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EN40: Dynamics and Vibrations
Solutions to Differential Equations of Motion for Vibrating Systems
Here, we summarize the solutions to the most important differential equations of motion that we
encounter when analyzing single degree of freedom linear systems.
CASE I:
2
2 2
n
d x
x
dt
CASE II
2
2 2
d x
x
dt
CASE III
2
2 2
n n
d x dx
x
dt dt
CASE IV
2
2 2
n n
d x dx
x KF t
dt dt
with 0
F t ( ) F sin t
CASE V
2
2 2
n n n
d x dx dy
x K y
dt dt dt
with 0
y Y sin t
CASE VI
2 2
2 2 2
n n
d x dx d y
x K
dt dt dt
with 0
y Y sin t
SOLUTION 1:
The equation
2
2 2
n
d x
x
dt
with initial conditions
0 0
dx
x x v t
dt
has solution
0
2 2 2 1 0
0 0 0
0
sin
/ tan
n
n
n
x X t
x
X x v
v
or, equivalently
0
0
( ) cos sin n n
n
v
x t x t t
SOLUTION 2
The equation
2
2 2
d x
x
dt
with initial conditions
0 0
dx
x x v t
dt
has solution
0 0
0 0
( ) exp exp
v v
x t x t x t
SOLUTION 4:
2
2 2
n n
d x dx
x KF t
dt dt
with 0
F t ( ) F sin t
and initial conditions
0 0
dx
x x v t
dt
has solution of the form
h p
x t x t x t
where the steady state solution (or particular integral) is
0
1 0
0 1/2 2 2
2
2 2 2
( ) sin
tan
p
n
n
n n
x t X t
KF
X
while the transient solution (or homogeneous solution) is:
Case I: Overdamped System 1
0 0 0 0
( ) exp( ) exp( ) exp( )
h h h h
n d n d
h n d d
d d
v x v x
x t t t t
where
2
d n
Case II: Critically Damped System 1
( ) exp( )
h h h
h n n
x t x v x t t
Case III: Underdamped System 1
0 0
0
( ) exp( ) cos sin
h h
h n
h n d d
d
v x
x t t x t t
where
2
d n
In all three preceding cases, we have set
0 0 0 0
0 0 0 0
0
(0) sin
cos
h
p
p h
t
x x x x X
dx
v v v X
dt
Observe that for large time, the transient solution always decays to zero.
The graphs below plot the amplitude of the steady state vibration and the steady state phase lead.
(a) (b)
Steady state response of a forced spring—mass system (a) amplitude and (b) phase
SOLUTION 5
The equation
2
2 2
n n n
d x dx dy
x K y
dt dt dt
with 0
y t ( ) Y sin t
and initial conditions
0 0
dx
x x v t
dt
has solution of the form
h p
x t x t x t
where the steady state solution (or particular integral) is
0
1/
2
3 3 0
1
0 1/2 2 2 2
2
2 2 2
( ) sin
tan
p
n
n
n
n n
x t X t
KY
X
while the transient solution (or homogeneous solution) is:
Case I: Overdamped System 1
0 0 0 0
( ) exp( ) exp( ) exp( )
h h h h
n d n d
h n d d
d d
v x v x
x t t t t
where
2
d n
SOLUTION 6
The equation
2 2
2 2 2 2
n n n
d x dx K d y
x
dt dt dt
with 0
y Y sin t
and initial conditions
0 0
dx
x x v t
dt
has solution of the form
h p
x t x t x t
where the steady state solution (or particular integral) is
0
2 2
0 1
0 1/2 2 2
2 2 2 2
( ) sin
tan
p
n n
n
n n
x t X t
KY
X
while the transient solution (or homogeneous solution) is:
Case I: Overdamped System 1
0 0 0 0
( ) exp( ) exp( ) exp( )
h h h h
n d n d
h n d d
d d
v x v x
x t t t t
where
2
d n
Case II: Critically Damped System 1
0 0 0
( ) exp( )
h h h
h n n
x t x v x t t
Case III: Underdamped System 1
0 0
0
( ) exp( ) cos sin
h h
h n
h n d d
d
v x
x t t x t t
where
2
d n
In all three preceding cases, we have set
0 0 0 0
0 0 0 0
0
(0) sin
cos
h
p
p h
t
x x x x X
dx
v v v X
dt
Observe that for large time, the transient solution always decays to zero.
The graphs below show the steady state amplitude and phase lead for Case 6.
(a) (b)
Steady state response of a rotor excited spring—mass system (a) Amplitude; (b) Phase
