Free (Unforced) Vibrations
Consider a mass resting on a frictionless, horizontal surface, connected by a linear spring
and a linear damper to a fixed support. X is measured from the un-stretched position of the
spring.
 .
Consider a free-body diagram (FBD) of the mass located at X with a velocity X
  ma  m X

FX   k X  c X
X
(23.1)
  c X
 kX
0  mX
(23.2)
   c  X
   k  X
0X
m
m
(23.3)
This is a linear, ordinary differential equation with constant coefficients.
Regardless of what someone in the mathematics department may have told you, this general
class of differential equation is essentially the only one of importance to engineers and very
probably the only one any of you will ever be required to solve.
Because the highest derivative is a second derivative, this is referred to as a second-order
differential equation. Because there are no terms on the other side of the equal sign, this is
referred to as a homogeneous differential equation.
MAE3501 - 23 - 1
Make the following substitutions that demonstrate that I have previously solved a bunch
of these differential equations.
c
(23.4)
 2
m
k
(23.5)
 n 2
m
Note that  n is the natural angular frequency of the un-damped oscillator.
  2  X
  2 X
0X
n
(23.6)
The method of solution is always the same.
X  A e t
(23.7)
Substitute this general solution into the differential equation.
0  A 2 e  t  2  A  e  t   n 2 A e  t
t
 A e (  2     n )
2
2
(23.8)
After simplifying, equation (23.8) is a quadratic equation.

 2   4  2  4 n 2
     2  n 2
(23.9)
2
There are three cases, depending on the value of the discriminant, the quantity under the
square root.
I.
(23.10)
 2  n 2  0
In this case, there are two distinct, real roots to the quadratic equation.
1    
 2  n 2
(23.11)
2    
 2  n 2
(23.12)
The solution of the differential equation.
X(t )  A1 e 1 t  A 2 e  2 t
(23.13)
Case I is referred to as over-damped, corresponding to a large value of the damping
coefficient. A mass moving in a tub full of molasses. If the mass is displaced from its
equilibrium position and released from rest, it never crosses the equilibrium position. This case
is of no interest to us.
MAE3501 - 23 - 2
II.
 2  n 2  0
(23.14)
In this case, there is one repeated root to the quadratic equation.

(23.15)
It is a theorem of mathematics that the solution of a homogeneous, second-order
differential equation must consist of exactly two linearly independent terms. I will not bore you
with the method used to generate the second term.
The solution of the differential equation.
X(t )  A e  t  B t e t
(23.16)
Case II is referred to as critically damped, corresponding to a large value of the damping
coefficient, but not as large as in case I. A mass moving in a tub full of vegetable oil. If the mass
is displaced from its equilibrium position and released from rest, it crosses the equilibrium
position exactly once. This case is of no interest to us, except to comment that automobile shock
absorbers are designed to be close to critically damped.
MAE3501 - 23 - 3
III.
 2  n 2  0
(23.17)
In this case, there are two complex-conjugate roots to the quadratic equation.
 1     j n 2   2
(23.18)
 2     j n 2   2
(23.19)
Make the following substitution.
  n 2   2
(23.20)
The solution of the differential equation.
X(t )  e   t ( A1e j t  A 2e  j t )
(23.21)
Case III is referred to as under-damped, corresponding to a small value of the damping
coefficient. A mass moving in air. If the mass is displaced from its equilibrium position and
released from rest, it oscillates back and forth, with ever-decreasing amplitude. This case is of
interest to us.
Substitute Euler’s Identity into equation (23.21).
X(t )  e   t ( A 1 cos t  A 1 j sin t  A 2 cos t  A 2 j sin t )
 e   t ( A 1  A 2 ) cos t  ( A 1  A 2 ) j sin t 
(23.22)
Rename the constants in equation (23.22).
X(t )  e   t  A cos t  B sin t 
(23.23)
Note that the frequency of oscillation of the damped oscillator is less than the natural
frequency of the un-damped oscillator.
MAE3501 - 23 - 4
The constants in equation (23.23) are determined through boundary conditions,
frequently initial conditions.
X(t  0)  X 0  A
(23.24)
 (t  0)  V    A  B 
X
0
(23.25)
B
V0   A V0   X 0



(23.26)
The general solution of the free (unforced) under-damped oscillator.


 V   X0 
X(t )  e   t  X 0 cos t   0
 sin t 





MAE3501 - 23 - 5
(23.27)
Forced (Driven) Vibrations
Consider a mass resting on a frictionless, horizontal surface, connected by a linear spring
and a linear damper to a fixed support. X is measured from the un-stretched position of the
spring. The mass is subjected to an external force that is a function of time.
It is straightforward to develop the differential equation of motion.
 FX   k X  c X  F(t)  m a X  m X
(23.28)
  c X
 kX
F(t )  m X
(23.29)
F(t )   c    k 
 X  X  X
m
m
m
(23.30)
Make the same substitutions that we made in the unforced case.
c
 2
m
k
 n 2
m
F(t ) 
  2 X
 X  2 X
n
m
(23.31)
(23.32)
(23.33)
The method of solution of a differential equation of this form is always the same. First
determine the homogeneous, or transient, solution, the solution if there were zero on the left of
the equal sign. This is, of course, the solution we obtained for the unforced case. We will
consider only the under-damped case.
MAE3501 - 23 - 6
Make the following substitution.
  n 2   2
(23.34)
X H (t )  e   t  A cos t  B sin t 
(23.35)
Because the differential equation is second order, there must be exactly two linearly
independent terms in the homogeneous solution.
The next step is to determine the particular, or steady-state, solution. The steady-state
solution is what remains after the transient solution has decayed to zero, or essentially zero. The
steady-state solution is a solution of the complete differential equation that does not contain the
homogeneous solution. It is a theorem of mathematics that there is one, and only one, particular
solution of a differential equation.
If the forcing function has a finite number of derivatives, it is straightforward to use the
method of undetermined coefficients to determine the particular solution.
Example
  2 X
  3X
3t2  X
(E.1)
F(t )  3 t 2
(E.2)
d F( t )
 6t
dt
(E.3)
d 2 F( t )
6
(E.4)
dt 2
All subsequent derivatives are equal to zero. The particular solution is a linear
combination of the forcing function and all its derivatives, each multiplied by an undetermined
coefficient.
X P (t )  C t 2  D t  E
(E.5)
 (t )  2 C t  D
X
P
(E.6)
 (t )  2 C
X
P
(E.7)
Substitute equations (E.5) through (E.7) into equation (E.1).
3 t 2  2 C  2 (2 C t  D)  3 (C t 2  D t  E)
MAE3501 - 23 - 7
(E.8)
Collect terms in powers of t.
t2 :
3  3C
(E.9)
t1 :
0  4C  3 D
(E.10)
t0 :
0  2C  2 D  3E
(E.11)
C1
D
E
(E.12)
4
3
(E.13)
2
9
(E.14)
 4
 2
X P (t )  t 2    t   
 3
 9
MAE3501 - 23 - 8
(E.15)
Example
  2 X
  3X
2 ea t  X
(E.1)
F(t )  2 e a t
(E.2)
All derivatives of the forcing function are multiples of each other. Realize that a is a
known value.
X P (t )  C e a t
(E.3)
 (t )  C a e a t
X
P
(E.4)
 (t )  C a 2 e a t
X
P
(E.5)
Substitute equations (E.3) through (E.5) into equation (E.1).
2 ea t  C a 2 ea t  2 C a ea t  3 C ea t
C
2
(E.7)
( a2  2a  3 )
XP 
2
( a  2a  3 )
2
(E.6)
e at
(E.8)
a has units of inverse time. Therefore, it would appear that we are adding terms with
different units in the denominator. However, examine the differential equation. The “2” on the
left side must have the units of force. The “3” on the right side must have the units of force
divided by length. The “2” on the right side must have the units of force multiplied by time
divided by length. The “1” on the right side, the multiplier of the acceleration term, must have
the units of force multiplied by time squared divided by length.
The method of undetermined coefficients will not work for some forcing functions, such
1
as log t and . Each of these functions possesses an infinite number of non-zero derivatives.
t
MAE3501 - 23 - 9
Return to forced vibrations.
F(t ) 
  2 X
 X  2 X
n
m
(23.36)
The most important class of forcing functions for lightly damped vibrations is sinusoidal,
either sine or cosine. Realize that F0 and  are known values.
F(t )  F0 cos  t
(23.37)
d F( t )
   F0 sin  t
dt
(23.38)
All subsequent derivatives have one of the above two functional dependences on time.
X P (t )  C sin t  D cos  t
(23.39)
 (t )  C  cos  t  D  sin t
X
P
(23.40)
 (t )   C  2 sin t  D  2 cos  t
X
P
(23.41)
Substitute equations (23.39) through (23.41) into equation (23.36).
 F0 
  cos  t  (  C  2 sin  t  D  2 cos  t )
m
 2  ( C  cos  t  D  sin  t )
(23.42)
  n 2 (C sin  t  D cos  t )
Sin:
0   C  2  2  D   n 2 C
(23.43)
Cos:
F0
  D  2  2  C   n 2 D
m
(23.44)
MAE3501 - 23 - 10
Solve equation (23.43) for D in terms of C.
D
C ( n 2   2 )
2 
(23.45)
Substitute equation (23.45) into equation (23.44).

F0
(  2   2 )2 
 C 2    n
C
m
2 


 ( 2   ) 2  ( n 2   2 ) 2 


2 



2 
F  
C 0

 m   ( 2   ) 2  (  n 2   2 ) 2 
(23.46)
(23.47)
Substitute equation (23.47) into equation (23.45).

( n 2   2 )
 F0  
D 

 m   ( 2   ) 2  (  n 2   2 ) 2 
(23.48)
Recall equations (23.37) and (23.39).
F(t )  F0 cos  t
(23.49)
X P (t )  C sin t  D cos  t
(23.50)
MAE3501 - 23 - 11
There are three cases to consider, depending on the relative magnitude of  compared to
n .
Consider   n .
C0
(23.51)
F  1  F
D 0 2 0
 m    n  k
(23.52)
The steady state response is essentially in phase with the driving force. The amplitude of
the steady-state response is equal to the amplitude of the driving force divided by the spring
constant. Neither of these results is affected by the magnitude of the weak damping coefficient.
Consider   n .
 F   2 
C 0 3
 m  
(23.53)
F   1 
D 0 2
 m  
(23.54)
The steady state response is essentially 180 out of phase with the driving force. This
result is weakly affected by the magnitude of the weak damping coefficient. The amplitude of
the steady-state response is very small. This result is not affected by the magnitude of the weak
damping coefficient.
Consider   n .
 F   1  F0
 
C   0  
 m   2   c
(23.55)
D0
(23.56)
The steady state response lags the driving force by 90 . This result is not affected by the
magnitude of the weak damping coefficient. The amplitude of the steady-state response is large,
and is inversely proportional to the magnitude of the weak damping coefficient.
MAE3501 - 23 - 12
Un-damped, Forced Vibrations
Set the damping coefficient equal to zero in equations (23.47) and (23.48).

2 
F  
C 0
0
 m   ( 2   ) 2  (  n 2   2 ) 2 
(23.57)
  F0  

( n 2   2 )
1
F  
D 0






 m   ( 2   ) 2  (  n 2   2 ) 2   m   (  n 2   2 ) 
(23.58)
Consider   n .
D0
(23.59)
The steady state response is in phase with the driving force. The amplitude of the steadystate response increases without limit as the driving frequency approaches the natural frequency
of the system.
Consider   n
D0
(23.60)
The steady state response is 180 out of phase with the driving force. The amplitude of
the steady-state response increases without limit as the driving frequency approaches the natural
frequency of the system.
MAE3501 - 23 - 13
Homework
Read sections 7-1 through 7-3.
Consider a free oscillator with the following characteristics and initial conditions.
m  14.0 kg
N
k  7.0  
m
N s
c  1.0 

 m 
X 0  0.5 m
m
V0   0.25  
 s
Determine the equation of motion of the oscillator, and the time at which it first returns to
the un-stretched position of the spring.
MAE3501 - 23 - 14