Free Damped Vibrations

The equation of motion for the linear displacement:

The equation of motion for the angular

displacement:

All real systems dissipate energy when they

vibrate, and the damping must be included in
analysis, particularly when the amplitude of
vibration is required. Energy is dissipated by
frictional effects. The general expression of the
viscous damping force is given by:

The motion stops when the distance is x n ≤ x s . The

The most common types of damping are as follows: number of steps n for which the body vibrates back
Dry damping, when and forth is determined from:
Viscous damping, when
where c is the coefficient of viscous damping
and v= ẋ The amplitudes of motion decrease linearly in time.
Arbitrary damping, where The zone x=± x s is called the dead zone.
This damping leads to a nonlinear differential
equation of motion. FREE DAMPED VIBRATIONS WITH VISCOUS
FREE DAMPED VIBRATIONS WITH DRY DAMPING
DAMPING This type of vibration appears for the case of
In this case, the viscous damping force R is given motion in a liquid environment with low viscosity or
by Coulomb's law of friction, in the case of motion in air with speed under 1m/s.
Figure below shows a mechanical model for a free
where N is the normal force and μ is the coefficient vibration with linear damping and free vibration with
of friction. torsional damping.
The resistance force has a constant value and it is
opposed to the direction of motion. A simple
example is shown below.

The differential equation of motion is:

The minus sign is for the case of positive motion

along the x axis. Consider the single degree of freedom model with
The general solutions of the above equation are: viscous damping shown below.

Assume the body of mass m at rest and the spring

is compressed (or stretched) such that its initial
displacement is With the initial
conditions The differential equation of motion is

c
where ∝= =c ω
2m r
The characteristic equation in r is

Oscillation decay, shown below, for free damped

vibration with dry damping: with the solutions
The expression of the critical damping coefficient is

C
The damping ratio is c r =
C cr
One can classify the vibrations with respect to
critical damping coefficient as follows:
 In this case the displacement measured at equal
time intervals of one quasiperiod decreases in
this is also known as underdamped geometric progression.
To characterize this decay, the logarithmic
Complex Conjugate Roots decrement δ is introduced
The term is the quasicircular
frequency also known as damped natural
frequency.
The oscillation continues until the amplitude of
The roots are motion is so small that the maximum spring force is
The solution of the differential equation is unable to overcome the friction force.

this is also known as overdamped

If we use the initial condition
Real and Distinct Roots
The roots of the characteristic equation are
and the derivative with respect to time, negative

Damping ratio
the constants are C
C r=
C cr
The solution of in this case is

The roots are

C2 λ 1=−C r ω−ω √ C r −1
2
tan φ=
C1 λ 2=−C r ω+ω √ C 2r −1
The solution is
the solution is then
( C 1 e−ω √ C −1+ C2 e ω √C −1 )
2 2
−C r ωt
x=e r r

or
which represents a nonoscillatory response, C 1∧C2
are arbitrary constants of integration to be
determine by initial conditions.
or
x= A e
−αt
sin ( βt+ φ) −v o + (−Cr + √C 2r −1 ) ω x o
C 1=
The exponential decay, A e−αt , decreases in time, 2 ω √C r −1
2

so one can obtain a motion that is modulated in

v o + ( Cr + √ C r −1 ) ω x o
2
amplitude. The damped natural (or quasiangular) C 2=
frequency β is 2ω √ C r −1
2

The quasiperiod of motion is

The rate of decay of oscillation is

Oscillation decay, shown below, for free damped

vibration with viscous damping: The motion is a periodic and tends asymptotically
to the rest position ( x → 0 when t → ∞ ).

this is also known as critically damped

Real Identical Roots
The roots of the characteristic equation are

The solution of the differential equation, namely the

law of motion, in this case becomes

C 1∧C2 are arbitrary constants of integration to be

determine by initial conditions.
C 1=ω x o +v o and C 2=x o
 ∞
When t → ∞ , one obtains the undeterminate x= .
∞
The L'Hospital rule is applied in this case:

namely, the motion stops a periodically.

Solved Problem:
1. A small 30 mm long, welded to a stationary
table so that it is fixed at one point of contact,
with a 12 mm bolt welded to the other end,
which is free to move. The mass of system is
about 49.2 x 10-3 kg and the spring constant is
857.8 N/m. The damping rate of the spring is
measured to be 0.11 kg/s. Calculate the
damping ratio and determine if the free motion
of the spring-bolt system is overdamped,
underdamped or critically damped. ans: 0.85%
2. The average natural frequency of human leg is
measured to be 20 Hz when in its rigid position
in the longitudinal direction, with a damping
ration of 0.224. Calculate the response of the
tip of the leg bone to an initial velocity of 0.6
m/s and zero initial displacement (this would
correspond to the vibration induced while
landing on your feet, with your knees locked
from a height of 18 mm) and plot the response.
What is the maximum acceleration experienced
by the leg assuming no damping? ans: 7.68g

Assignment:
For a damped system, the mass, damped
coefficient and spring constant are known to be 1
kg, 2 kg/s and 10 N/m respectively. Determine the
damping ratio, natural frequency and determine if
the free motion of the system is overdamped,
underdamped or critically damped. Plot the
response.