Маятник состоит из жёсткого стержня длиной 𝑙, несущего массу 𝑚 на своём конце. К стержню прикреплены две пружины жёсткости 𝑐 на расстоянии 𝑎 от его
верхнего конца; противоположные концы пружин закреплены. Пренебрегая массой стержня, найти период малых колебаний маятника.
T
=
m
v
2
2
{\displaystyle T={\frac {mv^{2}}{2}}}
V
=
2
k
(
Δ
x
)
2
2
−
m
g
y
{\displaystyle V=2{\frac {k(\Delta x)^{2}}{2}}-mgy}
Δ
x
{\displaystyle \Delta x}
v
=
ϕ
˙
{\displaystyle v={\dot {\phi }}}
y
=
l
cos
(
ϕ
)
{\displaystyle y=l\cos(\phi )}
Δ
x
=
a
sin
(
ϕ
)
{\displaystyle \Delta x=a\sin(\phi )}
L
=
T
−
V
=
1
2
m
l
2
ϕ
˙
2
+
m
g
l
cos
ϕ
−
k
a
2
sin
2
ϕ
{\displaystyle L=T-V={\frac {1}{2}}ml^{2}{\dot {\phi }}^{2}+mgl\cos \phi -ka^{2}\sin ^{2}\phi }
L
~
=
1
2
m
l
2
ϕ
˙
2
−
1
2
m
g
l
ϕ
2
−
k
a
2
ϕ
2
{\displaystyle {\tilde {L}}={\frac {1}{2}}ml^{2}{\dot {\phi }}^{2}-{\frac {1}{2}}mgl\phi ^{2}-ka^{2}\phi ^{2}}
ω
2
=
1
2
m
g
l
+
k
a
2
1
2
m
l
2
=
g
l
+
2
k
a
2
m
l
2
{\displaystyle \omega ^{2}={\frac {{\frac {1}{2}}mgl+ka^{2}}{{\frac {1}{2}}ml^{2}}}={\frac {g}{l}}+{\frac {2ka^{2}}{ml^{2}}}}
ω
=
2
π
ν
=
2
π
T
{\displaystyle \omega =2\pi \nu ={\frac {2\pi }{T}}}
T
=
2
π
ω
=
2
π
g
l
+
2
k
a
2
m
l
2
{\displaystyle T={\frac {2\pi }{\omega }}={\frac {2\pi }{\sqrt {{\frac {g}{l}}+{\frac {2ka^{2}}{ml^{2}}}}}}}
Ответ:
T
=
2
π
g
l
+
2
k
a
2
m
l
2
{\displaystyle T={\frac {2\pi }{\sqrt {{\frac {g}{l}}+{\frac {2ka^{2}}{ml^{2}}}}}}}
Предполагая, что маятник, описанный в предыдущей задаче, установлен так,что масса 𝑚 расположена выше точки подвеса, определить условие, при котором вертикальное положение равновесия маятника устойчиво, и вычислить период малых колебаний маятника.
y
=
−
l
cos
(
ϕ
)
{\displaystyle y=-l\cos(\phi )}
y
=
l
cos
(
ϕ
)
{\displaystyle y=l\cos(\phi )}
V
=
m
g
l
cos
ϕ
+
k
a
2
sin
2
ϕ
{\displaystyle V=mgl\cos \phi +ka^{2}\sin ^{2}\phi }
L
~
=
1
2
m
l
2
ϕ
˙
2
−
1
2
m
g
l
ϕ
2
+
k
a
2
ϕ
2
{\displaystyle {\tilde {L}}={\frac {1}{2}}ml^{2}{\dot {\phi }}^{2}-{\frac {1}{2}}mgl\phi ^{2}+ka^{2}\phi ^{2}}
ω
2
=
−
g
l
+
2
k
a
2
m
l
2
{\displaystyle \omega ^{2}=-{\frac {g}{l}}+{\frac {2ka^{2}}{ml^{2}}}}
T
=
2
π
−
g
l
+
2
k
a
2
m
l
2
{\displaystyle T={\frac {2\pi }{\sqrt {-{\frac {g}{l}}+{\frac {2ka^{2}}{ml^{2}}}}}}}
Ответ:
T
=
2
π
−
g
l
+
2
k
a
2
m
l
2
{\displaystyle T={\frac {2\pi }{\sqrt {-{\frac {g}{l}}+{\frac {2ka^{2}}{ml^{2}}}}}}}
Цилиндр диаметром d и массой т может катиться без скольжения по горизонтальной плоскости. Две одинаковые пружины жёсткости с прикреплены посередине его длины на расстоянии а от оси цилиндра; противоположные концы пружин закреплены. Определить период малых колебаний цилиндра.
V
=
d
2
φ
˙
{\displaystyle V={\frac {d}{2}}{\dot {\varphi }}}
I
=
m
d
2
8
{\displaystyle I={\frac {md^{2}}{8}}}
T
=
m
2
v
2
+
I
φ
˙
2
2
=
m
2
⋅
4
d
2
φ
˙
2
+
m
d
2
φ
˙
2
16
=
3
16
m
d
2
φ
˙
z
{\displaystyle T={\frac {m}{2}}v^{2}+{\frac {I{\dot {\varphi }}^{2}}{2}}={\frac {m}{2\cdot 4}}d^{2}{\dot {\varphi }}^{2}+{\frac {md^{2}{\dot {\varphi }}^{2}}{16}}={\frac {3}{16}}md^{2}{\dot {\varphi }}^{z}}
Δ
ℓ
=
Δ
x
cos
α
=
(
d
2
+
a
)
φ
{\displaystyle \Delta \ell =\Delta x\cos \alpha =\left({\frac {d}{2}}+a\right)\varphi }
Π
=
2
⋅
c
Δ
ℓ
2
2
=
c
φ
2
(
d
2
+
a
)
2
{\displaystyle \Pi =2\cdot {\frac {c{\Delta \ell }^{2}}{2}}=c\varphi ^{2}\left({\frac {d}{2}}+a\right)^{2}}
L
=
3
16
m
d
2
φ
˙
2
−
c
φ
2
(
d
2
+
a
)
2
{\displaystyle L={\frac {3}{16}}md^{2}{\dot {\varphi }}^{2}-c\varphi ^{2}\left({\frac {d}{2}}+a\right)^{2}}
ω
2
=
c
(
d
2
+
a
)
2
3
16
m
d
2
=
16
c
(
1
2
+
a
d
)
2
3
m
{\displaystyle \omega ^{2}={\frac {c\left({\frac {d}{2}}+a\right)^{2}}{{\frac {3}{16}}md^{2}}}={\frac {16c\left({\frac {1}{2}}+{\frac {a}{d}}\right)^{2}}{3m}}}
ω
=
4
3
c
m
d
/
2
+
a
d
=
4
c
3
m
(
1
2
+
a
d
)
=
4
c
3
m
(
1
+
2
a
d
)
{\displaystyle \omega ={\frac {4}{\sqrt {3}}}{\sqrt {\frac {c}{m}}}{\frac {d/2+a}{d}}=4{\sqrt {\frac {c}{3m}}}\left({\frac {1}{2}}+{\frac {a}{d}}\right)={\sqrt {\frac {4c}{3m}}}\left(1+{\frac {2a}{d}}\right)}
τ
=
2
π
/
ω
=
π
3
m
c
1
1
+
2
a
/
d
{\displaystyle \tau =2\pi /\omega =\pi {\sqrt {\frac {3m}{c}}}{\frac {1}{1+2a/d}}}
τ
=
π
3
m
c
1
1
+
2
a
/
d
{\displaystyle \tau =\pi {\sqrt {\frac {3m}{c}}}{\frac {1}{1+2a/d}}}
Неоднородный диск радиуса R и массы М, центр масс С которого расположен на расстоянии а от его геометрического центра О , может катиться без проскальзывания по горизонтальной направляющей. Момент инерции диска относительно оси, перпендикулярной его плоскости и проходящей через центр масс равен J. Найти малые колебания системы около устойчивого положения равновесия.
K
=
m
2
(
R
−
a
)
2
φ
˙
2
+
J
φ
˙
2
2
{\displaystyle K={\frac {m}{2}}(R-a)^{2}{\dot {\varphi }}^{2}+{\frac {J{\dot {\varphi }}^{2}}{2}}}
Π
=
mga
(
1
−
cos
φ
)
≈
mga
φ
2
2
{\displaystyle \Pi =\operatorname {mga} (1-\cos \varphi )\approx \operatorname {mga} {\frac {\varphi ^{2}}{2}}}
ω
=
m
g
a
m
(
R
−
a
)
2
+
J
{\displaystyle \omega ={\sqrt {\frac {mga}{m(R-a)^{2}+J}}}}
φ
(
t
)
=
c
1
cos
m
g
a
m
(
R
−
a
)
2
+
J
t
+
c
2
sin
m
g
a
m
(
R
−
a
)
2
+
J
t
{\displaystyle \varphi (t)=c_{1}\cos {\sqrt {\frac {mga}{m(R-a)^{2}+J}}}t+c_{2}\sin {\sqrt {\frac {mga}{m(R-a)^{2}+J}}}t}
T
=
2
π
m
(
R
−
a
)
2
+
y
m
g
a
{\displaystyle T=2\pi {\sqrt {\frac {m(R-a)^{2}+y}{mga}}}}
R, находящегося на негладкой горизонтальной плоскости, по которой он может катиться без скольжения.
Ищем центр масс
править
{
x
c
=
r
cos
φ
y
c
=
r
sin
φ
{\displaystyle {\begin{cases}x_{c}=r\cos \varphi \\y_{c}=r\sin \varphi \end{cases}}}
x
c
=
0
{\displaystyle x_{c}=0}
y
c
=
∫
y
d
m
m
=
∫
y
d
S
S
=
∫
0
R
r
2
d
r
∫
0
π
sin
φ
d
φ
π
R
2
2
=
4
R
3
π
{\displaystyle y_{c}={\frac {\int ydm}{m}}={\frac {\int ydS}{S}}={\frac {\int _{0}^{R}r^{2}dr\int _{0}^{\pi }\sin \varphi d\varphi }{\pi {\frac {R^{2}}{2}}}}={\frac {4R}{3\pi }}}
p
=
4
R
3
π
{\displaystyle p={\frac {4R}{3\pi }}}
Моменты инерции
править
J
=
J
T
+
m
d
2
{\displaystyle J=J_{T}+md^{2}}
J
S
=
1
2
m
R
2
{\displaystyle J_{S}={\frac {1}{2}}mR^{2}}
J
S
=
J
T
+
m
p
2
{\displaystyle J_{S}=J_{T}+mp^{2}}
J
H
=
J
T
+
m
(
r
−
p
)
2
{\displaystyle J_{H}=J_{T}+m(r-p)^{2}}
J
H
=
J
S
−
m
p
2
+
m
r
2
−
2
m
r
p
+
m
p
2
J
H
=
J
S
+
m
r
2
−
2
m
r
p
=
1
4
m
r
2
+
m
r
2
−
2
m
r
4
r
3
π
J
H
=
m
r
2
(
3
2
−
8
3
π
)
{\displaystyle {\begin{array}{c}{J_{H}=J_{S}-mp^{2}+mr^{2}-2mrp+mp^{2}}\\{J_{H}=J_{S}+mr^{2}-2mrp={\frac {1}{4}}mr^{2}+mr^{2}-2mr{\frac {4r}{3\pi }}}\\{J_{H}=mr^{2}\left({\frac {3}{2}}-{\frac {8}{3\pi }}\right)}\end{array}}}
h
=
p
−
p
cos
α
0
≐
p
(
1
−
(
1
−
α
2
2
)
)
=
2
r
α
2
3
π
{\displaystyle h=p-p\cos \alpha _{0}\doteq p\left(1-\left(1-{\frac {\alpha ^{2}}{2}}\right)\right)={\frac {2r\alpha ^{2}}{3\pi }}}
V
=
m
g
h
=
2
m
g
r
α
2
3
π
{\displaystyle V=mgh={\frac {2mgr\alpha ^{2}}{3\pi }}}
T
=
1
2
J
H
ω
2
=
1
2
m
r
2
(
3
2
−
8
3
π
)
ω
2
{\displaystyle T={\frac {1}{2}}J_{H}\omega ^{2}={\frac {1}{2}}mr^{2}\left({\frac {3}{2}}-{\frac {8}{3\pi }}\right)\omega ^{2}}
1
2
m
r
2
(
3
2
−
8
3
π
)
ω
2
=
m
g
2
r
3
π
{\displaystyle {\frac {1}{2}}mr^{2}\left({\frac {3}{2}}-{\frac {8}{3\pi }}\right)\omega ^{2}=mg{\frac {2r}{3\pi }}}
ω
2
=
4
g
3
π
r
(
3
2
−
8
3
π
)
ω
2
=
8
g
r
(
9
π
−
16
)
{\displaystyle {\begin{array}{c}{\omega ^{2}={\frac {4g}{3\pi r\left({\frac {3}{2}}-{\frac {8}{3\pi }}\right)}}}\\{\omega ^{2}={\frac {8g}{r(9\pi -16)}}}\end{array}}}
T
=
2
π
r
(
9
π
−
16
)
8
g
T
=
π
r
(
9
π
−
16
)
2
g
{\displaystyle {\begin{array}{l}{T=2\pi {\sqrt {\frac {r(9\pi -16)}{8g}}}}\\{T=\pi {\sqrt {\frac {r(9\pi -16)}{2g}}}}\end{array}}}
Три нити связаны в точке С: две из них перекинуты через небольшие неподвижные блоки А и В, расположенные на одной горизонтали, и несут на концах равные грузы m 1 ; к концу третьей нити подвешен груз с массой m 2 , причём m2 <
2m 1 . Найти период малых колебаний этой
системы около положения равновесия, если СА = СВ = а, считая, что грузы могут двигаться только по вертикали.
V
=
2
m
1
g
a
sin
α
0
(
sin
α
0
−
sin
α
)
−
m
2
g
a
sin
α
0
(
sin
α
0
−
sin
α
)
{\displaystyle V=2m_{1}ga\sin \alpha _{0}(\sin \alpha _{0}-\sin \alpha )-m_{2}ga\sin \alpha _{0}\left(\sin \alpha _{0}-\sin \alpha \right)}
V
=
(
2
m
1
−
m
2
)
g
⋅
a
sin
α
0
(
sin
α
0
−
sin
α
)
{\displaystyle V=\left(2m_{1}-m_{2}\right)g\cdot a\sin \alpha _{0}\left(\sin \alpha _{0}-\sin \alpha \right)}
∂
V
∂
q
=
∂
V
∂
α
=
−
(
2
m
1
−
m
2
)
g
⋅
a
sin
α
0
cos
α
=
0
{\displaystyle {\frac {\partial V}{\partial q}}={\frac {\partial V}{\partial \alpha }}=-\left(2m_{1}-m_{2}\right)g\cdot a\sin \alpha _{0}\cos \alpha =0}
α
=
{
π
/
2
- уст.
−
π
/
2
- неуст.
{\displaystyle \alpha =\left\{{\begin{array}{c}{\pi /2}{\text{- уст.}}\\{-\pi /2}{\text{- неуст.}}\end{array}}\right.}
∂
2
V
∂
q
2
=
∂
2
V
∂
α
2
=
(
2
m
1
−
m
2
)
g
⋅
a
⋅
sin
α
0
⋅
sin
α
{\displaystyle {\frac {\partial ^{2}V}{\partial q^{2}}}={\frac {\partial ^{2}V}{\partial \alpha ^{2}}}=\left(2m_{1}-m_{2}\right)g\cdot a\cdot \sin \alpha _{0}\cdot \sin \alpha }
∂
2
V
∂
α
2
(
π
2
)
>
0
{\displaystyle {\frac {\partial ^{2}V}{\partial \alpha ^{2}}}\left({\frac {\pi }{2}}\right)>0}
- устойчивое положение равновесия;
∂
2
V
∂
α
2
(
−
π
2
)
<
0
{\displaystyle {\frac {\partial ^{2}V}{\partial \alpha ^{2}}}\left(-{\frac {\pi }{2}}\right)<0}
- неустойчивое.
T
=
1
2
J
1
w
2
+
1
2
J
2
w
2
{\displaystyle T={\frac {1}{2}}J_{1}w^{2}+{\frac {1}{2}}J_{2}w^{2}}
ω
→
=
α
˙
e
z
→
{\displaystyle {\vec {\omega }}={\dot {\alpha }}{\vec {e_{z}}}}
T
=
1
2
m
2
a
2
α
˙
2
+
2
m
1
2
a
2
α
˙
2
=
1
2
(
2
m
1
+
m
2
)
a
2
α
˙
2
{\displaystyle T={\frac {1}{2}}m_{2}a^{2}{\dot {\alpha }}^{2}+{\frac {2m_{1}}{2}}a^{2}{\dot {\alpha }}^{2}={\frac {1}{2}}\left(2m_{1}+m_{2}\right)a^{2}{\dot {\alpha }}^{2}}
ω
=
(
2
m
1
−
m
2
)
g
a
sin
α
0
2
m
2
1
2
a
2
=
(
2
m
1
−
m
2
)
g
sin
α
0
m
2
a
{\displaystyle \omega ={\sqrt {\frac {\left(2m_{1}-m_{2}\right)ga\sin \alpha _{0}}{2m_{2}{\frac {1}{2}}a^{2}}}}={\sqrt {\frac {\left(2m_{1}-m_{2}\right)g\sin \alpha _{0}}{m_{2}a}}}}
T
=
2
π
m
2
a
(
2
m
1
−
m
2
)
g
sin
α
0
{\displaystyle T=2\pi {\sqrt {\frac {m_{2}a}{\left(2m_{1}-m_{2}\right)g\sin \alpha _{0}}}}}
Уголок, составленный из тонких однородных стержней длин l и 2l с углом
π
2
{\displaystyle {\frac {\pi }{2}}}
О . Определить период малых колебаний системы около положения равновесия.
T
=
J
1
ω
2
2
+
J
2
ω
2
2
=
1
2
(
J
1
+
J
2
)
φ
˙
2
{\displaystyle T={\frac {J_{1}\omega ^{2}}{2}}+{\frac {J_{2}\omega ^{2}}{2}}={\frac {1}{2}}\left(J_{1}+J_{2}\right){\dot {\varphi }}^{2}}
J
1
=
1
3
m
l
2
{\displaystyle J_{1}={\frac {1}{3}}ml^{2}}
J
2
=
2
m
(
2
l
)
2
3
=
8
3
m
l
2
{\displaystyle J_{2}={\frac {2m(2l)^{2}}{3}}={\frac {8}{3}}ml^{2}}
T
=
3
2
m
l
2
φ
˙
2
{\displaystyle T={\frac {3}{2}}ml^{2}{\dot {\varphi }}^{2}}
∂
Π
∂
q
=
c
q
{\displaystyle {\frac {\partial \Pi }{\partial q}}=cq}
y
1
=
−
l
2
cos
ψ
{\displaystyle y_{1}=-{\frac {l}{2}}\cos \psi }
y
2
=
l
sin
ψ
{\displaystyle y_{2}=l\sin \psi }
x
E
=
l
6
(
sin
ψ
+
4
cos
ψ
)
{\displaystyle x_{E}={\frac {l}{6}}(\sin \psi +4\cos \psi )}
y
E
=
l
6
(
4
sin
ψ
−
cos
ψ
)
=
0
{\displaystyle y_{E}={\frac {l}{6}}(4\sin \psi -\cos \psi )=0}
cos
ψ
=
4
sin
ψ
{\displaystyle \cos \psi =4\sin \psi }
1
−
sin
2
ψ
=
16
sin
2
ψ
{\displaystyle 1-\sin ^{2}\psi =16\sin ^{2}\psi }
sin
ψ
=
1
17
{\displaystyle \sin \psi ={\frac {1}{\sqrt {17}}}}
x
E
=
17
6
l
sin
ψ
=
17
6
l
{\displaystyle x_{E}={\frac {17}{6}}l\sin \psi ={\frac {\sqrt {17}}{6}}l}
Π
=
−
3
m
g
x
E
cos
φ
{\displaystyle \Pi =-3mgx_{E}\cos \varphi }
∂
Π
∂
φ
=
3
m
g
x
E
sin
φ
≈
3
m
g
x
E
φ
{\displaystyle {\frac {\partial \Pi }{\partial \varphi }}=3mgx_{E}\sin \varphi \approx 3mgx_{E}\varphi }
τ
=
2
π
6
l
g
17
=
2
π
6
17
4
l
g
{\displaystyle \tau =2\pi {\sqrt {\frac {6l}{g{\sqrt {17}}}}}=2\pi {\frac {\sqrt {6}}{\sqrt[{4}]{17}}}{\sqrt {\frac {l}{g}}}}
Найти период малых колебаний около устойчивого положения равновесия тяжёлого однородного стержня ОА длины I и массы т, который может свободно вращаться вокруг неподвижной горизонтальной оси О и к концу А которого прикреплена пружина АВ. Пружина закреплена в точке В, расположенной на расстоянии I над точкой О. Длина нерастянутой пружины I, коэффициент её жёсткости с.
l
+
Δ
l
sin
φ
=
l
cos
φ
2
{\displaystyle {\frac {l+\Delta l}{\sin \varphi }}={\frac {l}{\cos {\frac {\varphi }{2}}}}}
l
+
Δ
l
=
2
sin
φ
2
⋅
l
{\displaystyle l+\Delta l=2\sin {\frac {\varphi }{2}}\cdot l}
Δ
l
=
(
2
sin
φ
2
−
1
)
⋅
l
{\displaystyle \Delta l=\left(2\sin {\frac {\varphi }{2}}-1\right)\cdot l}
V
=
m
g
l
2
cos
φ
+
1
2
c
l
2
(
2
sin
φ
2
−
1
)
2
{\displaystyle V=mg{\frac {l}{2}}\cos \varphi +{\frac {1}{2}}cl^{2}\left(2\sin {\frac {\varphi }{2}}-1\right)^{2}}
∂
V
∂
φ
=
−
m
g
l
2
sin
φ
+
1
2
c
l
2
⋅
2
(
2
sin
φ
2
−
1
)
2
⋅
1
2
cos
φ
2
=
−
m
g
l
2
sin
φ
+
c
l
2
sin
φ
−
c
l
2
cos
φ
2
{\displaystyle {\frac {\partial V}{\partial \varphi }}=-mg{\frac {l}{2}}\sin \varphi +{\frac {1}{2}}cl^{2}\cdot 2\left(2\sin {\frac {\varphi }{2}}-1\right)2\cdot {\frac {1}{2}}\cos {\frac {\varphi }{2}}=-mg{\frac {l}{2}}\sin \varphi +cl^{2}\sin \varphi -cl^{2}\cos {\frac {\varphi }{2}}}
∂
V
∂
φ
=
0
{\displaystyle {\frac {\partial V}{\partial \varphi }}=0}
∂
V
∂
φ
=
−
m
g
l
2
2
sin
φ
2
cos
φ
2
+
c
l
(
2
sin
φ
2
−
1
)
cos
φ
2
=
0
{\displaystyle {\frac {\partial V}{\partial \varphi }}=-mg{\frac {l}{2}}2\sin {\frac {\varphi }{2}}\cos {\frac {\varphi }{2}}+cl\left(2\sin {\frac {\varphi }{2}}-1\right)\cos {\frac {\varphi }{2}}=0}
cos
φ
2
(
−
m
g
l
sin
φ
2
+
2
c
l
sin
φ
2
−
c
l
)
=
0
{\displaystyle \cos {\frac {\varphi }{2}}\left(-mgl\sin {\frac {\varphi }{2}}+2cl\sin {\frac {\varphi }{2}}-cl\right)=0}
cos
α
2
=
0
φ
2
=
π
2
φ
=
π
{\displaystyle \cos {\frac {\alpha }{2}}=0\quad {\frac {\varphi }{2}}={\frac {\pi }{2}}\quad \varphi =\pi }
∂
2
V
∂
φ
2
=
l
2
(
(
2
c
l
−
m
g
)
cos
φ
+
c
l
2
⋅
1
2
sin
φ
2
=
l
2
(
(
2
c
l
−
m
g
)
cos
φ
+
c
l
sin
φ
2
)
=
0
{\displaystyle {\frac {\partial ^{2}V}{\partial \varphi ^{2}}}={\frac {l}{2}}((2cl-mg)\cos \varphi +cl^{2}\cdot {\frac {1}{2}}\sin {\frac {\varphi }{2}}={\frac {l}{2}}\left((2cl-mg)\cos \varphi +cl\sin {\frac {\varphi }{2}}\right)=0}
∂
2
V
∂
φ
2
=
l
2
(
2
c
l
−
m
g
+
c
l
)
=
l
2
(
m
g
−
c
l
)
{\displaystyle {\frac {\partial ^{2}V}{\partial \varphi ^{2}}}={\frac {l}{2}}\left(2cl-mg+cl\right)={\frac {l}{2}}(mg-cl)}
m
g
>
c
l
{\displaystyle mg>cl}
T
=
1
2
m
l
2
3
ϕ
˙
2
{\displaystyle T={\frac {1}{2}}{\frac {ml^{2}}{3}}{\dot {\phi }}^{2}}
ω
2
=
l
2
(
m
g
−
c
l
)
1
3
m
l
2
=
3
(
m
g
−
c
l
)
2
m
l
{\displaystyle \omega ^{2}={\frac {{\frac {l}{2}}(mg-cl)}{{\frac {1}{3}}ml^{2}}}={\frac {3(mg-cl)}{2ml}}}
τ
=
2
π
2
m
l
3
(
m
g
−
c
l
)
{\displaystyle \tau =2\pi {\sqrt {\frac {2ml}{3(mg-cl)}}}}
Другой случай:
∂
2
V
∂
φ
2
=
l
2
(
(
2
c
l
−
m
g
)
(
1
−
2
c
2
l
2
(
2
c
l
−
m
g
)
2
)
+
c
2
l
2
2
c
l
−
m
g
)
=
l
2
(
2
c
l
−
m
g
−
c
2
l
2
2
c
l
−
m
g
)
=
1
2
(
2
c
l
−
m
g
−
c
l
)
(
2
c
l
−
m
g
+
c
l
)
2
c
l
−
m
g
=
{\displaystyle {\frac {\partial ^{2}V}{\partial \varphi ^{2}}}={\frac {l}{2}}\left((2cl-mg)\left(1-2{\frac {c^{2}l^{2}}{(2cl-mg)^{2}}}\right)+{\frac {c^{2}l^{2}}{2cl-mg}}\right)={\frac {l}{2}}(2cl-mg-{\frac {c^{2}l^{2}}{2cl-mg}})={\frac {1}{2}}{\frac {(2cl-mg-cl)(2cl-mg+cl)}{2cl-mg}}=}
=
1
2
(
c
l
−
m
g
)
(
3
c
l
−
m
q
)
2
c
l
−
m
g
{\displaystyle ={\frac {1}{2}}{\frac {(cl-mg)(3cl-mq)}{2cl-mg}}}
ω
2
=
l
2
(
c
l
−
m
g
)
(
3
c
l
−
m
q
)
2
c
l
−
m
g
m
l
2
3
=
3
(
c
l
−
m
g
)
(
3
c
l
−
m
g
)
2
m
l
(
2
c
l
−
m
g
)
{\displaystyle \omega ^{2}={\frac {{\frac {l}{2}}{\frac {(cl-mg)(3cl-mq)}{2cl-mg}}}{\frac {ml^{2}}{3}}}={\frac {3(cl-mg)(3cl-mg)}{2ml(2cl-mg)}}}
τ
=
2
π
2
m
l
(
2
c
l
−
m
g
)
3
(
c
l
−
m
g
)
(
3
c
l
−
m
g
)
{\displaystyle \tau =2\pi {\sqrt {\frac {2ml(2cl-mg)}{3(cl-mg)(3cl-mg)}}}}
m
g
<
c
l
{\displaystyle mg<cl}
Тяжёлая квадратная платформа ABCD массы М подвешена на четырёх упругих канатах, жёсткости с каждый, к неподвижной точке О, отстоящей в положении равновесия системы на расстоянии I по вертикали от центра Е платформы. Длина диагонали платформы а. Определить период вертикальных колебаний системы (точки А, В, С, D перемещаются только вертикально).
O
E
=
ℓ
{\displaystyle OE=\ell }
A
C
=
a
{\displaystyle AC=a}
l
0
=
l
2
+
a
2
4
{\displaystyle l_{0}={\sqrt {l^{2}+{\frac {a^{2}}{4}}}}}
V
=
−
M
g
y
+
4
ℓ
2
(
y
2
+
a
2
4
−
ℓ
~
)
2
=
−
M
g
y
+
2
C
(
y
2
+
a
2
4
−
2
ℓ
~
y
2
+
a
2
4
+
ℓ
~
2
)
{\displaystyle V=-Mgy+{\frac {4\ell }{2}}\left({\sqrt {y^{2}+{\frac {a^{2}}{4}}}}-{\tilde {\ell }}\right)^{2}=-Mgy+2C(y^{2}+{\frac {a^{2}}{4}}-2{\tilde {\ell }}{\sqrt {y^{2}+{\frac {a^{2}}{4}}}}+{\tilde {\ell }}^{2})}
V
0
′
=
−
M
g
+
4
c
y
−
4
c
ℓ
~
ℓ
ℓ
0
=
0
{\displaystyle V'_{0}=-Mg+4cy-{\frac {4c{\tilde {\ell }}\ell }{\ell _{0}}}=0}
ℓ
~
=
(
4
c
ℓ
−
M
g
)
l
0
4
c
ℓ
{\displaystyle {\tilde {\ell }}={\frac {(4c\ell -Mg)l_{0}}{4c\ell }}}
V
″
=
4
ℓ
−
4
c
ℓ
~
(
y
2
+
a
2
4
−
y
y
y
2
+
a
2
4
y
2
+
a
2
4
)
=
4
ℓ
−
4
c
ℓ
~
a
2
4
1
(
y
2
+
a
2
4
)
3
2
=
4
ℓ
−
c
ℓ
~
a
2
1
(
y
2
+
a
2
4
)
3
2
=
{\displaystyle V''=4\ell -4c{\tilde {\ell }}\left({\frac {{\sqrt {y^{2}+{\frac {a^{2}}{4}}}}-y{\frac {y}{\sqrt {y^{2}+{\frac {a^{2}}{4}}}}}}{y^{2}+{\frac {a^{2}}{4}}}}\right)=4\ell -4c{\tilde {\ell }}{\frac {a^{2}}{4}}{\frac {1}{\left(y^{2}+{\frac {a^{2}}{4}}\right)^{\frac {3}{2}}}}=4\ell -c{\tilde {\ell }}a^{2}{\frac {1}{\left(y^{2}+{\frac {a^{2}}{4}}\right)^{\frac {3}{2}}}}=}
V
″
(
0
)
=
4
ℓ
−
c
ℓ
~
a
2
1
ℓ
0
3
=
c
(
4
−
a
2
ℓ
0
2
(
1
−
M
g
4
c
ℓ
)
)
=
4
c
−
(
c
−
M
g
4
ℓ
)
a
2
l
0
2
=
16
c
l
l
0
2
−
4
a
2
c
l
+
M
g
a
2
4
l
l
0
2
=
4
c
l
(
4
l
2
+
a
2
)
−
4
a
2
c
l
+
M
g
a
2
4
l
l
0
2
=
16
c
l
3
+
M
g
a
2
4
l
l
0
2
=
16
c
l
3
+
M
g
a
2
l
(
4
l
2
+
a
2
)
{\displaystyle V''(0)=4\ell -c{\tilde {\ell }}a^{2}{\frac {1}{\ell _{0}^{3}}}=c\left(4-{\frac {a^{2}}{\ell _{0}^{2}}}\left(1-{\frac {Mg}{4c\ell }}\right)\right)=4c-\left(c-{\frac {Mg}{4\ell }}\right){\frac {a^{2}}{l_{0}^{2}}}={\frac {16cll_{0}^{2}-4a^{2}cl+Mga^{2}}{4ll_{0}^{2}}}={\frac {4cl\left(4l^{2}+a^{2}\right)-4a^{2}cl+Mga^{2}}{4ll_{0}^{2}}}={\frac {16cl^{3}+Mga^{2}}{4ll_{0}^{2}}}={\frac {16cl^{3}+Mga^{2}}{l\left(4l^{2}+a^{2}\right)}}}
T
=
1
2
M
y
˙
2
{\displaystyle T={\frac {1}{2}}M{\dot {y}}^{2}}
τ
=
2
π
M
l
(
4
l
2
+
a
2
)
16
c
l
3
+
M
g
a
2
{\displaystyle \tau =2\pi {\sqrt {\frac {Ml\left(4l^{2}+a^{2}\right)}{16cl^{3}+Mga^{2}}}}}
Точки подвеса двух одинаковых математических маятников с массой т и длиной I расположены на одной горизонтальной прямой. Точки этих маятников, отстоящие от точек подвеса на расстоянии h (0 < h < /), соединены между собой пружиной жёсткости с; пружина не натянута, когда маятники занимают вертикальное положение. Найти малые колебания системы в вертикальной плоскости, проходящей через точки подвеса маятников, если в начальный момент один из них отклонён от вертикали на угол а, а начальные скорости равны нулю.
V
=
m
g
y
A
1
+
m
g
y
A
2
+
c
2
(
L
−
L
0
)
2
{\displaystyle V=mgy_{A_{1}}+mgy_{A_{2}}+{\frac {c}{2}}\left(L-L_{0}\right)^{2}}
V
=
−
m
g
l
(
cos
φ
1
+
cos
φ
2
)
+
c
2
(
(
l
0
−
h
sin
φ
1
+
h
sin
φ
2
)
2
+
(
h
cos
φ
1
−
h
cos
φ
2
)
2
−
L
0
)
2
{\displaystyle V=-mgl\left(\cos \varphi _{1}+\cos \varphi _{2}\right)+{\frac {c}{2}}\left({\sqrt {(l_{0}-h\sin \varphi _{1}+h\sin \varphi _{2})^{2}+\left(h\cos \varphi _{1}-h\cos \varphi _{2}\right)^{2}}}-L_{0}\right)^{2}}
(
φ
1
,
φ
2
)
=
(
0
,
0
)
{\displaystyle (\varphi _{1},\varphi _{2})=(0,0)}
(
L
0
−
h
φ
1
+
h
φ
2
)
2
+
h
2
(
−
φ
1
2
2
+
φ
2
2
2
)
2
−
L
0
≈
L
0
−
h
φ
1
+
h
φ
1
−
L
0
=
−
h
φ
1
+
h
φ
2
{\displaystyle {\sqrt {(L_{0}-h\varphi _{1}+h\varphi _{2})^{2}+h^{2}\left(-{\frac {\varphi _{1}^{2}}{2}}+{\frac {\varphi _{2}^{2}}{2}}\right)^{2}}}-L_{0}\approx L_{0}-h\varphi _{1}+h\varphi _{1}-L_{0}=-h\varphi _{1}+h\varphi _{2}}
V
=
−
m
g
l
(
1
−
φ
1
2
2
+
1
−
φ
2
2
2
)
+
c
2
h
2
(
φ
2
−
φ
1
)
2
+
o
(
φ
1
2
+
φ
2
2
)
{\displaystyle V=-mgl\left(1-{\frac {\varphi _{1}^{2}}{2}}+1-{\frac {\varphi _{2}^{2}}{2}}\right)+{\frac {c}{2}}h^{2}\left(\varphi _{2}-\varphi _{1}\right)^{2}+o\left(\varphi _{1}^{2}+\varphi _{2}^{2}\right)}
V
(
φ
1
,
φ
2
)
=
V
(
0
,
0
)
+
∂
V
∂
φ
1
(
0
,
0
)
φ
1
+
∂
V
∂
φ
2
(
0
,
0
)
φ
2
+
1
2
(
∂
2
V
∂
φ
1
2
(
0
,
0
)
φ
1
2
+
2
∂
V
∂
φ
1
∂
φ
2
(
0
,
0
)
φ
1
φ
2
+
∂
2
V
∂
φ
2
2
(
0
,
0
)
φ
2
2
)
{\displaystyle V\left(\varphi _{1},\varphi _{2}\right)=V(0,0)+{\frac {\partial V}{\partial \varphi _{1}}}(0,0)\varphi _{1}+{\frac {\partial V}{\partial \varphi _{2}}}(0,0)\varphi _{2}+{\frac {1}{2}}\left({\frac {\partial ^{2}V}{\partial \varphi _{1}^{2}}}(0,0)\varphi _{1}^{2}+2{\frac {\partial V}{\partial \varphi _{1}\partial \varphi _{2}}}\left(0,0\right)\varphi _{1}\varphi _{2}+{\frac {\partial ^{2}V}{\partial \varphi _{2}^{2}}}(0,0)\varphi _{2}^{2}\right)}
B
(
φ
1
,
φ
2
)
=
(
m
g
l
+
c
h
2
−
c
h
2
−
c
h
2
m
g
l
+
c
h
2
)
{\displaystyle B\left(\varphi _{1},\varphi _{2}\right)=\left({\begin{array}{lc}mgl+ch^{2}&-ch^{2}\\-ch^{2}&mgl+ch^{2}\end{array}}\right)}
T
=
T
1
+
T
2
=
m
l
2
2
(
φ
˙
1
2
+
φ
˙
2
2
)
{\displaystyle T=T_{1}+T_{2}={\frac {ml^{2}}{2}}\left({\dot {\varphi }}_{1}^{2}+{\dot {\varphi }}_{2}^{2}\right)}
A
=
(
m
l
2
0
0
m
l
2
)
{\displaystyle A=\left({\begin{array}{cc}ml^{2}&0\\0&ml^{2}\end{array}}\right)}
det
(
B
−
A
ω
2
)
=
|
m
g
l
+
c
h
2
−
m
l
2
w
2
−
c
h
2
−
c
h
2
m
g
l
+
c
h
2
−
m
l
2
ω
2
|
=
m
g
l
+
c
h
2
−
m
l
2
ω
2
−
(
c
h
2
)
2
=
(
m
g
l
+
c
h
2
−
m
l
2
w
2
)
2
−
(
c
h
2
)
2
=
0
{\displaystyle \det \left(B-A\omega ^{2}\right)=\left|{\begin{array}{lc}mgl+ch^{2}-ml^{2}w^{2}&-ch^{2}\\-ch^{2}&mgl+ch^{2}-ml^{2}\omega ^{2}\end{array}}\right|=mgl+ch^{2}-ml^{2}\omega ^{2}-\left(ch^{2}\right)^{2}=\left(mgl+ch^{2}-ml^{2}w^{2}\right)^{2}-\left(ch^{2}\right)^{2}=0}
w
1
2
=
m
g
l
+
2
c
h
2
m
l
2
=
g
e
+
2
c
h
2
m
l
2
{\displaystyle w_{1}^{2}={\frac {mgl+2ch^{2}}{ml^{2}}}={\frac {g}{e}}+{\frac {2ch^{2}}{ml^{2}}}}
w
2
2
=
g
l
{\displaystyle w_{2}^{2}={\frac {g}{l}}}
B
(
φ
1
,
φ
2
)
=
(
−
c
h
2
−
c
h
2
−
c
h
2
−
c
h
2
)
(
u
1
u
2
)
{\displaystyle B\left(\varphi _{1},\varphi _{2}\right)=\left({\begin{array}{lc}-ch^{2}&-ch^{2}\\-ch^{2}&-ch^{2}\end{array}}\right)\left({\begin{array}{l}u_{1}\\u_{2}\end{array}}\right)}
u
1
→
=
(
1
−
1
)
{\displaystyle {\vec {u_{1}}}=\left({\begin{array}{l}1\\-1\end{array}}\right)}
u
2
→
=
(
1
1
)
{\displaystyle {\vec {u_{2}}}=\left({\begin{array}{l}1\\1\end{array}}\right)}
(
φ
1
φ
2
)
=
(
1
−
1
)
C
1
sin
g
l
+
2
c
h
2
m
l
2
t
+
D
1
cos
g
l
+
2
c
h
2
m
l
2
t
+
(
1
1
)
(
C
2
sin
g
l
t
+
D
2
cos
g
l
t
)
{\displaystyle \left({\begin{array}{l}\varphi _{1}\\\varphi _{2}\end{array}}\right)=\left({\begin{array}{c}1\\-1\end{array}}\right)C_{1}\sin {\sqrt {{\frac {g}{l}}+{\frac {2ch^{2}}{ml^{2}}}}}t+D_{1}\cos {\sqrt {{\frac {g}{l}}+{\frac {2ch^{2}}{ml^{2}}}}}t+\left({\begin{array}{l}1\\1\end{array}}\right)(C_{2}\sin {\sqrt {\frac {g}{l}}}t+D_{2}\cos {\sqrt {\frac {g}{l}}}t)}
φ
1
(
0
)
=
φ
0
{\displaystyle \varphi _{1}(0)=\varphi _{0}}
φ
2
(
0
)
=
φ
0
{\displaystyle \varphi _{2}(0)=\varphi _{0}}
φ
˙
1
(
0
)
=
0
{\displaystyle {\dot {\varphi }}_{1}(0)=0}
φ
˙
2
(
0
)
=
0
{\displaystyle {\dot {\varphi }}_{2}(0)=0}
(
φ
0
−
φ
0
)
=
(
1
−
1
)
D
1
+
(
1
1
)
D
2
{\displaystyle \left({\begin{array}{c}\varphi _{0}\\-\varphi _{0}\end{array}}\right)=\left({\begin{array}{l}1\\-1\end{array}}\right)D_{1}+\left({\begin{array}{l}1\\1\end{array}}\right)D_{2}}
{
D
1
+
D
2
=
φ
0
−
D
1
+
D
2
=
−
φ
0
{\displaystyle \left\{{\begin{array}{l}D_{1}+D_{2}=\varphi _{0}\\-D_{1}+D_{2}=-\varphi _{0}\end{array}}\right.}
2
D
2
=
0
{\displaystyle 2D_{2}=0}
D
2
=
0
{\displaystyle D_{2}=0}
D
1
=
φ
0
{\displaystyle D_{1}=\varphi _{0}}
(
0
0
)
=
(
1
−
1
)
c
1
w
1
+
(
1
1
)
c
2
w
2
{\displaystyle \left({\begin{array}{l}0\\0\end{array}}\right)=\left({\begin{array}{c}1\\-1\end{array}}\right)c_{1}w_{1}+\left({\begin{array}{l}1\\1\end{array}}\right)c_{2}w_{2}}
{
c
1
w
1
+
c
2
w
2
=
0
−
c
w
1
+
c
2
w
2
=
0
{\displaystyle \left\{{\begin{array}{c}{c_{1}w_{1}+c_{2}w_{2}=0}\\{-cw_{1}+c_{2}w_{2}=0}\end{array}}\right.}
C
1
=
C
2
=
0
{\displaystyle C_{1}=C_{2}=0}
Два груза массы т каждый, соединённые между собой пружиной жёсткости с, а с неподвижными стенками пружинами жёсткости 2с каждая, могут скользить по гладкой горизонтальной направляющей. К каждому грузу подвешен математический маятник массы
m
2
{\displaystyle {\frac {m}{2}}}
l. Найти малые колебания системы. При вычислениях положить
c
=
m
g
2
l
{\displaystyle c={\frac {mg}{2l}}}
x
1
−
l
sin
φ
1
≈
x
1
−
l
φ
1
{\displaystyle x_{1}-l\sin \varphi _{1}\approx x_{1}-l\varphi _{1}}
v
=
x
˙
1
−
l
φ
˙
1
{\displaystyle v={\dot {x}}_{1}-l{\dot {\varphi }}_{1}}
K
=
m
x
˙
1
2
2
+
m
4
(
x
˙
1
−
l
φ
˙
1
)
2
+
m
x
˙
2
2
2
+
m
4
(
x
˙
2
−
ℓ
φ
˙
2
)
2
{\displaystyle K={\frac {m{\dot {x}}_{1}^{2}}{2}}+{\frac {m}{4}}\left({\dot {x}}_{1}-l{\dot {\varphi }}_{1}\right)^{2}+{\frac {m{\dot {x}}_{2}^{2}}{2}}+{\frac {m}{4}}\left({\dot {x}}_{2}-\ell {\dot {\varphi }}_{2}\right)^{2}}
Π
=
2
c
2
x
1
2
+
c
2
(
x
2
−
x
1
)
2
+
2
c
2
x
2
2
+
m
4
g
ℓ
(
φ
1
2
+
φ
2
2
)
=
3
c
2
(
x
1
2
+
x
2
2
)
−
c
x
1
x
2
+
m
4
g
l
(
φ
1
2
+
φ
2
2
)
{\displaystyle \Pi ={\frac {2c}{2}}x_{1}^{2}+{\frac {c}{2}}\left(x_{2}-x_{1}\right)^{2}+{\frac {2c}{2}}x_{2}^{2}+{\frac {m}{4}}g\ell \left(\varphi _{1}^{2}+\varphi _{2}^{2}\right)={\frac {3c}{2}}\left(x_{1}^{2}+x_{2}^{2}\right)-cx_{1}x_{2}+{\frac {m}{4}}gl\left(\varphi _{1}^{2}+\varphi _{2}^{2}\right)}
L
=
K
−
Π
{\displaystyle L=K-\Pi }
∂
L
∂
x
˙
1
=
3
m
x
˙
1
2
−
m
2
l
φ
˙
1
∂
L
∂
x
1
=
−
3
c
x
1
+
c
x
2
∂
L
∂
x
˙
2
=
3
m
x
˙
2
2
−
m
2
l
φ
˙
2
{\displaystyle {\begin{array}{l}{{\frac {\partial L}{\partial {\dot {x}}_{1}}}={\frac {3m{\dot {x}}_{1}}{2}}-{\frac {m}{2}}l{\dot {\varphi }}_{1}}\\{{\frac {\partial L}{\partial x_{1}}}=-3cx_{1}+cx_{2}}\\{{\frac {\partial L}{\partial {\dot {x}}_{2}}}={\frac {3m{\dot {x}}_{2}}{2}}-{\frac {m}{2}}l{\dot {\varphi }}_{2}}\end{array}}}
∂
L
∂
x
2
=
−
3
c
x
2
+
c
x
1
{\displaystyle {\frac {\partial L}{\partial {x_{2}}}}=-3cx_{2}+cx_{1}}
∂
L
∂
φ
˙
1
=
m
l
2
(
ℓ
φ
˙
1
−
x
˙
1
)
{\displaystyle {\frac {\partial L}{\partial {\dot {\varphi }}_{1}}}={\frac {ml}{2}}\left(\ell {\dot {\varphi }}_{1}-{\dot {x}}_{1}\right)}
∂
L
∂
φ
1
=
−
m
2
g
l
φ
1
{\displaystyle {\frac {\partial L}{\partial \varphi _{1}}}=-{\frac {m}{2}}gl\varphi _{1}}
∂
L
∂
φ
˙
2
=
m
ℓ
2
(
ℓ
φ
˙
2
−
x
˙
2
)
∂
L
∂
φ
2
=
−
m
2
g
l
φ
2
{\displaystyle {\begin{array}{l}{{\frac {\partial L}{\partial {\dot {\varphi }}_{2}}}={\frac {m\ell }{2}}\left(\ell {\dot {\varphi }}_{2}-{\dot {x}}_{2}\right)}\\{{\frac {\partial L}{\partial \varphi _{2}}}=-{\frac {m}{2}}gl\varphi _{2}}\end{array}}}
d
d
t
(
∂
L
∂
q
˙
)
−
∂
L
∂
q
=
0
{\displaystyle {\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {q}}}}\right)-{\frac {\partial L}{\partial q}}=0}
c
=
m
g
2
l
{\displaystyle c={\frac {mg}{2l}}}
3
x
¨
1
−
l
φ
¨
1
+
3
g
l
x
1
−
g
l
x
2
=
0
{\displaystyle 3{\ddot {x}}_{1}-l{\ddot {\varphi }}_{1}+{\frac {3g}{l}}x_{1}-{\frac {g}{l}}x_{2}=0}
3
x
2
−
l
φ
¨
2
+
3
g
l
x
2
−
g
l
x
1
=
0
{\displaystyle 3x_{2}-l{\ddot {\varphi }}_{2}+{\frac {3g}{l}}x_{2}-{\frac {g}{l}}x_{1}=0}
ℓ
φ
¨
1
−
x
¨
1
+
g
ℓ
⋅
l
φ
1
=
0
{\displaystyle \ell {\ddot {\varphi }}_{1}-{\ddot {x}}_{1}+{\frac {g}{\ell }}\cdot l\varphi _{1}=0}
ℓ
φ
¨
2
−
x
¨
2
+
g
ℓ
⋅
l
φ
2
=
0
{\displaystyle \ell {\ddot {\varphi }}_{2}-{\ddot {x}}_{2}+{\frac {g}{\ell }}\cdot l\varphi _{2}=0}
A
q
¨
+
C
q
=
0
{\displaystyle A{\ddot {q}}+Cq=0}
q
=
(
x
1
l
φ
1
x
2
l
φ
2
)
{\displaystyle q=\left({\begin{array}{l}{x_{1}}\\{l\varphi _{1}}\\{x_{2}}\\{l\varphi _{2}}\end{array}}\right)}
A
=
(
3
−
1
0
0
−
1
1
0
0
0
0
3
−
1
0
0
−
1
1
)
{\displaystyle A=\left({\begin{array}{cccc}{3}&{-1}&{0}&{0}\\{-1}&{1}&{0}&{0}\\{0}&{0}&{3}&{-1}\\{0}&{0}&{-1}&{1}\end{array}}\right)}
C
=
g
l
(
3
0
−
1
0
0
1
0
0
−
1
0
3
0
0
0
0
1
)
{\displaystyle C={\frac {g}{l}}\left({\begin{array}{cccc}{3}&{0}&{-1}&{0}\\{0}&{1}&{0}&{0}\\{-1}&{0}&{3}&{0}\\{0}&{0}&{0}&{1}\end{array}}\right)}
det
(
3
−
3
λ
λ
−
1
0
λ
1
−
λ
0
0
−
1
0
3
−
3
λ
λ
0
0
λ
1
−
λ
)
=
0
{\displaystyle \det \left({\begin{array}{cccc}{3-3\lambda }&{\lambda }&{-1}&{0}\\{\lambda }&{1-\lambda }&{0}&{0}\\{-1}&{0}&{3-3\lambda }&{\lambda }\\{0}&{0}&{\lambda }&{1-\lambda }\end{array}}\right)=0}
λ
1
=
g
2
l
,
λ
2
=
2
g
ℓ
{\displaystyle \lambda _{1}={\frac {g}{2l}},\quad \lambda _{2}={\frac {2g}{\ell }}}
λ
3
,
4
=
7
±
17
4
g
l
{\displaystyle \lambda _{3,4}={\frac {7\pm {\sqrt {17}}}{4}}{\frac {g}{l}}}
u
1
=
(
1
1
1
1
)
{\displaystyle u_{1}=\left({\begin{array}{l}{1}\\{1}\\{1}\\{1}\end{array}}\right)}
u
2
=
(
1
−
2
1
−
2
)
{\displaystyle u_{2}=\left({\begin{array}{c}{1}\\{-2}\\{1}\\{-2}\end{array}}\right)}
u
3
=
(
2
17
+
1
−
2
−
17
−
1
)
{\displaystyle u_{3}=\left({\begin{array}{c}{2}\\{{\sqrt {17}}+1}\\{-2}\\{-{\sqrt {17}}-1}\end{array}}\right)}
u
4
=
(
2
1
−
17
−
2
17
−
1
)
{\displaystyle u_{4}=\left({\begin{array}{c}{2}\\{1-{\sqrt {17}}}\\{-2}\\{{\sqrt {17}}-1}\end{array}}\right)}
(
x
1
l
φ
1
x
2
ℓ
φ
2
)
=
(
1
1
1
1
)
(
C
1
cos
g
2
l
t
+
C
2
sin
g
2
l
t
)
+
(
1
−
2
1
−
2
)
(
C
3
cos
2
g
ℓ
t
+
C
4
sin
2
g
ℓ
t
)
+
(
2
1
+
17
−
2
−
17
−
1
)
(
C
5
cos
7
−
17
4
g
ℓ
t
+
C
6
sin
7
17
4
g
ℓ
t
)
+
(
2
1
−
17
−
2
17
−
1
)
(
C
7
cos
7
+
17
4
q
ℓ
t
+
C
8
sin
7
+
17
4
q
ℓ
t
)
{\displaystyle {\begin{aligned}\left({\begin{array}{c}{x_{1}}\\{l\varphi _{1}}\\{x_{2}}\\{\ell \varphi _{2}}\end{array}}\right)=&\left({\begin{array}{r}{1}\\{1}\\{1}\\{1}\end{array}}\right)\left(C_{1}\cos {\sqrt {\frac {g}{2l}}}t+C_{2}\sin {\sqrt {\frac {g}{2l}}}t\right)+\left({\begin{array}{r}{1}\\{-2}\\{1}\\{-2}\end{array}}\right)\left(C_{3}\cos {\sqrt {\frac {2g}{\ell }}}t+C_{4}\sin {\sqrt {\frac {2g}{\ell }}}t\right)+\\&\left({\begin{array}{c}{2}\\{1+{\sqrt {17}}}\\{-2}\\{-{\sqrt {17}}-1}\end{array}}\right)\left(C_{5}\cos {\sqrt {{\frac {7-{\sqrt {17}}}{4}}{\frac {g}{\ell }}}}t+C_{6}\sin {\sqrt {{\frac {7{\sqrt {17}}}{4}}{\frac {g}{\ell }}}}t\right)+\\&\left({\begin{array}{c}{2}\\{1-{\sqrt {17}}}\\-2\\{{\sqrt {17}}-1}\end{array}}\right)\left(C_{7}\cos {\sqrt {{\frac {7+{\sqrt {17}}}{4}}{\frac {q}{\ell }}}}t+C_{8}\sin {\sqrt {{\frac {7+{\sqrt {17}}}{4}}{\frac {q}{\ell }}}}t\right)\end{aligned}}}
Двойной физический маятник состоит из однородного прямолинейного стержня
O
1
O
2
{\displaystyle O_{1}O_{2}}
а и массы т 1 , вращающегося вокруг неподвижной горизонтальной оси О1 , и из однородного прямолинейного стержня АВ массы т 2 , шарнирно соединённого в своём центре масс с концом O2 первого стержня. Определить движение системы, если в начальный момент стержень О1 O2 отклонён на угол φ0 от вертикали, а стержень АВ занимает вертикальное положение и имеет начальную угловую скорость ω 0 .
K
1
=
I
φ
˙
2
2
=
m
1
⋅
(
2
a
)
2
3
⋅
φ
˙
2
2
=
2
3
m
a
2
φ
˙
2
{\displaystyle K_{1}={\frac {I{\dot {\varphi }}^{2}}{2}}={\frac {m_{1}\cdot (2a)^{2}}{3}}\cdot {\frac {{\dot {\varphi }}^{2}}{2}}={\frac {2}{3}}ma^{2}{\dot {\varphi }}^{2}}
K
2
=
m
2
2
(
2
a
φ
˙
)
2
+
1
12
m
2
(
2
l
)
2
⋅
ψ
˙
2
2
{\displaystyle K_{2}={\frac {m_{2}}{2}}(2a{\dot {\varphi }})^{2}+{\frac {1}{12}}m_{2}(2l)^{2}\cdot {\frac {{\dot {\psi }}^{2}}{2}}}
Π
=
m
1
g
a
(
1
−
cos
φ
)
+
m
2
g
⋅
2
a
(
1
−
cos
φ
)
≈
(
m
1
+
2
m
2
)
g
a
⋅
φ
2
2
{\displaystyle \Pi =m_{1}ga(1-\cos \varphi )+m_{2}g\cdot 2a(1-\cos \varphi )\approx \left(m_{1}+2m_{2}\right)ga\cdot {\frac {\varphi ^{2}}{2}}}
L
=
(
2
3
m
1
a
2
+
2
m
2
a
2
)
φ
˙
2
+
1
6
m
2
ℓ
2
ψ
˙
2
−
(
m
1
+
2
m
2
)
g
a
⋅
φ
2
2
{\displaystyle L=\left({\frac {2}{3}}m_{1}a^{2}+2m_{2}a^{2}\right){\dot {\varphi }}^{2}+{\frac {1}{6}}m_{2}\ell ^{2}{\dot {\psi }}^{2}-\left(m_{1}+2m_{2}\right)ga\cdot {\frac {\varphi ^{2}}{2}}}
∂
L
∂
φ
˙
=
(
4
3
m
1
+
4
m
2
)
a
2
φ
˙
{\displaystyle {\frac {\partial L}{\partial {\dot {\varphi }}}}=\left({\frac {4}{3}}m_{1}+4m_{2}\right)a^{2}{\dot {\varphi }}}
∂
L
∂
φ
=
−
(
m
1
+
2
m
2
)
g
a
φ
{\displaystyle {\frac {\partial L}{\partial \varphi }}=-\left(m_{1}+2m_{2}\right)ga\varphi }
∂
L
∂
ψ
˙
=
1
3
m
2
l
2
ψ
˙
{\displaystyle {\frac {\partial L}{\partial {\dot {\psi }}}}={\frac {1}{3}}m_{2}l^{2}{\dot {\psi }}}
ψ
¨
=
0
{\displaystyle {\ddot {\psi }}=0}
ψ
(
t
)
=
ψ
0
+
ω
0
t
=
ω
0
t
{\displaystyle \psi (t)=\psi _{0}+\omega _{0}t=\omega _{0}t}
(
4
3
m
1
+
4
m
2
)
a
2
φ
¨
+
(
m
1
+
2
m
2
)
g
a
φ
=
0
{\displaystyle \left({\frac {4}{3}}m_{1}+4m_{2}\right)a^{2}{\ddot {\varphi }}+\left(m_{1}+2m_{2}\right)ga\varphi =0}
φ
¨
+
m
1
+
2
m
2
m
1
+
3
m
2
⋅
3
g
4
a
⋅
φ
=
0
{\displaystyle {\ddot {\varphi }}+{\frac {m_{1}+2m_{2}}{m_{1}+3m_{2}}}\cdot {\frac {3g}{4a}}\cdot \varphi =0}
φ
(
t
)
=
A
cos
m
1
+
2
m
2
m
1
+
3
m
2
⋅
3
g
4
a
t
+
B
sin
m
1
+
2
m
2
m
1
+
3
m
2
3
g
4
a
t
{\displaystyle \varphi (t)=A\cos {\sqrt {{\frac {m_{1}+2m_{2}}{m_{1}+3m_{2}}}\cdot {\frac {3g}{4a}}}}t+B\sin {\sqrt {{\frac {m_{1}+2m_{2}}{m_{1}+3m_{2}}}{\frac {3g}{4a}}}}t}
При
φ
(
0
)
=
φ
0
,
φ
˙
(
0
)
=
0
{\displaystyle \varphi (0)=\varphi _{0},\quad {\dot {\varphi }}(0)=0}
φ
(
t
)
=
φ
0
cos
m
1
+
2
m
2
m
1
+
3
m
2
⋅
3
g
4
a
t
{\displaystyle \varphi (t)=\varphi _{0}\cos {\sqrt {{\frac {m_{1}+2m_{2}}{m_{1}+3m_{2}}}\cdot {\frac {3g}{4a}}}}t}
К бруску массы М, который может двигаться по гладкой горизонтальной направляющей, подвешен двойной маятник, причём
m
1
=
m
2
=
M
2
,
l
1
=
l
2
=
l
{\displaystyle m_{1}=m_{2}={\frac {M}{2}},\quad l_{1}=l_{2}=l}
Найти малые колебания системы в окрестности стационарного движения, если при 𝑡 = 0
x
˙
=
v
0
,
φ
1
=
φ
2
=
0
{\displaystyle {\dot {x}}=v_{0},\quad \varphi _{1}=\varphi _{2}=0}
v
1
=
x
˙
+
l
φ
˙
1
{\displaystyle v_{1}={\dot {x}}+l{\dot {\varphi }}_{1}}
v
2
=
x
˙
+
ℓ
φ
˙
1
+
l
φ
˙
2
{\displaystyle v_{2}={\dot {x}}+\ell {\dot {\varphi }}_{1}+l{\dot {\varphi }}_{2}}
h
1
=
ℓ
(
1
−
cos
φ
1
)
{\displaystyle h_{1}=\ell \left(1-\cos \varphi _{1}\right)}
h
2
=
ℓ
(
1
−
cos
φ
1
)
+
ℓ
(
1
−
cos
φ
2
)
{\displaystyle h_{2}=\ell \left(1-\cos \varphi _{1}\right)+\ell \left(1-\cos \varphi _{2}\right)}
L
=
K
−
Π
=
M
2
x
˙
2
+
M
4
(
x
˙
+
φ
˙
1
)
2
+
M
4
(
x
˙
+
ℓ
φ
˙
1
+
ℓ
φ
˙
2
)
2
−
M
4
g
ℓ
φ
1
2
−
M
4
g
ℓ
(
φ
1
2
+
φ
2
2
)
{\displaystyle L=K-\Pi ={\frac {M}{2}}{\dot {x}}^{2}+{\frac {M}{4}}\left({\dot {x}}+{\dot {\varphi }}_{1}\right)^{2}+{\frac {M}{4}}\left({\dot {x}}+\ell {\dot {\varphi }}_{1}+\ell {\dot {\varphi }}_{2}\right)^{2}-{\frac {M}{4}}g\ell \varphi _{1}^{2}-{\frac {M}{4}}g\ell \left(\varphi _{1}^{2}+\varphi _{2}^{2}\right)}
∂
L
∂
x
˙
=
2
M
x
˙
+
M
ℓ
φ
1
˙
+
M
2
ℓ
φ
2
˙
{\displaystyle {\frac {\partial L}{\partial {\dot {x}}}}=2M{\dot {x}}+M\ell {\dot {\varphi _{1}}}+{\frac {M}{2}}\ell {\dot {\varphi _{2}}}}
∂
L
∂
φ
˙
1
=
M
ℓ
2
(
x
˙
+
ℓ
φ
˙
1
)
+
M
ℓ
2
(
x
˙
+
ℓ
φ
˙
1
+
ℓ
φ
˙
2
)
{\displaystyle {\frac {\partial L}{\partial {\dot {\varphi }}_{1}}}={\frac {M\ell }{2}}\left({\dot {x}}+\ell {\dot {\varphi }}_{1}\right)+{\frac {M\ell }{2}}\left({\dot {x}}+\ell {\dot {\varphi }}_{1}+\ell {\dot {\varphi }}_{2}\right)}
∂
L
∂
φ
1
=
−
M
g
ℓ
φ
1
{\displaystyle {\frac {\partial L}{\partial \varphi _{1}}}=-Mg\ell \varphi _{1}}
∂
L
∂
φ
˙
2
=
M
ℓ
2
(
x
˙
+
ℓ
φ
˙
1
+
ℓ
φ
˙
2
)
{\displaystyle {\frac {\partial L}{\partial {\dot {\varphi }}_{2}}}={\frac {M\ell }{2}}\left({\dot {x}}+\ell {\dot {\varphi }}_{1}+\ell {\dot {\varphi }}_{2}\right)}
∂
L
∂
φ
2
=
−
1
2
M
g
ℓ
φ
2
{\displaystyle {\frac {\partial L}{\partial \varphi _{2}}}=-{\frac {1}{2}}Mg\ell \varphi _{2}}
2
x
¨
+
ℓ
φ
¨
1
+
1
2
ℓ
φ
¨
2
=
0
{\displaystyle 2{\ddot {x}}+\ell {\ddot {\varphi }}_{1}+{\frac {1}{2}}\ell {\ddot {\varphi }}_{2}=0}
x
¨
+
ℓ
φ
¨
1
+
1
2
ℓ
φ
¨
2
+
g
ℓ
⋅
ℓ
φ
1
=
0
{\displaystyle {\ddot {x}}+\ell {\ddot {\varphi }}_{1}+{\frac {1}{2}}\ell {\ddot {\varphi }}_{2}+{\frac {g}{\ell }}\cdot \ell {\varphi }_{1}=0}
x
¨
+
ℓ
φ
¨
1
+
ℓ
φ
¨
2
+
g
ℓ
⋅
ℓ
φ
2
=
0
{\displaystyle {\ddot {x}}+\ell {\ddot {\varphi }}_{1}+\ell {\ddot {\varphi }}_{2}+{\frac {g}{\ell }}\cdot \ell {\varphi }_{2}=0}
A
q
¨
+
C
q
=
0
{\displaystyle A{\ddot {q}}+Cq=0}
q
=
(
x
ℓ
φ
1
ℓ
φ
2
)
{\displaystyle q=\left({\begin{array}{c}{x}\\{\ell {\varphi }_{1}}\\{\ell {\varphi }_{2}}\end{array}}\right)}
A
=
(
2
1
1
/
2
1
1
1
/
2
1
1
1
)
{\displaystyle A=\left({\begin{array}{lll}{2}&{1}&{1/2}\\{1}&{1}&{1/2}\\{1}&{1}&{1}\end{array}}\right)}
C
=
g
ℓ
(
0
0
0
0
1
0
0
0
1
)
{\displaystyle C={\frac {g}{\ell }}\left({\begin{array}{lll}{0}&{0}&{0}\\{0}&{1}&{0}\\{0}&{0}&{1}\end{array}}\right)}
det
(
C
−
λ
A
)
=
|
−
2
λ
−
λ
−
λ
/
2
−
λ
g
ℓ
−
λ
−
λ
/
2
−
λ
−
λ
g
ℓ
−
λ
|
=
λ
(
−
2
g
2
ℓ
2
+
5
2
g
ℓ
λ
−
1
2
λ
2
)
=
0
{\displaystyle \operatorname {det} (C-\lambda A)=\left|{\begin{array}{ccc}{-2\lambda }&{-\lambda }&{-\lambda /2}\\{-\lambda }&{{\frac {g}{\ell }}-\lambda }&{-\lambda /2}\\{-\lambda }&{-\lambda }&{{\frac {g}{\ell }}-\lambda }\end{array}}\right|=\lambda \left(-2{\frac {g^{2}}{\ell ^{2}}}+{\frac {5}{2}}{\frac {g}{\ell }}\lambda -{\frac {1}{2}}\lambda ^{2}\right)=0}
λ
1
=
0
,
λ
2
=
g
ℓ
{\displaystyle \lambda _{1}=0,\quad \lambda _{2}={\frac {g}{\ell }}}
λ
3
=
4
g
ℓ
{\displaystyle \lambda _{3}={\frac {4g}{\ell }}}
u
1
=
(
1
0
0
)
{\displaystyle u_{1}=\left({\begin{array}{l}{1}\\{0}\\{0}\end{array}}\right)}
u
2
=
(
1
−
3
2
)
{\displaystyle u_{2}=\left({\begin{array}{c}{1}\\{-3}\\{2}\end{array}}\right)}
u
3
=
(
1
−
4
4
)
{\displaystyle u_{3}=\left({\begin{array}{c}{1}\\{-4}\\{4}\end{array}}\right)}
(
x
ℓ
φ
1
ℓ
φ
2
)
=
(
1
0
0
)
(
C
1
t
+
C
2
)
+
(
1
−
3
2
)
(
C
3
cos
g
ℓ
t
+
C
4
sin
g
ℓ
t
)
+
(
1
−
4
4
)
(
C
5
cos
2
g
ℓ
t
+
C
6
sin
2
g
ℓ
t
)
$
{\displaystyle \left({\begin{array}{c}{x}\\{\ell \varphi _{1}}\\{\ell \varphi _{2}}\end{array}}\right)=\left({\begin{array}{l}{1}\\{0}\\{0}\end{array}}\right)\left(C_{1}t+C_{2}\right)+\left({\begin{array}{c}{1}\\{-3}\\{2}\end{array}}\right)\left(C_{3}\cos {\sqrt {\frac {g}{\ell }}}t+C_{4}\sin {\sqrt {\frac {g}{\ell }}}t\right)+\left({\begin{array}{c}{1}\\{-4}\\{4}\end{array}}\right)\left(C_{5}\cos 2{\sqrt {\frac {g}{\ell }}}t+C_{6}\sin 2{\sqrt {\frac {g}{\ell }}}t\right)\$}
$
x
˙
=
v
0
,
y
1
=
φ
2
=
0
$
{\displaystyle \${\dot {x}}=v_{0},\quad y_{1}=\varphi _{2}=0\quad \$}
$
t
=
0
{\displaystyle \$\quad t=0}
(
x
ℓ
φ
1
ℓ
φ
2
)
=
(
1
0
0
)
(
v
0
t
+
x
0
)
+
(
1
−
3
2
)
C
4
sin
g
ℓ
t
+
(
1
−
4
4
)
C
6
sin
2
g
ℓ
t
{\displaystyle \left({\begin{array}{c}{x}\\{\ell \varphi _{1}}\\{\ell \varphi _{2}}\end{array}}\right)=\left({\begin{array}{l}{1}\\{0}\\{0}\end{array}}\right)\left(v_{0}t+x_{0}\right)+\left({\begin{array}{c}{1}\\{-3}\\{2}\end{array}}\right)C_{4}\sin {\sqrt {\frac {g}{\ell }}}t+\left({\begin{array}{c}{1}\\{-4}\\{4}\end{array}}\right)C_{6}\sin 2{\sqrt {\frac {g}{\ell }}}t}
По гладкой горизонтальной трубке, вращающейся с постоянной угловой скоростью ω вокруг вертикальной оси, может двигаться точка массы m. 1) Найти общее решение канонических уравнений движения точки. 2) Найти полный интеграл уравнения Гамильтона — Якоби. Получить решение методом Якоби. За обобщённую координату принять х — расстояние точки до оси вращения.
ω
→
=
ω
e
z
→
{\displaystyle {\vec {\omega }}=\omega {\vec {e_{z}}}}
V
a
b
s
.
2
=
x
˙
2
+
w
2
x
2
{\displaystyle V_{abs.}^{2}={\dot {x}}^{2}+w^{2}x^{2}}
T
=
1
2
m
(
x
˙
2
+
w
2
x
2
)
,
V
=
0
{\displaystyle T={\frac {1}{2}}m\left({\dot {x}}^{2}+w^{2}x^{2}\right),\quad V=0}
L
=
T
−
V
=
T
=
1
2
m
x
2
+
1
2
m
(
w
x
)
2
{\displaystyle L=T-V=T={\frac {1}{2}}mx^{2}+{\frac {1}{2}}m(wx)^{2}}
H
=
p
2
2
m
−
1
2
m
(
ω
x
)
2
{\displaystyle H={\frac {p^{2}}{2m}}-{\frac {1}{2}}m(\omega x)^{2}}
{
p
˙
=
−
∂
H
∂
q
q
˙
=
∂
H
∂
p
{\displaystyle \left\{{\begin{array}{l}{{\dot {p}}=-{\frac {\partial H}{\partial q}}}\\{{\dot {q}}={\frac {\partial H}{\partial p}}}\end{array}}\right.}
{
p
˙
=
∂
H
∂
x
x
˙
=
p
m
{\displaystyle \left\{{\begin{array}{l}{{\dot {p}}={\frac {\partial H}{\partial x}}}\\{{\dot {x}}={\frac {p}{m}}}\end{array}}\right.}
{
p
=
m
ω
2
x
x
˙
=
p
m
{\displaystyle \left\{{\begin{array}{l}{p=m\omega ^{2}x}\\{{\dot {x}}={\frac {p}{m}}}\end{array}}\right.}
x
¨
=
ω
2
x
{\displaystyle {\ddot {x}}=\omega ^{2}x}
λ
2
−
w
2
=
0
{\displaystyle \lambda ^{2}-w^{2}=0}
λ
=
±
w
{\displaystyle \lambda =\pm w}
x
(
t
)
=
c
1
e
ω
t
+
c
2
e
−
ω
t
{\displaystyle x(t)=c_{1}e^{\omega t}+c_{2}e^{-\omega t}}
p
(
t
)
=
m
w
(
c
1
e
w
t
−
c
2
e
−
w
t
)
{\displaystyle p(t)=mw\left(c_{1}e^{wt}-c_{2}e^{-wt}\right)}
∂
W
∂
t
+
H
(
t
,
∂
W
∂
x
,
x
)
=
0
{\displaystyle {\frac {\partial W}{\partial t}}+H\left(t,{\frac {\partial W}{\partial x}},x\right)=0}
∂
W
∂
t
+
1
2
m
(
∂
W
∂
t
)
2
−
1
2
m
ω
2
x
2
=
0
{\displaystyle {\frac {\partial W}{\partial t}}+{\frac {1}{2m}}\left({\frac {\partial W}{\partial t}}\right)^{2}-{\frac {1}{2}}m\omega ^{2}x^{2}=0}
d
H
d
t
=
0
{\displaystyle {\frac {dH}{dt}}=0}
W
=
−
h
t
+
S
(
x
,
h
)
{\displaystyle W=-ht+S(x,h)}
H
(
∂
S
∂
x
,
x
)
=
h
{\displaystyle H\left({\frac {\partial S}{\partial x}},x\right)=h}
1
2
m
(
∂
S
∂
x
)
2
−
1
2
m
ω
2
x
2
=
h
{\displaystyle {\frac {1}{2m}}\left({\frac {\partial S}{\partial x}}\right)^{2}-{\frac {1}{2}}m\omega ^{2}x^{2}=h}
(
∂
S
∂
x
)
2
=
(
h
−
1
2
m
ω
2
x
2
)
2
m
{\displaystyle \left({\frac {\partial S}{\partial x}}\right)^{2}=\left(h-{\frac {1}{2}}m\omega ^{2}x^{2}\right)2m}
∂
S
∂
x
=
2
m
h
−
m
2
w
2
x
2
{\displaystyle {\frac {\partial S}{\partial x}}={\sqrt {2mh-m^{2}w^{2}x^{2}}}}
S
=
∫
2
m
h
−
m
2
w
2
x
2
d
x
{\displaystyle S=\int {\sqrt {2mh-m^{2}w^{2}x^{2}}}dx}
W
=
−
h
t
+
∫
2
m
h
−
m
2
w
2
x
2
d
x
{\displaystyle W=-ht+\int {\sqrt {2mh-m^{2}w^{2}x^{2}}}dx}
Гладкая трубка АВ вращается с постоянной угловой скоростью ω вокруг вертикальной оси CD, составляя с ней неизменный угол
π
4
{\displaystyle {\frac {\pi }{4}}}
т. 1) Решить задачу Коши для канонических уравнений движения шарика, если его начальная относительная скорость равна нулю и начальное расстояние от точки О равно а.
2) Найти полный интеграл уравнения Гамильтона — Якоби и найти решение канонических уравнений методом Якоби.
H
=
T
+
V
=
p
2
2
m
+
m
g
q
cos
α
+
m
2
ω
2
q
2
sin
2
α
{\displaystyle H=T+V={\frac {p^{2}}{2m}}+mgq\cos \alpha +{\frac {m}{2}}\omega ^{2}q^{2}\sin ^{2}\alpha }
d
p
d
t
=
−
∂
H
∂
q
=
−
m
g
cos
α
−
m
ω
2
q
sin
2
α
{\displaystyle {\frac {dp}{dt}}=-{\frac {\partial H}{\partial q}}=-mg\cos \alpha -m\omega ^{2}q\sin ^{2}\alpha }
d
q
d
t
=
∂
H
∂
p
=
p
m
{\displaystyle {\frac {dq}{dt}}={\frac {\partial H}{\partial p}}={\frac {p}{m}}}
(
p
˙
q
˙
)
=
(
0
−
m
ω
2
sin
2
α
1
/
m
0
)
(
p
q
)
+
(
−
m
g
cos
α
0
)
{\displaystyle \left({\begin{array}{l}{\dot {p}}\\{\dot {q}}\end{array}}\right)=\left({\begin{array}{cc}{0}&{-m\omega ^{2}\sin ^{2}\alpha }\\{1/m}&{0}\end{array}}\right)\left({\begin{array}{l}{p}\\{q}\end{array}}\right)+\left({\begin{array}{c}{-mg\cos \alpha }\\{0}\end{array}}\right)}
(
p
˙
q
˙
)
=
(
0
−
m
ω
2
sin
2
α
1
/
m
0
)
(
p
q
)
{\displaystyle \left({\begin{array}{l}{\dot {p}}\\{\dot {q}}\end{array}}\right)=\left({\begin{array}{cc}{0}&{-m\omega ^{2}\sin ^{2}\alpha }\\{1/m}&{0}\end{array}}\right)\left({\begin{array}{l}{p}\\{q}\end{array}}\right)}
det
(
A
−
λ
E
)
=
|
−
λ
−
m
ω
2
sin
2
α
1
/
m
−
λ
|
=
λ
2
+
ω
2
sin
2
α
{\displaystyle \det(A-\lambda E)=\left|{\begin{array}{cc}{-\lambda }&{-m\omega ^{2}\sin ^{2}\alpha }\\{1/m}&{-\lambda }\end{array}}\right|=\lambda ^{2}+\omega ^{2}\sin ^{2}\alpha }
λ
1
=
ω
i
sin
α
=
ω
sin
α
e
i
π
/
2
{\displaystyle \lambda _{1}=\omega i\sin \alpha =\omega \sin \alpha e^{i\pi /2}}
(
p
q
)
=
C
1
(
1
m
ω
sin
α
e
i
π
/
2
)
exp
(
i
ω
sin
α
t
)
+
C
2
(
1
m
ω
sin
α
e
−
i
π
/
2
)
exp
(
−
i
ω
sin
α
t
)
{\displaystyle \left({\begin{array}{l}{p}\\{q}\end{array}}\right)=C_{1}\left({\begin{array}{c}{1}\\{m\omega \sin \alpha e^{i\pi /2}}\end{array}}\right)\exp(i\omega \sin \alpha t)+C_{2}\left({\begin{array}{c}{1}\\{m\omega \sin \alpha e^{-i\pi /2}}\end{array}}\right)\exp(-i\omega \sin \alpha t)}
(
p
q
)
1
=
(
0
−
g
cos
α
ω
2
sin
2
α
)
{\displaystyle \left({\begin{array}{l}{p}\\{q}\end{array}}\right)_{1}=\left({\begin{array}{c}{0}\\{-{\frac {g\cos \alpha }{\omega ^{2}\sin ^{2}\alpha }}}\end{array}}\right)}
(
p
q
)
=
(
C
1
exp
(
i
ω
sin
α
t
)
+
C
2
exp
(
−
i
ω
sin
α
t
)
−
g
cos
α
ω
2
sin
2
α
+
C
1
m
ω
sin
α
exp
(
i
π
2
+
i
ω
sin
α
t
)
+
C
2
m
ω
sin
α
exp
(
−
i
π
2
−
i
ω
sin
α
t
)
)
{\displaystyle \left({\begin{array}{l}{p}\\{q}\end{array}}\right)=\left({\begin{array}{l}C_{1}\exp(i\omega \sin \alpha t)+C_{2}\exp(-i\omega \sin \alpha t)\\-g{\frac {\cos \alpha }{\omega ^{2}\sin ^{2}\alpha }}+C_{1}m\omega \sin \alpha \exp \left({\frac {i\pi }{2}}+i\omega \sin \alpha t\right)+C_{2}m\omega \sin \alpha \exp \left(-{\frac {i\pi }{2}}-i\omega \sin \alpha t\right)\end{array}}\right)}
(
p
q
)
0
=
(
0
a
)
{\displaystyle \left({\begin{array}{l}{p}\\{q}\end{array}}\right)_{0}=\left({\begin{array}{l}{0}\\{a}\end{array}}\right)}
(
p
q
)
|
t
=
0
=
(
C
1
+
C
2
−
g
cos
α
ω
2
sin
2
α
+
i
(
C
1
−
C
2
)
ω
sin
α
)
{\displaystyle \left.\left({\begin{array}{l}{p}\\{q}\end{array}}\right)\right|{}_{t=0}=\left({\begin{array}{l}{C_{1}+C_{2}}\\{-g{\frac {\cos \alpha }{\omega ^{2}\sin ^{2}\alpha }}+i\left(C_{1}-C_{2}\right)\omega \sin \alpha }\end{array}}\right)}
C
1
+
C
2
=
0
,
C
2
=
−
C
1
=
−
C
{\displaystyle C_{1}+C_{2}=0,\quad C_{2}=-C_{1}=-C}
q
|
t
=
0
=
−
g
cos
α
ω
2
sin
2
α
+
i
⋅
2
c
m
ω
sin
α
=
a
{\displaystyle \left.q\right|{}_{t=0}=-g{\frac {\cos \alpha }{\omega ^{2}\sin ^{2}\alpha }}+i\cdot 2cm\omega \sin \alpha =a}
C
=
a
+
g
cos
α
ω
2
sin
2
α
2
i
m
ω
sin
α
=
−
i
2
(
a
+
g
cos
α
ω
2
sin
2
α
m
ω
sin
α
)
{\displaystyle C={\frac {a+g{\frac {\cos \alpha }{\omega ^{2}\sin ^{2}\alpha }}}{2im\omega \sin \alpha }}=-{\frac {i}{2}}\left({\frac {a+{\frac {g\cos \alpha }{\omega ^{2}\sin ^{2}\alpha }}}{m\omega \sin \alpha }}\right)}
(
p
q
)
=
(
a
+
g
cos
α
ω
2
sin
2
α
2
i
m
ω
sin
α
(
exp
(
−
π
i
2
)
exp
(
i
ω
sin
α
t
)
+
exp
(
π
i
2
)
exp
(
−
i
ω
sin
α
t
)
)
−
g
cos
α
ω
2
sin
2
α
+
a
+
g
cos
α
ω
2
sin
2
α
2
i
m
ω
sin
α
e
−
i
π
/
2
e
i
π
/
2
e
i
ω
sin
α
t
+
e
i
π
/
2
e
−
i
π
/
2
e
−
i
ω
sin
α
t
)
=
(
a
+
g
cos
α
ω
2
sin
2
α
m
ω
sin
α
sin
(
ω
sin
α
t
)
−
g
cos
α
ω
2
sin
2
α
+
a
+
g
cos
α
ω
2
sin
2
α
2
i
m
ω
sin
α
cos
(
ω
sin
α
t
)
)
{\displaystyle \left({\begin{array}{l}{p}\\{q}\end{array}}\right)=\left({\begin{array}{l}{\frac {a+g{\frac {\cos \alpha }{\omega ^{2}\sin ^{2}\alpha }}}{2im\omega \sin \alpha }}\left(\exp \left(-{\frac {\pi i}{2}}\right)\exp(i\omega \sin \alpha t)+\exp \left({\frac {\pi i}{2}}\right)\exp(-i\omega \sin \alpha t)\right)\\-g{\frac {\cos \alpha }{\omega ^{2}\sin ^{2}\alpha }}+{\frac {a+g{\frac {\cos \alpha }{\omega ^{2}\sin ^{2}\alpha }}}{2im\omega \sin \alpha }}e^{-i\pi /2}e^{i\pi /2}e^{i\omega \sin \alpha t}+e^{i\pi /2}e^{-i\pi /2}e^{-i\omega \sin \alpha t}\end{array}}\right)=\left({\begin{array}{l}{\frac {a+{\frac {g\cos \alpha }{\omega ^{2}\sin ^{2}\alpha }}}{m\omega \sin \alpha }}\sin(\omega \sin \alpha t)\\-g{\frac {\cos \alpha }{\omega ^{2}\sin ^{2}\alpha }}+{\frac {a+g{\frac {\cos \alpha }{\omega ^{2}\sin ^{2}\alpha }}}{2im\omega \sin \alpha }}\cos(\omega \sin \alpha t)\end{array}}\right)}
⊐
α
=
45
∘
{\displaystyle \sqsupset \alpha =45^{\circ }}
(
p
q
)
=
(
a
+
2
g
ω
2
m
ω
2
sin
(
ω
t
2
)
a
+
2
g
ω
2
m
ω
2
cos
(
ω
t
2
)
−
2
g
ω
2
)
{\displaystyle \left({\begin{array}{l}{p}\\{q}\end{array}}\right)=\left({\begin{array}{l}{\frac {a+{\frac {{\sqrt {2}}g}{\omega ^{2}}}}{m\omega }}{\sqrt {2}}\sin \left({\frac {\omega t}{\sqrt {2}}}\right)\\{\frac {a+{\frac {{\sqrt {2}}g}{\omega ^{2}}}}{m\omega }}{\sqrt {2}}\cos \left({\frac {\omega t}{\sqrt {2}}}\right)-{\frac {{\sqrt {2}}g}{\omega ^{2}}}\end{array}}\right)}
H
(
p
,
q
)
=
p
2
2
m
+
m
g
q
cos
α
+
m
ω
2
q
2
sin
2
α
2
{\displaystyle H(p,q)={\frac {p^{2}}{2m}}+mgq\cos \alpha +{\frac {m\omega ^{2}q^{2}\sin ^{2}\alpha }{2}}}
W
=
−
h
t
+
S
(
q
)
{\displaystyle W=-ht+S(q)}
H
(
∂
S
∂
q
,
q
)
=
h
{\displaystyle H\left({\frac {\partial S}{\partial q}},q\right)=h}
1
2
m
(
∂
S
∂
q
)
2
+
m
g
q
cos
α
+
m
2
ω
2
q
2
sin
2
α
=
h
{\displaystyle {\frac {1}{2m}}\left({\frac {\partial S}{\partial q}}\right)^{2}+mgq\cos \alpha +{\frac {m}{2}}\omega ^{2}q^{2}\sin ^{2}\alpha =h}
∂
S
∂
q
=
2
m
⋅
h
−
m
g
q
2
−
m
w
2
q
2
4
{\displaystyle {\frac {\partial S}{\partial q}}={\sqrt {2m}}\cdot {\sqrt {h-{\frac {mgq}{\sqrt {2}}}-{\frac {mw^{2}q^{2}}{4}}}}}
S
(
q
)
=
∫
0
q
2
m
⋅
h
−
m
g
t
2
−
m
w
2
t
2
4
d
t
{\displaystyle S(q)=\int _{0}^{q}{\sqrt {2m}}\cdot {\sqrt {h-{\frac {mgt}{\sqrt {2}}}-{\frac {mw^{2}t^{2}}{4}}}}dt}
W
=
−
h
t
+
∫
0
q
2
m
⋅
h
−
m
g
t
2
−
m
w
2
t
2
4
d
t
{\displaystyle W=-ht+\int _{0}^{q}{\sqrt {2m}}\cdot {\sqrt {h-{\frac {mgt}{\sqrt {2}}}-{\frac {mw^{2}t^{2}}{4}}}}dt}
Шарик массы m, привязанный к нерастяжимой нити, скользит по гладкой горизонтальной плоскости; другой конец нити втягивают с постоянной скоростью а в отверстие, сделанное на плоскости. 1) Решить задачу Коши для канонических уравнений движения шарика, если в начальный момент нить расположена по прямой, расстояние между шариком и отверстием равно R, а проекция начальной скорости шарика на перпендикуляр к направлению нити равна ν 0 . 2) Найти полный интеграл уравнения Гамильтона — Якоби и общее решение канонических уравнений движения методом Якоби.
T
−
m
2
(
x
˙
2
+
y
˙
2
)
=
m
2
(
a
2
+
(
R
−
a
t
)
2
φ
˙
2
)
{\displaystyle T-{\frac {m}{2}}\left({\dot {x}}^{2}+{\dot {y}}^{2}\right)={\frac {m}{2}}\left(a^{2}+(R-at)^{2}{\dot {\varphi }}^{2}\right)}
L
=
T
{\displaystyle L=T}
p
φ
=
∂
L
∂
φ
˙
=
m
(
R
−
a
t
)
2
φ
˙
{\displaystyle p_{\varphi }={\frac {\partial L}{\partial {\dot {\varphi }}}}=m(R-at)^{2}{\dot {\varphi }}}
φ
˙
=
p
ψ
m
(
R
−
a
t
)
2
{\displaystyle {\dot {\varphi }}={\frac {p\psi }{m(R-at)^{2}}}}
H
=
p
φ
2
2
m
(
R
−
a
t
)
2
−
m
a
2
2
{\displaystyle H={\frac {p_{\varphi }^{2}}{2m(R-at)^{2}}}-{\frac {ma^{2}}{2}}}
φ
˙
=
p
φ
m
(
R
−
a
t
)
2
{\displaystyle {\dot {\varphi }}={\frac {p_{\varphi }}{m(R-at)^{2}}}}
200
R
φ
˙
=
v
0
=
p
φ
m
R
{\displaystyle 200R{\dot {\varphi }}=v_{0}={\frac {p_{\varphi }}{mR}}}
p
φ
=
m
R
v
0
{\displaystyle p_{\varphi }=mRv_{0}}
φ
˙
=
R
ν
0
(
R
−
a
t
)
2
{\displaystyle {\dot {\varphi }}={\frac {R\nu _{0}}{(R-at)^{2}}}}
φ
=
R
v
0
a
(
R
−
a
t
)
+
C
{\displaystyle \varphi ={\frac {Rv_{0}}{a(R-at)}}+C}
c
=
−
v
0
a
{\displaystyle c=-{\frac {v_{0}}{a}}}
φ
=
R
v
0
a
(
R
−
a
t
)
−
v
0
a
=
v
0
t
R
−
a
t
{\displaystyle \varphi ={\frac {Rv_{0}}{a(R-at)}}-{\frac {v_{0}}{a}}={\frac {v_{0}t}{R-at}}}
Точка массы т под действием собственного веса движется по циклоиде
x
=
r
(
θ
+
sin
θ
)
y
=
−
r
(
1
+
cos
θ
)
{\displaystyle {\begin{array}{l}{x=r(\theta +\sin \theta )}\\{y=-r(1+\cos \theta )}\end{array}}}
,
расположенной в вертикальной плоскости. 1) Решить задачу Коши для канонических уравнений движения точки, если при t = 0 Θ = 0, v = v0 .В качестве обобщённой координаты взять дугу циклоиды s =
⌣
M
0
M
{\displaystyle \smile M_{0}M}
{
x
=
r
(
θ
+
sin
θ
)
y
=
−
r
(
1
+
cos
θ
)
{\displaystyle \left\{{\begin{array}{l}{x=r(\theta +\sin \theta )}\\{y=-r(1+\cos \theta )}\end{array}}\right.}
d
x
d
θ
=
r
(
1
+
cos
θ
)
d
y
d
θ
=
r
sin
θ
{\displaystyle {\begin{array}{l}{{\frac {dx}{d\theta }}=r(1+\cos \theta )}\\{{\frac {dy}{d\theta }}=r\sin \theta }\end{array}}}
S
=
∫
0
θ
(
d
x
d
θ
)
2
+
(
d
y
d
θ
)
2
d
θ
=
∫
0
θ
2
r
1
+
cos
θ
d
θ
=
2
2
r
1
+
cos
θ
tg
(
θ
2
)
{\displaystyle S=\int _{0}^{\theta }{\sqrt {\left({\frac {dx}{d\theta }}\right)^{2}+\left({\frac {dy}{d\theta }}\right)^{2}}}d\theta =\int _{0}^{\theta }{\sqrt {2}}r{\sqrt {1+\cos \theta }}d\theta =2{\sqrt {2}}r{\sqrt {1+\cos \theta }}\operatorname {tg} \left({\frac {\theta }{2}}\right)}
t
=
tg
(
θ
2
)
{\displaystyle t=\operatorname {tg} \left({\frac {\theta }{2}}\right)}
s
=
2
2
r
1
+
1
−
t
2
1
+
t
2
t
=
2
2
r
2
1
+
t
2
=
4
r
t
1
+
t
2
{\displaystyle s=2{\sqrt {2}}r{\sqrt {1+{\frac {1-t^{2}}{1+t^{2}}}}}t=2{\sqrt {2}}r{\sqrt[{}]{\frac {2}{1+t^{2}}}}={\frac {4rt}{\sqrt {1+t^{2}}}}}
(
1
+
t
2
)
s
2
=
16
r
2
t
2
{\displaystyle \left(1+t^{2}\right)s^{2}=16r^{2}t^{2}}
t
2
=
s
2
16
r
2
−
s
2
{\displaystyle t^{2}={\frac {s^{2}}{16r^{2}-s^{2}}}}
y
=
−
r
(
1
+
cos
θ
)
=
−
r
(
1
+
1
−
t
2
1
+
t
2
)
=
−
2
r
1
+
t
2
=
s
2
−
16
r
2
8
r
{\displaystyle y=-r(1+\cos \theta )=-r\left(1+{\frac {1-t^{2}}{1+t^{2}}}\right)={\frac {-2r}{1+t^{2}}}={\frac {s^{2}-16r^{2}}{8r}}}
H
(
p
,
s
)
=
p
2
2
m
+
m
g
(
y
+
2
r
)
⋅
p
2
2
m
+
m
g
s
2
8
r
{\displaystyle H(p,s)={\frac {p^{2}}{2m}}+mg(y+2r)\cdot {\frac {p^{2}}{2m}}+{\frac {mgs^{2}}{8r}}}
p
=
−
∂
H
∂
s
=
−
m
g
s
4
r
s
˙
=
p
m
{\displaystyle p=-{\frac {\partial H}{\partial s}}={\frac {-mgs}{4r}}\quad {\dot {s}}={\frac {p}{m}}}
(
s
˙
p
˙
)
=
(
0
1
/
m
−
m
g
/
4
r
0
)
(
s
p
)
{\displaystyle \left({\begin{array}{c}{\dot {s}}\\{\dot {p}}\end{array}}\right)=\left({\begin{array}{cc}{0}&{1/m}\\{-mg/4r}&{0}\end{array}}\right)\left({\begin{array}{l}{s}\\{p}\end{array}}\right)}
det
(
A
−
λ
E
)
=
|
−
1
1
/
m
−
m
g
4
r
−
λ
|
=
λ
2
+
g
4
r
=
0
{\displaystyle \det(A-\lambda E)=\left|{\begin{array}{cc}{-1}&{1/m}\\{-{\frac {mg}{4r}}}&{-\lambda }\end{array}}\right|=\lambda ^{2}+{\frac {g}{4r}}=0}
λ
1
=
g
4
r
i
λ
2
=
−
g
4
r
i
{\displaystyle \lambda _{1}={\sqrt {\frac {g}{4r}}}i\quad \lambda _{2}=-{\sqrt {{\frac {g}{4r}}i}}}
u
1
=
(
−
i
m
4
r
g
1
)
u
2
=
(
+
i
m
4
r
g
1
)
{\displaystyle u_{1}=\left({\begin{array}{c}{-{\frac {i}{m}}{\sqrt {\frac {4r}{g}}}}\\{1}\end{array}}\right)\quad u_{2}=\left({\begin{array}{c}{+{\frac {i}{m}}{\sqrt {\frac {4r}{g}}}}\\{1}\end{array}}\right)}
(
s
p
)
=
C
1
(
−
i
m
4
r
g
1
)
exp
(
g
4
r
i
t
)
+
C
2
(
i
m
4
r
g
1
)
exp
(
−
g
4
r
i
t
)
{\displaystyle \left({\begin{array}{l}{s}\\{p}\end{array}}\right)=C_{1}\left({\begin{array}{l}-{\frac {i}{m}}{\sqrt {\frac {4r}{g}}}\\1\end{array}}\right)\exp \left({\sqrt {\frac {g}{4r}}}it\right)+C_{2}\left({\begin{array}{l}{\frac {i}{m}}{\sqrt {\frac {4r}{g}}}\\1\end{array}}\right)\exp \left(-{\sqrt {\frac {g}{4r}}}it\right)}
s
|
t
=
0
=
0
p
l
|
t
=
0
=
m
v
0
{\displaystyle \left.s\right|_{t=0}=0\quad \left.pl\right|_{t=0}=mv_{0}}
(
s
p
)
|
t
=
0
=
(
(
C
2
−
C
)
i
m
4
r
g
C
2
+
C
1
)
=
(
0
m
v
0
)
{\displaystyle \left.\left({\begin{array}{l}{s}\\{p}\end{array}}\right)\right|_{t=0}=\left({\begin{array}{c}{\left(C_{2}-C\right){\frac {i}{m}}{\sqrt {\frac {4r}{g}}}}\\{C_{2}+C_{1}}\end{array}}\right)=\left({\begin{array}{c}{0}\\{mv_{0}}\end{array}}\right)}
C
1
=
C
2
=
m
v
0
2
{\displaystyle C_{1}=C_{2}={\frac {mv_{0}}{2}}}
(
s
p
)
=
(
−
v
0
i
r
g
exp
(
g
4
r
i
t
)
+
v
0
i
r
g
exp
(
−
g
4
r
i
t
)
m
v
0
2
exp
(
g
4
r
i
t
)
+
m
v
0
2
exp
(
−
g
4
r
i
t
)
)
=
(
4
r
g
v
0
sin
g
4
r
t
m
v
0
cos
g
4
r
t
)
{\displaystyle \left({\begin{array}{l}{s}\\{p}\end{array}}\right)=\left({\begin{array}{l}-v_{0}i{\sqrt {\frac {r}{g}}}\exp \left({\sqrt {\frac {g}{4r}}}it\right)+v_{0}i{\sqrt {\frac {r}{g}}}\exp \left(-{\sqrt {\frac {g}{4r}}}it\right)\\{\frac {mv_{0}}{2}}\exp \left({\sqrt {\frac {g}{4r}}}it\right)+{\frac {mv_{0}}{2}}\exp \left(-{\sqrt {\frac {g}{4r}}}it\right)\end{array}}\right)=\left({\begin{array}{l}{{\sqrt {\frac {4r}{g}}}v_{0}\sin {\sqrt {\frac {g}{4r}}}t}\\{mv_{0}\cos {\sqrt {\frac {g}{4r}}}t}\end{array}}\right)}
H
(
p
,
q
)
=
p
2
2
m
+
m
g
q
2
8
r
{\displaystyle H(p,q)={\frac {p^{2}}{2m}}+{\frac {mgq^{2}}{8r}}}
W
=
−
h
t
+
S
(
q
)
{\displaystyle W=-ht+S(q)}
H
(
∂
S
∂
q
,
q
)
=
h
{\displaystyle H\left({\frac {\partial S}{\partial q}},q\right)=h}
1
2
m
(
∂
S
∂
q
)
2
+
m
g
q
2
8
r
=
h
{\displaystyle {\frac {1}{2m}}\left({\frac {\partial S}{\partial q}}\right)^{2}+{\frac {mgq^{2}}{8r}}=h}
∂
S
∂
q
=
2
m
(
h
−
m
g
q
2
8
r
)
{\displaystyle {\frac {\partial S}{\partial q}}={\sqrt {2m\left(h-{\frac {mgq^{2}}{8r}}\right)}}}
S
(
q
)
=
∫
0
g
2
m
(
h
−
m
g
t
2
8
r
)
d
t
{\displaystyle S(q)=\int _{0}^{g}{\sqrt {2m\left(h-{\frac {mgt^{2}}{8r}}\right)}}dt}
W
(
t
,
q
,
h
)
=
−
h
t
+
∫
0
g
2
m
(
h
−
m
g
t
2
8
r
)
d
t
{\displaystyle W(t,q,h)=-ht+\int _{0}^{g}{\sqrt {2m\left(h-{\frac {mgt^{2}}{8r}}\right)}}dt}
−
β
=
∂
W
∂
h
=
−
t
+
∫
0
q
d
x
h
−
m
g
x
2
8
r
{\displaystyle -\beta ={\frac {\partial W}{\partial h}}=-t+\int _{0}^{q}{\frac {dx}{\sqrt {h-{\frac {mgx^{2}}{8r}}}}}}
−
β
+
t
=
8
r
m
g
arcsin
m
g
8
r
h
q
{\displaystyle -\beta +t={\sqrt {\frac {8r}{mg}}}\arcsin {\sqrt {\frac {mg}{8rh}}}q}
m
g
8
r
h
q
=
sin
m
g
8
r
(
t
−
β
)
{\displaystyle {\sqrt {\frac {mg}{8rh}}}q=\sin {\sqrt {\frac {mg}{8r}}}(t-\beta )}
q
=
8
r
h
m
g
sin
m
g
8
r
(
t
−
β
)
{\displaystyle q={\sqrt {\frac {8rh}{mg}}}\sin {\sqrt {\frac {mg}{8r}}}(t-\beta )}
Тяжёлая точка М массы т движется по поверхности круглого цилиндра радиуса r, ось которого вертикальна. 1) Решить задачу Коши для канонических уравнений движения точки, если её начальная скорость равна по величине v0 α с горизонтом. 2) Найти полный интеграл уравнения Гамильтона — Якоби и закон движения точки методом Якоби.
v
2
=
r
˙
2
+
r
2
φ
˙
2
+
h
˙
2
{\displaystyle v^{2}={\dot {r}}^{2}+r^{2}{\dot {\varphi }}^{2}+{\dot {h}}^{2}}
L
=
m
2
r
˙
2
+
m
r
˙
2
2
φ
2
+
m
2
h
˙
2
−
V
(
r
,
φ
,
h
)
{\displaystyle L={\frac {m}{2}}{\dot {r}}^{2}+{\frac {m{\dot {r}}^{2}}{2}}\varphi ^{2}+{\frac {m}{2}}{\dot {h}}^{2}-V(r,\varphi ,h)}
p
r
=
m
r
˙
p
φ
=
m
r
2
φ
˙
p
h
=
m
h
˙
{\displaystyle p_{r}=m{\dot {r}}\quad p_{\varphi }=mr^{2}{\dot {\varphi }}\quad p_{h}=m{\dot {h}}}
H
=
1
2
m
(
p
r
2
+
1
r
2
p
φ
2
+
p
h
2
)
+
V
(
r
,
φ
,
h
)
{\displaystyle H={\frac {1}{2m}}\left(p_{r}^{2}+{\frac {1}{r^{2}}}p_{\varphi }^{2}+p_{h}^{2}\right)+V(r,\varphi ,h)}
r
=
c
o
n
s
t
{\displaystyle r=const}
H
(
p
φ
,
p
h
,
φ
,
h
)
=
1
2
m
r
2
p
φ
2
+
p
h
2
2
m
+
m
g
h
{\displaystyle H\left(p_{\varphi },p_{h},\varphi ,h\right)={\frac {1}{2mr^{2}}}p_{\varphi }^{2}+{\frac {p_{h}^{2}}{2m}}+mgh}
h
˙
=
∂
H
∂
ρ
h
=
p
h
m
p
h
˙
=
−
∂
H
∂
h
=
−
m
g
{\displaystyle {\dot {h}}={\frac {\partial H}{\partial \rho _{h}}}={\frac {p_{h}}{m}}\quad {\dot {p_{h}}}=-{\frac {\partial H}{\partial h}}=-mg}
φ
˙
=
∂
H
∂
p
φ
=
p
φ
m
r
2
p
φ
˙
=
−
∂
H
∂
φ
=
0
{\displaystyle {\dot {\varphi }}={\frac {\partial H}{\partial p_{\varphi }}}={\frac {p_{\varphi }}{mr^{2}}}\quad {\dot {p_{\varphi }}}=-{\frac {\partial H}{\partial \varphi }}=0}
d
d
t
(
φ
p
φ
h
p
h
)
=
(
0
1
/
m
r
2
0
0
0
0
0
0
0
0
0
1
/
m
0
0
0
0
)
(
φ
p
φ
h
p
h
)
+
(
0
0
0
−
m
g
)
{\displaystyle {\frac {d}{dt}}\left({\begin{array}{l}{\varphi }\\p_{\varphi }\\{h}\\{p_{h}}\end{array}}\right)=\left({\begin{array}{llll}{0}&{1/mr^{2}}&{0}&{0}\\{0}&{0}&{0}&{0}\\{0}&{0}&{0}&{1/m}\\{0}&{0}&{0}&{0}\end{array}}\right)\left({\begin{array}{l}{\varphi }\\p_{\varphi }\\{h}\\{p_{h}}\end{array}}\right)+\left({\begin{array}{c}{0}\\{0}\\{0}\\{-mg}\end{array}}\right)}
(
φ
p
φ
h
p
h
)
f
=
(
0
0
−
g
t
2
/
2
−
m
g
t
)
{\displaystyle \left({\begin{array}{l}{\varphi }\\p_{\varphi }\\{h}\\{p_{h}}\end{array}}\right)_{f}=\left({\begin{array}{c}{0}\\{0}\\{-gt^{2}/2}\\{-mgt}\end{array}}\right)}
det
(
−
λ
ln
r
2
0
0
0
−
λ
0
0
0
0
−
λ
1
/
m
0
0
0
−
λ
)
=
λ
4
=
0
{\displaystyle \det \left({\begin{array}{cccc}{-\lambda }&{\ln r^{2}}&{0}&{0}\\{0}&{-\lambda }&{0}&{0}\\{0}&{0}&{-\lambda }&{1/m}\\{0}&{0}&{0}&{-\lambda }\end{array}}\right)=\lambda ^{4}=0}
A
u
=
(
0
1
/
m
r
2
0
0
0
0
0
0
0
0
0
1
/
m
0
0
0
0
)
(
a
b
c
d
)
=
(
b
/
m
r
2
0
d
/
m
0
)
=
0
{\displaystyle Au=\left({\begin{array}{llll}{0}&{1/mr^{2}}&{0}&{0}\\{0}&{0}&{0}&{0}\\{0}&{0}&{0}&{1/m}\\{0}&{0}&{0}&{0}\end{array}}\right)\left({\begin{array}{l}{a}\\{b}\\{c}\\{d}\end{array}}\right)=\left({\begin{array}{c}{b/mr^{2}}\\{0}\\{d/m}\\{0}\end{array}}\right)=0}
u
1
=
(
1
0
0
0
)
u
2
=
(
0
0
1
0
)
{\displaystyle u_{1}=\left({\begin{array}{l}{1}\\{0}\\{0}\\0\end{array}}\right)\quad u_{2}=\left({\begin{array}{l}{0}\\{0}\\{1}\\{0}\end{array}}\right)}
A
u
=
u
1
{\displaystyle Au=u_{1}}
(
b
/
m
r
2
0
d
/
m
0
)
=
(
1
0
0
0
)
{\displaystyle \left({\begin{array}{c}{b/mr^{2}}\\{0}\\{d/m}\\{0}\end{array}}\right)=\left({\begin{array}{l}{1}\\{0}\\{0}\\{0}\end{array}}\right)}
u
3
=
(
0
m
r
2
0
0
)
{\displaystyle u_{3}=\left({\begin{array}{c}{0}\\{mr^{2}}\\{0}\\{0}\end{array}}\right)}
A
u
=
u
2
{\displaystyle Au=u_{2}}
(
b
/
m
r
2
0
d
/
m
0
)
=
(
0
0
1
0
)
{\displaystyle \left({\begin{array}{c}{b/mr^{2}}\\{0}\\{d/m}\\{0}\end{array}}\right)=\left({\begin{array}{l}{0}\\{0}\\{1}\\{0}\end{array}}\right)}
u
4
=
(
0
0
0
m
)
{\displaystyle u_{4}=\left({\begin{array}{c}{0}\\{0}\\{0}\\{m}\end{array}}\right)}
(
φ
p
φ
h
p
h
)
=
C
1
(
1
0
0
0
)
+
C
2
(
t
(
1
0
0
0
)
+
(
0
m
r
2
0
0
)
)
+
C
3
(
0
0
1
0
)
+
C
4
(
t
(
0
0
1
0
)
+
(
0
0
0
m
)
)
{\displaystyle \left({\begin{array}{l}{\varphi }\\p_{\varphi }\\{h}\\{p_{h}}\end{array}}\right)=C_{1}\left({\begin{array}{l}{1}\\{0}\\{0}\\{0}\end{array}}\right)+C_{2}\left(t\left({\begin{array}{l}{1}\\{0}\\{0}\\{0}\end{array}}\right)+\left({\begin{array}{c}{0}\\{mr^{2}}\\{0}\\{0}\end{array}}\right)\right)+C_{3}\left({\begin{array}{l}{0}\\{0}\\{1}\\{0}\end{array}}\right)+C_{4}\left(t\left({\begin{array}{l}{0}\\{0}\\{1}\\{0}\end{array}}\right)+\left({\begin{array}{c}{0}\\{0}\\{0}\\{m}\end{array}}\right)\right)}
(
φ
p
φ
h
p
h
)
|
t
=
0
=
(
C
1
m
r
2
C
2
C
3
m
C
4
)
=
(
0
m
r
v
cos
α
0
m
v
sin
α
)
{\displaystyle \left.\left({\begin{array}{l}{\varphi }\\p_{\varphi }\\{h}\\{p_{h}}\end{array}}\right)\right|_{t=0}=\left({\begin{array}{l}{C_{1}}\\{mr^{2}C_{2}}\\{C_{3}}\\{mC_{4}}\end{array}}\right)=\left({\begin{array}{c}{0}\\{mrv\cos \alpha }\\{0}\\{mv\sin \alpha }\end{array}}\right)}
(
φ
p
φ
h
p
h
)
=
(
V
r
cos
α
t
m
r
v
cos
α
v
sin
α
t
−
g
t
2
/
2
m
v
sin
α
−
m
g
t
)
{\displaystyle \left({\begin{array}{l}{\varphi }\\p_{\varphi }\\{h}\\{p_{h}}\end{array}}\right)=\left({\begin{array}{l}{{\frac {V}{r}}\cos \alpha t}\\mrv\cos \alpha \\{v\sin \alpha t-gt^{2}/2}\\{mv\sin \alpha -mgt}\end{array}}\right)}
H
=
p
h
2
2
m
+
p
φ
2
2
m
r
2
+
m
g
h
{\displaystyle H={\frac {p_{h}^{2}}{2m}}+{\frac {p_{\varphi }^{2}}{2mr^{2}}}+mgh}
V
=
W
(
t
,
φ
,
h
,
α
1
,
α
2
)
{\displaystyle V=W\left(t,\varphi ,h,\alpha _{1},\alpha _{2}\right)}
−
t
(
α
1
+
α
2
)
+
S
1
(
φ
)
+
S
2
(
h
)
{\displaystyle -t\left(\alpha _{1}+\alpha _{2}\right)+S_{1}(\varphi )+S_{2}(h)}
1
2
m
r
2
(
∂
S
1
∂
φ
)
2
=
α
1
{\displaystyle {\frac {1}{2mr^{2}}}\left({\frac {\partial S_{1}}{\partial \varphi }}\right)^{2}=\alpha _{1}}
S
1
(
φ
)
=
2
m
r
2
α
1
φ
{\displaystyle S_{1}(\varphi )={\sqrt {2mr^{2}\alpha _{1}}}\varphi }
1
2
m
(
∂
S
2
∂
h
)
2
+
m
g
h
=
α
2
{\displaystyle {\frac {1}{2m}}\left({\frac {\partial S_{2}}{\partial h}}\right)^{2}+mgh=\alpha _{2}}
S
2
(
h
)
=
2
m
∫
0
h
α
2
−
m
g
x
d
x
=
2
2
⋅
α
2
m
α
2
3
m
g
(
1
−
(
1
−
m
g
h
α
2
)
3
/
2
)
{\displaystyle S_{2}(h)={\sqrt {2m}}\int _{0}^{h}{\sqrt {\alpha _{2}-mgx}}dx={\frac {2{\sqrt {2}}\cdot \alpha _{2}{\sqrt {m\alpha _{2}}}}{3mg}}\left(1-\left(1-{\frac {mgh}{\alpha _{2}}}\right)^{3/2}\right)}
W
(
t
,
φ
,
h
,
α
1
,
α
2
)
=
−
t
(
α
1
+
α
2
)
+
2
m
r
2
α
1
φ
+
2
2
3
⋅
α
2
m
α
2
m
g
(
1
−
(
1
−
m
g
h
α
2
)
3
/
2
)
{\displaystyle W\left(t,\varphi ,h,\alpha _{1},\alpha _{2}\right)=-t\left(\alpha _{1}+\alpha _{2}\right)+{\sqrt {2mr^{2}\alpha _{1}}}\varphi +{\frac {2{\sqrt {2}}}{3}}\cdot {\frac {\alpha _{2}{\sqrt {m\alpha _{2}}}}{mg}}\left(1-\left(1-{\frac {mgh}{\alpha _{2}}}\right)^{3/2}\right)}
β
1
=
∂
W
∂
α
1
,
β
2
=
∂
W
∂
α
2
{\displaystyle \beta _{1}={\frac {\partial W}{\partial \alpha _{1}}},\quad \beta _{2}={\frac {\partial W}{\partial \alpha _{2}}}}
β
1
=
∂
W
∂
α
1
=
−
t
+
2
m
r
2
φ
⋅
1
2
α
1
{\displaystyle \beta _{1}={\frac {\partial W}{\partial \alpha _{1}}}=-t+{\sqrt {2mr^{2}}}\varphi \cdot {\frac {1}{2{\sqrt {\alpha _{1}}}}}}
φ
(
t
)
=
2
α
1
m
r
2
(
t
+
β
1
)
{\displaystyle \varphi (t)={\sqrt {\frac {2\alpha _{1}}{mr^{2}}}}(t+\beta _{1})}
β
2
=
∂
W
∂
α
2
=
2
α
2
m
g
2
−
2
(
α
2
−
m
g
h
)
m
g
2
−
t
{\displaystyle \beta _{2}={\frac {\partial W}{\partial \alpha _{2}}}={\sqrt {\frac {2\alpha _{2}}{mg^{2}}}}-{\sqrt {\frac {2\left(\alpha _{2}-mgh\right)}{mg^{2}}}}-t}
2
α
2
−
g
(
t
+
β
2
)
m
=
2
(
α
2
−
m
g
h
)
{\displaystyle {\sqrt {2\alpha _{2}}}-g\left(t+\beta _{2}\right){\sqrt {m}}={\sqrt {2\left(\alpha _{2}-mgh\right)}}}
2
α
2
−
2
2
α
2
g
(
t
+
β
2
)
m
+
g
2
(
t
+
β
2
)
2
m
=
2
α
2
−
2
m
g
h
{\displaystyle 2\alpha _{2}-2{\sqrt {2\alpha _{2}}}g\left(t+\beta _{2}\right){\sqrt {m}}+g^{2}\left(t+\beta _{2}\right)^{2}m=2\alpha _{2}-2mgh}
h
(
t
)
=
2
α
2
m
(
t
+
β
2
)
−
g
(
t
+
β
2
)
2
2
{\displaystyle h(t)={\sqrt {\frac {2\alpha _{2}}{m}}}\left(t+\beta _{2}\right)-{\frac {g\left(t+\beta _{2}\right)^{2}}{2}}}
Два одинаковых шарика массы т, связанные между собой пружиной жёсткости с (длина пружины в недеформированном состоянии равна /), могут скользить без трения по трубке, вращающейся с постоянной угловой скоростью ω вокруг вертикальной оси. 1) Найти переменные у 1 , у 2 , в которых полный интеграл уравнения Гамильтона — Якоби нашёлся бы методом разделения переменных.
2) Найти полный интеграл и решение канонических уравнений Гамильтона методом Якоби.
v
→
1
=
w
x
1
e
→
y
+
x
˙
1
e
→
x
{\displaystyle {\vec {v}}_{1}=wx_{1}{\vec {e}}_{y}+{\dot {x}}_{1}{\vec {e}}_{x}}
V
→
=
w
x
2
e
¯
y
+
x
˙
1
e
¯
x
{\displaystyle {\vec {V}}=wx_{2}{\overline {e}}_{y}+{\dot {x}}_{1}{\overline {e}}_{x}}
T
=
T
1
+
T
2
=
m
2
(
w
2
(
x
1
2
+
x
2
2
)
+
x
˙
2
+
y
˙
2
)
{\displaystyle T=T_{1}+T_{2}={\frac {m}{2}}(w^{2}\left(x_{1}^{2}+x_{2}^{2}\right)+{\dot {x}}^{2}+{\dot {y}}^{2})}
V
=
c
(
x
2
−
x
1
−
l
)
2
2
{\displaystyle V={\frac {c\left(x_{2}-x_{1}-l\right)^{2}}{2}}}
p
1
=
m
x
˙
1
{\displaystyle p_{1}=m{\dot {x}}_{1}}
p
2
=
m
x
˙
2
{\displaystyle p_{2}=m{\dot {x}}_{2}}
H
=
m
2
(
−
w
2
(
x
1
2
+
x
2
2
)
+
x
˙
1
2
+
x
1
2
˙
)
+
c
2
(
x
2
−
x
1
−
l
)
2
=
p
1
2
2
m
+
p
2
2
2
m
−
m
w
2
2
(
x
1
2
+
x
2
2
)
+
c
2
(
x
2
−
x
1
−
1
)
2
{\displaystyle H={\frac {m}{2}}\left(-w^{2}\left(x_{1}^{2}+x_{2}^{2}\right)+{\dot {x}}_{1}^{2}+{\dot {x_{1}^{2}}}\right)+{\frac {c}{2}}\left(x_{2}-x_{1}-l\right)^{2}={\frac {p_{1}^{2}}{2m}}+{\frac {p_{2}^{2}}{2m}}-{\frac {mw^{2}}{2}}\left(x_{1}^{2}+x_{2}^{2}\right)+{\frac {c}{2}}\left(x_{2}-x_{1}-1\right)^{2}}
p
˙
1
=
m
w
2
x
1
+
c
(
x
2
−
x
1
−
l
)
{\displaystyle {\dot {p}}_{1}=mw^{2}x_{1}+c\left(x_{2}-x_{1}-l\right)}
p
˙
2
=
m
w
2
x
2
−
c
(
x
2
−
x
1
−
l
)
{\displaystyle {\dot {p}}_{2}=mw^{2}x_{2}-c\left(x_{2}-x_{1}-l\right)}
x
˙
1
=
p
1
m
{\displaystyle {\dot {x}}_{1}={\frac {p_{1}}{m}}}
x
˙
2
=
p
2
m
{\displaystyle {\dot {x}}_{2}={\frac {p_{2}}{m}}}
m
x
¨
1
=
m
ω
2
x
1
+
c
(
x
2
−
x
1
−
l
)
{\displaystyle m{\ddot {x}}_{1}=m\omega ^{2}x_{1}+c\left(x_{2}-x_{1}-l\right)}
m
x
¨
2
=
m
w
2
x
2
−
c
(
x
2
−
x
1
−
l
)
{\displaystyle m{\ddot {x}}_{2}=mw^{2}x_{2}-c\left(x_{2}-x_{1}-l\right)}
y
1
=
x
2
−
x
1
l
{\displaystyle y_{1}={\frac {x_{2}-x_{1}}{l}}}
y
2
=
x
2
+
x
1
l
{\displaystyle y_{2}={\frac {x_{2}+x_{1}}{l}}}
2
m
y
¨
2
=
2
m
w
2
y
2
{\displaystyle 2m{\ddot {y}}_{2}=2mw^{2}y_{2}}
−
2
m
y
¨
1
=
−
2
m
w
2
y
1
+
4
c
(
y
1
−
l
)
{\displaystyle -2m{\ddot {y}}_{1}=-2mw^{2}y_{1}+4c\left(y_{1}-l\right)}
Стержень вращается с постоянной угловой скоростью ω в вертикальной плоскости вокруг горизонтальной оси (которая проходит через некоторую точку стержня). Найти общее решение канонических уравнений движения колечка массы m, насаженного на стержень
1) непосредственным интегрированием, и 2) методом Якоби; выписать полный интеграл уравнения Гамильтона — Якоби.
T
=
1
2
m
(
x
˙
2
+
ω
2
x
2
)
{\displaystyle T={\frac {1}{2}}m\left({\dot {x}}^{2}+\omega ^{2}x^{2}\right)}
V
=
−
m
g
x
cos
ω
t
{\displaystyle V=-mgx\cos \omega t}
L
=
1
2
⋅
m
(
x
˙
2
+
ω
2
x
2
)
+
m
g
x
cos
ω
t
{\displaystyle L={\frac {1}{2}}\cdot m\left({\dot {x}}^{2}+\omega ^{2}x^{2}\right)+mgx\cos \omega t}
p
=
∂
L
∂
x
=
m
x
˙
{\displaystyle p={\frac {\partial L}{\partial x}}=m{\dot {x}}}
H
=
1
2
m
p
2
−
1
2
m
⋅
ω
2
x
2
−
m
g
x
cos
ω
t
{\displaystyle H={\frac {1}{2m}}p^{2}-{\frac {1}{2}}m\cdot \omega ^{2}x^{2}-mgx\cos \omega t}
d
S
d
t
+
[
1
2
m
(
d
S
∂
x
)
2
−
1
2
m
ω
2
x
2
]
−
x
m
g
cos
ω
t
=
0
{\displaystyle {\frac {dS}{dt}}+\left[{\frac {1}{2m}}\left({\frac {dS}{\partial x}}\right)^{2}-{\frac {1}{2}}m\omega ^{2}x^{2}\right]-xmg\cos \omega t=0}
S
=
A
x
2
2
+
B
x
+
C
{\displaystyle S={\frac {Ax^{2}}{2}}+Bx+C}
A
˙
x
2
2
+
B
˙
x
+
C
+
1
2
m
(
A
x
+
B
)
2
−
1
2
m
ω
2
x
2
−
x
m
g
cos
ω
t
=
0
{\displaystyle {\frac {{\dot {A}}x^{2}}{2}}+{\dot {B}}x+C+{\frac {1}{2m}}(Ax+B)^{2}-{\frac {1}{2}}m\omega ^{2}x^{2}-xmg\cos \omega t=0}
{
A
˙
2
+
A
2
2
m
−
1
2
m
ω
2
=
0
B
˙
+
1
m
A
B
−
m
g
cos
ω
t
=
0
C
˙
+
1
2
m
B
2
=
0
<
m
a
t
h
>
A
˙
=
m
ω
2
−
A
2
m
{\displaystyle {\begin{cases}{\frac {\dot {A}}{2}}+{\frac {A^{2}}{2m}}-{\frac {1}{2}}m\omega ^{2}=0\\{\dot {B}}+{\frac {1}{m}}AB-mg\cos \omega t=0\\{\dot {C}}+{\frac {1}{2m}}B^{2}=0\end{cases}}<math>{\dot {A}}=m\omega ^{2}-{\frac {A^{2}}{m}}}
∫
d
A
m
ω
2
−
A
2
m
=
m
∫
d
A
m
ω
2
−
A
2
=
1
2
ω
ln
|
m
ω
+
A
m
ω
−
A
|
=
t
+
C
{\displaystyle \int {\frac {dA}{m\omega ^{2}-{\frac {A^{2}}{m}}}}=m\int {\frac {dA}{m\omega ^{2}-A^{2}}}={\frac {1}{2\omega }}\ln \left|{\frac {m\omega +A}{m\omega -A}}\right|=t+C}
P
=
∂
L
∂
x
˙
{\displaystyle P={\frac {\partial L}{\partial {\dot {x}}}}}
x
˙
=
p
m
{\displaystyle {\dot {x}}={\frac {p}{m}}}
p
=
−
∂
H
∂
x
=
m
ω
2
x
+
m
g
cos
ω
t
{\displaystyle p=-{\frac {\partial H}{\partial x}}=m\omega ^{2}x+mg\cos \omega t}
(
x
˙
p
˙
)
=
(
0
1
m
m
ω
2
0
)
(
x
p
)
{\displaystyle \left({\begin{array}{l}{\dot {x}}\\{\dot {p}}\end{array}}\right)=\left({\begin{array}{cc}0&{\frac {1}{m}}\\m\omega ^{2}&0\end{array}}\right)\left({\begin{array}{l}x\\p\end{array}}\right)}
λ
2
−
ω
2
=
0
{\displaystyle \lambda ^{2}-\omega ^{2}=0}
λ
1
,
2
=
±
ω
{\displaystyle \lambda _{1,2}=\pm \omega }
x
(
t
)
=
C
1
e
ω
t
+
C
2
e
−
ω
t
{\displaystyle x(t)=C_{1}e^{\omega t}+C_{2}e^{-\omega t}}
p
(
t
)
=
A
1
e
ω
t
+
A
2
e
−
ω
t
{\displaystyle p(t)=A_{1}e^{\omega t}+A_{2}e^{-\omega t}}
ω
C
1
e
ω
t
−
ω
C
2
e
−
ω
t
=
A
1
m
e
ω
t
+
A
2
m
e
−
ω
t
{\displaystyle \omega C_{1}e^{\omega t}-\omega C_{2}e^{-\omega t}={\frac {A_{1}}{m}}e^{\omega t}+{\frac {A_{2}}{m}}e^{-\omega t}}
{
A
1
=
m
ω
C
1
A
2
=
−
m
ω
C
2
{\displaystyle \left\{{\begin{array}{l}{A_{1}=m\omega C_{1}}\\{A_{2}=-m\omega C_{2}}\end{array}}\right.}
A
1
=
m
ω
C
1
A
2
=
−
m
ω
C
2
{\displaystyle {\begin{array}{l}{A_{1}=m\omega C_{1}}\\{A_{2}=-m\omega C_{2}}\end{array}}}
{
A
1
=
m
ω
C
1
A
2
=
−
m
ω
C
2
{\displaystyle {\begin{cases}{\begin{array}{l}{A_{1}=m\omega C_{1}}\\{A_{2}=-m\omega C_{2}}\end{array}}\end{cases}}}
{
λ
(
t
)
=
C
1
e
ω
t
+
C
2
e
−
ω
t
+
C
(
t
)
p
(
t
)
=
m
ω
(
C
1
e
ω
t
−
C
2
e
−
ω
t
)
+
B
(
t
)
{\displaystyle {\begin{array}{l}{\begin{cases}\lambda (t)=C_{1}e^{\omega t}+C_{2}e^{-\omega t}+C(t)\\{p(t)=m\omega \left(C_{1}e^{\omega t}-C{}_{2}e^{-\omega t}\right)+B(t)}\end{cases}}\end{array}}}
⊐
C
=
−
ρ
2
ω
2
cos
ω
t
{\displaystyle \sqsupset C=-{\frac {\rho }{2\omega ^{2}}}\cos \omega t}
B
=
m
g
2
ω
sin
ω
t
{\displaystyle B={\frac {mg}{2\omega }}\sin \omega t}
m
g
2
cos
ω
t
=
−
m
ω
2
g
2
ω
2
cos
w
t
+
m
g
cos
ω
t
{\displaystyle {\frac {mg}{2}}\cos \omega t=-m\omega ^{2}{\frac {g}{2\omega ^{2}}}\cos wt+mg\cos \omega t}
{
x
(
t
)
=
C
1
e
ω
t
+
c
2
e
−
∞
−
g
2
ω
2
cos
ω
t
p
(
t
)
=
m
ω
(
c
1
e
ω
t
−
C
2
e
−
ω
t
)
+
g
2
ω
sin
ω
t
)
{\displaystyle {\begin{cases}x(t)=C_{1}e^{\omega t}+c_{2}e^{-\infty }-{\frac {g}{2\omega ^{2}}}\cos \omega t\\p(t)=m\omega (c_{1}e^{\omega t}-C_{2}e^{-\omega t})+{\frac {g}{2\omega }}\sin \omega t)\end{cases}}}
B
˙
+
1
m
⋅
ω
m
B
=
m
g
cos
ω
t
{\displaystyle {\dot {B}}+{\frac {1}{m}}\cdot \omega mB=mg\cos \omega t}
B
˙
+
ω
B
=
m
g
cos
ω
t
{\displaystyle {\dot {B}}+\omega B=mg\cos \omega t}
B
0
=
α
e
−
ω
t
{\displaystyle B_{0}=\alpha e^{-\omega t}}
B
1
=
α
sin
ω
t
+
β
cos
ω
t
{\displaystyle B_{1}=\alpha \sin \omega t+\beta \cos \omega t}
B
˙
1
=
−
ω
α
sin
ω
t
+
ω
β
cos
ω
t
{\displaystyle {\dot {B}}_{1}=-\omega \alpha \sin \omega t+\omega \beta \cos \omega t}
−
ω
α
cos
ω
t
+
ω
β
sin
ω
t
+
ω
α
sin
ω
t
+
ω
β
cos
ω
t
=
m
g
cos
ω
t
{\displaystyle -\omega \alpha \cos \omega t+\omega \beta \sin \omega t+\omega \alpha \sin \omega t+\omega \beta \cos \omega t=mg\cos \omega t}
{
−
w
α
+
ω
β
=
m
g
ω
β
=
−
ω
α
{\displaystyle {\begin{cases}-w\alpha +\omega \beta =mg\\\omega \beta =-\omega \alpha \end{cases}}}
β
=
m
g
2
ω
{\displaystyle \beta ={\frac {mg}{2\omega }}}
α
=
−
m
g
2
ω
{\displaystyle \alpha =-{\frac {mg}{2\omega }}}
B
=
α
e
−
ω
t
+
β
(
cos
ω
t
−
sin
ω
t
)
{\displaystyle B=\alpha e^{-\omega t}+\beta (\cos \omega t-\sin \omega t)}
C
=
∫
−
B
2
2
m
d
t
{\displaystyle C=\int -{\frac {B^{2}}{2m}}dt}
Интегрирование...
C
=
α
2
4
ω
e
−
2
ω
t
−
α
m
g
4
ω
2
(
e
−
ω
t
sin
ω
t
)
−
m
2
g
2
4
ω
2
(
t
+
1
ω
cos
2
ω
t
)
{\displaystyle C={\frac {\alpha ^{2}}{4\omega }}e^{-2\omega t}-\alpha {\frac {mg}{4\omega ^{2}}}\left(e^{-\omega t}\sin \omega t\right)-{\frac {m^{2}g^{2}}{4\omega ^{2}}}\left(t+{\frac {1}{\omega }}\cos 2\omega t\right)}
S
=
ω
m
x
2
2
+
x
(
α
e
−
ω
t
+
m
g
2
ω
(
cos
ω
t
−
sin
ω
t
)
+
C
{\displaystyle S={\frac {\omega mx^{2}}{2}}+x(\alpha e^{-\omega t}+{\frac {mg}{2\omega }}(\cos \omega t-\sin \omega t)+C}
Материальная точка М массы т движется под действием двух ньютоновских сил притяжения, центры С1 и С2 которых расположены на неподвижной оси, расстояние между ними равно 2с. Найти интегралы канонических уравнений Гамильтона методом Якоби.
За обобщённые координаты принять угол поворота плоскости, определяемой точками М, С1 и С2 и эллиптические координаты в этой плоскости
λ
=
r
1
+
r
2
2
c
{\displaystyle \lambda ={\frac {r_{1}+r_{2}}{2c}}}
μ
=
r
1
−
r
2
2
c
{\displaystyle \mu ={\frac {r_{1}-r_{2}}{2c}}}
где r1 = С1 М, r2 = С2 М.
λ
=
r
1
+
r
2
2
c
{\displaystyle \lambda ={\frac {r_{1}+r_{2}}{2c}}}
μ
=
r
1
−
r
2
2
c
{\displaystyle \mu ={\frac {r_{1}-r_{2}}{2c}}}
z
=
c
λ
μ
{\displaystyle z=c\lambda \mu }
r
=
c
λ
2
−
1
1
−
μ
2
{\displaystyle r=c{\sqrt {\lambda ^{2}-1}}{\sqrt {1-\mu ^{2}}}}
r
1
=
c
(
λ
+
μ
)
{\displaystyle r_{1}=c(\lambda +\mu )}
r
2
=
c
(
λ
−
μ
)
{\displaystyle r_{2}=c(\lambda -\mu )}
Π
=
−
f
1
r
1
−
f
2
r
2
=
−
1
c
(
λ
2
−
μ
2
)
[
(
f
1
+
f
2
)
λ
+
(
f
2
−
f
1
)
μ
]
{\displaystyle \Pi =-{\frac {f_{1}}{r_{1}}}-{\frac {f_{2}}{r_{2}}}={\frac {-1}{c\left(\lambda ^{2}-\mu ^{2}\right)}}[\left(f_{1}+f_{2}\right)\lambda +\left(f_{2}-f_{1}\right)\mu ]}
T
=
1
2
(
z
˙
2
+
r
˙
2
+
r
2
φ
˙
2
)
=
1
2
c
2
(
(
λ
2
−
μ
2
)
(
λ
˙
2
λ
2
−
1
+
μ
˙
2
1
−
μ
2
)
+
φ
2
(
λ
2
−
1
)
(
1
−
μ
2
)
{\displaystyle T={\frac {1}{2}}\left({\dot {z}}^{2}+{\dot {r}}^{2}+r^{2}{\dot {\varphi }}^{2}\right)={\frac {1}{2}}c^{2}\left(\left(\lambda ^{2}-\mu ^{2}\right)\right.\left({\frac {{\dot {\lambda }}^{2}}{\lambda ^{2}-1}}+{\frac {{\dot {\mu }}^{2}}{1-\mu ^{2}}}\right)+\varphi ^{2}\left(\lambda ^{2}-1\right)\left(1-\mu ^{2}\right)}
H
=
1
2
c
2
[
p
λ
2
(
λ
2
−
1
)
λ
2
−
μ
2
+
p
μ
2
(
1
−
μ
2
)
λ
2
−
μ
2
+
p
φ
2
(
λ
2
−
1
)
(
1
−
μ
2
)
]
−
1
c
(
λ
2
−
μ
2
)
[
(
f
1
+
f
2
)
λ
+
(
f
2
−
f
1
)
μ
]
{\displaystyle H={\frac {1}{2c^{2}}}\left[{\frac {p_{\lambda }^{2}\left(\lambda ^{2}-1\right)}{\lambda ^{2}-\mu ^{2}}}+{\frac {p_{\mu }^{2}\left(1-\mu ^{2}\right)}{\lambda ^{2}-\mu ^{2}}}+{\frac {p_{\varphi }^{2}}{(\lambda ^{2}-1)\left(1-\mu ^{2}\right)}}\right]-{\frac {1}{c\left(\lambda ^{2}-\mu ^{2}\right)}}[\left(f_{1}+f_{2}\right)\lambda +\left(f_{2}-f_{1}\right)\mu ]}
W
=
β
φ
φ
+
W
∗
(
λ
,
μ
)
{\displaystyle W=\beta _{\varphi }\varphi +W^{*}(\lambda ,\mu )}
(
λ
2
−
1
)
(
∂
W
∗
∂
r
)
2
+
(
1
−
μ
2
)
(
∂
W
∗
∂
q
μ
)
2
+
β
φ
2
λ
2
−
1
+
β
φ
2
1
−
μ
2
−
2
c
[
(
f
1
+
f
2
)
λ
+
(
f
2
−
f
1
)
μ
]
=
2
h
c
2
(
λ
2
−
μ
2
)
{\displaystyle \left(\lambda ^{2}-1\right)\left({\frac {\partial W^{*}}{\partial r}}\right)^{2}+\left(1-\mu ^{2}\right)\left({\frac {\partial W^{*}}{\partial q\mu }}\right)^{2}+{\frac {\beta _{\varphi }^{2}}{\lambda ^{2}-1}}+{\frac {\beta _{\varphi }^{2}}{1-\mu ^{2}}}-2c\left[\left(f_{1}+f_{2}\right)\lambda +\left(f_{2}-f_{1}\right)\mu \right]=2hc^{2}\left(\lambda ^{2}-\mu ^{2}\right)}
W
∗
(
λ
,
μ
)
=
W
1
(
λ
)
+
W
2
(
μ
)
{\displaystyle W^{*}(\lambda ,\mu )=W_{1}(\lambda )+W_{2}(\mu )}
(
λ
2
−
1
)
(
∂
W
1
∂
μ
)
2
+
β
φ
2
λ
2
−
1
−
2
c
(
f
1
+
f
2
)
λ
−
2
h
c
2
λ
2
=
−
[
(
1
−
μ
2
)
(
d
W
2
d
μ
)
2
+
β
φ
2
1
−
μ
2
−
2
c
(
f
2
−
f
1
)
μ
+
2
h
c
2
μ
2
]
{\displaystyle \left(\lambda ^{2}-1\right)\left({\frac {\partial W_{1}}{\partial \mu }}\right)^{2}+{\frac {\beta _{\varphi }^{2}}{\lambda ^{2}-1}}-2c\left(f_{1}+f_{2}\right)\lambda -2hc^{2}\lambda ^{2}=-[\left(1-\mu ^{2}\right)\left({\frac {dW_{2}}{d\mu }}\right)^{2}+{\frac {\beta _{\varphi }^{2}}{1-\mu ^{2}}}-2c\left(f_{2}-f_{1}\right)\mu +2hc^{2}\mu ^{2}]}
(
λ
2
−
1
)
(
∂
W
1
∂
μ
)
2
=
−
β
−
β
φ
2
λ
2
−
1
+
2
c
(
f
1
+
f
2
)
λ
+
2
h
c
2
λ
2
{\displaystyle \left(\lambda ^{2}-1\right)\left({\frac {\partial W_{1}}{\partial \mu }}\right)^{2}=-\beta -{\frac {\beta _{\varphi }^{2}}{\lambda ^{2}-1}}+2c\left(f_{1}+f_{2}\right)\lambda +2hc^{2}\lambda ^{2}}
(
1
−
μ
2
)
(
d
W
2
d
μ
)
2
=
β
−
β
φ
2
1
−
μ
2
+
2
c
(
f
2
−
f
1
)
μ
−
2
h
c
2
μ
2
{\displaystyle \left(1-\mu ^{2}\right)\left({\frac {dW_{2}}{d\mu }}\right)^{2}=\beta -{\frac {\beta \varphi ^{2}}{1-\mu ^{2}}}+2c\left(f_{2}-f_{1}\right)\mu -2hc^{2}\mu ^{2}}
W
=
β
φ
φ
+
∫
F
1
(
λ
)
d
λ
λ
2
−
1
+
∫
F
2
(
μ
)
d
μ
μ
2
−
1
{\displaystyle W=\beta _{\varphi }\varphi +\int {\sqrt {F_{1}(\lambda )}}{\frac {d\lambda }{\lambda ^{2}-1}}+\int {\sqrt {F_{2}(\mu )}}{\frac {d\mu }{\mu ^{2}-1}}}
F
1
(
λ
)
=
(
λ
2
−
1
)
(
2
h
c
2
λ
2
+
2
c
(
f
1
+
f
2
)
λ
−
β
)
−
β
φ
2
{\displaystyle F_{1}(\lambda )=\left(\lambda ^{2}-1\right)\left(2hc^{2}\lambda ^{2}+2c\left(f_{1}+f_{2}\right)\lambda -\beta \right)-\beta _{\varphi }^{2}}
F
2
(
μ
)
=
(
1
−
μ
2
)
(
−
2
h
c
2
μ
2
+
2
c
(
k
2
−
k
1
)
μ
+
α
)
−
α
φ
2
{\displaystyle F_{2}(\mu )=\left(1-\mu ^{2}\right)\left(-2hc^{2}\mu ^{2}+2c\left(k_{2}-k_{1}\right)\mu +\alpha \right)-\alpha _{\varphi }^{2}}
∂
W
∂
β
φ
=
φ
−
β
φ
∫
d
λ
(
λ
2
−
1
)
F
1
(
λ
)
−
β
φ
∫
d
μ
(
1
−
μ
2
)
F
2
(
μ
)
=
α
φ
{\displaystyle {\frac {\partial W}{\partial \beta _{\varphi }}}=\varphi -\beta _{\varphi }\int {\frac {d\lambda }{\left(\lambda ^{2}-1\right){\sqrt {F_{1}(\lambda )}}}}-\beta _{\varphi }\int {\frac {d\mu }{\left(1-\mu ^{2}\right){\sqrt {F_{2}(\mu )}}}}=\alpha _{\varphi }}
∂
W
∂
β
=
−
∫
d
λ
F
1
(
λ
)
+
∫
d
μ
F
2
(
μ
)
=
2
α
{\displaystyle {\frac {\partial W}{\partial \beta }}=-\int {\frac {d\lambda }{\sqrt {F_{1}(\lambda )}}}+\int {\frac {d\mu }{\sqrt {F_{2}(\mu )}}}=2\alpha }
∂
W
∂
h
=
c
2
∫
λ
2
d
λ
F
1
(
λ
)
−
c
2
∫
μ
2
d
μ
F
2
(
μ
)
=
t
−
t
0
{\displaystyle {\frac {\partial W}{\partial h}}=c^{2}\int {\frac {\lambda ^{2}d\lambda }{\sqrt {F_{1}(\lambda )}}}-c^{2}\int {\frac {\mu ^{2}d\mu }{\sqrt {F_{2}(\mu )}}}=t-t_{0}}
p
φ
=
β
φ
{\displaystyle p_{\varphi }=\beta _{\varphi }}
p
λ
=
F
1
λ
2
−
1
{\displaystyle p_{\lambda }={\frac {\sqrt {F_{1}}}{\lambda ^{2}-1}}}
p
μ
=
F
2
μ
2
−
1
{\displaystyle p_{\mu }={\frac {\sqrt {F_{2}}}{\mu ^{2}-1}}}
Доказать свойства скобок Пуассона...
1.
{
F
,
G
}
=
−
{
G
,
F
}
{\displaystyle \{F,G\}=-\{G,F\}}
править
{
F
,
G
}
=
∑
j
=
1
n
(
∂
F
∂
q
j
⋅
∂
G
∂
p
j
−
∂
F
∂
p
j
⋅
∂
G
∂
q
j
)
{\displaystyle \{F,G\}=\sum _{j=1}^{n}\left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}\right)}
−
{
G
,
F
}
=
∑
j
=
1
n
(
−
∂
G
∂
q
j
⋅
∂
F
∂
p
j
+
∂
G
∂
p
j
⋅
∂
F
∂
q
j
)
{\displaystyle -\{G,F\}=\sum _{j=1}^{n}\left(-{\frac {\partial G}{\partial q_{j}}}\cdot {\frac {\partial F}{\partial p_{j}}}+{\frac {\partial G}{\partial p_{j}}}\cdot {\frac {\partial F}{\partial q_{j}}}\right)}
2. Если
F
=
φ
(
f
1
,
…
,
f
m
)
{\displaystyle F=\varphi (f_{1,}\ldots ,f_{m})}
{
F
,
G
}
=
∑
i
=
1
m
∂
φ
∂
f
i
{
f
i
,
G
}
{\displaystyle \{F,G\}=\sum _{i=1}^{m}{\frac {\partial \varphi }{\partial f_{i}}}\{f_{i},G\}}
править
{
F
,
G
}
=
∑
s
=
1
n
(
∂
F
∂
q
s
⋅
∂
G
∂
p
s
−
∂
F
∂
p
s
⋅
∂
G
∂
q
s
)
{\displaystyle \{F,G\}=\sum _{s=1}^{n}\left({\frac {\partial F}{\partial q_{s}}}\cdot {\frac {\partial G}{\partial p_{s}}}-{\frac {\partial F}{\partial p_{s}}}\cdot {\frac {\partial G}{\partial q_{s}}}\right)}
∂
F
∂
q
s
=
∑
i
=
1
m
∂
φ
∂
f
i
∂
f
i
∂
q
s
∂
F
∂
p
s
=
∑
i
=
1
m
∂
φ
∂
f
i
∂
f
i
∂
p
s
{\displaystyle {\frac {\partial F}{\partial q_{s}}}=\sum _{i=1}^{m}{\frac {\partial \varphi }{\partial f_{i}}}{\frac {\partial f_{i}}{\partial q_{s}}}\quad {\frac {\partial F}{\partial p_{s}}}=\sum _{i=1}^{m}{\frac {\partial \varphi }{\partial f_{i}}}{\frac {\partial f_{i}}{\partial p_{s}}}}
{
F
,
G
}
=
∑
s
=
1
n
(
∑
i
=
1
m
∂
φ
∂
f
i
∂
f
i
∂
q
s
⋅
∂
G
∂
p
s
−
∑
i
=
1
m
∂
φ
∂
f
i
∂
f
i
∂
p
s
⋅
∂
G
∂
q
s
)
=
∑
s
=
1
n
[
∂
φ
∂
f
i
(
∑
i
=
1
m
∂
f
i
∂
q
s
⋅
∂
G
∂
p
s
−
∑
i
=
1
m
∂
f
i
∂
p
s
⋅
∂
G
∂
q
s
)
]
=
∑
i
=
1
m
∂
φ
∂
f
i
{
f
i
,
G
}
{\displaystyle \{F,G\}=\sum _{s=1}^{n}\left(\sum _{i=1}^{m}{\frac {\partial \varphi }{\partial f_{i}}}{\frac {\partial f_{i}}{\partial q_{s}}}\cdot {\frac {\partial G}{\partial p_{s}}}-\sum _{i=1}^{m}{\frac {\partial \varphi }{\partial f_{i}}}{\frac {\partial f_{i}}{\partial p_{s}}}\cdot {\frac {\partial G}{\partial q_{s}}}\right)=\sum _{s=1}^{n}\left[{\frac {\partial \varphi }{\partial f_{i}}}\left(\sum _{i=1}^{m}{\frac {\partial f_{i}}{\partial q_{s}}}\cdot {\frac {\partial G}{\partial p_{s}}}-\sum _{i=1}^{m}{\frac {\partial f_{i}}{\partial p_{s}}}\cdot {\frac {\partial G}{\partial q_{s}}}\right)\right]=\sum _{i=1}^{m}{\frac {\partial \varphi }{\partial f_{i}}}\{f_{i},G\}}
3.
{
{
F
,
G
}
,
H
}
+
{
{
H
,
F
}
,
G
}
+
{
{
G
,
H
}
,
F
}
=
0
{\displaystyle \{\{F,G\},H\}+\{\{H,F\},G\}+\{\{G,H\},F\}=0}
править
{
F
,
G
}
=
∑
s
=
1
n
(
∂
F
∂
q
s
⋅
∂
G
∂
p
s
−
∂
F
∂
p
s
⋅
∂
G
∂
q
s
)
{\displaystyle \{F,G\}=\sum _{s=1}^{n}\left({\frac {\partial F}{\partial q_{s}}}\cdot {\frac {\partial G}{\partial p_{s}}}-{\frac {\partial F}{\partial p_{s}}}\cdot {\frac {\partial G}{\partial q_{s}}}\right)}
{
{
F
,
G
}
,
H
}
=
∑
s
=
1
n
(
∂
{
F
,
G
}
∂
q
s
⋅
∂
H
∂
p
s
−
∂
{
F
,
G
}
∂
p
s
⋅
∂
H
∂
q
s
)
=
∑
s
=
1
n
(
∂
[
∑
j
=
1
n
(
∂
F
∂
q
j
⋅
∂
G
∂
p
j
−
∂
F
∂
p
j
⋅
∂
G
∂
q
j
)
]
∂
q
s
⋅
∂
H
∂
p
s
−
∂
[
∑
j
=
1
n
(
∂
F
∂
q
j
⋅
∂
G
∂
p
j
−
∂
F
∂
p
j
⋅
∂
G
∂
q
j
)
]
∂
p
s
⋅
∂
H
∂
q
s
)
{\displaystyle \{\{F,G\},H\}=\sum _{s=1}^{n}\left({\frac {\partial \{F,G\}}{\partial q_{s}}}\cdot {\frac {\partial H}{\partial p_{s}}}-{\frac {\partial \{F,G\}}{\partial p_{s}}}\cdot {\frac {\partial H}{\partial q_{s}}}\right)=\sum _{s=1}^{n}\left({\frac {\partial \left[\sum _{j=1}^{n}\left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}\right)\right]}{\partial q_{s}}}\cdot {\frac {\partial H}{\partial p_{s}}}-{\frac {\partial \left[\sum _{j=1}^{n}\left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}\right)\right]}{\partial p_{s}}}\cdot {\frac {\partial H}{\partial q_{s}}}\right)}
∂
[
∑
j
=
1
n
(
∂
F
∂
q
j
⋅
∂
G
∂
p
j
−
∂
F
∂
p
j
⋅
∂
G
∂
q
j
)
]
∂
x
=
∑
j
=
1
n
∂
(
∂
F
∂
q
j
⋅
∂
G
∂
p
j
−
∂
F
∂
p
j
⋅
∂
G
∂
q
j
)
∂
x
=
∑
j
=
1
n
[
∂
(
∂
F
∂
q
j
⋅
∂
G
∂
p
j
)
∂
x
−
∂
(
∂
F
∂
p
j
⋅
∂
G
∂
q
j
)
∂
x
]
{\displaystyle {\frac {\partial \left[\sum _{j=1}^{n}\left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}\right)\right]}{\partial x}}=\sum _{j=1}^{n}{\frac {\partial \left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}\right)}{\partial x}}=\sum _{j=1}^{n}\left[{\frac {\partial \left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}\right)}{\partial x}}-{\frac {\partial \left({\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}\right)}{\partial x}}\right]}
∂
(
∂
F
∂
q
j
⋅
∂
G
∂
p
j
)
∂
x
=
∂
2
F
∂
x
∂
q
j
⋅
∂
G
∂
p
j
+
∂
F
∂
q
j
⋅
∂
2
G
∂
x
∂
p
j
{\displaystyle {\frac {\partial \left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}\right)}{\partial x}}={\frac {\partial ^{2}F}{\partial x\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}+{\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial ^{2}G}{\partial x\partial p_{j}}}}
∑
j
=
1
n
[
∂
(
∂
F
∂
q
j
⋅
∂
G
∂
p
j
)
∂
x
−
∂
(
∂
F
∂
p
j
⋅
∂
G
∂
q
j
)
∂
x
]
=
∑
j
=
1
n
(
∂
2
F
∂
x
∂
q
j
⋅
∂
G
∂
p
j
+
∂
F
∂
q
j
⋅
∂
2
G
∂
x
∂
p
j
−
∂
2
F
∂
x
∂
p
j
⋅
∂
G
∂
q
j
−
∂
F
∂
p
j
⋅
∂
2
G
∂
x
∂
q
j
)
{\displaystyle \sum _{j=1}^{n}\left[{\frac {\partial \left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}\right)}{\partial x}}-{\frac {\partial \left({\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}\right)}{\partial x}}\right]=\sum _{j=1}^{n}\left({\frac {\partial ^{2}F}{\partial x\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}+{\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial ^{2}G}{\partial x\partial p_{j}}}-{\frac {\partial ^{2}F}{\partial x\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial ^{2}G}{\partial x\partial q_{j}}}\right)}
∑
j
=
1
n
(
∂
2
F
∂
x
∂
q
j
⋅
∂
G
∂
p
j
+
∂
F
∂
q
j
⋅
∂
2
G
∂
x
∂
p
j
−
∂
2
F
∂
x
∂
p
j
⋅
∂
G
∂
q
j
−
∂
F
∂
p
j
⋅
∂
2
G
∂
x
∂
q
j
)
{\displaystyle \sum _{j=1}^{n}\left({\frac {\partial ^{2}F}{\partial x\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}+{\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial ^{2}G}{\partial x\partial p_{j}}}-{\frac {\partial ^{2}F}{\partial x\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial ^{2}G}{\partial x\partial q_{j}}}\right)}
∑
j
=
1
n
(
∂
2
F
∂
q
s
∂
q
j
⋅
∂
G
∂
p
j
+
∂
F
∂
q
j
⋅
∂
2
G
∂
q
s
∂
p
j
−
∂
2
F
∂
q
s
∂
p
j
⋅
∂
G
∂
q
j
−
∂
F
∂
p
j
⋅
∂
2
G
∂
q
s
∂
q
j
)
{\displaystyle \sum _{j=1}^{n}\left({\frac {\partial ^{2}F}{\partial q_{s}\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}+{\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial ^{2}G}{\partial q_{s}\partial p_{j}}}-{\frac {\partial ^{2}F}{\partial q_{s}\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial ^{2}G}{\partial q_{s}\partial q_{j}}}\right)}
∑
j
=
1
n
(
∂
2
F
∂
p
s
∂
q
j
⋅
∂
G
∂
p
j
+
∂
F
∂
q
j
⋅
∂
2
G
∂
p
s
∂
p
j
−
∂
2
F
∂
p
s
∂
p
j
⋅
∂
G
∂
q
j
−
∂
F
∂
p
j
⋅
∂
2
G
∂
p
s
∂
q
j
)
{\displaystyle \sum _{j=1}^{n}\left({\frac {\partial ^{2}F}{\partial p_{s}\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}+{\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial ^{2}G}{\partial p_{s}\partial p_{j}}}-{\frac {\partial ^{2}F}{\partial p_{s}\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial ^{2}G}{\partial p_{s}\partial q_{j}}}\right)}
{
{
F
,
G
}
,
H
}
=
{\displaystyle \{\{F,G\},H\}=}
=
∑
s
=
1
n
∑
j
=
1
n
(
[
∂
2
F
∂
q
s
∂
q
j
⋅
∂
G
∂
p
j
+
∂
F
∂
q
j
⋅
∂
2
G
∂
q
s
∂
p
j
−
∂
2
F
∂
q
s
∂
p
j
⋅
∂
G
∂
q
j
−
∂
F
∂
p
j
⋅
∂
2
G
∂
q
s
∂
q
j
]
⋅
∂
H
∂
p
s
−
[
∂
2
F
∂
p
s
∂
q
j
⋅
∂
G
∂
p
j
+
∂
F
∂
q
j
⋅
∂
2
G
∂
p
s
∂
p
j
−
∂
2
F
∂
p
s
∂
p
j
⋅
∂
G
∂
q
j
−
∂
F
∂
p
j
⋅
∂
2
G
∂
p
s
∂
q
j
]
⋅
∂
H
∂
q
s
)
{\displaystyle =\sum _{s=1}^{n}\sum _{j=1}^{n}\left(\left[{\frac {\partial ^{2}F}{\partial q_{s}\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}+{\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial ^{2}G}{\partial q_{s}\partial p_{j}}}-{\frac {\partial ^{2}F}{\partial q_{s}\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial ^{2}G}{\partial q_{s}\partial q_{j}}}\right]\cdot {\frac {\partial H}{\partial p_{s}}}-\left[{\frac {\partial ^{2}F}{\partial p_{s}\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}+{\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial ^{2}G}{\partial p_{s}\partial p_{j}}}-{\frac {\partial ^{2}F}{\partial p_{s}\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial ^{2}G}{\partial p_{s}\partial q_{j}}}\right]\cdot {\frac {\partial H}{\partial q_{s}}}\right)}
{
{
F
,
G
}
,
H
}
+
{
{
H
,
F
}
,
G
}
+
{
{
G
,
H
}
,
F
}
=
{\displaystyle \{\{F,G\},H\}+\{\{H,F\},G\}+\{\{G,H\},F\}=}
=
∑
s
=
1
n
∑
j
=
1
n
(
[
∂
2
F
∂
q
s
∂
q
j
⋅
∂
G
∂
p
j
+
∂
F
∂
q
j
⋅
∂
2
G
∂
q
s
∂
p
j
−
∂
2
F
∂
q
s
∂
p
j
⋅
∂
G
∂
q
j
−
∂
F
∂
p
j
⋅
∂
2
G
∂
q
s
∂
q
j
]
⋅
∂
H
∂
p
s
−
[
∂
2
F
∂
p
s
∂
q
j
⋅
∂
G
∂
p
j
+
∂
F
∂
q
j
⋅
∂
2
G
∂
p
s
∂
p
j
−
∂
2
F
∂
p
s
∂
p
j
⋅
∂
G
∂
q
j
−
∂
F
∂
p
j
⋅
∂
2
G
∂
p
s
∂
q
j
]
⋅
∂
H
∂
q
s
)
+
{\displaystyle =\sum _{s=1}^{n}\sum _{j=1}^{n}\left(\left[{\frac {\partial ^{2}F}{\partial q_{s}\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}+{\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial ^{2}G}{\partial q_{s}\partial p_{j}}}-{\frac {\partial ^{2}F}{\partial q_{s}\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial ^{2}G}{\partial q_{s}\partial q_{j}}}\right]\cdot {\frac {\partial H}{\partial p_{s}}}-\left[{\frac {\partial ^{2}F}{\partial p_{s}\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}+{\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial ^{2}G}{\partial p_{s}\partial p_{j}}}-{\frac {\partial ^{2}F}{\partial p_{s}\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial ^{2}G}{\partial p_{s}\partial q_{j}}}\right]\cdot {\frac {\partial H}{\partial q_{s}}}\right)+}
+
∑
s
=
1
n
∑
j
=
1
n
(
[
∂
2
H
∂
q
s
∂
q
j
⋅
∂
F
∂
p
j
+
∂
H
∂
q
j
⋅
∂
2
F
∂
q
s
∂
p
j
−
∂
2
H
∂
q
s
∂
p
j
⋅
∂
F
∂
q
j
−
∂
H
∂
p
j
⋅
∂
2
F
∂
q
s
∂
q
j
]
⋅
∂
G
∂
p
s
−
[
∂
2
H
∂
p
s
∂
q
j
⋅
∂
F
∂
p
j
+
∂
H
∂
q
j
⋅
∂
2
F
∂
p
s
∂
p
j
−
∂
2
H
∂
p
s
∂
p
j
⋅
∂
F
∂
q
j
−
∂
H
∂
p
j
⋅
∂
2
F
∂
p
s
∂
q
j
]
⋅
∂
G
∂
q
s
)
+
{\displaystyle +\sum _{s=1}^{n}\sum _{j=1}^{n}\left(\left[{\frac {\partial ^{2}H}{\partial q_{s}\partial q_{j}}}\cdot {\frac {\partial F}{\partial p_{j}}}+{\frac {\partial H}{\partial q_{j}}}\cdot {\frac {\partial ^{2}F}{\partial q_{s}\partial p_{j}}}-{\frac {\partial ^{2}H}{\partial q_{s}\partial p_{j}}}\cdot {\frac {\partial F}{\partial q_{j}}}-{\frac {\partial H}{\partial p_{j}}}\cdot {\frac {\partial ^{2}F}{\partial q_{s}\partial q_{j}}}\right]\cdot {\frac {\partial G}{\partial p_{s}}}-\left[{\frac {\partial ^{2}H}{\partial p_{s}\partial q_{j}}}\cdot {\frac {\partial F}{\partial p_{j}}}+{\frac {\partial H}{\partial q_{j}}}\cdot {\frac {\partial ^{2}F}{\partial p_{s}\partial p_{j}}}-{\frac {\partial ^{2}H}{\partial p_{s}\partial p_{j}}}\cdot {\frac {\partial F}{\partial q_{j}}}-{\frac {\partial H}{\partial p_{j}}}\cdot {\frac {\partial ^{2}F}{\partial p_{s}\partial q_{j}}}\right]\cdot {\frac {\partial G}{\partial q_{s}}}\right)+}
+
∑
s
=
1
n
∑
j
=
1
n
(
[
∂
2
G
∂
q
s
∂
q
j
⋅
∂
H
∂
p
j
+
∂
G
∂
q
j
⋅
∂
2
H
∂
q
s
∂
p
j
−
∂
2
G
∂
q
s
∂
p
j
⋅
∂
H
∂
q
j
−
∂
G
∂
p
j
⋅
∂
2
H
∂
q
s
∂
q
j
]
⋅
∂
F
∂
p
s
−
[
∂
2
G
∂
p
s
∂
q
j
⋅
∂
H
∂
p
j
+
∂
G
∂
q
j
⋅
∂
2
H
∂
p
s
∂
p
j
−
∂
2
G
∂
p
s
∂
p
j
⋅
∂
H
∂
q
j
−
∂
G
∂
p
j
⋅
∂
2
H
∂
p
s
∂
q
j
]
⋅
∂
F
∂
q
s
)
=
0
{\displaystyle +\sum _{s=1}^{n}\sum _{j=1}^{n}\left(\left[{\frac {\partial ^{2}G}{\partial q_{s}\partial q_{j}}}\cdot {\frac {\partial H}{\partial p_{j}}}+{\frac {\partial G}{\partial q_{j}}}\cdot {\frac {\partial ^{2}H}{\partial q_{s}\partial p_{j}}}-{\frac {\partial ^{2}G}{\partial q_{s}\partial p_{j}}}\cdot {\frac {\partial H}{\partial q_{j}}}-{\frac {\partial G}{\partial p_{j}}}\cdot {\frac {\partial ^{2}H}{\partial q_{s}\partial q_{j}}}\right]\cdot {\frac {\partial F}{\partial p_{s}}}-\left[{\frac {\partial ^{2}G}{\partial p_{s}\partial q_{j}}}\cdot {\frac {\partial H}{\partial p_{j}}}+{\frac {\partial G}{\partial q_{j}}}\cdot {\frac {\partial ^{2}H}{\partial p_{s}\partial p_{j}}}-{\frac {\partial ^{2}G}{\partial p_{s}\partial p_{j}}}\cdot {\frac {\partial H}{\partial q_{j}}}-{\frac {\partial G}{\partial p_{j}}}\cdot {\frac {\partial ^{2}H}{\partial p_{s}\partial q_{j}}}\right]\cdot {\frac {\partial F}{\partial q_{s}}}\right)=0}
4.
∂
{
F
,
G
}
∂
x
=
{
∂
F
∂
x
,
G
}
+
{
F
,
∂
G
∂
x
}
{\displaystyle {\frac {\partial \{F,G\}}{\partial x}}=\left\{{\frac {\partial F}{\partial x}},G\right\}+\left\{F,{\frac {\partial G}{\partial x}}\right\}}
править
∂
[
∑
j
=
1
n
(
∂
F
∂
q
j
⋅
∂
G
∂
p
j
−
∂
F
∂
p
j
⋅
∂
G
∂
q
j
)
]
∂
x
=
∑
j
=
1
n
∂
(
∂
F
∂
q
j
⋅
∂
G
∂
p
j
−
∂
F
∂
p
j
⋅
∂
G
∂
q
j
)
∂
x
=
∑
j
=
1
n
[
∂
(
∂
F
∂
q
j
⋅
∂
G
∂
p
j
)
∂
x
−
∂
(
∂
F
∂
p
j
⋅
∂
G
∂
q
j
)
∂
x
]
{\displaystyle {\frac {\partial \left[\sum _{j=1}^{n}\left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}\right)\right]}{\partial x}}=\sum _{j=1}^{n}{\frac {\partial \left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}\right)}{\partial x}}=\sum _{j=1}^{n}\left[{\frac {\partial \left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}\right)}{\partial x}}-{\frac {\partial \left({\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}\right)}{\partial x}}\right]}
∂
(
∂
F
∂
q
j
⋅
∂
G
∂
p
j
)
∂
x
=
∂
2
F
∂
x
∂
q
j
⋅
∂
G
∂
p
j
+
∂
F
∂
q
j
⋅
∂
2
G
∂
x
∂
p
j
{\displaystyle {\frac {\partial \left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}\right)}{\partial x}}={\frac {\partial ^{2}F}{\partial x\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}+{\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial ^{2}G}{\partial x\partial p_{j}}}}
∑
j
=
1
n
[
∂
(
∂
F
∂
q
j
⋅
∂
G
∂
p
j
)
∂
x
−
∂
(
∂
F
∂
p
j
⋅
∂
G
∂
q
j
)
∂
x
]
=
∑
j
=
1
n
(
∂
2
F
∂
x
∂
q
j
⋅
∂
G
∂
p
j
+
∂
F
∂
q
j
⋅
∂
2
G
∂
x
∂
p
j
−
∂
2
F
∂
x
∂
p
j
⋅
∂
G
∂
q
j
−
∂
F
∂
p
j
⋅
∂
2
G
∂
x
∂
q
j
)
=
{\displaystyle \sum _{j=1}^{n}\left[{\frac {\partial \left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}\right)}{\partial x}}-{\frac {\partial \left({\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}\right)}{\partial x}}\right]=\sum _{j=1}^{n}\left({\frac {\partial ^{2}F}{\partial x\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}+{\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial ^{2}G}{\partial x\partial p_{j}}}-{\frac {\partial ^{2}F}{\partial x\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial ^{2}G}{\partial x\partial q_{j}}}\right)=}
=
∑
j
=
1
n
(
∂
(
∂
F
∂
x
)
∂
q
j
⋅
∂
G
∂
p
j
−
∂
(
∂
F
∂
x
)
∂
p
j
⋅
∂
G
∂
q
j
+
∂
F
∂
q
j
⋅
∂
(
∂
G
∂
x
)
∂
p
j
−
∂
F
∂
p
j
⋅
∂
(
∂
G
∂
x
)
∂
q
j
)
=
{
∂
F
∂
x
,
G
}
+
{
F
,
∂
G
∂
x
}
{\displaystyle =\sum _{j=1}^{n}\left({\frac {\partial \left({\frac {\partial F}{\partial x}}\right)}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}-{\frac {\partial \left({\frac {\partial F}{\partial x}}\right)}{\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}+{\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial \left({\frac {\partial G}{\partial x}}\right)}{\partial p_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial \left({\frac {\partial G}{\partial x}}\right)}{\partial q_{j}}}\right)=\left\{{\frac {\partial F}{\partial x}},G\right\}+\left\{F,{\frac {\partial G}{\partial x}}\right\}}
H
=
F
{
f
n
[
…
f
2
(
f
1
(
q
1
,
p
1
)
,
q
2
,
p
2
)
]
,
t
}
{\displaystyle H=F\{f_{n}[\ldots f_{2}(f_{1}\left(q_{1},p_{1}\right),q_{2},p_{2})],t\}}
∑
j
=
1
n
(
∂
f
i
∂
q
j
⋅
∂
F
∂
p
j
−
∂
f
i
∂
p
j
⋅
∂
F
∂
q
j
)
=
{\displaystyle \sum _{j=1}^{n}\left({\frac {\partial f_{i}}{\partial q_{j}}}\cdot {\frac {\partial F}{\partial p_{j}}}-{\frac {\partial f_{i}}{\partial p_{j}}}\cdot {\frac {\partial F}{\partial q_{j}}}\right)=}
=
∑
j
=
1
n
(
∂
f
i
∂
q
j
⋅
∂
F
∂
f
n
⋅
∂
f
n
∂
f
n
−
1
⋯
∂
f
j
∂
p
j
−
∂
f
i
∂
p
j
⋅
∂
F
∂
f
n
⋅
∂
f
n
∂
f
n
−
1
⋯
∂
f
j
∂
q
j
)
=
{\displaystyle =\sum _{j=1}^{n}\left({\frac {\partial f_{i}}{\partial q_{j}}}\cdot {\frac {\partial F}{\partial f_{n}}}\cdot {\frac {\partial f_{n}}{\partial f_{n-1}}}\cdots {\frac {\partial f_{j}}{\partial p_{j}}}-{\frac {\partial f_{i}}{\partial p_{j}}}\cdot {\frac {\partial F}{\partial f_{n}}}\cdot {\frac {\partial f_{n}}{\partial f_{n-1}}}\cdots {\frac {\partial f_{j}}{\partial q_{j}}}\right)=}
=
∑
j
=
1
n
(
∂
F
∂
f
n
⋅
∂
f
n
∂
f
n
−
1
⋯
(
∂
f
i
∂
q
j
⋅
∂
f
j
∂
p
j
−
∂
f
i
∂
p
j
⋅
∂
f
j
∂
q
j
)
)
{\displaystyle =\sum _{j=1}^{n}\left({\frac {\partial F}{\partial f_{n}}}\cdot {\frac {\partial f_{n}}{\partial f_{n-1}}}\cdots \left({\frac {\partial f_{i}}{\partial q_{j}}}\cdot {\frac {\partial f_{j}}{\partial p_{j}}}-{\frac {\partial f_{i}}{\partial p_{j}}}\cdot {\frac {\partial f_{j}}{\partial q_{j}}}\right)\right)}
Это выражение равно
0
{\displaystyle 0}
i
<
j
{\displaystyle i<j}
f
i
{\displaystyle f_{i}}
p
j
,
q
j
{\displaystyle p_{j},q_{j}}
Если же
i
⩾
j
{\displaystyle i\geqslant j}
j
{\displaystyle j}
![{\displaystyle \tau =2\pi {\sqrt {\frac {6l}{g{\sqrt {17}}}}}=2\pi {\frac {\sqrt {6}}{\sqrt[{4}]{17}}}{\sqrt {\frac {l}{g}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48ad06c6444206c1d824c9178f63ec28dc8951d0)
![{\displaystyle s=2{\sqrt {2}}r{\sqrt {1+{\frac {1-t^{2}}{1+t^{2}}}}}t=2{\sqrt {2}}r{\sqrt[{}]{\frac {2}{1+t^{2}}}}={\frac {4rt}{\sqrt {1+t^{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fc799bcec841ba678963f6f51da2c8f2c4a70bc)
![{\displaystyle {\frac {dS}{dt}}+\left[{\frac {1}{2m}}\left({\frac {dS}{\partial x}}\right)^{2}-{\frac {1}{2}}m\omega ^{2}x^{2}\right]-xmg\cos \omega t=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6cac5ab28b937767bea5360fea864b97d6a25e9b)
![{\displaystyle \Pi =-{\frac {f_{1}}{r_{1}}}-{\frac {f_{2}}{r_{2}}}={\frac {-1}{c\left(\lambda ^{2}-\mu ^{2}\right)}}[\left(f_{1}+f_{2}\right)\lambda +\left(f_{2}-f_{1}\right)\mu ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/57da84062bc9d6f9190da0c24ac21733c31432a2)
![{\displaystyle H={\frac {1}{2c^{2}}}\left[{\frac {p_{\lambda }^{2}\left(\lambda ^{2}-1\right)}{\lambda ^{2}-\mu ^{2}}}+{\frac {p_{\mu }^{2}\left(1-\mu ^{2}\right)}{\lambda ^{2}-\mu ^{2}}}+{\frac {p_{\varphi }^{2}}{(\lambda ^{2}-1)\left(1-\mu ^{2}\right)}}\right]-{\frac {1}{c\left(\lambda ^{2}-\mu ^{2}\right)}}[\left(f_{1}+f_{2}\right)\lambda +\left(f_{2}-f_{1}\right)\mu ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/25e4a66c8517eeb0991c965f95ef95e582db026d)
![{\displaystyle \left(\lambda ^{2}-1\right)\left({\frac {\partial W^{*}}{\partial r}}\right)^{2}+\left(1-\mu ^{2}\right)\left({\frac {\partial W^{*}}{\partial q\mu }}\right)^{2}+{\frac {\beta _{\varphi }^{2}}{\lambda ^{2}-1}}+{\frac {\beta _{\varphi }^{2}}{1-\mu ^{2}}}-2c\left[\left(f_{1}+f_{2}\right)\lambda +\left(f_{2}-f_{1}\right)\mu \right]=2hc^{2}\left(\lambda ^{2}-\mu ^{2}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5336b3e334ff8e7576086f545ea004b00aae5242)
![{\displaystyle \left(\lambda ^{2}-1\right)\left({\frac {\partial W_{1}}{\partial \mu }}\right)^{2}+{\frac {\beta _{\varphi }^{2}}{\lambda ^{2}-1}}-2c\left(f_{1}+f_{2}\right)\lambda -2hc^{2}\lambda ^{2}=-[\left(1-\mu ^{2}\right)\left({\frac {dW_{2}}{d\mu }}\right)^{2}+{\frac {\beta _{\varphi }^{2}}{1-\mu ^{2}}}-2c\left(f_{2}-f_{1}\right)\mu +2hc^{2}\mu ^{2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b6ee9dfacb0ca4f6007d21acd08012b3476d5695)
![{\displaystyle \{F,G\}=\sum _{s=1}^{n}\left(\sum _{i=1}^{m}{\frac {\partial \varphi }{\partial f_{i}}}{\frac {\partial f_{i}}{\partial q_{s}}}\cdot {\frac {\partial G}{\partial p_{s}}}-\sum _{i=1}^{m}{\frac {\partial \varphi }{\partial f_{i}}}{\frac {\partial f_{i}}{\partial p_{s}}}\cdot {\frac {\partial G}{\partial q_{s}}}\right)=\sum _{s=1}^{n}\left[{\frac {\partial \varphi }{\partial f_{i}}}\left(\sum _{i=1}^{m}{\frac {\partial f_{i}}{\partial q_{s}}}\cdot {\frac {\partial G}{\partial p_{s}}}-\sum _{i=1}^{m}{\frac {\partial f_{i}}{\partial p_{s}}}\cdot {\frac {\partial G}{\partial q_{s}}}\right)\right]=\sum _{i=1}^{m}{\frac {\partial \varphi }{\partial f_{i}}}\{f_{i},G\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/481806823590551f691e92104513604661f0de91)
![{\displaystyle \{\{F,G\},H\}=\sum _{s=1}^{n}\left({\frac {\partial \{F,G\}}{\partial q_{s}}}\cdot {\frac {\partial H}{\partial p_{s}}}-{\frac {\partial \{F,G\}}{\partial p_{s}}}\cdot {\frac {\partial H}{\partial q_{s}}}\right)=\sum _{s=1}^{n}\left({\frac {\partial \left[\sum _{j=1}^{n}\left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}\right)\right]}{\partial q_{s}}}\cdot {\frac {\partial H}{\partial p_{s}}}-{\frac {\partial \left[\sum _{j=1}^{n}\left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}\right)\right]}{\partial p_{s}}}\cdot {\frac {\partial H}{\partial q_{s}}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c6c28c4b5f3bda51d0f73384268f4eed963716f)
![{\displaystyle {\frac {\partial \left[\sum _{j=1}^{n}\left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}\right)\right]}{\partial x}}=\sum _{j=1}^{n}{\frac {\partial \left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}\right)}{\partial x}}=\sum _{j=1}^{n}\left[{\frac {\partial \left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}\right)}{\partial x}}-{\frac {\partial \left({\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}\right)}{\partial x}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31d7a5f975d4bd293ede9e07b208992edc3ee997)
![{\displaystyle \sum _{j=1}^{n}\left[{\frac {\partial \left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}\right)}{\partial x}}-{\frac {\partial \left({\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}\right)}{\partial x}}\right]=\sum _{j=1}^{n}\left({\frac {\partial ^{2}F}{\partial x\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}+{\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial ^{2}G}{\partial x\partial p_{j}}}-{\frac {\partial ^{2}F}{\partial x\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial ^{2}G}{\partial x\partial q_{j}}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e9598c6f5c4bbbe7398784f25045cc6785abe39)
![{\displaystyle =\sum _{s=1}^{n}\sum _{j=1}^{n}\left(\left[{\frac {\partial ^{2}F}{\partial q_{s}\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}+{\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial ^{2}G}{\partial q_{s}\partial p_{j}}}-{\frac {\partial ^{2}F}{\partial q_{s}\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial ^{2}G}{\partial q_{s}\partial q_{j}}}\right]\cdot {\frac {\partial H}{\partial p_{s}}}-\left[{\frac {\partial ^{2}F}{\partial p_{s}\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}+{\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial ^{2}G}{\partial p_{s}\partial p_{j}}}-{\frac {\partial ^{2}F}{\partial p_{s}\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial ^{2}G}{\partial p_{s}\partial q_{j}}}\right]\cdot {\frac {\partial H}{\partial q_{s}}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/acc8a0d756f57b2e0fc217e3f98c3cd7c1bef655)
![{\displaystyle =\sum _{s=1}^{n}\sum _{j=1}^{n}\left(\left[{\frac {\partial ^{2}F}{\partial q_{s}\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}+{\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial ^{2}G}{\partial q_{s}\partial p_{j}}}-{\frac {\partial ^{2}F}{\partial q_{s}\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial ^{2}G}{\partial q_{s}\partial q_{j}}}\right]\cdot {\frac {\partial H}{\partial p_{s}}}-\left[{\frac {\partial ^{2}F}{\partial p_{s}\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}+{\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial ^{2}G}{\partial p_{s}\partial p_{j}}}-{\frac {\partial ^{2}F}{\partial p_{s}\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial ^{2}G}{\partial p_{s}\partial q_{j}}}\right]\cdot {\frac {\partial H}{\partial q_{s}}}\right)+}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3194d3839375314e34111204e9afbe7528ea749d)
![{\displaystyle +\sum _{s=1}^{n}\sum _{j=1}^{n}\left(\left[{\frac {\partial ^{2}H}{\partial q_{s}\partial q_{j}}}\cdot {\frac {\partial F}{\partial p_{j}}}+{\frac {\partial H}{\partial q_{j}}}\cdot {\frac {\partial ^{2}F}{\partial q_{s}\partial p_{j}}}-{\frac {\partial ^{2}H}{\partial q_{s}\partial p_{j}}}\cdot {\frac {\partial F}{\partial q_{j}}}-{\frac {\partial H}{\partial p_{j}}}\cdot {\frac {\partial ^{2}F}{\partial q_{s}\partial q_{j}}}\right]\cdot {\frac {\partial G}{\partial p_{s}}}-\left[{\frac {\partial ^{2}H}{\partial p_{s}\partial q_{j}}}\cdot {\frac {\partial F}{\partial p_{j}}}+{\frac {\partial H}{\partial q_{j}}}\cdot {\frac {\partial ^{2}F}{\partial p_{s}\partial p_{j}}}-{\frac {\partial ^{2}H}{\partial p_{s}\partial p_{j}}}\cdot {\frac {\partial F}{\partial q_{j}}}-{\frac {\partial H}{\partial p_{j}}}\cdot {\frac {\partial ^{2}F}{\partial p_{s}\partial q_{j}}}\right]\cdot {\frac {\partial G}{\partial q_{s}}}\right)+}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ade955bb3af70c220ef0a526a81ffc80b9dcd7b0)
![{\displaystyle +\sum _{s=1}^{n}\sum _{j=1}^{n}\left(\left[{\frac {\partial ^{2}G}{\partial q_{s}\partial q_{j}}}\cdot {\frac {\partial H}{\partial p_{j}}}+{\frac {\partial G}{\partial q_{j}}}\cdot {\frac {\partial ^{2}H}{\partial q_{s}\partial p_{j}}}-{\frac {\partial ^{2}G}{\partial q_{s}\partial p_{j}}}\cdot {\frac {\partial H}{\partial q_{j}}}-{\frac {\partial G}{\partial p_{j}}}\cdot {\frac {\partial ^{2}H}{\partial q_{s}\partial q_{j}}}\right]\cdot {\frac {\partial F}{\partial p_{s}}}-\left[{\frac {\partial ^{2}G}{\partial p_{s}\partial q_{j}}}\cdot {\frac {\partial H}{\partial p_{j}}}+{\frac {\partial G}{\partial q_{j}}}\cdot {\frac {\partial ^{2}H}{\partial p_{s}\partial p_{j}}}-{\frac {\partial ^{2}G}{\partial p_{s}\partial p_{j}}}\cdot {\frac {\partial H}{\partial q_{j}}}-{\frac {\partial G}{\partial p_{j}}}\cdot {\frac {\partial ^{2}H}{\partial p_{s}\partial q_{j}}}\right]\cdot {\frac {\partial F}{\partial q_{s}}}\right)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e3884e18fc145365b8cccfe270761886230a479)
![{\displaystyle \sum _{j=1}^{n}\left[{\frac {\partial \left({\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}\right)}{\partial x}}-{\frac {\partial \left({\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}\right)}{\partial x}}\right]=\sum _{j=1}^{n}\left({\frac {\partial ^{2}F}{\partial x\partial q_{j}}}\cdot {\frac {\partial G}{\partial p_{j}}}+{\frac {\partial F}{\partial q_{j}}}\cdot {\frac {\partial ^{2}G}{\partial x\partial p_{j}}}-{\frac {\partial ^{2}F}{\partial x\partial p_{j}}}\cdot {\frac {\partial G}{\partial q_{j}}}-{\frac {\partial F}{\partial p_{j}}}\cdot {\frac {\partial ^{2}G}{\partial x\partial q_{j}}}\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2fbe0b0ee476b21f07670d692f2a3dfb9c6ea1f9)
![{\displaystyle H=F\{f_{n}[\ldots f_{2}(f_{1}\left(q_{1},p_{1}\right),q_{2},p_{2})],t\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/09a9254eaf8a96e7dbd9f4564ba13745ccf99805)
