MAE 524 course notes – Spring 2002, Copyrighted L. Silverberg
- -
16
C. Free Vibration
C.1 Undamped vibration
 (1.34) mx(t )  kx(t )  f (t ),
where m is mass, k is stiffness and it shall be assumed for
now that the external force acting on the system is zero.
The equation governing the free vibration is
(1.35) x(t )   02 x(t )  0,  02  k / m
Where  0 is the system’s natural frequency of
oscillation. The general solution of Eq. (1.35) is
(1.37) x(t )  A sin( 0 t )  B cos( 0 t )
 Let’s now return to the solution of the single degree of
freedom system, Eq. (1.37), and determine the constants
A and B in that equation. Letting x0 denote the system’s
initial position, and letting v0 denote it’s initial velocity,
we get from Eq. (1.35)
(1.40) x(0)  x0  B, x (0)  v0   0 A
 Solving for A and B in Eq. (1.40) and substituting the
result into Eq. (1.37), yields the free response of the
undamped system
v
(1.41) x(t )  x0 cos( 0 t )  0 sin( 0 t )
0
FREE VIBRATION
Lecture notes prepared by L.Silverberg and J.Morton
MAE 524 course notes – Spring 2002, Copyrighted L. Silverberg
- -
17
C.2 Damped vibration (stable cases only)
 (1.42) mx(t )  cx (t )  kx(t )  f (t ), f (t )  0
where c is the damping constant. Dividing Eq. (1.42) by
m, yields Eq. (1.43).
c
k
x(t )  2x (t )   02 x (t )  0,  
,  02  .
2m
m
 This time, let’s determine the solution by trying a
solution in the complex form
(1.44) x  e st
where s is a complex number. Substituting Eq. (1.44)
st
into Eq. (1.43), and dividing the result by e , yields
(1.45) s 2  2s   02  0
 Equation (1.45) is a quadratic equation with real
coefficients. The solutions can be pairs of numbers that
are complex, real multiples, or real distinct.
FREE VIBRATION
Lecture notes prepared by L.Silverberg and J.Morton
MAE 524 course notes – Spring 2002, Copyrighted L. Silverberg
- -
18
Underdamped (0)
 When 0, the quadratic equation (1.45) admits the
two solutions
2
2
2
(1.46) s    i,    0  
 Substituting Eq. (1.46) into Eq. (1.44), and invoking the
principle of linear superposition
(1.47) x(t )  A1e s1t  A2 e s2t  e t [ A1e it  A2 e it ]
 e t [( A1  A2 ) cos(t )  i( A1  A2 ) sin( t )]
 e t [ A cos(t )  B sin( t )]
 where represents the system’s natural rate of decay,
and  denotes the system’s damped natural frequency
of oscillation. The arbitrary constants in Eq. (1.47) are
determined from the initial conditions. From Eq. (1.47)
(1.48) x0  A, v0  A  B
 The underdamped free response is
(1.49) x(t )  e t [ x0 cos(t ) 
v0  x0

sin(t )].
FREE VIBRATION
Lecture notes prepared by L.Silverberg and J.Morton
MAE 524 course notes – Spring 2002, Copyrighted L. Silverberg
- -
Critically damped (   0 )
 When 0, the quadratic equation (1.45) admits the
two repeated solutions
(1.50) s1  s 2  
t
(1.51) x1  e t , x2  te
 By the principle of linear superposition, the general
solution is
(1.52) x(t )  e t [ A  Bt ]
 where A and B are arbitrary constants. Substituting the
initial conditions into Eq. (1.52) yields the critically
damped free vibration response
(1.53) x  e t [ x0  (x0  v0 )t ]
FREE VIBRATION
Lecture notes prepared by L.Silverberg and J.Morton
19
MAE 524 course notes – Spring 2002, Copyrighted L. Silverberg
- -
20
Overdamped (   0 )
 When    0 , the quadratic equation (1.45) admits two
real solutions.
2
2
2
(1.54) s1     , s2    ,      0
 The general overdamped free vibration response is
(1.55) x  Ae
s1t
 Be s2t  e t [ Ae t  Be t ]
  (   ) x0  v0 t (   ) x 0  v0 t 
 e t 
e 
e 




2
2


C.3 Comparison between limiting cases
 The undamped, underdamped, critically damped and
overdamped responses, Eqs. (1.41), (1.49), (1.53), and
(1.55), are identical to each other in limiting cases.
 The limits are
lim
(1.56)
 0
sin( t )

t cos(t )
t
 0
1
 lim
and
(1.57) lim
0
et  e t

tet  te t
 lim
 2t.
 0
1
FREE VIBRATION
Lecture notes prepared by L.Silverberg and J.Morton