Applications of Second-Order
Differential Equations
ymy/2013
Building Intuition
• Even though there are an infinite number of differential
equations, they all share common characteristics that allow
intuition to be developed:
– Particular and complementary solutions
– Effects of initial conditions
Lect12
EEE 202
2
 Mechanical Vibrations
• The study of motion of objects and the effect of forces acting
on those objects.
Spring:
Support Structure
Example mass – spring system,
cantilever, pendulum, …
 Electric Circuits
Mass:
Mass of the bridge
structure
Spring-Mass Systems
We consider the motion of an object with mass at the end of a
spring that is either vertical (as in Figure 1) or horizontal on a level
surface (as in Figure 2).
Figure 2
Figure 1
An object of mass m is suspended from
the spring and stretches it a length s
when the spring comes to rest in an
equilibrium position.
According to Hooke’s Law, the tension
force in the spring is ks , where k is the
spring constant. The force due to gravity
pulling down on the spring is mg , and
equilibrium requires that
ks
mg
(1)
Hooke’s Law: The spring exerts a restoring force F opposite to the direction of
elongation and proportional to the amount of elongation
Forces acting on the object :
Fp
mg
Fs
k(s
Fr
dy
d
dt
the propulsion force due to
gravity
y) the restoring force of the
spring’s tension
a frictional force assumed
proportional to velocity
The propulsion force (weight), Fp pulls
the mass downward, but the spring
restoring force Fs and frictional force Fr
pull the mass upward.
The motion starts at y y0 with the
mass vibrating up and down.
The frictional force tends to retard the motion of the object. The
resultant of these forces is
F
Fp
and by Newton’s second law F
d 2y
m 2
dt
By Equation (1), mg
ks
Fs
Fr
ma , we then have
mg
ks
ky
dy
d
dt
0 , so this last equation becomes
d 2y
m 2
dt
subject to the initial conditions
dy
d
dt
y(0)
ky
0
(2)
y0 , y (0)
0
Remark
•
•
The nature of the vibrations/oscillations described by the
differential equation depends on the constants m, d , and k.
d is known as the damping constant.
•
The damping may be due to a damper such as a dashpot,
internal damping such as friction within the spring, or
external damping such as aerodynamic drag.
•
The displacement, y, at any time, t, of the mass is the
output from the system.
Free, Undamped Vibrations : Simple
Harmonic Motion
• Suppose there is no retarding frictional force. Then d 0 and
there is no damping. In this case, once motion has started it will
continue for ever.
• The motion is governed by
d 2y
m 2
dt
ky
0
(3)
 Second –order linear homogeneous constant-coefficient equation
If w
2
k
, then the second-order equation (3) becomes
m
d 2y
2
w
y 0 with y(0) y0 , y (0) 0
2
dt
The auxiliary equation is r 2
roots
r
w2
0 having the imaginary
wi
The general solution to the differential equation in (3) is
y(t )
C1 cos(wt )
Substituting the conditions yields C1
y(t )
C2 sin(wt )
y0 and C2
y0 cos(wt )
0
(4)
describes the motion of the object. Equation (4) represents
simple harmonic motion of amplitude y0 and period
2p
T
w
Note
The general solution to the differential equation (3), that is
y(t )
C1 cos(wt )
y(t )
A cos(wt
C2 sin(wt )
can also be written as
q)
where
w
k
m
A
C12
cos q
C1
A
frequency
C 22
sin q
frequency at which the system “wants
to oscillate” without external
interference.
amplitude
C2
q phase angle
A
Example 1
A spring with a mass of 2 kg has natural length 0.5 m. A force of
25.6 N is required to maintain it stretched to a length of 0.7 m. If
the spring is stretched to a length of 0.7 m and then released with
initial velocity 0, find the position of the mass at any time.
Solution
d 2y
The motion is governed by m 2
dt
ky
0
(3)
From Hooke’s Law, the force required to stretch the spring is
k(0.2)
so the spring constant,
k
25.6
0.2
25.6
128
Using this value of k together with m = 2 in Equation (3), we have
d 2y
2 2
dt
The solution of this equation is
y(t )
128y
0
C1 cos 8t
C2 sin 8t (*)
Given the initial condition y(0) 0.2 . But, from Equation (*)
y(0) C1
C1 0.2
Differentiating Equation (*), we get
y (t )
8C1 sin 8t
8C2 cos 8t
Since the initial velocity is y (0) 0 , we have C 2
the solution is
1
y(t )
cos 8t
5
0 and so
Damped Motion
The motion of a spring is subject to a frictional force (in the case of
the horizontal spring of Figure 2) or a damping force (in the case
where a vertical spring moves through a fluid as in Figure 3).
Figure 2
Coil spring
Dashpot for damping
Figure 3
Example: the damping force supplied by
“coilovers” in motorcycle suspension
So d
0 in Equation (2),
d 2y
m 2
dt
If we substitute
w
2
dy
d
dt
ky
k
and 2b
m
0
(2)
d
m
then the differential equation (2) is
d 2y
dt 2
dy
2b
dt
2
r
The auxiliary equation is
r
w 2y
2br
b
0
w2
b2
(5)
0 with roots
w2
We now have 3 cases depending upon the relative sizes of b and w
Case 1: b w
The double root of the auxiliary equation is real and equals r
The general solution to Equation (5) is
y(t )
(C1
C 2t )e
wt
This situation of motion is called critical damping and is not
oscillatory.
w
Case 2: b w
The roots of the auxiliary equation are real and unequal, given by
r1
b
b2
w2 and r2
b
b2
w2
The general solution to Equation (5) is
y(t )
C1e
b
b 2 w2 t
C 2e
b
b 2 w2 t
Here again the motion is not oscillatory and both r1 and r2 are
negative. Thus y approaches zero as time goes on. This motion is
referred to as overdamping.
Case 3: b w
The roots of the auxiliary equation are complex and given by
r
b
i w2
b2
The general solution to Equation (5) is
y(t )
e
bt
C1 cos w2
b2 t
C 2 sin w2
b2 t
This situation, called underdamping, represents damped oscillatory
motion.
2p
 It is analogous to simple harmonic motion of period T
2
2
w
b
but the amplitude is not constant but damped by the factor e bt .
Therefore, the motion tends to zero as t increases, so the vibrations
tend to die out as time goes on.
Three examples of damped vibratory motion for a spring system
with friction, d 0
(a) damping is just sufficient to suppress vibrations.
(b) strong damping force (high-viscosity oil or grease) compared with a weak
spring or small mass
(c) the motion decays to 0 as time increases.
Example 2
Suppose that the spring of Example 1 is immersed in a fluid with
damping constant d 40 . Find the position of the mass at any
time t if it starts from the equilibrium position and is given a push
to start it with an initial velocity of 0.6 m/s.
Solution
From Example 1 the mass is m = 2 and the spring constant is
k = 128 , so the differential equation (5) becomes
d 2y
dt 2
dy
20
dt
64y
0
2
r
The auxiliary equation is
having roots
r
20r
(r
64
4 and r
4)(r
16)
16
so the motion is overdamped and the solution is
y(t )
We are given that y(0)
we get
y (t )
so
Since C 2
Therefore
y (0)
C1e
4t
C 2e
0 so C1
4C 1e
C2
4t
4C1
0.05(e
0 . Differentiating,
16C 2e
16C 2
C1 , this gives 12C1
y(t )
16t
4t
0.6
C1
0.6
e
16t
16t
)
0.05
0
Exercise
1. Find the damping parameter and natural frequency of the
system governed by
d 2y
dy
6
9y 0
2
dt
dt
2. A spring with a mass of 2 kg has damping constant 14, and a
force of 6 N is required to keep the spring stretched 0.5 m
beyond its natural length. The spring is stretched 1 m beyond its
natural length and then released with zero velocity.
(i) Find the position of the mass at any time t.
(ii) Find the mass that would produce critical damping.
2
d 2y
dt 2
14
dy
dt
12y
0,
y(0)
1, y (0)
0
d2
4mk
0
m
d2
4k
142
4(12)
49
kg
12
3. A spring has a mass of 1 kg and its spring constant is k = 100 .
The spring is released at a point 0.1 m above its equilibrium
position. Graph the position function for the following values of
the damping constant d : 10, 15, 20, 25, 30. What type of
damping occurs in each case?
d
10 & 15 , underdamped
d
20 , critically damped
d
25 & 30 , overdamping
Forced Vibrations
Suppose that, in addition to the restoring force and the damping
force, the motion of the spring is affected by an external force F(t) .
Then Newton’s Second Law gives
d 2y
m 2
dt
restoring force + damping force + external force
= ky
dy
d
dt
F (t )
Thus, the motion of the spring is now governed by the following
non-homogeneous differential equation:
d 2y
m 2
dt
dy
d
dt
ky
F (t )
(6)
The general solution to Equation (6) is
y(t )
yp (t )
yc (t )
where the complementary solution will be the solution to the free,
damped case and the particular solution will be found using
undetermined coefficients or variation of parameters.
Note
• The complementary solution will approach zero as t increases.
Because of this the complementary solution is often called the
transient solution.
• The particular solution is often called the steady state solution
or forced response.
Remark
A commonly occurring type of external force is a periodic force
function
F (t )
F0 cos w0t
where w0
In the absence of a damping force ( d
y(t )
C1 cos wt
C 2 sin wt
k
m
w
0)
F0
m(w2
w02 )
cos w02t
If w0 w , then the applied frequency reinforces the natural
frequency and the result is vibrations of large amplitude. This is
the phenomenon of resonance.
Example
Suppose a machine is attached to a 1 kilogram mass and the
machine exerts a force of sin t newtons on the mass at time t. In
addition the mass is attached to a spring having spring constant
4. The mass slides along a frictionless horizontal surface. The
equation of motion is
for t > 0.
d 2x
dt 2
4x (t )
sin t
Solution
Here x(t) is the displacement from the equilibrium position (which
is taken to be 0) at time t.
• General solution of the corresponding homogeneous
equation:
• Particular integral of the non-homogeneous equation:
y p (t )
1
sin t
3
• The general solution of the original non-homogeneous
equation:
y(t )
1
sin t
3
C1 cos 2t
C 2 sin 2t
• Determine the constants C1 and C2 from the initial conditions:
Simple Electric Circuits
The LCR circuit contains a resistor R , an inductor L, and a
capacitor C, connected in series with a switch and battery (d.c.
voltage source).
V
IR potential difference across resistor
Q
V
potential difference across capacitor
C
dI
V L
potential difference across inductor
dt
The total potential difference around the circuit must be zero and
dI
Q
we have
L
RI
0
dt
C
The principle of conservation of charge tells us that the current is
equal to the rate of change of charge, that is we have :
I
dQ
dt
We use this to eliminate Q, we obtain
dI
L
dt
RI
1
C
I dt
0
or to eliminate I, we obtain
d 2Q
L
dt
dQ
R
dt
Differentiating either equations, we obtain
d 2I
dI
1
L
R
I
dt
dt C
1
Q
C
0
0
(7)
Note
• We have differential equations for one of the variables: either
the charge on the capacitor in the circuit or the current in the
circuit.
• The equations are equivalent, but equation (7) is the most
usual and occurs widely in engineering applications.
Example
A circuit contains a 1 henry inductor, a 2000 ohm resistor, and a 1
microfarad capacitor. Initially there is no charge in the circuit and
the initial current is 1 ampere. Find the charge in the circuit as a
function of time.
Solution
The equation governing the charge is
d 2Q
dt
dQ
2000
dt
1000000Q(t )
0,
Q(0)
0 , Q (0)
1
The characteristic equation has a repeated real root at −1000. So
the general solution is
Q(t )
(C1
C 2t )e
1000t
Since Q(0)
0 , Q (0)
Q(t )
1 , 0 = C1 and 1 = C2. Hence
te
1000t
Exercise
Suppose in the previous example the resistance of the resistor is
(i) 1990 ohms
(ii) 2010 ohms
Find the charge in the circuit.
(i)
(ii)
a periodic solution with exponentially decaying amplitude
distinct real roots that leads to rapidly decaying charge
Oscillations in Electrical Circuits
If an alternating voltage signal is applied to a simple LCR
electrical circuit, the equation governing the resulting oscillations
is also a second-order linear ODE.
Suppose the circuit contains an electromotive
force (supplied by a battery or generator),
a resistor , an inductor , and a capacitor , in
series.
V
V
V
IR
voltage drop across resistor
I
dt voltage drop across capacitor
C
dI
L
voltage drop across inductor
dt
Kirchhoff’s voltage law says that the sum of these voltage drops
is equal to the supplied voltage:
dI
1
L
RI
I dt E (t )
dt
C
Since I
dQ
, this equation becomes
dt
d 2Q
dQ
1
L 2
R
Q E (t )
dt
C
dt
(8)
which is a second-order linear differential equation with
constant coefficients. If the charge Q0 and the current I0 are
known at time 0, then we have the initial conditions
Q(0)
Q0
Q (0)
I (0)
I0
Note the similarity between the differential equations of the two
quite different systems given above.
As L, R, and C are all positive, this system behaves just like the
mass and spring system.
The position of the mass is replaced by the current. Mass is
replaced by the inductance, damping is replaced by resistance
and the spring constant is replaced by one over the capacitance.
The change in voltage becomes the forcing function. Hence for
constant voltage this is an unforced motion.
This similarity is used in analogue systems in which a
mechanical system can be simulated by the equivalent electrical
system.
Example
Find the charge and current at time in the circuit if R
L
1 H,C
16 10
4
F, E(t )
40
,
100 cos 10t and the initial
charge and current are both 0.
Solution
With the given values of L, R, C, and E(t), Equation (8) becomes
d 2Q
dt 2
dQ
40
dt
625Q
100 cos 10t
The characteristic equation have complex roots
the complementary function is
Qc (t )
e
20t
(C1 cos15t
20
C 2 sin15t )
15i . So
Using the method of undetermined coefficients we try the trial
solution
Qp (t )
A cos10t
B sin10t
Substituting into the DE and equating coefficients, we have (DIY)
the particular integral
Qp (t )
4
(21cos10t
697
16 sin10t )
and the general solution is
Q(t )
4
(21cos10t 16sin10t ) e
697
20t
(C1 cos15t C 2 sin15t )
Imposing the initial conditions Q(0)
formula for the charge is
0 , I (0)
0 , we get the
4
e 20t
Q(t )
(21cos10t 16sin10t )
( 63cos15t 116sin15t )
697
3
and the expression for the current is
I (t )
1
120( 21sin10t 16cos10t ) e
2091
20t
( 1920cos15t 13060sin15t )
Exercise
The equation for an LCR circuit with applied voltage E is
dI
L
dt
RI
1
Q
C
E
By differentiating this equation find the solution for Q(t) and I(t) if
L = 1, R = 100, C = 10−4 and E = 1000 given that Q = 0 and I = 0
at t = 0.