

m d2 x = kx αẋ α > 0 (3.5 dt2 ( de dt = d dt ( 1 2 mẋ kx2 = mẍẋ + kxẋ = (mẍ + kxẋ = αẋẋ = αẋ 2 < 0 (3.6 Joule Joule 1843 Joule ( A B (> A ( 3-2


II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R

A大扉・騒音振動.qxd

DE-resume

x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x

M3 x y f(x, y) (= x) (= y) x + y f(x, y) = x + y + *. f(x, y) π y f(x, y) x f(x + x, y) f(x, y) lim x x () f(x,y) x 3 -

2014 S hara/lectures/lectures-j.html r 1 S phone: ,

1 1 u m (t) u m () exp [ (cπm + (πm κ)t (5). u m (), U(x, ) f(x) m,, (4) U(x, t) Re u k () u m () [ u k () exp(πkx), u k () exp(πkx). f(x) exp[ πmxdx

S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt

DVIOUT

高等学校学習指導要領

高等学校学習指導要領

, x R, f (x),, df dx : R R,, f : R R, f(x) ( ).,, f (a) d f dx (a), f (a) d3 f dx 3 (a),, f (n) (a) dn f dx n (a), f d f dx, f d3 f dx 3,, f (n) dn f

入試の軌跡

W u = u(x, t) u tt = a 2 u xx, a > 0 (1) D := {(x, t) : 0 x l, t 0} u (0, t) = 0, u (l, t) = 0, t 0 (2)

6. Euler x

S I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d

熊本県数学問題正解

r III... IV.. grad, div, rot. grad, div, rot 3., B grad, div, rot I, II ɛ-δ web page (

I II III 28 29

生活設計レジメ

44 4 I (1) ( ) (10 15 ) ( 17 ) ( 3 1 ) (2)


I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT

Chap11.dvi

x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ dt iωζ = ẍ + ω2 x (2.1) ζ ζ = Aωe iωt = Aω cos ωt + iaω sin

I ( ) 2019

K E N Z OU

i 18 2H 2 + O 2 2H 2 + ( ) 3K

m dv = mg + kv2 dt m dv dt = mg k v v m dv dt = mg + kv2 α = mg k v = α 1 e rt 1 + e rt m dv dt = mg + kv2 dv mg + kv 2 = dt m dv α 2 + v 2 = k m dt d

18 ( ) I II III A B C(100 ) 1, 2, 3, 5 I II A B (100 ) 1, 2, 3 I II A B (80 ) 6 8 I II III A B C(80 ) 1 n (1 + x) n (1) n C 1 + n C

II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re

2 Chapter 4 (f4a). 2. (f4cone) ( θ) () g M. 2. (f4b) T M L P a θ (f4eki) ρ H A a g. v ( ) 2. H(t) ( )

Korteweg-de Vries

[ ] x f(x) F = f(x) F(x) f(x) f(x) f(x)dx A p.2/29

dynamics-solution2.dvi

2011de.dvi

(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0

A

6 6.1 B A: Γ d Q S(B) S(A) = S (6.1) T (e) Γ (6.2) : Γ B A R (reversible) 6-1

1

(1) (2) (3) (4) HB B ( ) (5) (6) (7) 40 (8) (9) (10)

50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq

5.. z = f(x, y) y y = b f x x g(x) f(x, b) g x ( ) A = lim h 0 g(a + h) g(a) h g(x) a A = g (a) = f x (a, b)

<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>

(1) D = [0, 1] [1, 2], (2x y)dxdy = D = = (2) D = [1, 2] [2, 3], (x 2 y + y 2 )dxdy = D = = (3) D = [0, 1] [ 1, 2], 1 {

() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y

1.2 y + P (x)y + Q(x)y = 0 (1) y 1 (x), y 2 (x) y 1 (x), y 2 (x) (1) y(x) c 1, c 2 y(x) = c 1 y 1 (x) + c 2 y 2 (x) 3 y 1 (x) y 1 (x) e R P (x)dx y 2

( ) 2.1. C. (1) x 4 dx = 1 5 x5 + C 1 (2) x dx = x 2 dx = x 1 + C = 1 2 x + C xdx (3) = x dx = 3 x C (4) (x + 1) 3 dx = (x 3 + 3x 2 + 3x +

i

08-Note2-web

5. [1 ] 1 [], u(x, t) t c u(x, t) x (5.3) ξ x + ct, η x ct (5.4),u(x, t) ξ, η u(ξ, η), ξ t,, ( u(ξ,η) ξ η u(x, t) t ) u(x, t) { ( u(ξ, η) c t ξ ξ { (

振動と波動


mugensho.dvi

1 nakayama/print/ Def (Definition ) Thm (Theorem ) Prop (Proposition ) Lem (Lemma ) Cor (Corollary ) 1. (1) A, B (2) ABC

微分積分 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 初版 1 刷発行時のものです.

x ( ) x dx = ax


meiji_resume_1.PDF

untitled

II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka )

(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y

, 3, 6 = 3, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 43, 4, 3, 9, 6, 6,, 0 p, p, p 3,..., p n N = p p p 3 p n + N p n N p p p, p 3,..., p n p, p,..., p n N, 3,,,,

2.2 h h l L h L = l cot h (1) (1) L l L l l = L tan h (2) (2) L l 2 l 3 h 2.3 a h a h (a, h)

5.. z = f(x, y) y y = b f x x g(x) f(x, b) g x ( ) A = lim h g(a + h) g(a) h g(x) a A = g (a) = f x (a, b)

( ) a, b c a 2 + b 2 = c : 2 2 = p q, p, q 2q 2 = p 2. p 2 p q 2 p, q (QED)

( )

[1.1] r 1 =10e j(ωt+π/4), r 2 =5e j(ωt+π/3), r 3 =3e j(ωt+π/6) ~r = ~r 1 + ~r 2 + ~r 3 = re j(ωt+φ) =(10e π 4 j +5e π 3 j +3e π 6 j )e jωt

) ] [ h m x + y + + V x) φ = Eφ 1) z E = i h t 13) x << 1) N n n= = N N + 1) 14) N n n= = N N + 1)N + 1) 6 15) N n 3 n= = 1 4 N N + 1) 16) N n 4

A 99% MS-Free Presentation

<4D F736F F D B B83578B6594BB2D834A836F815B82D082C88C60202E646F63>


Transcription:

2{3. 53 2{3 [ ] 4 2 1 2 10,15 m 10,10 m 2 2

54 2 III 1{I U 2.4 U r (2.16 F U F =, du dt du dr > 0 du dr < 0 O r 0 r 2.4: 1 m =1:00 10 kg 1:20 10 kgf 8:0 kgf g =9:8 m=s 2 (a) x N mg 2.5: N

2{3. 55 (b) x 1 kgf 1 g = 9.80665 m=s 2 1kg 1 kgf 1 kgf = 1 kg 9:80665 m=s 2 =9:80665 N 2 3 N kgf 2 3 (i) 2.5 mg N N 1 kgw kg. 2 kg

56 2 (ii a ma = N, mg (iii) a a = N m, g (a) ; N =1:20 10 kgf = 1:20 10 kg g; m =1:00 10 kg N ( )+( ) 1:20 10 kg g a =, g 1:00 10 kg =0:20g =0:20 9:8 m=s 2 2:0m=s 2 (b) ; N =8:0 kgf=8:0 kg g; m =1:00 10 kg a = 8:0 kg g 1:00 10 kg, g =,0:20g,2:0m=s 2 a <0 a =0 N = mg a =,g N =0 0 kgf 1 a C M m N C N C g (1)(a) (b) (2)(a) (b) (3) N ; N C M;m;g;a 3 3 1 (1)(a) N (b) mg (2)(a) N C (b) N ;Mg (3) N = m(g+a); N C =(m+m)(g+a)

2{3. 57 a b c C D 2.6: 2 2 2.4 0 2.6 a (a) (b) b c C 2.7: a b C c D 1 2 2.7 (a) 2 (b) (a) 2 (b)

58 2 0 4 1 N R 05 R< 0 N (2.29) 0 N N 2 2 R R = N (2.30) 2 6 I{2{1 3 < 0 (2.31) 4 5 F F Reibung 6

2{3. 59 2 2.8 m N = c R g (1) =0 mg sin (2) < c mg cos mg (3) 2.8: (4) > c 4 mg N R 2.8 x y x +mg sin y,mg cos (1),(2) 0 ( x :0=mg sin, R (2:32) y :0=N, mg cos (2:33) (2.32) (2) R = mg sin =0 (1) R =0 (3) 0 (2.33) N = mg cos (2.29) R 0 N; mg sin <mgcos! c mg sin c! mg cos c 0 =tan c (4) R = N = mg cos x ma = mg sin, mg cos

60 2 a = g(sin, cos ) tan >tan c (3) sin cos > 0 (2.31) sin > 0 cos >cos a>0 2 l = 176 cm h =93cm s = 168 cm t =2:0 s 7 2, 1 2 2 2.9: 2.9 1m 2 10 8 N 2.9 7 2s 2 = tan, =0:521 gt 2 cos sin = h l

2{3. 61 3 2.10 m m F F F g t =0 1 C D D E C D E 2.11 m g F N F C N C m g E N C x C 2.10: T CD N F m g D N C C N DF T DE F E D E N E y N E m g T DE T CD 2.11:

62 2 N E C N C CD D C T CD E N E D E T DE 4 1 D T CD T DE N DF C D E 2 2.10 x y O t =0 t x y y = x dy dx dt = dt d2 y dt 2 = d2 x dt 2 8 F v a C D E v a D 4 3 C D E F (x ) m a = N C (2.34) (y ) m 0=N F, m g (2.35) m a = m g, N E (2.36) C 0 a = T CD, N C (2.37) D(x ) 0 a cos 4 = N DF cos 4, T CD (2.38) D(y ) 0 a sin 4 = T DE, N DF sin 4 (2.39) E 0 a = N E, T DE (2.40) C E D 8

2{3. 63 4 4 (2.34) (2.40) (2.35) (m + m )a = m g (2.41) 1 a = m m + m g (2.42) (2.42) a (2.34)(2.40) N F = m g; N C = T CD = T DE = N E = p 1 N DF = 2 m m m + m 5 t =0 t t 0 t =0 x = y =0 v =0 t v = at; x = y = 1 2 at2 6 : v t x y I{1{2 a = dv (2.41) dt (m + m ) dv dt = m g v v = dy dt dy dt (m + m )v dv dt = m g dy dt

64 2 t v y t =0 t y =0 y (m + m ) Z v 0 vdv = m g 1 Z y 0 dy 2 (m + m )v 2, m gy =0 (2.43) y =0; v =0 t =0 0 U y =0 U =0 U =,m gy N C W W = N C x N F N F 0 N E W W =,N E y (2.37) (2.40) N E = N C W + W =0 C D E 1 3 2.12 m m 2.12: g t =0 1 [ ] L = 1 (2.41) 2 (m +m )v 2 +m gy d dt @L @v, @L @y =0

2{3. 65 y v 2 4 2.13 m m P P m C C g l 1 + l 2 C l 2 t =0 C m = M m =3M m C =4M C 2 t 1 C P 2.14 C P T 1 P P P 3 P l 1 + l 2 m g P l 2 P Q 2.13: T 1 T 2 T 1 T 1 P 2 2.14: Q T 2 T 1 P C Q Q 2 2.15 O x Q 2l 1 PC 2l 2 2 T 1 m g C C T 2 m C g 2 3 a = m, m sin m + m g; 1 2 (m + m )v 2, (m, m sin )gy =0

66 2 (x, x P )+(x, x P )=2l 1 (2.44) x C + x P =2l 2 (2.45) C P 4 4,2 =2P x C O C x P P x r = x,x P x P x r (2.24) (2.25) x P x P x = x P + x r (2.46) x =2l 1 + x P, (x, x P )=2l 1 + x P, x r (2.47) x C =2l 2, x P (2.48) x x C P v v v C v P ; a a a C a P (2.25) t 0 2.15: x C dt + x P dt =0 v C + v P =0 C v C P v P =,v C (2.24) t x dt, x P x + dt dt, x P =0 dt P v r = v, v P P,v r = v, v P P v r,v r v = v P + v r (2.49) v = v P, v r (2.50) v C =,v P (2.51) (2.49) (2.50) (2.51) t a = a P + a r (2.52) a = a P, a r (2.53) a C =,a P (2.54) 3 P P 0

2{3. 67 0 a P =0 g +2T 1, T 2 P C T 2 =2T 1 C m a = m g, T 1 (2.55) m a = m g, T 1 (2.56) C m C a C = m C g, 2T 1 (2.57) 4 (2.55) (2.56) (2.57) (2.52) (2.53) (2.54) m (a p + a r )=m g, T 1 (2.58) m (a p, a r )=m g, T 1 (2.59) C m C (,a P )=m C g, 2T 1 (2.60) T 1 (2.58),(2.59) (2.58)+(2:59), (2:60) (m, m )a P +(m + m )a r =(m, m )g (2.61) (m + m + m C )a P +(m, m )a r =(m + m, m C )g (2.62) 3 m = M m =3M m C =4M (2.61),(2.62) (,2Ma P +4Ma r =,2Mg 8Ma P, 2Ma r =0 a P =, 1 7 g; a r =, 4 7 g (2.52) (2.53) (2.54) a =, 5 7 g; a = 3 7 g; a C = 1 7 g 3 1 [ ] L = 2 m (v p + v r ) 2 + m (v p, v r ) 2 + 1 2 m Cv 2 P + f(m + m, m C )x P +(m, m )x r gg d @L, @L d @L =0;, @L =0 dt @v P @x P dt @v r @x r (2.61),(2.62)

68 2 C C (2.55) T 1 = 1 2 T 2 = m (g, a )= 12 7 Mg 5 t =0 x = x = l 1 + l 2 x C = l 2 v = v = v C =0 C v =, 5 7 gt; v = 3 7 gt; v C = 1 7 gt C t x = l 1 + l 2, 5 14 gt2 ; x = l 1 + l 2 + 3 14 gt2 ; x C = l 2 + 1 7 gt2 6 :(2.61) (2.62) a P = dv P dt a r = dv r dt v r v P t (v r =) dx r dt (v P =) dx P (2.61) 2 (2.62) 2 dt Z vr (m + m )v P 1dv r =(m + m )v P v r 0 Z vp (m + m )v r 1dv P =(m + m )v P v r 0 v P v r (2.61)+(2.62) Z vp Z vr v r (m + m + m C ) v P dv P +2(m + m )v P v r +(m, m ) 0 0 dv r Z xp Z xr =(m + m, m C )g 1dx P +(m, m )g l 2 0 x P = l 2 x r = x, x P =(l 1 + l 2 ), l 2 1 2 (m + m + m C )v 2 P +2(m + m )v P v r + 1 2 (m, m )v 2 r l 1 1dx r

2{3. 69 =(m + m, m C )g x P, l 2 )+(m, m )g(x r, l 1 ) 1 2 m (v P + v r ) 2 + 1 2 m (v P, v r ) 2 + 1 2 m Cv 2 P 1 2 m (v P + v r ) 2 + 1 2 m (v P, v r ) 2 + 1 2 m Cv 2 P, m g(x P + x r ), m g(2l 1 + x P, x r ), m C (2l 2, x P ) =,m g(l 1 + l 2 ), m g(l 1 + l 2 ), m C l 2 (2.63) (2.46)(2.51) 1 2 m v 2 + 1 2 m v 2 + 1 2 m Cv 2,m C gx,m gx,m C gx C =,m g(l 1 + l 2 ), m g(l 1 + l 2 ), m C gl 2 C (2.23) x =0 4 2.16 m P m P Q g 4 P Q 2.16: N N R R 3 4 5 C m m 0 2.17: 2.17 N 0 g 4 4 a = 1 2 a = 2m, m g; T =2T = 6m m g m +4m m +4m N C

70 2 1 2.18 N R N N m g N R C 2.18: C N C R N N m g 2 x y x v v ; a a R a R v = v ; a = a b R v 6= v ; a 6= a 3 a R a = a = a N C (x ) m a = N, R (2.63) (y ) 0=N, m g (2.64) (x ) m a = R (2.65) (y ) 0=N C, N, m g (2.66) b R (2.30) R = N y x (x ) m a = N N (2.67) (x ) m a = N (2.68) 4 a R (2.63)+(2.65) (m + m )a = N (2.69)

2{3. 71 N a = N m + m (2.70) N (2.64) (2.66) N = m g (2.71) N C = N + m g =(m + m )g (2.72) (2.65) (2.71) R = m a = m m + m N (2.73) (2.29) R < 0 N (2.70) R < 0 m g (2.74) (2.71) (2.73) m m + m N< 0 m g N< m (m + m ) 0 g m N b R m N < (m + m ) = 0 g m (2.71) R = m g (2.75) (2.67) (2.68) a = N m, g; a = m m g

72 2 5 O t =0 x =0 x =0 v = v =0 a b 6 a R (2.69) v dx dt x a = dv dt (m + m )v dv dt = N dx dt t v x 0 v 0 x (m + m ) Z v 0 vdv = N N =const. 1 Z x 0 1dx 2 (m + m )v 2, 0=Nx (2.76) N 0 ) b R (2.67) v dx dt (2.68) v dx dt m v dv dt + m v dv dt =(N, N ) dx dt + N dx dt (2.71) N = m g 1 2 m v 2 + 1 2 m v 2, 0=Nx, m g(x, x ) (2.77) x >x 2 N 100% 2 5 C m m N C 2.19:

2{3. 73 0 2.19 N 0 N N l g 5 3 4 2.4 3 4 2.20 f l 1 x 1 l 2 x 2 2.21 f > 0 5 5 N = 0 (m + m )g; m gl x 1 > 0 x 1 +x 2 > 0 f>0 l 1 l 2 f<0 x 1 < 0 x 1 +x 2 < 0 2.20:

74 2 f x 2 l 2 K f x 1 l 1 f = K x 1 l 1 = K x 2 l 2 n l 1 l 2 l n x 1 x 2 x n l = l 1 + l 2 + + l n f K x 2 l 2 x =x 1 +x 2 + +x n K x 1 l 1 f K x 1 = K x 2 = = K x n l 1 l 2 l n 2.21: = K x 1 +x 2 + +x n l 1 + l 2 + + l n K = k l x x >0 x <0 = K x l,t x T x,! T T x,kx (2.78),,! T,T y (,kx < 0 ;,kx > 0 ) 2.22: 6 k 6

2{3. 75 2.22 T 0 6 2.23 m x O x 2.23: k 1 O x x>0 x <0 x,kx 2 a x ma =,kx (2.79) a = d2 x dt 2 m d2 x dt 2 =,kx II 3 t =0 t =0 x v =0

76 2 4 (2.79) v dx dt a = dv dt mv dv dt =,kxdx dt t v x 0 v x m Z v 0 vdv =,k 1 Z x xdx 2 mv2 =, 1 2 kx2 + 1 2 k2 1 2 mv2 1 2 kx2 = 1 2 k2 (2.80) U (x) = 1 2 kx2 (2.81) x =0( ) ( 2.4 (2.80) v x v xy! x 2 a 2 + y2 b 2 =1, O x x = a cos ; y = b sin,! k m =!2 (2.80) x 2 + v2 2 (!) =1 2 2.24: x, v x v 2.24!

2{3. 77 t t =0 x = cos 0 = ; v =! sin0=0 x v (i) (ii) (iii) j,j (iv) (v) 7 =!t x = cos!t (2.82) T x = cos 2 = cos!t T = 2! =2 r m k (2.83), p v =! 2, x 2 =! sin!t!t < v<0, v =,!sin!t (2.84) 8 (2.80) a = k m x =,!2 x =,! 2 cos!t a, t v, t x, t 6 2.25 k m g (1) (2) O 1 2 kx2 (3) 9 7 8 v = dx dt = d cos!t =,! sin!t dt 9 6 (1)2r m mg mg,(2) (3) k k k 2.25: