Коткин Г.Л., Сербо В.Г., Черных А.И. Лекции по аналитической механике
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ω = ω
0
+ δω δω
ω = ω
0
+ κb
2
, κ =
3β
8ω
0
−
5α
2
12ω
3
0
. (30.4)
cos γt = cos(γt + δ − δ) = cos (γt + δ) cos δ + sin (γt + δ) sin δ
cos(γt + δ) sin(γt + δ)
(ω
2
− γ
2
) b =
f
m
cos δ , −2λγb =
f
m
sin δ .
b δ b
ω
2
0
→ ω
2
= (ω
0
+ κb
2
)
2
≈ ω
2
0
+ 2κω
0
b
2
(30.5)
b
2
b
2
h
ω
2
0
+ 2κω
0
b
2
− γ
2
2
+ 4λ
2
γ
2
i
=
f
2
m
2
. (30.6)
b(γ)
κ
κ
b(γ) λ
κ
0 0.5 1 1.5 2
γ/ω
0
b
b γ
λ
λ κ = 0
λ
γ
b(γ)
λ ≪ ω
0
ǫ = γ − ω
0
b
2
[(ǫ − κb
2
)
2
+ λ
2
] =
f
2
4m
2
ω
2
0
. (30.7)
b
2
ǫ b < b
max
≈ f/(2mω
0
λ)
ǫ
ǫ = κb
2
±
s
f
2mω
0
b
2
− λ
2
. (30.8)
(κ = 0)
κ
f κ > 0
b
ǫ κb
2
b
max
κ b(ǫ)
κ
ǫ
1
< ǫ < ǫ
2
ǫ
κ
ǫ(b
2
)
b
2
b
2
= 2λ/(
√
3κ)
κ = κ ≡
32
3
√
3
m
2
ω
2
0
λ
3
f
2
. ( 30.9)
κ
“
κ < −κ
b(ǫ) f
f ≪ f
k
f < f
k
f > f
k
b
b
b
ǫ
ǫ
ǫ
A
B
C
D
E
F
а)
б)
в)
ǫ
1
ǫ
2
b
ǫ = γ − ω
0
κ
κ “
f > f =
s
32
3
√
3
m
2
ω
2
0
λ
3
|κ|
. (30.10)
ǫ
ABCDEF
ǫ
ABC C
E
DEF “ ǫ
F ED
B
“
ǫ
1
< ǫ < ǫ
2
ABC F ED
CD
l ϕ = x/l
U = mgl( 1 −cos ϕ) ≈ mgl
1
2
ϕ
2
−
1
24
ϕ
4
=
m
2
ω
2
0
x
2
+
m
4
βx
4
.
α = 0, β = −
ω
2
0
6l
3
, λ = 0, 095 ω
0
, f = 0, 26 mω
2
0
l .
−0.3 −0.2 −0.1 0 0.1
0
0.3
0.6
0.9
ǫ/ω
0
b
D
B
C
E
ǫ/ω
0
b
D
B
C
E
ǫ/ω
0
b
D
B
C
E
b
D
B
C
E
+“
γ o“
β
ǫ
“
x(t)
“
x ∼ l
m l(t)
x
L =
ml
2
(t)
2
˙x
2
−
mgl(t)
2
x
2
d
dt
l
2
(t)
dx
dt
+ gl(t)x = 0.
“ t
′
dt
′
= dt/l
2
(t)
ω
2
(t
′
) = gl
3
(t)
d
2
x
dt
2
+ ω
2
(t) x = 0 , (31.1)
ω(t)
ω
2
(t)
ω
2
(t) = ω
2
0
(1 + h cos γt) . ( 31.2)
d
2
x
dt
2
+ ω
2
0
(1 + h cos γt) x = 0 , (31.3)
γ ≈ 2ω
0
h ≪ 1
γ =
2ω
0
+ǫ ǫ |ǫ| ≪ ω
0
d
2
x
dt
2
+ ω
2
0
x = hω
2
0
x cos γt . (3 1 .4 )
h
x = a cos ω
0
t
hω
2
0
x cos γt =
1
2
hω
2
0
a [cos (ω
0
+ ǫ)t + cos (3ω
0
+ ǫ)t] .
h
x(t) = a(t) cos (γt/2) + b(t) sin (γt/2) ,
a(t) b(t)
¨a
¨
b
4 ˙a + (2ǫ + hω
0
) b = 0 ,
4
˙
b + (2ǫ − hω
0
) a = 0 .
|ǫ| < hω
0
/2
a(t) = α
1
C
1
e
−st
+ C
2
e
st
,
b(t) = α
2
C
1
e
−st
− C
2
e
st
,
s =
1
4
p
(hω
0
)
2
− 4ǫ
2
, α
1,2
=
p
hω
0
± 2ǫ . (31.5)
x(t) = Ce
st
cos
γ
2
t − ϕ
+ De
−st
cos
γ
2
t + ϕ
, (31.6)
x
t
x
t
tg ϕ = α
1
/α
2
s
|ǫ| > hω
0
/2 s = ±(i/4)
p
4ǫ
2
− (hω
0
)
2
γ ≈ 2ω
0
γ ≈ ω
0
/n n ≥ 1
γ ≈ 2ω
0
h