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3.4. Compact of dynamics problems (resonance) 61
If contains procedures 4 and 5: it is an element of the ideal solving of a
problem.
Figure 3.10
Who can write an equation of motion for problem shown in Fig. 3.7–3.9?
Nobody can. Well. Since we want to master out the procedure “to find the
solution of a problem without solving it” we’ll write the equation of motion for
no one problem. Thus, we have used procedure 1.
The peculiarity of this procedure is that it proposes not to solve the whole
problem but only its main part.
We must find the equation for a natural frequency.
Procedure 3 suggests to simplify our problem in such a way that our simple
problem would have a simple solution. Let’s simplify problem 1, since we had
a simple solution in the previous paragraph (the so-called procedure “to begin
from the beginning”, see Fig. 3.3.
l =const,m=const, sin ϕ ≈ ϕ.
2
d(ml
ϕ)
+ mgl sin ϕ =0, (3.14)
dt
2
ml
¨ϕ + mglϕ =0, (3.15)
2
ϕ =0, (3.16)
¨ϕ + p
g
2
p
. (3.17)
=
l
62 Chapter 3
A
ω
Is expression (1.17) correct or not? I am waiting for your proposals. Why
is it correct? Did want you to say that the expression is not correct? You used
an intuition by an inversion principle. Is it true? Why is it not correct? Thus,
the same answer,
g
2
,
=
p
l
is correct for the particular problem (Fig. 3.13) and is not right for our problem.
The correct answer is
mgl
2
p
=
. (3.18)
2
ml
Thus, the premature cancellation of mlwas a hard mistake for our problem.
What is the difference between our problem number one and the problem
shown in Fig. 3.3? This is because of presence of the elastic force moment.
Will we find the value of this moment? If we look for it then we’ll violate
the procedure “to find the solution of a problem without solving it”. At first
we’ll find a place where we must write down our information. There are two
variants in formula (3.18): numerator and denominator. Can it be written in the
denominator? It cannot? Why? Because a momentum deals with the moment of
a gravity force and the moment of an elastic force.
η
n
00=
p
n
1,00=
p
B
1
0
1
p
0
Figure 3.11
3.4. Compact of dynamics problems (resonance) 63
1
X
Thus, the place of a new information is in the numerator:
mgl +
2
=
p
. (3.19)
2
ml
e
X
X
2
m
x
0
T
y
-nt
t
Figure 3.12 Figure 3.13
We have two parameters: stiffness (c) and length (l). How can they be
connected? Let’s look at the left term of our sum. mgl is the moment of weight,
and ca(the product of stiffness by length) is a force. In order that the dimension
in the numerator will not change,it is necessary to take the unknown expression
2
as ca
.
Do you like the answer:
ml
2
2
? (3.20)
2
p
mgl + ca
=
But we need the answer for each problem, therefore let’s substitute ± instead of +:
2
p
=
± mgl ± ca
ml
2
. (3.21)
2
Let’s presume our results. We solved a simplified problem (see Fig. 3.13),
used the above complicated solution, but have written the solution of all three
problems. If we know integrating mechanics, then we can find a mistake in
solution (3.20). But we’ll not look for it. Let’s try to obtain the solution by
another way. The procedure “to find the solution of a problem without solving
it” in mathematics has a classical example of theorem about finding the number
64 Chapter 3
of the algebraic equation roots by changing the symbols of the coefficients of an
algebraic equation.
Thus, instead of writing the equations of motion let’s examine the sign in
the numerator of expression (3.20). Let’s form table 1.1. Why are the problems
placed in order one, two, three?
Tab le 3.1
No problem sign mgl sign c · a
2
1 + +
2 0 +
3 − +
Possibly we need a number of problems, we must remember procedure 5.
We need in the dialectical “fork”. It is safe to say that the first and the third
problems are contrary ones. The third problem is directly opposite to the first
problem. There is a geometrical inversion. If we solve problems in order 1, 2,
3, then we’ll not be able to use the dialectical “fork”. That is why we solve the
first, the third, and only then the second problem.
In the first problem: the moment of weight and the moment of the elastic
force are of the same sign. In the third problem: they are of different sign. What
entity should have “+”? It is obvious that the elastic force moment is a positive
quantity, it is greater than the moment of weight, since ca
2
cannot be negative.
If the moment of weight is more than the elastic force moment then then the
vibrations cannot occur. But we are interested in a vibration process (we use the
procedure introducing an additional information). Therefore sign “+” should be
referred to ca
Now let’s examine the second problem. Since the systems comes back to
the initial position, the form ca
2
, and sign “−”—tomgl.
2
will have sign “+” as in problems 1.3. What is
the sign of mgl ? In the first problem it was “+”,in the third – “-”. Am waiting
for your proposals. Let’s look on procedure 5. We find null. Right! We find the
right answer intuitively, using procedure 5. But we have no physical reasons to
prove it.
The final answer:
(+; 0; −) mgl (+;+;+) ca
2
p
=
ml
2
2
, problems −−1, 2, 3. (3.22)
The problem is solved. Let’s now come back and form the equations of
motion to find the physical understanding of our problems.
3.4. Compact of dynamics problems (resonance) 65
ω
Problem 1 (see Fig. 3.14). The equation of motion:
2
¨ϕ +(mgl+ ca2) ϕ =0. (3.23)
ml
2
ml
¨ϕ +(− mgl + ca2) ϕ =0. (3.24)
2
¨ϕ + ca2ϕ =0. (3.25)
ml
c
1
m tP
1
x
1
sin
Figure 3.14
Why the moment of weight is not presented in the latter equation? We
return to the physical sense of Newton’s law:
2
d (ml
dt
˙ϕ)
e
= M
, (3.26)
the time rate of change the moment of momentum of a system equals the
external moment acting on the system.
The right writing of equation 3.22:
1.ml
3.ml
2.ml
2
¨ϕ =0 − mgl sin ϕ − ca2tan ϕ,
2
¨ϕ =0 + mgl sin ϕ − ca2tan ϕ,
2
¨ϕ =0 + mgl − mgl cos ϕ
=0
− ca2tan ϕ.
Resonance, damping of vibration
The problem of the forces vibration of the linear spring-mass vibrating
system with harmonic excitation (see Fig. 3.4) had been discussed before.
The physical equation:
m¨x + cx= F sinwt, (3.27)
66 Chapter 3
where c – spring constant (stiffness), m — mass of a block, F — magnitude of
exciting force.
The mathematical equation:
2
¨x + p
x = h sin wt, (3.28)
where p =
c
— natural circular frequency, h =
m
F
— amplitude of the
m
excitation acceleration.
The solution (3.28):
x = C
cos pt+ C2sin pt+ A sin wt;
1
A =
h
p2− w
;
2
initial conditions:
x = x
cos pt+
0
˙x
p
free vibration
t =0; x = x
0
sin pt
−
p (p2− w2)
accompanying vibration
0
hw
;˙x =˙x0;
sin pt
h
+
p2− w
explicit forced vibration
sin pt
2
. (3.29)
We also have introduced the magnification factor:
A
where A
A
dyn
st
|=
η =|
— amplitude of the forced vibration, Ast— static deflection of a
dyn
| 1 − (
1
, (3.30)
w
)2|
p
system that would be produced by the amplitude F of the harmonic force.
So, the engineer notion of a resonance is: when the value of a frequency of
a forced vibration coincides with one of a free vibration, the growth of displacement amplitude occurs.
Hence, the notion of a resonance is connected with the notion of a free
vibration. Will the resonance occur if the free vibration is absent? The not right
answer: no. The right answer: the frequencies of both free and forced vibration are of importance for the resonance phenomenon: but the accompanying
vibration have a frequency of the free vibration. This form of vibration does not
depend on an initial condition unlike the free vibration.
3.4. Compact of dynamics problems (resonance) 67
Figure 3.15
h
w
P'
,
P
1
,
P
2
w
P''
P
Figure 3.16
The physical notion of resonance:
lim
w→ p
(−
p (p2− w2)
= lim
w→ p
hw
(
sin pt+
h (−w sin pt+ p sin wt)
p (p2− w2)
When w closely tends to p, the expression has the form
h
p2− w
sin wt)=
2
).
0
.
0
L’Hospital’s rule — let’s differentiate the numerator and denominator with
68 Chapter 3
Figure 3.17
respect to w:
h (− sin pt+ p (cos wt)t)
(
lim
w→ p
− p 2w
Because of the term
ht
a resonance amplitude increases with time.
2p
)=
h
2p
sin pt−
2
ht
cos pt. (3.31)
2p
How many resonances does the one – degree – of – freedom system possess? The most of authors say that the system possesses only one resonance. We
can correct this opinion: if w → p, then the system has
one mathematical resonance but infinite number of physical resonances.
Indeed,
x = x
cos pt+
0
0
sin pt−
p
hw
p (p2− w2)
sin pt+
h
p2− w
sin pt.
2
˙x
Resonance for x, resonance for ˙x, resonance for ¨x etc. All these resonances
have different amplitudes, signs etc.
3.4. Compact of dynamics problems (resonance) 69
The damping of vibration
There are three ways to avoid or minimize the damage or discomfort of the
resonance:
• to operate far from the critical frequency (resonance) in the high-frequency
band (point A in Fig. 3.11);
• to operate far from the critical frequency in the low-frequency band
(point B in Fig. 3.11);
• to operate in the zone of critical frequency (in the zone of resonance,
point C in Fig. 3.11); we use principle “to turn harm into benefit”;
• damping: viscous damping, Coulomb friction;
• damping material (for example, use of the material providing high internal
friction in springs).
Let’s consider the damping proportional to velocity (viscous damping).
A free damping vibration (see Fig. 3.12):
where F
m¨x + cx= −F
= α ˙x represents a dissipative force, or viscous damping force.
x
= − α ˙x, (3.32)
x
m¨x + α ˙x + cx=0
— physical equation
— mathematical equation
where n =
,
2
¨x +2n ˙x + p
x =0, (3.33)
,
α
— is the damping coefficient, p =
2m
c
— natural circular
m
frequency.
The dimension of the damping coefficient is the same as one of the natural
circular frequency.
The characteristic equation for equation (1.32) is :
2
r
+2nr+ p2=0,
r
= −n ±n2− p2,n2∠∠p2,p
1,2
= − n ± p1i, i =√−1.
r
1,2
2
= p2− n2;
1
70 Chapter 3
Here r
are the roots of the characteristic equation, “iis an imaginary
1,2
unit.
The general solution of equation (1.32) then is
x = C
(−n+p1i)t
e
1
−nt
= e
(C1cos p1t + C2sin p1t).
+ x = C2e
(−n−p1i)t
The ration of two successive maximum displacements equals
−nt
e
sin p1t
sin p1(t + T )
1
= enT. (3.35)
2
or
y
1
=
y
−n(t+T )
2
e
y
y
Then the quantity
y
δ =ln
1
= nT (3.36)
y
2
is called the logarithmic decrement (see Fig. 3.13).
The period of the damped vibration equals:
T =
2π
p
=
1
2π
p
1 −
0
= T0(1 + 0,5
2
n
2
p
0
=
, (3.34)
2
n
). (3.37)
2
p
0
Here the factor p
vibration, p
and T0— frequency and period of free undamped vibration, re-
0
=p2− n2is called the frequency of the free damped
1
spectively.
Using Newton’s binomial, we obtain
(1 + x)
n
=1−nx; x 1.
The isochronism of vibration: there are two processes with different ve-
locities in one problem, the period of vibration changes slightly (i. e., less
than 0.6%), but an amplitude changes significantly
y
1
(
=2;p1=0, 96p0).
y
2
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