Updated 12-15-2010
Vibration Analysis Theory
Governing Equations for a Damped Harmonic Oscillator
Figure 1 shows a detailed and simplified lumped-parameter model of the mechanical side of the
Dynamics Module.
Figure 1: The mechanical-vibration experiment implements a simple mass-spring-damper system. A linear actuator
creates the forcing function, f(t).
A mass, m, which can be varied by attaching different masses to the mass platform, is acted
upon by a force, f(t), which is produced by an electromagnetic linear actuator. The force, which
is linearly related to the current passing through the actuator, is directly controlled by the
amplifier of the actuator. LabVIEW software controls the amplifier, so any type of forcing
function (sinusoidal, impulsive, etc.) can be created and applied to the mechanical system. The
spring supporting the mass has a spring constant k which can be varied by changing the
mounting point of the spring, i.e., changing the number of active coils of the spring. The viscous
damping of the system is provided by a variable damper and is characterized by the damping
coefficient, c. There is also damping from Coulomb friction [1]. However, the effect of the
Coulomb friction is minimized by applying an equal and opposite force as the friction force using
the linear actuator. The minimization of the friction force is done in an attempt to make the
system behave more like the ideal spring-mass-damper system. Therefore, from a force-balance
procedure, we obtain the following differential equation for this damped, forced, single-degreeof-freedom system:
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.. .
m x  c x  kx  f (t ) .
(1)
The forcing function, f(t), is controlled by the LabVIEW software, and the equation to describe a
sinusoidal forcing function with amplitude F0 and frequency  (radians/s) is shown below.
f (t )  F0 sin( t )
(2)
The solution to Eq. (1) can be written as [1]:
x(t )exp  n t X 1sin(  d t   ) 
F0 k
 
1  
   n



2
2
  
  2 
    n
sin( t   )



2
(3)
where:
ωn 
ζ 
k
 undamped natural frequency
m
c
 damping ratio
cc
cc  2ωn m  2 km  critical damping coefficient
X 1  constant that depends on the initial c onditions
ωd  ωn 1  ζ 2  damped natural frequency
 and θ  pha se shifts between forcing function and system response
The first term on the right hand side of Eq. (3) is called the homogeneous or transient part of
the solution. The contribution of this term to the complete solution becomes negligible after
the time t becomes sufficiently large. The transient solution describes the vibrations of the
system when an impulse function is used to apply a quick force to the mass platform. After
applying a quick force, the mass platform is allowed to vibrate freely, with a frequency equal to
the damped natural frequency, and come to rest at its equilibrium position.
The second term of the general solution in Eq. (3) is called the particular or steady-state
solution. It is the response of the system to the forcing function after the effects of the
transient solution have faded away, and the frequency of the steady-state vibration will be
equal to the frequency of the forcing function, .
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Transient (Homogeneous) Solution
The transient (or homogeneous) solution assumes the following three different forms
depending upon the value of  :
For  1: The system is over-damped and will never return to the equilibrium position
once it is disturbed.
For  =1: The system is critically damped (i.e., c  2 km ) and will return to its
equilibrium without oscillation.
For  1: The system is underdamped and will undergo damped oscillatory motion.
In this experiment, the damping is variable, but the system is always underdamped. Therefore,
this lab will focus solely on underdamped systems. The response of underdamped systems when
excited with an impulsive force is similar to the waveform shown in Figure 2.
(x1,t1)
(x2,t1+τ)
x(t) [in]
(x3,t1+2τ)
t [sec]
Figure 2: Amplitude ratios of the transient response of an under-damped harmonic oscillator, ζ<1.0, can be used to
calculate the damping ratio.
The damped natural frequency, d, can be found by measuring the frequency of oscillation of
the transient response. The damped natural frequency is related to the time required to
complete one cycle of oscillation, i.e., the period, τ. The relationship between damped natural
frequency and period is
d 
2

.
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(4)
Updated 12-15-2010
The damping ratio, , can be found using the logarithmic decrement method which is based on
the ratio of successive peaks in the time waveform. Implementing the transient-response
portion of Eq. (3), the ratio of the maximum amplitudes of two successive cycles of the transient
response is
xt1    exp  n t1   X 1 sin( d (t   )   ) exp  n t1  
 n


exp  .
 n t1 
 n t1  exp
xt1 
exp
X 1 sin( d t   )
exp
(5)
If the natural logarithm of both sides of Eq. (5) is taken and simplified, the logarithmic
decrement, , is expressed as
 x( t ) 
1
 .
  ln
x
(
t

 1  )
(6)
The logarithmic decrement can also be found using the ratio of two amplitudes that are n cycles
apart with the following equation
1 

x( t )
1
.
  ln
n  x( t1  n ) 
(7)
From Eq. (5), the logarithmic decrement can also be defined in terms of the damping ratio,
   n .
(8)
After considering the definition of d in Eq. (3) and (4) and substituting Eq. (4) into (8), the
logarithmic decrement can be defined as
 2

1 2
.
(9)
Eq. (9) can be solved for the damping ratio, , in terms of the logarithmic decrement (which can
be found experimentally using Eq. (6) or (7)),


4 2   2
.
(10)
1. Find the time constant, , with a linear regression between the times of the max (or min)
positions and the corresponding cycle number, n (refer to Figure 2). Use  to determine the
damped natural frequency, d.
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 xt1
2. Determine  using a linear regression between ln 
 x(t  n )
 1

 and n (see Eq. 7). Use  to


calculate . Note that the oscillation may not be about x=0; the time waveform may have a
DC offset that must be removed prior to finding . Include this regression and the data
points in your report.
3. To what physical property of the system is  related, and why is the natural logarithm
function used in the calculation of ?
4. Use your estimate of the damped natural frequency and the damping ratio to estimate the
natural frequency of the system.
5. Using the information calculated in the above questions and the mass of the system, m, find
the values of k and c.
6. Estimate the statistical uncertainty in , n, and d using the three data sets collected.
Steady-State (Particular) Solution
From Eq. (3), at a large time, t, the transient solution is negligible and the steady-state
(particular) solution can be written as
x p (t )  X sin( t   )
(11)
where
X 
F0 k
 
1   
  n 
2 2
    
  2  
   n 
.
(12)
2
The maximum displacement, X, for a given forcing frequency is found by evaluating xp(t) for
t- = /2 (positive displacement) or t- = 3/2 (negative displacement). Figure 3 shows the
normalized maximum displacement of the particular solution for different frequency and
damping ratios. X is very dependent on frequency, and the largest X is obtained when the
frequency ratio is near one, especially for low levels of damping. The frequency, , at which X is
largest is the resonant frequency, r.
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As mentioned previously, the angle  in Eq. (3) and (11) indicates the phase relationship
between the mass-platform displacement and the excitation force and is often called, simply,
the phase angle [1]. The theoretical expression for  is given by


 2    
  

1
 tan   n 2 
   
 1     
  n 
(13)
Figure 3 shows the phase response as a function of the forcing frequency for various values of
damping ratio. When excited at a low frequency relative to the natural frequency, the phase
shift is small, so the output is nearly in phase with the input. When excited at the undamped
natural frequency, the system output lags the input (forcing function) by 90°. When excited at
high frequencies, the input and output will trend toward being 180° out of phase.
Figure 3: The normalized maximum displacement of a system changes with forcing frequency [1].
The quantities X and  can also be determined experimentally by means of Lissajous diagrams,
as shown in Figure 4. If the displacement of the mass platform is graphed on the y-axis and the
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angular position of the forcing function is plotted on the x-axis then the maximum massplatform deflection, X, is equal to A.
Figure 4: The maximum displacement and phase angle of the system can be found from a Lissajous diagram.
To find , the first step is to determine the direction of rotation of the of the pattern (clockwise
or counterclockwise) and the direction of the slope (positive or negative) of the major axis of the
pattern. The direction of rotation and the slope will dictate if the phase angle is located in
quadrant III or IV in the diagram of Figure 5. After determining the quadrant in which  is
located, the following equations are used to calculate :
if X is in quadrant IV:
 
(14)
if X is in quadrant III:
  180   
(15)
where
B
 A
  sin 1   .
(16)
The value B in Eq. (16) is the distance between y = 0 and the value of y at x = 0 [2].
The need to use the direction of rotation and the slope of the Lissajous pattern to determine the
phase angle stems from the arcsin function being defined only for angles between -90° and 90°.
The phase of the vibrating system, however, spans between 0° and 180°, depending on the
excitation frequency (see Eq.(13)), so Eq. (14) and (15) must be used to appropriately find the
phase angle.
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Figure 5: The slope and direction of rotation of the Lissajous pattern help determine the correct phase angle.
Figure adapted from http://www.tedpavlic.com/teaching/osu/ece209/ .
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7. Use n, , k, and m from the transient-analysis section to predict the maximum deflection of
the system for a range of frequencies (see Eq. 12). These will be referred to as your
theoretical results. Normalize the frequencies with respect to n when plotting, and
normalize X with respect to k/F0. The final plot should look similar to that in Figure 3.
8. Plot the theoretical curves for  vs. (/n) for the same range of forcing frequencies as the
previous question (see Eq. 13). Again, use the n and  calculated in the transient analysis.
9. Make a Lissajous plot for each forcing frequency used during the experiment. Determine A
and B for each plot and the uncertainty associated with each.
10. Calculate the phase angle between the forcing function and the system response, , from
the values of A and B found in the previous question. Based on the phase angles for
different frequencies, what does it mean when the ellipse “leans” to the left vs. the right?
11. Using the data from the two previous questions, create a plot of amplitude (normalized) and
phase angle similar to that shown in Figure 3.
12. Estimate the uncertainty in the phase angle, , calculation in both Questions 8 and 10.
13. Find r from the plots created in Questions 11. Determine the uncertainty associated with
this estimate.
14. To compare the “theoretical” and experimental results, combine the plots from Questions 7
and 11 together on the same graph. Display the experimental data as discrete points (not
connected with lines). Do the same for the phase-angle plots from Questions 8 and 11.
15. Compare r with n and d (from your transient analysis). Explain why r is different, or the
same, as d. Be sure to consider the uncertainty in each estimate.
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Summary of Useful Equations

2
wd
  2

c = cC

cc  2 km
2
1 2
d  n 1  

n  k m
4 2
 2
References
1. W.T. Thomson, Theory of Vibration with Applications, Prentice-Hall, Englewood Cliffs,
NJ, 1972.
2. J.P. Holman, Experimental Methods for Engineers, 5th Ed., McGraw-Hill, 1989, p. 417.
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