2nd Order ODE: Engineering Applications with
Second-Order Differential Equations
Hishammudin Afifi Bin Huspi
Faculty of Engineering
Universiti Malaysia Sarawak
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Engineering Applications with 2nd Order DE
Learning Objective:
1. Analyse engineering problems (spring and
series/parallel electronic circuits) using 2nd
order ODE
Engineering Applications with 2nd Order DE
Chapter Contents:
Linear Models: Initial Value Problems
 Spring/Mass Systems
 Series Circuit Analogue
SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION
HOOKE’S LAW
ks
Suppose that a flexible spring is
suspended vertically from a rigid support
and then mass m is attached to its free
end.
The amount of elongation, s of the
spring will depend on the mass.
mg
By Hooke’s Law, the spring itself exerts a
restoring force F opposite to the
direction of elongation and proportional
to the amount of elongation s.
F = ks , where k is the spring constant
SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION
NEWTON’S SECOND LAW
After a mass m is attached to a spring, it
stretches the spring by an amount s and
attains a position of equilibrium at
which weight W is balanced by the
restoring force ks.
Weight, W = mg
At equilibrium, the condition can be
represented by:
mg – ks = 0
If the mass is displaced by an amount x
from its equilibrium position, the
restoring force of the spring is then
k(x + s)
SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION
NEWTON’S SECOND LAW
Let’s assume that:
1. There is no retarding forces acting on the system
2. The mass vibrates free of other external forces, i.e. free motion
By Newton’s 2nd Law:
At equilibrium,
mg  ks  0
Restoring Force
of the spring.
F  ma
d 2x
m 2  k ( s  x)  mg
dt
d 2x
m 2  ks  mg  kx
dt
d 2x
m 2  kx
dt
Weight of the
mass
Negative sign indicates
that the restoring force of
the spring acts opposite
to the direction of
motion.
SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION
FREE UNDAMPED MOTION
From equation
Divide by the mass m,
Or,
d 2x
m 2  kx
dt
d 2x
k
 ( )x  0
2
dt
m
d 2x
2


x0
2
dt
where
k
 
m
2
This equation is said to describe simple harmonic motion or free
undamped motion.
SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION
FREE UNDAMPED MOTION
d 2x
2


x0
2
dt
2 initial conditions can be associated with the differential equation:
x(0)  x0 - the initial displacement of the mass
x' (0)  x1 - the initial velocity of the mass
For example:
x0 > 0, x1 < 0
x0 < 0, x1 = 0
The mass starts from a point below the equilibrium position with
an upward velocity.
The mass is released from rest from a point x0 units above the
equlibrium position.
SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION
FREE UNDAMPED MOTION
d 2x
2


x0
2
dt
Given above second order equation, let’s solve the equation.
The auxiliary equations will be
m2   2  0
The roots will found to be complex number: m1  i, m2  i
From constant coefficients we know that for Case III,
y  c1ex cos x  c2ex sin x
Thus, in this problem
x(t )  c1 cos t  c2 sin t
SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION
FREE UNDAMPED MOTION
From equation
x(t )  c1 cos t  c2 sin t
The period of motion, T  2 /  where T represents the time (secs)
it takes the mass to execute one cycle of motion.
One complete oscillation cycle of a mass is when the mass move from,
say, the lowest point below the equilibrium position to the highest point
and back to the lowest point.
Extreme displacement of the mass means:
 maximum positive displacement, x(t) below the equilibrium position
 minimum negative displacement, x(t) above the equilibrium position
SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION
FREE UNDAMPED MOTION
From equation
x(t )  c1 cos t  c2 sin t
The frequency of motion, f 
completed every one second.
1 

is the number of cycles
T 2
After the constants c1 and c2 are determined, the resulting particular
solution is called the equation of motion.
SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION
FREE UNDAMPED MOTION
Example 1:
A mass weighing 2 pounds stretches a spring 6 inches. At t = 0 the mass is released
from a point 8 inches below the equilibrium position with an upward velocity of
4/3 ft/s. Determine the equation of motion.
Solution:
Standardized units,
Weight, W = mg
Hooke’s Law
6 inches = ½ ft, 8 inches = 2/3 ft
m = W/g = 2/32 = 1/16 slug.
F = ks
2 = k(1/2)
k = 4 lb /ft
SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION
FREE UNDAMPED MOTION
Solution (continue):
Newton’s 2nd Law
d 2x
m 2  kx
dt
1 d 2x
 4 x
2
16 dt
d 2x
 64 x  0
2
dt
SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION
FREE UNDAMPED MOTION
Solution (continue):
The initial conditions are, x(0) = 2/3, x’(0) = -4/3
d 2x
 64 x  0
2
dt
Second order DE
d 2x
2
Simple harmonic motion or free undamped motion equation is


x0
2
dt
Thus,   64 or   8 .
2
So, the general solution of the differential equation is
x(t )  c1 cos 8t  c2 sin 8t
SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION
FREE UNDAMPED MOTION
Solution (continue):
The initial conditions are, x(0) = 2/3, x’(0) = -4/3
x(t )  c1 cos 8t  c2 sin 8t
2
x(0)  :
3
2
3
2
c1 
3
x(0)  c1 cos 8(0)  c2 sin 8(0) 
4
x' (t )  c1 sin 8t  8c2 cos 8t
x' (0)   :
3 x' (0)  c sin 8(0)  8c cos 8(0)   4
1
2
3
c2  
1
6
SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION
FREE UNDAMPED MOTION
Solution (continue):
Thus, the equation of motion is
2
1
x(t )  cos 8t  sin 8t
3
6
SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION
FREE UNDAMPED MOTION
Equation of motion
x(t )  c1 cos t  c2 sin t
Alternatively written as
where
A  c1  c2
c1
tan  
c2
2
2
x(t )  A sin(t   )
is the amplitude
is the phase angle
c1  c2
2
2

c2
c1
SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION
FREE UNDAMPED MOTION
Using Example 1: Write in the form of
2
A  c1  c2
2
2
x(t )  A sin(t   )
2
17
2  1
      
6
3  6
c1
tan    4
c2
  tan 1 (4)  1.326rad
We must take
Thus,

to be the second-quadrant angle, so     (1.326)  1.816rad
17
x(t ) 
sin(8t  1.816)
6
Week 10
SPRING/MASS SYSTEMS: FREE UNDAMPED MOTION
FREE UNDAMPED MOTION
T
2

SPRING/MASS SYSTEMS: FREE DAMPED MOTION
FREE DAMPED MOTION
The concept of free harmonic motion is unrealistic, since the motion assumes no
retarding forces acting on the moving mass.
Unless the mass is suspended in a perfect vacuum, there will be at least a resisting
force due to the surrounding medium.
Mass
suspended in
viscous
medium
Mass connected to a
dashpot damping
device
SPRING/MASS SYSTEMS: FREE DAMPED MOTION
FREE DAMPED MOTION
When no other external forces are impressed on the system, from Newton's
second law:
2
d x
dx
m 2  kx  
dt
dt
where  is a positive damping constant and the negative sign shows that the
damping force acts in a direction opposite to the motion.
Divide by m, the differential equation of free damped motion is:
d 2 x    dx  k 
     x  0
2
dt
 m  dt  m 
SPRING/MASS SYSTEMS: FREE DAMPED MOTION
FREE DAMPED MOTION
d 2 x    dx  k 
     x  0
2
dt
 m  dt  m 
From equation
d 2x
dx
2

2



x0
2
dt
dt
Rewrite as
where
2 

m
and
k
 
m
2
The symbol 2λ is used only for algebraic convenience, so that we can write the
2
2
auxiliary equation as m  2m    0 and the corresponding roots are
then
m1    2   2 and m2    2   2
SPRING/MASS SYSTEMS: FREE DAMPED MOTION
FREE DAMPED MOTION
Based on the algebraic sign of
2   2
, we can distinguish 3 possible cases:
d 2x
dx
2

2



x0
2
dt
dt
CASE I:     0
2
2
x(t )  e
t
(c1e
2  2 t
 c2e
 2  2 t
)
Overdamped system
CASE II: 2   2  0
x(t )  et (c1  c2t )
Critically damped system
CASE III: 2   2  0
Underdamped system
x(t )  e t (c1 cos  2  2 t  c2 sin  2  2 t )
SPRING/MASS SYSTEMS: FREE DAMPED MOTION
FREE DAMPED MOTION
CASE I
Overdamped
system
CASE II
Critically
damped
system
CASE III
Underdamped
system
- The damping coefficient β is larger than k.
- Smooth and non-oscillatory motion
- Any slight decrease in damping force would
result in oscillatory motion.
- Motion is similar to overdamped system.
- Mass can pass through the equilibrium
position at most one time.
- The damping coefficient β is smaller than k.
- The motion is oscillatory but the amplitudes
of vibration  0 as t  ∞.
SPRING/MASS SYSTEMS: DRIVEN MOTION
DRIVEN MOTION WITH DAMPING
Suppose there is an external force f(t) acting on a vibrating mass on
a spring.
For example, f(t) could represent a driving force causing an
oscillatory vertical motion of the support of the spring.
Thus, if we include f(t) in the formulation of Newton’s second law,
the differential equation of driven motion/forced motion is,
d 2x
dx
m 2  kx  
 f (t )
dt
dt
SPRING/MASS SYSTEMS: DRIVEN MOTION
DRIVEN MOTION WITH DAMPING
From
d 2x
dx
m 2  kx  
 f (t )
dt
dt
d 2x
dx
2

2



x  F (t )
2
Divide by m gives,
dt
dt

f (t )
k
where F (t ) 
, 2 
and  2 
m
m
m
The above nonhomogeneous equation can be solved using the
method of undetermined coefficients or variation of parameters.
SPRING/MASS SYSTEMS: DRIVEN MOTION
Example: Solve the initial value problem
1 d 2x
dx

1
.
2
 2 x  5 cos 4t ,
2
5 dt
dt
1
x(0)  ,
2
x' (0)  0
Solution:
From driven motion/forced motion DE,
d 2x
dx
m 2  kx  
 f (t )
dt
dt
Thus, m = 1/5 kg, k = 2 N/m.
From initial conditions, known that the mass is released from rest at ½ meter
below the equilibrium position.
The motion is damped with β = 1.2 and is being driven by an external periodic
force beginning at t = 0.
SPRING/MASS SYSTEMS: DRIVEN MOTION
Solution (continue):
1 d 2x
dx

1
.
2
 2 x  5 cos 4t ,
2
5 dt
dt
1
x(0)  ,
2
x' (0)  0
Solve for the homogeneous part first, multiply by 5:
d 2x
dx

6
 10 x  0
2
dt
dt
Solve and found, m1  3  i and m2  3  i
Thus, xc (t )  e3t (c1 cos t  c2 sin t )
To solve the nonhomogeneous part, we use undetermined coefficient method
to assume that the particular solution take the form A cos 4t  B sin 4t
SPRING/MASS SYSTEMS: DRIVEN MOTION
Solution (continue):
Assuming
x p (t )  A cos 4t  B sin 4t
Differentiating x p (t ) twice and substitute into the DE gives
x' ' p 6 x' p 10 x p  (6 A  24B) cos 4t  (24 A  6B) sin 4t  25 cos 4t
25
50
and B 
102
51
25
50
cos 4t  sin 4t
Thus, x(t )  e 3t (c1 cos t  c2 sin t ) 
102
51
Solving the above equation gives A  
At t = 0, x = ½ :
At t = 0, x’ = 0:
38
51
86
c2  
51
c1 
38
86
25
50
cos 4t  sin 4t
Thus, equation of motion is x(t )  e ( cos t  sin t ) 
51
51
102
51
3t
SPRING/MASS SYSTEMS: DRIVEN MOTION
TRANSIENT AND STEADY-STATE TERMS
d 2x
dx
2

2



x  F (t )
2
dt
dt
When F is a periodic function such as F (t )  F0 sin t or
F (t )  F0 cos t , the general solution for λ >0 is the sum of a
nonperiodic function xc (t ) and a periodic function x p (t ).
Also, xc (t ) becomes zero as time increases (t  ∞).
Thus, for large values of time, the displacements of the mass are
closely approximated by the particular solution x p (t ) .
SPRING/MASS SYSTEMS: DRIVEN MOTION
TRANSIENT AND STEADY-STATE TERMS
The complementary function xc (t ) is said to be a transient term
or transient solution, and the particular solution x p (t ) is called the
steady-state term or steady-state solution.
SPRING/MASS SYSTEMS: DRIVEN MOTION
TRANSIENT AND STEADY-STATE TERMS
From previous example,
x(t )  e 3t (
38
86
25
50
cos t  sin t ) 
cos 4t  sin 4t
51
51
102
51
Transient
Steady-state
SPRING/MASS SYSTEMS: DRIVEN MOTION
DRIVEN MOTION WITHOUT DAMPING
With a periodic force applied and NO damping force, there is no
transient term in the solution of the problem.
Example: (Undamped Forced Motion)
Solve the initial value problem
d 2x
2


x  F0 sin t
2
dt
Where F0 is a constant and γ ≠ω.
x(0)  0, x' (0)  0
SPRING/MASS SYSTEMS: DRIVEN MOTION
DRIVEN MOTION WITHOUT DAMPING
Solution:
The complementary function is xc (t )  c1 cos t  c2 sin t
To find the particular solution, we assume that x p (t )  A cos t  B sin t
x' ' p  2 x p  A( 2   2 ) cos t  B( 2   2 ) sin t  F0 sin t
which gives A = 0 and B  F0 /( 2   2 ) .
F0
Therefore, x p (t )  2
sin t
2
 
SPRING/MASS SYSTEMS: DRIVEN MOTION
DRIVEN MOTION WITHOUT DAMPING
Solution (continue):
Thus, the general solution is
F0
x(t )  c1 cos t  c2 sin t  2
sin t
2
 
With initial conditions x(0)  0, x' (0)  0
 F0
Found c1  0 and c2 
 ( 2   2 )
F0
Thus, the solution is x(t ) 
( sin t   sin t ) ,
2
2
 (   )
 
SERIES CIRCUIT ANALOGUE
LRC SERIES CIRCUITS
The charge q(t) on the capacitor is
related to the current i(t) by
i  dq / dt
From
LRC series circuit
di
1
L  Ri  q  E (t )
dt
C
d 2q
dq 1
L 2  R  q  E (t )
dt
dt C
Note that the nomenclature used in analysis of circuits is similar to that used
to describe spring/mass systems.
SERIES CIRCUIT ANALOGUE
LRC SERIES CIRCUITS
If E(t) = 0, the electrical vibrations of the circuit are said to be free.
d 2q
dq 1
1
From L 2  R
 q  E (t ) the auxiliary equation is Lm2  Rm   0
dt
dt C
C
Thus, there will be 3 forms of the solution with R ≠ 0, depending on the
value of the discriminant R 2  4L / C
R 2  4L / C  0
Overdamped
R 2  4L / C  0
Critically damped
R 2  4L / C  0
Underdamped
SERIES CIRCUIT ANALOGUE
LRC SERIES CIRCUITS
R 2  4L / C  0
R 2  4L / C  0
R 2  4L / C  0
Overdamped
Critically damped
Underdamped
In each THREE cases, the general solution contains the factor
so, q(t)  0 as t  ∞.
e  Rt / 2 L
In the underdamped case, when q(0)  q0, the capacitor is charging
and discharging as t ∞.
When E(t) = 0, R=0, the circuit is said to be undamped and the
electrical vibrations do not approach zero as t ∞; the response of
the circuit is simple harmonic.
SERIES CIRCUIT ANALOGUE
LRC SERIES CIRCUITS
Example:
Find the charge q(t) on the capacitor in an LRC series circuit when L = 0.25
henry (h), R = 10 ohms (Ω), C = 0.001 farad (f), E(t) = 0, q(0) =q0 and i(0)=0.
Solution:
d 2q
dq 1
Since 1/C = 1000, equation L 2  R
 q  E (t ) becomes
dt
dt C
1
q' '10q'1000q  0
4
q' '40q'4000q  0
Solve for the homogeneous equation, found R 2  4L / C  0 means the
20t
circuit is underdamped and q(t )  e (c1 cos 60t  c2 sin 60t )
SERIES CIRCUIT ANALOGUE
LRC SERIES CIRCUITS
Solution(continue):
With initial solutions q(0) =q0 and i(0)=0:
1
Found c1  q0 and c2  q0.
3
Thus,
1
q(t )  q0e 20t (cos 60t  sin 60t )
3
SPRING MASS SYTEM/SERIES ELECTRICAL CIRCUIT
We have identified the DE for spring/mass system and series
electrical circuits:
d 2x
dx
m 2 
 kx  f (t )
dt
dt
Driving function/
External force f(t)
d 2q
dq 1
L 2  R  q  E (t )
dt
dt C
Driving function/
Applied voltage E(t)
Previous cases that we have gone through in previous sections only
considers problems in which the functions f and E were continuous.
SPRING MASS SYTEM/SERIES ELECTRICAL CIRCUIT
Discontinuous driving functions are possible and not uncommon.
For example, the applied voltage on a circuit could be piecewise
continuous and periodic such as the “sawtooth” function described
by graph below:
SPRING MASS SYTEM/SERIES ELECTRICAL CIRCUIT
Given these types of differential equations, it is difficult to solve
the DE using the techniques that we have studied so far (for
second order DE).
The Laplace transform will be a useful tool that simplifies the
solution of problems such as these.
The End