An analysis of a vibrating system in real life must include effects of friction or damping to account for its true motion, otherwise the solutions obtained with negligence to these effects would only describe the system's theoretical motion.
A viscous damping force is caused
by a fluid drag on the system as it vibrates through it. This type of force is called
viscous damping force. If the motion is slow, this drag force is
then proportional to the velocity, that is, F = c
where c is the coefficient of viscous damping (also known as constant of proportionality).
Ordinarily, the viscous damping force, F, would be subscripted with the letter d
to signify damped response, but this designation has been omitted here.
Consider the following FBD illustrating 1 DOF system with viscous damping:

Thomson WT, Dahleh MD 1993, 'Free Vibration'
in Theory of Vibration with Applications, 5th edn, Prentice Hall, USA,
p27
| Notice that c is assigned with a dashpot |
|
Deformation from unstretched position
is labeled with delta and the displacement from equilibrium position is labeled
with x. In the equilibrium state,
k
= w = mg. Noting also that acceleration is directed downwards, we can sum all forces
(taking down as positive) about the equilibrium position in the following manner:

The differential equation is of a linear second-order homogeneous type with the following solution (linear because x and its derivatives appear without any composed functions {such as x3, ex, etc.}, second-order because x is differentiated twice {as denoted by two dots above it} and homogeneous because it equates to zero):
where s is a constant
[click
here for a simple proof]
Substituting this new value of x into the original differential equation, we obtain:

ms2 + cs + k = 0 is a characteristic equation (also known as an auxiliary equation). By the use of a quadratic formula, we can derive two solutions for this characteristic equation:
---(eq1)
Note that ms2 + cs + k was divided by m initially.
The general solution of eq1 can be written as:
... where A and B are constants to be evaluated
from initial conditions x(0) and
(0) --(eq2)
Substituting eq1 into eq2, we obtain:
---(eq3)
Eq3 looks very daunting but it's actually
quite simple. The first term in front of the parentheses is an exponentially decaying
function of time. The terms inside the parentheses describe the system's periodic
motion and will depend on whether the numerical value within the radical is positive,
negative or zero. At this point, we will introduce critical damping coefficient,
denoted by cc, and a nondimensional damping ratio, denoted by
(Greek letter zeta):
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--- (eq4) |
Now we can define three distinct cases of viscously damped free vibration:
Overdamped system
occurs when the c > cc and so
> 1. The roots s1 and s2 therefore become real roots (since
the radical is positive). Damping is so strong that as soon as the system is released
from its displaced position, it begins to creep back to its original, unstretched
position without oscillating (it portrays nonoscillatory motion).
Substituting (eq4) into the general solution (eq2), we get the solution for overdamped system and a graph of aperiodic motion.
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Critically damped system
will occur if c = cc (i.e.
=
1) making the roots s1 and s2 equal to each other (since radical
is zero). This is the limiting case between oscillatory and nonoscillatory motion
and it results in simple decaying motion (although it may overshoot the original
position once before it comes to rest).
s1 = s2 = -
n
{you can get this by substituting
= 1 into
(eq4)} which makes the two terms in eq2 combine into one single term. Through methods
of differential equations, it may be shown that the general solution becomes:
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Underdamped system
where c < cc and
< 1 produces
complex conjugate roots (because you cannot take a square root of an negative integer
{Z-}, so imaginary numbers need to be introduced). Underdamped systems
correspond to oscillatory motion with an exponential decay in amplitude. In other
words, the system continues to oscillate but with a smaller displacement from the
unstretched position with each completion of a full period.
As mentioned in the previous paragraph,
the radical in (eq4) must be greater than union, which produces the required complex
conjugates, as shown below. The following derivation of the general solution is
greatly simplified in that it omits the substitution of initial conditions into
the general solution to explain how arbitrary constants A, B, X and
were obtained.
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| Note in the above example that the term |
|
.
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Example 1
| The mass of the viscously damped system
shown to the right is released from rest at x(0) = 150mm when t = 0 seconds. Determine the displacement x(t) at t = 0.55s if c = 200N.s.m-1 and k = 200N.m-1 and assuming no friction. |
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Solution
First we need to find out whether the system is underdamped, critically damped or overdamped:
So we need to use the solution for critically damped case:
Example 2
A car weighing 3372lbf is supported by 4 springs and 4 dampers. If the static deflection of the car on the springs is 0.6562ft, due to its own weight, determine the required damping constant of each of the 4 shock absorbers in order to produce critical damping.
Solution
We will assume 1 DOF in vertical direction. Let's draw a FBD to visualise the problem:
Now with the help of our FBD, we can sum the forces in vertical direction:
Note that -mg and 4k
static have been included in force summation. In fact, the two terms equal each other at equilibrium position and don't need to be included in the force equation, but since the question doesn't give us a value for k, and it does for
static and mg, we can use that data to work out k: