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3.3. The basic information compact of dynamics problems 51
The aim is to rise a problem from the level of equations of motion to an
energy level.
An energy level
Equations of motion
The notion of a workless constraints is used in the energy level:
(Ri· δri)=0,
i
(Fi· δri+Ri· δri+Φi· δri)=0
i
— common dynamics equation, which itself is one of the equations of an energy
conservation.
Another approach: change in energy.
The change in the kinetic energy equals the change in the work done by
external forces:
ΔT =ΔA
e
.
More often the equations of motion are not formed by the last method, but
velocities and accelerations are determined from the energy equations.
Lagrange “treated” the common dynamics equation Σ(F · δr)=0goes
down on the level of forces, and obtains the equation which is called Lagrange’s
equation of second type:
dt
∂T
d
∂ ˙q
∂T
−
∂q
i
i
= −
∂V
∂q
+ Qi,
i
where q
ergy, Q
— generalized coordinate, T,V — kinetic energy, potential en-
i
— generalized active force.
i
Thus, the problem can be solved in both level of energy equations and level
of equations of motion.
There are also the direct and reverse problems in mechanics:
¨
= m
r →
F
direct problem — determining displacements using forces given; reverse problem — determining forces using displacements given.
Methods of composition of Newton’s equations: direct —F = m¨r,reverse — r = f(¨r,F ) (for bending vibration).
Besides listed there also are:
52 Chapter 3
• Newton’s equation of motion;
• Lagrange’s equation of motion;
• energy equation in terms of its change;
• energy equation (common dynamics equation);
• equation of motion obtained from energy equation;
• equation of displacements.
There exist:
• Lagrange’s equation of motion;
• Lagrange’s equation for the first type (the introduction of indefinite coeffi-
cient);
• system method for the composition of equations of motion [2].
Altogether there exist eight methods of solving besides variation methods.
Analysis of assumed equations of motion
Wehaddiscussedthetwosteps:
• the criterion of the applicability of classical mechanics basic motions,
• methods of composition of equations of motion.
The third step is the analysis of assumed equations (see Fig. 3.3).
l
j
Figure 3.3
mg
3.3. The basic information compact of dynamics problems 53
Let’s use Newton’s method for analysis of the particle motion about center O. The linear momentum is mv,wherev =˙ϕl, whence
mv= m ˙ϕl.
The moment of momentum about O is
2
.
Here ml
ml ˙ϕl = m ˙ϕl
2
– rotational mass of the particle. We’ll call ml2amomentof
inertia.
d
2
(ml
dt
˙ϕ)=−mgl sin ϕ.
The time rate of change of momentum of the system equals the change of
moment of external forces (the moment of a gravity force about point O).
Thus,
d
2
(ml
dt
˙ϕ)+mgl sin ϕ =0.
Example 1.
m =const; l =const; sinϕ ≈ ϕ;
2
ml
¨ϕ + mglϕ=0. ¨ϕ + p2ϕ =0,p2=
g
l
— these are ordinary linear vibration.
Example 2.
3
ϕ
m =const; l =const; sinϕ = ϕ −
.
6
We have a nonlinear vibration of pseudo-harmonic type (the parameters of a
system depends on ϕ ), which are obtained by the equation
2
2
ml
¨ϕ + mgl(1 −
ϕ
6
)ϕ =0.
This is Van-der-Paul’s equation used in radio engineering. He made a
presumption that some heart diseases are caused by vibrations of that sort. Naturally, the complete model of a heart is described by more complex equations.
But Van-der-Paul’s ideas turned valid.
54 Chapter 3
Example 3.
m =const; l =const; sinϕ ≈ ϕ;
ml
2
¨ϕ +2ml
dl
˙ϕ + mglϕ=0
dt
— quasi-harmonic vibration.
A parametric vibration is a particular case of quasi-harmonic vibration.
A phenomenon of parametric resonance is known in nuclear physics, and in
many other problems of physics and mechanics.
Example 4.
3
ϕ
m =const; l =const; sinϕ = ϕ −
2
ml
2
¨ϕ +2ml
dl
˙ϕ + mgl(1 −
dt
ϕ
6
.
6
) ϕ =0
— quasi- and pseudo-harmonic vibration acting simultaneously.
Example 5.
m =const; l =const; sinϕ ≈ ϕ.
ml
2
¨ϕ +
2
l
˙ϕ + mglϕ=0
dt
dm
— equation of motion of variable — mass particle, it is used in engineering of
space vehicles (for example, Meshersky’s equation, Tsiolkovsky’s formula).
Example 6.
3
ϕ
m =const; l =const; sinϕ = ϕ −
.
6
The assumed equation becomes very complicated, since the peculiarities of
pseudo-harmonic vibration are added to vibration of the variable-mass system:
2
ml
2
¨ϕ +
dm
dt
2
l
˙ϕ + mgl(1 −
ϕ
6
)ϕ =0.
Example 7.
m =const; l =const; sinϕ = ϕ.
ml
dm
2
¨ϕ +
dt
2
l
˙ϕ +2ml
dl
˙ϕ + mglϕ=0
dt
3.3. The basic information compact of dynamics problems 55
— the synthesis of vibration of a variable-mass system and pseudo-harmonic
vibration.
Example 8.
3
ϕ
2
ϕ
6
.
6
)ϕ =0
m =const; l =const; sinϕ = ϕ −
ml
2
¨ϕ +
dm
dt
2
l
˙ϕ +2ml
dl
˙ϕ + mgl(1 −
dt
— the synthesis of pseudo-harmonic, quasi-harmonic vibration, and vibration of
a variable — mass system.
It is very difficult to solve problem 8. We can easily solve equations of
problem 1 and problem 2.
Methods of solving assumed equations
We shall use only two methods: method taking into account the peculiarities
of our problem (Fourier’s method), see Fig. 3.4; method of the solving of a
problem in the most general form (method of variation of arbitrary constants),
see Fig. 3.5.
C
Fourier’s method
Assumed equations:
m
Figure 3.4 Figure 3.5
Ptsin w
x
m¨x + cx = F sin wt, (3.9)
x = x
+ x2,
1
m
x
Qt()
56 Chapter 3
x1— solution of the equation,
m¨x
+ cx1=0
1
— homogenous equation,
¨x
+ p2x1=0; p2=
1
x
— particular solution of the equation.
2
m¨x
+ cx2= F sin wt, ¨x2+ p2x2= hsin wt, h=
2
In the solution of a homogeneous equation
x
= C1cos pt+ C2sin pt,
1
constants C
The ratio
η =
and C2are determined from initial conditions:
1
;˙x =˙x0; x2= Asin wt.
0
=
m(p2− w2)F
−w
A
dyn
A
st
t =0; x = x
2
A sin wt + p2A sin wt= hA sin wt; A =
A
dyn
is called the magnification factor
A
st
=
hc
(p2− w2)F
x = C
cos pt+ C2sin pt+
1
t =0,x=0,x= x
x = x
˙x =˙x
0
= −pC1sin pt
0
cos pt+
free vibration
˙x
0
p
=0
=
C
2
sin pt
+pC2cos pt
˙x
0
p
−
p (p2− w2)
m
Fc
=
h
p2− w
,C1= x0.
0
+
=1
hw
−
p (p2− w2)
hw
sin pt
accompanying vibration
c
;
2
p
(p2− w
sin wt.
2
hw
p2− w
2
;
+
p2− w
explicit forced vibration
p2− w
=
2
1 − (
cos wt
=1
h
2
F
.
m
h
1
w
p
;
sin pt
.
2
.
2
)
.
3.3. The basic information compact of dynamics problems 57
Method of variation of arbitrary constants
All parameters of the problems coincide except for ones placed in the right
side
m¨x + cx= Q(t). (3.10)
+ p2x1=0; (3.11)
¨x
1
x
= C1cos pt+ C2sin pt; C1= C1(t); C2= C2(t);
1
˙x = −C
t =0; x = x
p sin pt+ pC2cosp t +˙C1cosp t +˙C2sin pt.
1
;˙x =˙x0;
0
The condition
˙
C
cos pt+˙C2sin pt=0
1
is chosen arbitrary for the convenience of solving.
¨x = −C
˙x = −pC
p2cos pt−C2p2sin pt−˙C1p sin pt+˙C2p cos pt.
1
sin pt+ pC2cos pt,
1
We substitute the latter expression into equation (3.10)
p2cos pt − C2p2sin pt−˙C1p sin pt +˙C2p cos pt+
−C
1
2
p
C1cos pt + p2C2sin pt =
˙
C
cos pt +˙C2sin pt =0;−˙C1p sin pt +˙C2p cos pt =
1
t
Q(t)
pm
= D2+
2
Q(τ)
sin pτdτ +sinpt
pm
t
1
pm
0
p((cos pt)
x = D
2
+(sinpt)2)=p;˙C1= −
t
C
= D1−
1
cos pt + D2sin pt − cos pt
1
x = D
Q(τ)
sin pτdτ; C
pm
0
cos pt + D2sin pt +
1
D
0
= x0; D2=
1
Q(t)
;
m
Q(t)
m
sin pt;˙C
t
Q(τ)
0
2
pm
Q(t)
=
pm
cos pτdτ;
t
Q(τ)
pm
0
Q(τ)sinp(t −τ)dτ ;
˙x
0
.
p
;
cos pt;
cos pτdτ.
58 Chapter 3
Sudden application of a force to vibrational system
If a constant force is suddenly exerted on the system being at rest, then the
system begins to vibrate about the position of statical equilibrium. The simplest
example is the motion of a linear oscillator caused by a force suddenly applied:
m¨x + cx = Q(t). (3.12)
The general solution
t
x = D
cos pt + D2sin pt +
1
1
pm
Q(τ)sinp(t −τ)dτ. (3.13)
0
In case of zero initial conditions and
x =
Q
pm
Since p
t
sin p (t −τ ) dτ =
0
c
2
=
,then
m
Q
2
mp
= Xst; x = Xst(1 − cos pt).
Q
cos p (t − τ ) |
2
mp
Q
t
=
0
mp
2
(1 − cos pt).
The parameter x changes between limits 0 and 2, if there is no damping in
the system vibrating under the action of a suddenly applied force. That is why
the minimum strength of the system equals 2 in the conditions of dynamical
loading. The part x(t) for (1.13) is given in Fig. 3.6.
3.4. Compact of dynamics problems (resonance)
Change of a natural frequency of a system due to the system orientation
Typical procedures for problems solving
Procedure 1.
Find the solution of a problem without solving it.
Procedure 2.
Do inversely.
Q =const,
Procedure 3.
3.4. Compact of dynamics problems (resonance) 59
x
x
ñò
t
Figure 3.6
a
c
mg
Figure 3.7. Problem No1
ϕ
l
Decomposition, to solve a problem with dividing it into a number of simple
ones.
Procedure 4.
Solve one general problem instead of several simple problems (use of the
inversion and decomposition procedures).
Procedure 5.
Use a dialectical “fork” (a positive theory, zero, a negative theory).
The most complex procedure is the first one.
60 Chapter 3
mg
l
Figure 3.8. Problem No2
mg
c
a
ϕ
l
ϕ
a
Figure 3.9. Problem No3
c
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