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- こおが そめや
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2 r(t) t r(t) O v(t) = dr(t) dt a(t) = dv(t) dt = d2 r(t) dt 2
3 r(t), v(t), a(t) t dr(t) dt r(t) =(x(t),y(t),z(t)) = d 2 r(t) dt 2 = ( dx(t) dt ( d 2 x(t) dt 2, dy(t), dz(t) dt dt ), d2 y(t) dt 2, d2 z(t) dt 2 = (ẋ(t), ẏ(t), ż(t)) ) = (ẍ(t), ÿ(t), z(t)) d dt r(t) 2 =2r(t) v(t) r(t) 2 = x(t) 2 + y(t) 2 + z(t) 2 d dt r(t) 2 = 2(x(t)ẋ(t)+y(t)ẏ(t)+z(t)ż(t)) =2r(t) v(t)
4 t = 0 (1, 0, 0) t = 1 ( 1, 1, 1) t = 1 ( 1, 1, 1) r(t) = (1 t)(1, 0, 0) + t( 1, 1, 1) = (1 2t, t, t) t = 0 (1, 0, 0) r(t) O v(t) =( 2, 1, 1) a(t) = (0, 0, 0)
5 t =0 (1, 0) ω v(t) a(t) O r(t) r(t) = (cos ωt, sin ωt) v(t) =( ω sin ωt, ω cos ωt) = ωr π/2 r(t) R θ = [ cos θ sin θ sin θ cos θ a(t) =( ω 2 cos ωt, ω 2 sin ωt) = ω 2 r(t) v(t) = ω a(t) = ω 2 ] θ
6 ma = f Isaac NEWTON ( ) m dv dt = f m d2 r dt 2 = f
7 y v 0 mg 0 x v 0 =(u 0,v 0 ) m d2 r dt 2 = f f = (0, mg) r(0) = 0 g dr dt (0) = v 0 { mẍ = 0 mÿ = mg { ẋ = u0 ẏ = v 0 gt x(0) = 0, ẋ(0) = u 0 y(0) = 0, ẏ(0) = v 0 x(t) = u 0 t y(t) = v 0 t g 2 t 2
8 x kv mg t = 0 v g v(t)
9 m dv dt = mg kv dy dx P(x) y = 0 log y = dy y = P(x) y = ce P(x)dx P(x)dx () dy dx αy = 0 y = ce αdx + c = ce αx
10 dy dx m dv dt P(x) y = Q(x) y = c(x)e = mg kv ( ) P(x )dx if Q(x) = 0 y = ce P(x)dx c(x) = e P(x )dx Q(x)dx + c y = e P(x )dx ( P(x )dx e Q(x)dx + c) dy dx αy = β y = β α + ce αx c:
11 x kv mg v = mg k t = 0 v (1 e k m t) m dv dt mg k v g = mg kv v = gt v(t) v(0) = 0 v = mg k t
12 v ( ) v --- dv/dt
13 du dt = ru(1 u/k) du/dt 0 < u < K u u > K u 0 K u dv dt = g k m v v < mg/k v > mg/k v v dv du/dt 0 mg/k uv
14 v v v(0) > mg/k mg k mg k v(0) = 0 t t dv dt = g k m v v < mg/k v > mg/k v v dv du/dt 0 mg/k uv
15 0 x(t) x x(t)
16 kx m m k x(t) d 2 x dt 2 + ω2 0x = 0 m d2 x dt 2 = kx ω 0 = k/m
17 d 2 x dt 2 + ω2 0x = 0 (1) x = cos ω 0 t x = sin ω 0 t A cos ω 0 t + B sin ω 0 t = 0 A = B = 0 x 2 (t) x 1 (t) x(t) = c 1 x 1 (t) + c 2 x 2 (t)
18 K k=0 a k (t) dk u dt k = f(t) f(t) =0 d 2 x dt 2 + ω2 0x = 0
19 d 2 x dt 2 + ω2 0x = 0 (1) A cos ω 0 t + B sin ω 0 t = 0 A = B = 0 t = 0 A = 0 t = π/(2ω 0 ) B = 0
20 d 2 x dt 2 + ω2 0x = 0 (1) x 2 (t) x 1 (t) x(t) = c 1 x 1 (t) + c 2 x 2 (t) d 2 x dt 2 + ω2 0x = d2 dt 2 (c 1x 1 + c 2 x 2 ) + ω0(c 2 1 x 1 + c 2 x 2 ) = c 1 ( d 2 x 1 dt 2 + ω2 0x 1 ) + c 2 ( d 2 x 2 dt 2 + ω2 0x 2 ) = 0
21 d 2 x dt 2 + ω2 0x = 0 (1) A, B x = A cos ω 0 t + B sin ω 0 t dx dt (0) = 0 x(0) = 1, dx dt (0) = 1 = C sin (ω 0 t + φ) x(0) = 0, dx x(0) = a, dt (0) = b
22 d 2 x dt 2 + ω2 0x = 0 (1) A, B x = A cos ω 0 t + B sin ω 0 t dx dt (0) = 0 = C sin (ω 0 t + φ) x(0) = 1, x = cos ω 0 t
23 d 2 x dt 2 + ω2 0x = 0 (1) A, B x = A cos ω 0 t + B sin ω 0 t = C sin (ω 0 t + φ) dx dt (0) = 1 x(0) = 0, x(t) = 1 ω 0 sin ω 0 t
24 d 2 x dt 2 + ω2 0x = 0 (1) A, B x = A cos ω 0 t + B sin ω 0 t = C sin (ω 0 t + φ) dx x(0) = a, dt (0) = b x(t) =a cos ω 0 t + b ω 0 sin ω 0 t
25 K k=0 a k (t) dk u dt k = f(t) f(t) =0
26 n n n n u 1,u 2,,,,u n α 1 u 1 + α 2 u 2 +,,,,+α n u n
27
28 K a k (t) d k k = 0 dt k L L(u) = f (t) L : L(u + v) = L(u) + L(v) L(αu) = αl(u) α :
29 m c dx dt x(t) m d2 x dt 2 = kx cdx dt
30 γ = c 2m, ω 0 = k m d 2 x dx + 2γ dt2 dt + ω2 0x = 0 x(t) = e λt x(t) = e λt λ λ 2 + 2γλ + ω 2 0 = 0
31 d 2 x dx + 2γ dt2 dt + ω2 0x = 0 = λ 2 + 2γλ + ω 2 0 = 0 γ 2 ω 2 0 γ = c/(2m) γ = 0 0 < γ < ω 0 γ = ω 0 γ > ω 0 x = A cos ω 0 t + B sin ω 0 t = C sin (ω 0 t + φ)
32 d 2 x dx + 2γ dt2 dt + ω2 0x = 0 γ > ω 0 λ 2 + 2γλ + ω 2 0 = 0 λ = γ ± γ 2 ω 2 0 x 1 = e ( γ+ γ 2 ω0 2)t x 2 = e ( γ γ 2 ω0 2)t x = Ae ( γ+ γ 2 ω 2 0 )t + Be ( γ γ 2 ω 2 0 )t
33 γ > ω 0 x = Ae ( γ+ γ 2 ω 2 0 )t + Be ( γ γ 2 ω 2 0 )t v γ = 5.0 ω 0 = 1.0 x x t
34 d 2 x dx + 2γ dt2 dt + ω2 0x = 0 0 < γ < ω 0 λ 2 + 2γλ + ω 2 0 = 0 λ = γ ± iω ω = ω 2 0 γ2 x 1 = e ( γ+iω)t = e γt e iωt = e γt (cos ωt + i sin ωt) x 2 = e ( γ iω)t = e γt e iωt = e γt (cos ωt i sin ωt) x 1, x 2 (x 1 + x 2 )/2 = e γt cos ωt (x 1 x 2 )/(2i) = e γt sin ωt x = e γt (A cos ωt + B sin ωt)
35 0 < γ < ω 0 x = e γt (A cos ωt + B sin ωt) v γ = 0.05 ω 0 = 1.0 x x t
36 d 2 x dx + 2γ dt2 dt + ω2 0x = 0 γ = ω 0 λ 2 + 2γλ + γ 2 = 0 λ = γ x 1 = e γt x 2 = te γt x = e γt (A + Bt)
37 γ = ω 0 x = e γt (A + Bt) v γ = 1.0 ω 0 = 1.0 x x t
38 d 2 x dx + 2γ dt2 dt + ω2 0x = 0 γ = 0 0 < γ < ω 0 γ = ω 0 x = A cos ω 0 t + B sin ω 0 t x = e γt (A cos ωt + B sin ωt) ω = ω 2 0 γ2 x = e γt (A + Bt) γ > ω 0 x = Ae ( γ+ γ 2 ω 2 0 )t + Be ( γ γ 2 ω 2 0 )t
39 γ = 0 v γ = 0.0 ω 0 = 1.0 x x t 0 < γ < ω 0 v γ = 0.05 ω 0 = 1.0 x x t
40 γ = ω 0 v γ = 1.0 ω 0 = 1.0 x x t γ > ω 0 v γ = 5.0 ω 0 = 1.0 x x t
41 0 γ < ω 0 γ > ω 0 γ = ω 0 x = e γt (A cos x = e γt (A + Bt) γ γ = ω 0 ) ω0 2 γ2 t + B sin ω0 2 γ2 t x = Ae ( γ+ γ 2 ω 2 0 )t + Be ( γ γ 2 ω 2 0 )t λ ω 0 0 ω 0 γ λ 2 + 2γλ + ω0 2 = 0 iω 0 iω 0
42 d 2 x dt 2 + 3dx dt + 2x = 0 d 2 x dt 2 + 2dx dt + 2x = 0 d 2 x dt 2 + 2dx dt + x = 0 d 2 x dt 2 + x = 0 d 2 x dt 2 + dx dt + x = 0 d 2 x dt 2 + 3dx dt + x = 0
43 x(0) = 1, ẋ(0) = 3 x(0) = 1, ẋ(0) = 0 x(0) = 1, ẋ(0) = 0 x(0) = 3, ẋ(0) = 1 x(0) = 2, ẋ(0) = 1 x(0) = 2, ẋ(0) = 3
44
45 ma cos ωt (A > 0, ω > 0) m d 2 x dx + 2γ dt2 dt + ω2 0x = A cos ωt 0 < γ < ω 0 ω
46 z(t) = x(t) + iy(t) d 2 z dz + 2γ dt2 dt + ω2 0z = Ae iωt z(t) x(t) d 2 x dx + 2γ dt2 dt + ω2 0x = A cos ωt y(t) d 2 y dy +2γ dt2 dt + ω2 0y = A sin ωt
47 =) d 2 z dz + 2γ dt2 dt + ω2 0z = Ae iωt z s (t) =R(ω)Ae iωt R(ω) {(ω 2 0 ω 2 )+2γωi}R(ω)Ae iωt = Ae iωt R(ω) = 1 (ω 2 0 ω2 )+2γωi
48 2 au 1 + bu 2 2 2
49 K a k (t) d k k = 0 dt k L(u) = f (t) L S 1 S 2 S 1 S 2 proof L(S 1 ) = f (t) L(S 2 ) = f (t) L(S 1 S 2 ) = L(S 1 ) L(S 2 ) = f (t) f (t) = 0
50 2 S 1 (= ) S 2 S 3 (= )
51 K a k (t) d k k = 0 dt k L(u) = 0 L u 1 u 2 αu 1 + βu 2 ( α,β Z) proof L(u 1 ) = 0 L(u 2 ) = 0 L(αu 1 + βu 2 ) = αl(u 1 ) + βl(u 2 ) L(αu 1 + βu 2 ) = αl(u 1 ) + βl(u 2 ) = 0
52 R(ω) R(ω) = 1 (ω 2 0 ω2 )+2γωi R(ω) =R 0 (ω)e iφ(ω) R 0 (ω) R 1 = R 1 0 eiφ R 0 (ω) = φ(ω) 0 < φ(ω) < π 1 (ω 2 0 ω 2 ) 2 + 4γ 2 ω 2 φ(ω) = Arg((ω 2 0 ω 2 ) + 2γωi) R 1 =(ω 2 0 ω 2 )+2γωi φ(ω)
53 x s (t) d 2 x dx + 2γ dt2 dt + ω2 0x = A cos ωt R 0 (ω), φ(ω) z s (t) =R 0 (ω)ae i(ωt φ(ω)) x s (t) = R 0 (ω)a cos (ωt φ(ω)) A cos ωt x s (t) t R 0 (ω) φ(ω)
54 d 2 x dx + 2γ dt2 dt + ω2 0x = 0 d 2 x dx + 2γ dt2 dt + ω2 0x = f(t) x s x = x s + y x s x y = x x s y
55 d 2 x dx + 2γ dt2 dt + ω2 0x = A cos ωt x s (t) = R 0 (ω)a cos (ωt φ(ω)) x(t) = x s (t) + e (B γt cos ω 2 0 γ2 t + C sin B, C ) ω0 2 γ2 t
56 γ > 0 d 2 x dx + 2γ dt2 dt + ω2 0x = A cos ωt x s (t) = R 0 (ω)a cos (ωt φ(ω)) x s (t) x(t) A cos ωt
57 R 0 (ω) ω > 0 R 0 (ω) ω R 0 ( ω) γ < ω 0 / 2 ω = ω0 2 2γ2 R 0 ( ω) = ω 0 = 1, γ = 0.1 ω R 0 ( ω) 1 2γ ω 2 0 γ2 ω ω = = = R 0 ( ω) = = 5 ω
58 ω R(ω) R 0 (ω) π 0 φ(ω) ω 0
59
60 ẍ +ẋ + x = cos t x(0) = 2, ẋ(0) = 0 z +ż + z = e it z s (t) =Re it ir =1 R = i z s (t) = ie it = i(cos t + i sin t) = sin t i cos t x s (t) = Re z s (t) = sin t 3 x(t) = sin t + e t/2 (A cos 2 t + B sin 3 x(t) = sin t +2e t/2 cos 2 t 3 2 t)
数学演習:微分方程式
( ) 1 / 21 1 2 3 4 ( ) 2 / 21 x(t)? ẋ + 5x = 0 ( ) 3 / 21 x(t)? ẋ + 5x = 0 x(t) = t 2? ẋ = 2t, ẋ + 5x = 2t + 5t 2 0 ( ) 3 / 21 x(t)? ẋ + 5x = 0 x(t) = t 2? ẋ = 2t, ẋ + 5x = 2t + 5t 2 0 x(t) = sin 5t? ẋ
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II 37 Wabash Avenue Bridge, Illinois 州 Winnipeg にある歩道橋 Esplanade Riel 橋6 6 斜張橋である必要は多分無いと思われる すぐ横に道路用桁橋有り しかも塔基部のレストランは 8 年には営業していなかった 9 9. 9.. () 97 [3] [5] k 9. m w(t) f (t) = f (t) + mg k w(t) Newton
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More information1 θ i (1) A B θ ( ) A = B = sin 3θ = sin θ (A B sin 2 θ) ( ) 1 2 π 3 < = θ < = 2 π 3 Ax Bx3 = 1 2 θ = π sin θ (2) a b c θ sin 5θ = sin θ f(sin 2 θ) 2
θ i ) AB θ ) A = B = sin θ = sin θ A B sin θ) ) < = θ < = Ax Bx = θ = sin θ ) abc θ sin 5θ = sin θ fsin θ) fx) = ax bx c ) cos 5 i sin 5 ) 5 ) αβ α iβ) 5 α 4 β α β β 5 ) a = b = c = ) fx) = 0 x x = x =
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1. 1 213 1 6 1 3 1: ( ) 2: 3: SF 1 2 3 1: 3 2 A m 2. 2 P M A 2 F = mmg AP AP 2 AP (G > : ) AP/ AP A P P j M j F = n j=1 mm j G AP j AP j 2 AP j 3 P ψ(p) j ψ(p j ) j (P j j ) A F = n j=1 mgψ(p j ) j AP
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m v = mg + kv m v = mg k v v m v = mg + kv α = mg k v = α e rt + e rt m v = mg + kv v mg + kv = m v α + v = k m v (v α (v + α = k m ˆ ( v α ˆ αk v = m v + α ln v α v + α = αk m t + C v α v + α = e αk m
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