ODE Lecture Notes
Section 3.8
Page 1 of 14
Section 3.8: Forced Vibrations
Big Idea: A damped spring-mass (or circuit) system described in section 3.7 that is driven by a
sinusoidal force will eventually settle into a steady oscillation with the same frequency as the
driving force. The amplitude of this steady state solution increases as the drive frequency gets
closer to the natural frequency of the system, and as the damping decreases. An undamped
spring-mass system that is driven by a sinusoidal force results in an oscillation that is the average
of the two frequencies and modulated by a sinusoidal envelope that is the difference of the
frequencies.
Big Skill: You should be able to solve sinusoidally-forced vibration problems.
Problems:
1. Find the solution of a spring-mass system IVP
u  u  1.25u  3cos  t  , u  0  2, u  0   3
ODE Lecture Notes
Section 3.8
Page 2 of 14
ODE Lecture Notes
Section 3.8
Page 3 of 14
Forced Vibrations with Damping
In this section, we will restrict our discussion to the case where the forcing function is a sinusoid.
Thus, we can make some general statements about the solution:
The equation of motion with damping will be given by:
mu   u  ku  F0 cos t 
Its solution will be of the form:
u  t   c1u1  t   c2u2  t   A cos t   B sin t 
homogeneous solution uh  t 
"transient solution"
particular solution u p  t 
"steady state solution"
Notes:
 The homogeneous solution uh  t   0 as t   , which is why it is called the “transient
solution.”
 The constants c1 and c2 of the transient solution are used to satisfy given initial
conditions.
 The particular solution u p  t  is all that remains after the transient solution dies away,
and is a steady oscillation at the same frequency of the driving function. That is why it is
called the “steady state solution,” or the “forced response.”
 The coefficients A and B must be determined by substitution into the differential
equation.
 If we replace u p  t   U  t   A cos t   B sin t  with u p  t   U  t   R cos t    ,

m 0 2   2 
2
F0

then R  , cos   
, sin   
,   m 2 0 2   2    2 2 , and



k
0 2  . (See scanned notes at end for derivation)
m
Note that as   0 , cos    1 and sin    0    0 .

Note that when   0 ,  



Note that as   ,    (mass is out of phase with drive).

2
The amplitude of the steady state solution can be written as a function of all the
parameters of the system:
ODE Lecture Notes
R

Page 4 of 14
F0
F0

2

m 2 0 2   2    2 2
F0
2
 2 
2
m 0 1  2    20 2 2
0
 0 
F0
2

Section 3.8
4
2
k2  2 
k 2
m 2 1  2    2
m  0 
m 0 2
2


F0
2
 2   2 2
k 1  2  
2
 0  mk 0
F0 / k
2
 2 
2
1  2    2
0
 0 
, 
2
mk
 k 
1
R  
2
 F0 
 2 
2
1




2 
0 2
 0 





 k 
Notice that R   is dimensionless (but proportional to the amplitude of the motion),
 F0 
F
since 0 is the distance a force of F0 would stretch a spring with spring constant k.
k
  mass 2 


  2    time  
2
Notice that  
is dimensionless…   
 1
mk
 mk   mass mass 

time2 

 k 
F
Note that as   0 , R    1  R  0 .
k
 F0 
Note that as   , R  0 (i.e., the drive is so fast that the system cannot respond to it
and so it remains stationary).
The frequency that generates the largest amplitude response is:
ODE Lecture Notes
Section 3.8
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d   k 
 R    0
d    F0  
1    2   2 
 
1




2
1


2



2




2
2
2
2   0   0 
0 2 
d    2 
2  
 1 
   2   
3
d    0 2 
0
2 2

 
   2 2



 1  2    2 
  0 
0 



    2 
 2   2  1  2      0
 0    0 

 2 
 2 1  2     0
 0 
2 
1 2 
0
2
  0,


2
max
 0 2  1 

2
max


2
max
k 2
 0 
m 2mk
2
 0 
2
2
2m 2
Plugging this value of the frequency into the amplitude formula gives us:
F0
Rmax 
0 1 




2

2
2
4mk
 1 , then the maximum value of R occurs for   0 .
4mk
Resonance is the name for the phenomenon when the amplitude grows very large
because the damping is relatively small and the drive frequency is close to the undriven
frequency of oscillation of the system.
If
ODE Lecture Notes
Section 3.8
Page 6 of 14
Practice:
2. Find the solution of u  0.125u  u  3cos t  , u  0  2, u  0  0 . Make graphs of the
solution for various values of .
ODE Lecture Notes
Section 3.8
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ODE Lecture Notes
Section 3.8
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Demonstration:
3. Compute the spring constant and resonant frequency of a given spring-mass system, and
then verify the resonant frequency calculation experimentally. Observe the following:
 pulling up and down on the spring very rapidly results in virtually no motion of
the mass.
 pulling up and down slowly results in the entire system simply moving up and
down with the pull.
 pulling up and down near the resonant frequency results in a large oscillation for a
pulling motion that is almost imperceptible to the human eye.
ODE Lecture Notes
Section 3.8
Forced Vibrations Without Damping
The equation of motion of an undamped forced oscillator is:
mu  ku  F0 cos t 
When   0 (non-resonant case), the solution is of the form:
u  t   c1 cos 0t   c2 sin 0t  
F0
k
cos t  , 0 
2
2
m
m 0   
When   0 (resonant case), the solution is of the form:
F
u  t   c1 cos 0t   c2 sin 0t   0 t sin 0t 
2m
Practice:
4. Derive both particular solutions above.
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ODE Lecture Notes
Section 3.8
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5. Show that for the non-resonant case with initial condition u  0  0, u  0  0 (i.e., starting
from rest at the equilibrium position), c1  
becomes u  t  
F0
, c 2  0 , and that the solution
m  0 2   2 
F0
 cos t   cos 0t   , which can be written as
m 0 2   2 

 0    t    0    t 
2 F0
u t   
sin

  sin 
 using the sum-to-product trig
2
2
2
2
 m 0    
  

u v  u v 
identity cos  u   cos  v   2sin 
 sin 
.
 2   2 
ODE Lecture Notes
Section 3.8
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Notice in the last problem that the solution can be considered to be a rapid-frequency oscillation
that is modulated by a low-frequency oscillation. In electronics, this is called amplitude
modulation. In acoustics, the low-frequency oscillation is called a beat because the sound wave
is perceived as a constant pitch whose loudness varies with a beat.
6. Solve the IVP u  u  0.5cos  0.8t  , u  0  0, u  0   0
ODE Lecture Notes
Section 3.8
7. Solve the IVP u  u  0.5cos  t  , u  0  0, u  0  0
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ODE Lecture Notes
Section 3.8
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ODE Lecture Notes
Section 3.8
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