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Ch 3.8: Mechanical & Electrical Vibrations
Two important areas of application for second order linear
equations with constant coefficients are in modeling
mechanical and electrical oscillations.
We will study the motion of a mass on a spring in detail.
An understanding of the behavior of this simple system is the
first step in investigation of more complex vibrating systems.
Spring – Mass System
Suppose a mass m hangs from vertical spring of original
length l. The mass causes an elongation L of the spring.
The force FG of gravity pulls mass down. This force has
magnitude mg, where g is acceleration due to gravity.
The force FS of spring stiffness pulls mass up. For small
elongations L, this force is proportional to L.
That is, Fs = kL (Hooke’s Law).
Since mass is in equilibrium, the forces balance each other:
mg пЂЅ kL
Spring Model
We will study motion of mass when it is acted on by an
external force (forcing function) or is initially displaced.
Let u(t) denote the displacement of the mass from its
equilibrium position at time t, measured downward.
Let f be the net force acting on mass. Newton’s 2nd Law:
m u п‚ўп‚ў ( t ) пЂЅ f ( t )
In determining f, there are four separate forces to consider:
Weight:
w = mg
(downward force)
Spring force: Fs = - k(L+ u) (up or down force, see next slide)
Damping force: Fd(t) = - пЃ§ uп‚ў (t) (up or down, see following slide)
External force: F (t)
(up or down force, see text)
Spring Model:
Spring Force Details
The spring force Fs acts to restore spring to natural position,
and is proportional to L + u. If L + u > 0, then spring is
extended and the spring force acts upward. In this case
Fs пЂЅ пЂ­ k ( L пЂ« u )
If L + u < 0, then spring is compressed a distance of |L + u|,
and the spring force acts downward. In this case
F s пЂЅ k L пЂ« u пЂЅ k пЃ›пЂ­ пЂЁ L пЂ« u пЂ©пЃќ пЂЅ пЂ­ k пЂЁ L пЂ« u пЂ©
In either case,
Fs пЂЅ пЂ­ k ( L пЂ« u )
Spring Model:
Damping Force Details
The damping or resistive force Fd acts in opposite direction as
motion of mass. Can be complicated to model.
Fd may be due to air resistance, internal energy dissipation
due to action of spring, friction between mass and guides, or a
mechanical device (dashpot) imparting resistive force to mass.
We keep it simple and assume Fd is proportional to velocity.
In particular, we find that
If uп‚ў > 0, then u is increasing, so mass is moving downward. Thus Fd
acts upward and hence Fd = - пЃ§ uп‚ў, where пЃ§ > 0.
If uп‚ў < 0, then u is decreasing, so mass is moving upward. Thus Fd
acts downward and hence Fd = - пЃ§ uп‚ў , пЃ§ > 0.
In either case,
Fd ( t ) пЂЅ пЂ­ пЃ§ u п‚ў( t ),
пЃ§ пЂѕ0
Spring Model:
Differential Equation
Taking into account these forces, Newton’s Law becomes:
m u п‚ўп‚ў( t ) пЂЅ mg пЂ« F s ( t ) пЂ« Fd ( t ) пЂ« F ( t )
пЂЅ mg пЂ­ k пЃ› L пЂ« u ( t ) пЃќ пЂ­ пЃ§ u п‚ў( t ) пЂ« F ( t )
Recalling that mg = kL, this equation reduces to
m u п‚ўп‚ў( t ) пЂ« пЃ§ u п‚ў( t ) пЂ« ku ( t ) пЂЅ F ( t )
where the constants m, пЃ§, and k are positive.
We can prescribe initial conditions also:
u ( 0 ) пЂЅ u 0 , u п‚ў( 0 ) пЂЅ v 0
It follows from Theorem 3.2.1 that there is a unique solution
to this initial value problem. Physically, if mass is set in
motion with a given initial displacement and velocity, then
its position is uniquely determined at all future times.
Example 1:
Find Coefficients
(1 of 2)
A 4 lb mass stretches a spring 2". The mass is displaced an
additional 6" and then released; and is in a medium that exerts
a viscous resistance of 6 lb when velocity of mass is 3 ft/sec.
Formulate the IVP that governs motion of this mass:
m u п‚ўп‚ў( t ) пЂ« пЃ§ u п‚ў( t ) пЂ« ku ( t ) пЂЅ F ( t ), u ( 0 ) пЂЅ u 0 , u п‚ў( 0 ) пЂЅ v 0
Find m:
w пЂЅ mg пѓћ m пЂЅ
w
пѓћ m пЂЅ
g
Find пЃ§ :
пЃ§ u п‚ў пЂЅ 6 lb пѓћ пЃ§ пЂЅ
4 lb
32 ft / sec
6 lb
пѓћ пЃ§ пЂЅ2
3 ft / sec
2
пѓћ m пЂЅ
1 lb sec
8
ft
lb sec
ft
Find k:
Fs пЂЅ пЂ­ k L пѓћ k пЂЅ
4 lb
2 in
пѓћ
k пЂЅ
4 lb
1 / 6 ft
пѓћ k пЂЅ 24
lb
ft
2
Example 1: Find IVP
(2 of 2)
Thus our differential equation becomes
1
u п‚ўп‚ў( t ) пЂ« 2 u п‚ў( t ) пЂ« 24 u ( t ) пЂЅ 0
8
and hence the initial value problem can be written as
u п‚ўп‚ў ( t ) пЂ« 16 u п‚ў ( t ) пЂ« 192 u ( t ) пЂЅ 0
u (0) пЂЅ
1
,
u п‚ў( 0 ) пЂЅ 0
2
This problem can be solved using
methods of Chapter 3.4. Given
on right is the graph of solution.
Spring Model:
Undamped Free Vibrations
(1 of 4)
Recall our differential equation for spring motion:
m u п‚ўп‚ў( t ) пЂ« пЃ§ u п‚ў( t ) пЂ« ku ( t ) пЂЅ F ( t )
Suppose there is no external driving force and no damping.
Then F(t) = 0 and пЃ§ = 0, and our equation becomes
m u п‚ўп‚ў( t ) пЂ« ku ( t ) пЂЅ 0
The general solution to this equation is
u ( t ) пЂЅ A cos пЃ· 0 t пЂ« B sin пЃ· 0 t ,
where
пЃ·0 пЂЅ k / m
2
Spring Model:
Undamped Free Vibrations
(2 of 4)
Using trigonometric identities, the solution
u ( t ) пЂЅ A cos пЃ· 0 t пЂ« B sin пЃ· 0 t , пЃ· 0 пЂЅ k / m
2
can be rewritten as follows:
u ( t ) пЂЅ A cos пЃ· 0 t пЂ« B sin пЃ· 0 t пѓ› u ( t ) пЂЅ R cos пЂЁпЃ· 0 t пЂ­ пЃ¤
пЂ©
пѓ› u ( t ) пЂЅ R cos пЃ¤ cos пЃ· 0 t пЂ« R sin пЃ¤ sin пЃ· 0 t ,
where
A пЂЅ R cos пЃ¤ , B пЂЅ R sin пЃ¤
пѓћ R пЂЅ
A пЂ« B , tan пЃ¤ пЂЅ
2
2
B
A
Note that in finding пЃ¤, we must be careful to choose correct
quadrant. This is done using the signs of cos пЃ¤ and sin пЃ¤.
Spring Model:
Undamped Free Vibrations
(3 of 4)
Thus our solution is
u ( t ) пЂЅ A cos пЃ· 0 t пЂ« B sin пЃ· 0 t пЂЅ R cos пЂЁпЃ· 0 t пЂ­ пЃ¤
пЂ©
where
пЃ·0 пЂЅ
k /m
The solution is a shifted cosine (or sine) curve, that describes
simple harmonic motion, with period
T пЂЅ
2пЃ°
пЃ·0
пЂЅ 2пЃ°
m
k
The circular frequency пЃ·0 (radians/time) is natural frequency
of the vibration, R is the amplitude of max displacement of
mass from equilibrium, and пЃ¤ is the phase (dimensionless).
Spring Model:
Undamped Free Vibrations
(4 of 4)
Note that our solution
u ( t ) пЂЅ A cos пЃ· 0 t пЂ« B sin пЃ· 0 t пЂЅ R cos пЂЁпЃ· 0 t пЂ­ пЃ¤ пЂ© ,
пЃ·0 пЂЅ
k /m
is a shifted cosine (or sine) curve with period
T пЂЅ 2пЃ°
m
k
Initial conditions determine A & B, hence also the amplitude R.
The system always vibrates with same frequency пЃ·0 , regardless
of initial conditions.
The period T increases as m increases, so larger masses vibrate
more slowly. However, T decreases as k increases, so stiffer
springs cause system to vibrate more rapidly.
Example 2: Find IVP
(1 of 3)
A 10 lb mass stretches a spring 2". The mass is displaced an
additional 2" and then set in motion with initial upward
velocity of 1 ft/sec. Determine position of mass at any later
time. Also find period, amplitude, and phase of the motion.
m u п‚ўп‚ў( t ) пЂ« ku ( t ) пЂЅ 0 , u ( 0 ) пЂЅ u 0 , u п‚ў( 0 ) пЂЅ v 0
Find m:
w пЂЅ mg пѓћ m пЂЅ
w
пѓћ m пЂЅ
g
Find k:
Fs пЂЅ пЂ­ k L пѓћ k пЂЅ
10 lb
32 ft / sec
10 lb
2 in
пѓћ k пЂЅ
2
10 lb
1 / 6 ft
пѓћ m пЂЅ
5 lb sec
16
пѓћ k пЂЅ 60
ft
lb
ft
Thus our IVP is
5 / 16 u п‚ўп‚ў( t ) пЂ« 60 u ( t ) пЂЅ 0 , u ( 0 ) пЂЅ 1 / 6 , u п‚ў( t ) пЂЅ пЂ­ 1
2
Example 2: Find Solution
(2 of 3)
Simplifying, we obtain
u п‚ўп‚ў( t ) пЂ« 192 u ( t ) пЂЅ 0 , u ( 0 ) пЂЅ 1 / 6 ,
u п‚ў( 0 ) пЂЅ пЂ­ 1
To solve, use methods of Ch 3.4 to obtain
u (t ) пЂЅ
1
cos
1
192 t пЂ­
6
sin
192 t
192
or
u (t ) пЂЅ
1
6
cos 8 3 t пЂ­
1
8 3
sin 8 3 t
u (t ) пЂЅ
1
cos 8 3 t пЂ­
6
Example 2:
Find Period, Amplitude, Phase (3 of 3)
1
sin 8 3 t
8 3
The natural frequency is
пЃ·0 пЂЅ
k /m пЂЅ
192 пЂЅ 8 3 пЃЂ 13 . 856 rad/sec
The period is
T пЂЅ 2пЃ° / пЃ· 0 пЃЂ 0 . 45345 sec
The amplitude is
R пЂЅ
A пЂ«B
2
2
пЃЂ 0 . 18162 ft
Next, determine the phase пЃ¤ :
A пЂЅ R cos пЃ¤ , B пЂЅ R sin пЃ¤ , tan пЃ¤ пЂЅ B / A
tan пЃ¤ пЂЅ
B
A
пѓћ tan пЃ¤ пЂЅ
пЂЁ
пЂ­
3
пѓћ пЃ¤ пЂЅ tan
4
Thus u ( t ) пЂЅ 0 . 182 cos 8 3 t пЂ« 0 . 409
пЂ©
пЂ­1
пѓ¦пЂ­ 3
пѓ§
пѓ§ 4
пѓЁ
пѓ¶
пѓ· пЃЂ пЂ­ 0 . 40864 rad
пѓ·
пѓё
Spring Model: Damped Free Vibrations
(1 of 8)
Suppose there is damping but no external driving force F(t):
m u п‚ўп‚ў( t ) пЂ« пЃ§ u п‚ў( t ) пЂ« ku ( t ) пЂЅ 0
What is effect of damping coefficient пЃ§ on system?
The characteristic equation is
r1 , r2 пЂЅ
пЂ­пЃ§ п‚±
пЃ§ пЂ­ 4 mk
пЃ§ пѓ©
2
пЂЅ
2m
пѓЄпЂ­ 1 п‚±
2m пѓ«
1пЂ­
4 mk пѓ№
пѓє
2
пЃ§ пѓ»
Three cases for the solution:
пЃ§ пЂ­ 4 mk пЂѕ 0 :
u ( t ) пЂЅ Ae
пЃ§ пЂ­ 4 mk пЂЅ 0 :
u ( t ) пЂЅ пЂЁ A пЂ« Bt пЂ©e
2
2
пЃ§ пЂ­ 4 mk пЂј 0 :
2
u (t ) пЂЅ e
Note : In all three cases,
r1 t
пЂ« Be
пЂ­пЃ§ t / 2 m
r2 t
, where r1 пЂј 0 , r2 пЂј 0 ;
пЂ­пЃ§ t / 2 m
пЂЁ A cos
, where пЃ§ / 2 m пЂѕ 0 ;
пЃ­ t пЂ« B sin пЃ­ t пЂ©, пЃ­ пЂЅ
lim u ( t ) пЂЅ 0 , as expected
tп‚® п‚Ґ
4 mk пЂ­ пЃ§
2
2m
from damping
term.
пЂѕ 0.
Damped Free Vibrations: Small Damping
(2 of 8)
Of the cases for solution form, the last is most important,
which occurs when the damping is small:
пЃ§ пЂ­ 4 mk пЂѕ 0 : u ( t ) пЂЅ Ae
2
r1 t
пЂ« Be
пЃ§ пЂ­ 4 mk пЂЅ 0 : u ( t ) пЂЅ пЂЁ A пЂ« Bt пЂ©e
2
пЃ§ пЂ­ 4 mk пЂј 0 : u ( t ) пЂЅ e
2
пЂ­пЃ§ t / 2 m
r2 t
, r1 пЂј 0 , r2 пЂј 0
пЂ­пЃ§ t / 2 m
пЂЁ A cos
Then
u (t ) пЂЅ R e
пЂ­пЃ§ t / 2 m
cos пЂЁ пЃ­ t пЂ­ пЃ¤
and hence
u (t ) п‚Ј R e
пЂ­пЃ§ t / 2m
(damped oscillation)
пЂ©
пЃ§ / 2m пЂѕ 0
пЃ­ t пЂ« B sin пЃ­ t пЂ©, пЃ­ пЂѕ 0
We examine this last case. Recall
A пЂЅ R cos пЃ¤ , B пЂЅ R sin пЃ¤
,
Damped Free Vibrations: Quasi Frequency
(3 of 8)
Thus we have damped oscillations:
u (t ) пЂЅ R e
пЂ­пЃ§ t / 2m
cos пЂЁ пЃ­ t пЂ­ пЃ¤
пЂ©
пѓћ
u (t ) п‚Ј R e
пЂ­пЃ§ t / 2m
Amplitude R depends on the initial conditions, since
u (t ) пЂЅ e
пЂ­пЃ§ t / 2 m
пЂЁ A cos пЃ­ t пЂ« B sin пЃ­ t пЂ©,
A пЂЅ R cos пЃ¤ , B пЂЅ R sin пЃ¤
Although the motion is not periodic, the parameter пЃ­
determines mass oscillation frequency.
Thus пЃ­ is called the quasi frequency.
Recall
пЃ­ пЂЅ
4 mk пЂ­ пЃ§
2m
2
Damped Free Vibrations: Quasi Period
(4 of 8)
Compare пЃ­ with пЃ·0 , the frequency of undamped motion:
пЃ­
пЃ·0
For small пЃ§
пЂЅ
4 km пЂ­ пЃ§
4 km пЂ­ пЃ§
2
пЂЅ
2m k / m
пЃЂ
1пЂ­
пЃ§
2
4 km
пЂ«
4m
пЃ§
2
64 k m
2
пЂЅ
4 km пЂ­ пЃ§
пЂЅ
2
пѓ¦
пЃ§
пѓ§1 пЂ­
пѓ§
8 km
пѓЁ
пЃ§
2
пЂЅ
1пЂ­
2
4 km
4 km
k /m
4
2
2
2
2
пѓ¶
пЃ§
пѓ· пЂЅ 1пЂ­
пѓ·
8 km
пѓё
Thus, small damping reduces oscillation frequency slightly.
Similarly, quasi period is defined as Td = 2пЃ°/пЃ­. Then
2
Td
пЃ·0 пѓ¦
2пЃ° / пЃ­
пЃ§
пЂЅ
пЂЅ
пЂЅ пѓ§пѓ§ 1 пЂ­
T
2пЃ° / пЃ· 0
пЃ­
4 km
пѓЁ
пѓ¶
пѓ·
пѓ·
пѓё
пЂ­1 / 2
2
пѓ¦
пЃ§
пЃЂ пѓ§пѓ§ 1 пЂ­
8 km
пѓЁ
пѓ¶
пѓ·
пѓ·
пѓё
пЂ­1
Thus, small damping increases quasi period.
пЃЂ 1пЂ«
пЃ§
2
8 km
Damped Free Vibrations:
Neglecting Damping for Small пЃ§ 2/4km
(5 of 8)
Consider again the comparisons between damped and
undamped frequency and period:
пѓ¦
пЃ§
пЂЅ пѓ§пѓ§ 1 пЂ­
пЃ·0 пѓЁ
4 km
пЃ­
2
пѓ¶
пѓ·
пѓ·
пѓё
1/ 2
пѓ¦
пЃ§
,
пЂЅ пѓ§пѓ§ 1 пЂ­
T
4 km
пѓЁ
Td
2
пѓ¶
пѓ·
пѓ·
пѓё
пЂ­1 / 2
Thus it turns out that a small пЃ§ is not as telling as a small
ratio пЃ§ 2/4km.
For small пЃ§ 2/4km, we can neglect effect of damping when
calculating quasi frequency and quasi period of motion. But
if we want a detailed description of motion of mass, then we
cannot neglect damping force, no matter how small.
Damped Free Vibrations:
Frequency, Period (6 of 8)
Ratios of damped and undamped frequency, period:
пѓ¦
пЃ§
пЂЅ пѓ§пѓ§ 1 пЂ­
пЃ·0 пѓЁ
4 km
пЃ­
2
пѓ¶
пѓ·
пѓ·
пѓё
1/ 2
пѓ¦
пЃ§
пЂЅ пѓ§пѓ§ 1 пЂ­
T
4 km
пѓЁ
2
Td
,
пѓ¶
пѓ·
пѓ·
пѓё
пЂ­1 / 2
Thus
пЃ­ пЂЅ 0 and
lim
пЃ§ п‚®2
km
lim
пЃ§ п‚®2
km
Td пЂЅ п‚Ґ
The importance of the relationship between пЃ§2 and 4km is
supported by our previous equations:
пЃ§ пЂ­ 4 mk пЂѕ 0 : u ( t ) пЂЅ Ae
2
r1 t
пЂ« Be
пЃ§ пЂ­ 4 mk пЂЅ 0 : u ( t ) пЂЅ пЂЁ A пЂ« Bt пЂ©e
2
пЃ§ пЂ­ 4 mk пЂј 0 : u ( t ) пЂЅ e
2
пЂ­пЃ§ t / 2 m
r2 t
, r1 пЂј 0 , r2 пЂј 0
пЂ­пЃ§ t / 2 m
пЂЁ A cos
,
пЃ§ / 2m пЂѕ 0
пЃ­ t пЂ« B sin пЃ­ t пЂ©, пЃ­ пЂѕ 0
Damped Free Vibrations:
Critical Damping Value (7 of 8)
Thus the nature of the solution changes as пЃ§ passes through
the value 2 km .
This value of пЃ§ is known as the critical damping value, and
for larger values of пЃ§ the motion is said to be overdamped.
Thus for the solutions given by these cases,
пЃ§ пЂ­ 4 mk пЂѕ 0 : u ( t ) пЂЅ Ae
2
r1 t
пЂ« Be
пЃ§ пЂ­ 4 mk пЂЅ 0 : u ( t ) пЂЅ пЂЁ A пЂ« Bt пЂ©e
2
пЃ§ пЂ­ 4 mk пЂј 0 : u ( t ) пЂЅ e
2
пЂ­пЃ§ t / 2 m
r2 t
, r1 пЂј 0 , r2 пЂј 0
пЂ­пЃ§ t / 2 m
пЂЁ A cos
,
пЃ§ / 2m пЂѕ 0
пЃ­ t пЂ« B sin пЃ­ t пЂ©, пЃ­ пЂѕ 0
(1)
(2)
(3)
we see that the mass creeps back to its equilibrium position
for solutions (1) and (2), but does not oscillate about it, as
for small пЃ§ in solution (3).
Soln (1) is overdamped and soln (2) is critically damped.
Damped Free Vibrations:
Characterization of Vibration
(8 of 8)
Mass creeps back to equilibrium position for solns (1) & (2),
but does not oscillate about it, as for small пЃ§ in solution (3).
пЃ§ пЂ­ 4 mk пЂѕ 0 : u ( t ) пЂЅ Ae
2
r1 t
пЂ« Be
пЃ§ пЂ­ 4 mk пЂЅ 0 : u ( t ) пЂЅ пЂЁ A пЂ« Bt пЂ©e
2
пЃ§ пЂ­ 4 mk пЂј 0 : u ( t ) пЂЅ e
2
пЂ­пЃ§ t / 2 m
r2 t
, r1 пЂј 0 , r2 пЂј 0
пЂ­пЃ§ t / 2 m
пЂЁ A cos
,
пЃ§ / 2m пЂѕ 0
пЃ­ t пЂ« B sin пЃ­ t пЂ©
(Green)
(1)
(Red, Black)
(2)
(Blue)
(3)
Soln (1) is overdamped and soln (2) is critically damped.
Example 3: Initial Value Problem
(1 of 4)
Suppose that the motion of a spring-mass system is governed
by the initial value problem
u п‚ўп‚ў пЂ« 0 . 125 u п‚ў пЂ« u пЂЅ 0 , u ( 0 ) пЂЅ 2 , u п‚ў( 0 ) пЂЅ 0
Find the following:
(a) quasi frequency and quasi period;
(b) time at which mass passes through equilibrium position;
(c) time пЃґ such that |u(t)| < 0.1 for all t > пЃґ.
For Part (a), using methods of this chapter we obtain:
u (t ) пЂЅ e
пЂ­ t / 16
пѓ¦
пѓ§ 2 cos
пѓ§
пѓЁ
255
16
tпЂ«
2
255
sin
255
16
пѓ¶
tпѓ· пЂЅ
пѓ·
пѓё
32
255
e
пЂ­ t / 16
пѓ¦ 255
cos пѓ§
t пЂ­пЃ¤
пѓ§ 16
пѓЁ
where
tan пЃ¤ пЂЅ
1
255
пѓћ пЃ¤ пЃЂ 0 . 06254
(recall A пЂЅ R cos пЃ¤ , B пЂЅ R sin пЃ¤ )
пѓ¶
пѓ·
пѓ·
пѓё
Example 3: Quasi Frequency & Period
(2 of 4)
The solution to the initial value problem is:
u (t ) пЂЅ e
пЂ­ t / 16
пѓ¦
пѓ§ 2 cos
пѓ§
пѓЁ
255
2
tпЂ«
16
255
sin
255
16
пѓ¶
tпѓ· пЂЅ
пѓ·
пѓё
32
255
e
пЂ­ t / 16
пѓ¦ 255
cos пѓ§
t пЂ­пЃ¤
пѓ§ 16
пѓЁ
The graph of this solution, along with solution to the
corresponding undamped problem, is given below.
The quasi frequency is
пЃ­ пЂЅ
255 / 16 пЃЂ 0 . 998
and quasi period
T d пЂЅ 2пЃ° / пЃ­ пЃЂ 6 . 295
For undamped case:
пЃ· 0 пЂЅ 1, T пЂЅ 2пЃ° пЃЂ 6 . 283
пѓ¶
пѓ·
пѓ·
пѓё
Example 3: Quasi Frequency & Period
(3 of 4)
The damping coefficient is пЃ§ = 0.125 = 1/8, and this is 1/16 of
the critical value 2 km пЂЅ 2
Thus damping is small relative to mass and spring stiffness.
Nevertheless the oscillation amplitude diminishes quickly.
Using a solver, we find that |u(t)| < 0.1 for t > пЃґ п‚» 47.515 sec
Example 3: Quasi Frequency & Period
(4 of 4)
To find the time at which the mass first passes through the
equilibrium position, we must solve
u (t ) пЂЅ
32
e
255
пЂ­ t / 16
пѓ¦ 255
cos пѓ§
t пЂ­пЃ¤
пѓ§ 16
пѓЁ
Or more simply, solve
255
t пЂ­пЃ¤ пЂЅ
16
пѓћ tпЂЅ
пЃ°
2
пѓ¦пЃ°
пѓ¶
пѓ§ пЂ« пЃ¤ пѓ· пЃЂ 1 . 637 sec
255 пѓЁ 2
пѓё
16
пѓ¶
пѓ·пЂЅ0
пѓ·
пѓё
Electric Circuits
The flow of current in certain basic electrical circuits is
modeled by second order linear ODEs with constant
coefficients:
L I п‚ўп‚ў ( t ) пЂ« R I п‚ў ( t ) пЂ«
1
I ( t ) пЂЅ E п‚ў( t )
C
I ( 0 ) пЂЅ I 0 , I п‚ў ( 0 ) пЂЅ I 0п‚ў
It is interesting that the flow of current in this circuit is
mathematically equivalent to motion of spring-mass system.
For more details, see text.
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