Манаков Н.Л., Некипелов А.А., Овсянников В.Д. Задачи по теоретической механике
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¨x + ω
0
2
x =
F
0
m
cos ωt.
ω 6= ω
0
B cos ωt
B =
F
0
m(ω
0
2
− ω
2
)
.
x(t) = A cos(ω
0
t + α) +
F
0
m(ω
0
2
− ω
2
)
cos ωt;
˙x(t) = −A sin(ω
0
t + α) −
F
0
ω
m(ω
0
2
− ω
2
)
sin ωt.
0 = A cos(α) +
F
0
m(ω
0
2
− ω
2
)
, 0 = −Aω
0
sin(α),
α = 0, A = −
F
0
m(ω
0
2
− ω
2
)
x(t) =
F
0
m(ω
0
2
− ω
2
)
[cos(ωt) − cos(ω
0
t)] =
=
2F
0
m(ω
0
2
− ω
2
)
sin
µ
ω
0
− ω
2
t
¶
sin
µ
ω
0
+ ω
2
t
¶
.
ω = ω
0
Bt sin(ω
0
t) ω → ω
0
x(t) =
F
0
t
2mω
0
sin(ω
0
t).
ω
T F
0
0 6 t 6 T
x(t) =
F
0
mω
2
(1 − cos ωt).
t > T
x(t) = A sin [ω(t − T ) + α] .
t = T
F
0
mω
2
(1 − cos ωT ) = A sin α;
F
0
mω
sin ωT = Aω cos α.
α =
ωT
2
, A =
2F
0
mω
2
sin
ωT
2
.
¨x + ω
2
x =
F (t)
m
dξ
dt
− iωξ =
F
m
,
ξ = ˙x + iωx x =
1
ω
Imξ ξ
ξ = e
iωt
t
Z
0
F (t
0
)
m
e
−iωt
0
dt
0
+ ξ
0
.
t > T
ξ = e
iωt
T
Z
0
F
0
m
e
−iωt
0
dt
0
, (ξ
0
= 0).
ξ = e
iωt
F
0
m
e
−iωT
− 1
−iω
= e
iωt
2F
0
mω
e
−
iωT
2
sin
ωT
2
.
x(t) =
1
ω
Imξ =
2F
0
mω
2
sin
ωT
2
sin
µ
ωt −
ωT
2
¶
ω F = F
0
t
T
T
A =
F
0
mT ω
3
p
ω
2
T
2
− 2ωT sin(ωT ) + 2(1 − cos(ωT )).
ω
F =
0 t < 0
F
0
t/T 0 < t < T
F
0
t > T,
A =
2F
0
mω
3
T
sin
ωT
2
.
L =
m
1
+ m
2
2
l
2
1
˙
ϕ
2
1
+
m
2
2
l
2
2
˙
ϕ
2
2
+
+ m
2
l
1
l
2
˙ϕ
1
˙ϕ
2
cos(ϕ
1
− ϕ
2
) + (m
1
+ m
2
)gl
1
cosϕ
1
+ m
2
gl
2
cos ϕ
2
.
(|ϕ
1
| ¿ 1, |ϕ
2
| ¿ 1)
ϕ
1
ϕ
2
L =
m
1
+ m
2
2
l
2
1
˙
ϕ
2
1
+
m
2
2
l
2
2
˙
ϕ
2
2
+ m
2
l
1
l
2
˙ϕ
1
˙ϕ
2
−
− (m
1
+ m
2
)gl
1
ϕ
2
1
2
− m
2
gl
2
ϕ
2
2
2
;
d
dt
∂L
∂ ˙ϕ
1
−
∂L
∂ϕ
1
= 0,
d
dt
∂L
∂ ˙ϕ
2
−
∂L
∂ϕ
2
= 0
(
(m
1
+ m
2
)l
1
¨ϕ
1
+ m
2
l
2
¨ϕ
2
+ (m
1
+ m
2
)gϕ
1
= 0;
l
1
¨ϕ
1
+ l
2
¨ϕ
2
+ gϕ
2
= 0.
ϕ
1
= A
1
e
iωt
; ϕ
2
= A
2
e
iωt
,
(
(m
1
+ m
2
)(g − l
1
ω
2
)A
1
− ω
2
m
2
l
2
A
2
= 0;
−l
1
ω
2
A
1
+ (g − l
2
ω
2
)A
2
= 0.
m
1
l
1
l
2
ω
4
− g(l
1
+ l
2
)(m
1
+ m
2
)ω
2
+ g
2
(m
1
+ m
2
) = 0.
ω
2
=
g(l
1
+ l
2
)(m
1
+ m
2
)
2l
1
l
2
m
1
"
1 ±
s
1 −
4l
1
l
2
m
1
(l
1
+ l
2
)
2
(m
1
+ m
2
)
#
.
m
1
→ ∞ : ω
2
1
∼
=
g
l
1
, ω
2
2
∼
=
g
l
2
.
m
2
→ ∞ : ω
2
1
∼
=
g
l
1
+ l
2
, ω
2
2
∼
=
g
l
1
+ l
2
l
1
l
2
m
1
m
2
À ω
2
1
.
m l l
0
κ
l
0
ϕ
1
= a cos(ω
1
t + α) + b cos(ω
2
t + β),
ϕ
2
= a cos(ω
1
t + α) − b cos(ω
2
t + β).
ω
1
=
r
g
l
, ω
2
=
r
g
l
+ 2
κ
m
F
0
cos γt
γ
U = −F
0
lϕ
2
cos γt
θ
1
=
1
2
(ϕ
1
+ ϕ
2
), θ
2
=
1
2
(ϕ
1
− ϕ
2
).
¨
θ
1
+ ω
2
1
θ
1
=
F
0
2ml
cos γt;
¨
θ
2
+ ω
2
2
θ
2
= −
F
0
2ml
cos γt.
ϕ
1
(t) =
F
0
2ml
·
1
ω
2
1
− γ
2
−
1
ω
2
2
− γ
2
¸
cos γt;
ϕ
2
(t) =
F
0
2ml
·
1
ω
2
1
− γ
2
+
1
ω
2
2
− γ
2
¸
cos γt,
ϕ
1
(t)
ϕ
2
(t)
=
ω
2
2
− ω
2
1
ω
2
2
+ ω
2
1
− 2γ
2
.
ω
2
2
− ω
2
1
¿ ω
2
2
+ ω
2
1
, γ ¿ ω
2
γ À ω
2
¯
¯
¯
¯
ϕ
1
(t)
ϕ
2
(t)
¯
¯
¯
¯
¿ 1,
ω
0
f = f
0
e
αt
cos γt.
¨x + 2λ ˙x + ω
2
0
x =
F (t)
m
¨z + 2λ ˙z + ω
2
0
z =
f
0
m
e
αt
e
iγt
,
x = Re z z z = Ae
(α+iγ)t
z A
A =
f
0
m
1
(α + iγ)
2
+ 2λ(α + iγ) + ω
2
0
.
x x = be
αt
cos(γt + δ)
b =
f
0
m
1
p
(α
2
+ ω
2
0
+ 2λα − γ
2
)
2
+ 4γ
2
(α + λ)
2
;
δ = −arctg
2γ(α + λ)
(α
2
+ ω
2
0
+ 2λα − γ
2
)
.
E
ω F (t) = F
0
e
−(t/τ)
2
t = −∞
F (t)
∆E =
1
2m
¯
¯
¯
¯
¯
¯
+∞
Z
−∞
F (t)e
−iωt
dt
¯
¯
¯
¯
¯
¯
2
.
∆E =
F
2
0
2m
¯
¯
¯
¯
¯
¯
¯
+∞
Z
−∞
e
−
t
2
τ
2
− iωt
dt
¯
¯
¯
¯
¯
¯
¯
2
=
F
2
0
2m
¯
¯
¯
¯
¯
¯
¯
¯
+∞
Z
−∞
e
−
µ
t
τ
+ i
τω
2
¶
2
−
τ
2
ω
2
4
dt
¯
¯
¯
¯
¯
¯
¯
¯
2
=
=
F
2
0
2m
πτ
2
e
−
τ
2
ω
2
2
.
t → −∞
x = a sin(ωt + α) a α
∆E =
m
2
£
|ξ(∞)|
2
− |ξ(−∞)|
2
¤
ξ = ˙x + iωx
ξ
∆E =
πF
2
0
τ
2
2m
e
−
1
2
(ωτ )
2
+
√
πF
0
aωτ e
−
1
4
(ωτ )
2
cos α.
mω
2
0
x
2
2
+
mα
3
x
3
+
mβ
4
x
4
.
L =
m ˙x
2
2
−
mω
2
0
x
2
2
−
mα
3
x
3
−
mβ
4
x
4
,
¨x + ω
2
0
x = −αx
2
− βx
3
.
¨x + ω
2
x = −αx
2
− βx
3
+ (ω
2
− ω
2
0
)x,
ω
x = x
(0)
+ x
(1)
+ x
(2)
+ . . . , ω = ω
0
+ ω
(1)
+ ω
(2)
+ . . . ,
¨x
(0)
+ ¨x
(1)
+ ¨x
(2)
+ . . . + ω
2
(x
(0)
+ x
(1)
+ x
(2)
+ . . . ) =
= −α(x
(0)
+ x
(1)
+ x
(2)
+ . . . )
2
− β(x
(0)
+ x
(1)
+ x
(2)
+ . . . )
3
+
+[(ω
0
+ ω
(1)
+ ω
(2)
+ . . . )
2
− ω
2
0
](x
(0)
+ x
(1)
+ x
(2)
+ . . . ).
¨x
(0)
+ ω
2
x
(0)
= 0;
¨x
(1)
+ ω
2
x
(1)
= −αx
(0)
2
+ 2ω
0
ω
(1)
x
(0)
;
¨x
(2)
+ ω
2
x
(2)
= −2αx
(0)
x
(1)
− βx
(0)
3
+ ω
(1)
2
x
(0)
+ 2ω
0
ω
(1)
x
(1)
+ 2ω
0
ω
(2)
x
(0)
.
x
(0)
= a cos ωt.
¨x
(1)
+ ω
2
x
(1)
= 2ω
0
ω
(1)
a cos ωt −
αa
2
2
(1 + cos 2ωt).
ω
2ω
0
ω
(1)
a cos ωt
∼ ω
(1)
t sin ωt
ω
(1)
= 0 x
(1)
x
(1)
= −
αa
2
2ω
2
0
+
αa
2
6ω
2
0
cos 2ωt.
x
(0)
, ω
(1)
, x
(1)
¨x
(2)
+ω
2
x
(2)
= −2αa cos ωt
µ
−
αa
2
2ω
2
0
+
αa
2
6ω
2
0
cos 2ωt
¶
−βa
3
cos
3
ωt+2ω
0
ω
(2)
a cos ωt.
cos ωt cos 2ωt =
1
2
cos ωt +
1
2
cos 3ωt, cos
3
ωt =
1
4
(cos 3ωt + 3 cos ωt),
x
(2)
¨x
(2)
+ ω
2
x
(2)
= −a
3
·
β
4
+
α
6ω
2
0
¸
cos 3ωt + a
·
2ω
(2)
+
5α
2
a
2
6ω
2
0
−
3
4
βa
2
¸
cos ωt.
cos ωt
ω
(2)
=
·
3
2
β −
5
3
α
2
ω
2
0
¸
a
2
4ω
0
,
x
(2)
x
(2)
=
a
3
16ω
2
0
µ
α
2
3ω
2
0
+
β
2
¶
cos 3ωt.
x = a cos ωt +
αa
2
2ω
2
0
µ
−1 +
1
3
cos 2ωt
¶
+
a
3
16ω
2
0
µ
α
2
3ω
2
0
+
β
2
¶
cos 3ωt + . . . ,
ω = ω
0
+
·
−
5
3
α
2
ω
2
0
+
3
2
β
¸
a
2
4ω
0
+ . . . .
−
αa
2
2ω
2
0
+ . . .
ω, 2ω, 3ω ω
q
α
p
α
α = s s
p
α
=
∂L
∂ ˙q
α
,
L = L(q, ˙q, t)
p
i
= mv
i
q
α
p
α
H
H(p, q, t) =
s
X
β=1
p
β
˙q
β
− L(q, ˙q, t),
˙q p q t
˙q p
α
(t) q
α
(t)
dp
α
dt
= −
∂H
∂q
α
,
dq
α
dt
=
∂H
∂p
α
, α = 1, . . . , s.
r
i
(q)
H(p, q, t) = T (p, q) + U(q, t),
T U
S
q(t)
S(q, t) =
Z
(t,q)
(t
0
q
0
)
L(q(t
0
), ˙q(t
0
), t
0
) dt
0
,
p
α
=
∂S
∂q
α
, α = 1, . . . , s,
∂S
∂t
= −H.
ϕ
1
(p, q, t) ϕ
2
(p, q, t),
{ϕ
1
, ϕ
2
} =
s
X
α=1
½
∂ϕ
1
∂p
α
∂ϕ
2
∂q
α
−
∂ϕ
1
∂q
α
∂ϕ
2
∂p
α
¾
.
{ϕ
1
, ϕ
2
} = −{ϕ
2
, ϕ
1
}.
c = const.
{ϕ, c} ≡ 0.
{c
1
ϕ
1
+ c
2
ϕ
2
, ψ} = c
1
{ϕ
1
, ψ} + c
2
{ϕ
2
, ψ},
c
1
c
2
{ϕ
1
· ϕ
2
, ψ} = ϕ
1
{ϕ
2
, ψ} + ϕ
2
{ϕ
1
, ψ}.
∂
∂t
{ϕ
1
, ϕ
2
} =
½
∂ϕ
1
∂t
, ϕ
2
¾
+
½
ϕ
1
,
∂ϕ
2
∂t
¾
.
ϕ
2
= p
α
{ϕ, p
α
} = −
∂ϕ
∂q
α
.
ϕ
2
= q
α
{ϕ, q
α
} =
∂ϕ
∂p
α
.
{p
α
, q
β
} = δ
αβ
, {p
α
, p
β
} = {q
α
, q
β
} = 0.
{ϕ
1
, {ϕ
2
, ϕ
3
}} + {ϕ
2
, {ϕ
3
, ϕ
1
}} + {ϕ
3
, {ϕ
1
, ϕ
2
}} = 0.
f(p, q, t)
df
dt
=
∂f
∂t
+ {H, f}, H