In the introductory tutorial in this
series we considered a free vibrating translational spring and mass
system with one degree of freedom with viscous damping as shown in Figure 1
below. Mass m connected to a spring with spring constant k is free
to move on a horizontal frictionless plane from the static
equilibrium position x = 0. A dashpot provides a damping force
F = c.v where **v** is velocity of mass **m** and
**c** the damping
coefficient.

From the dynamic* free body diagrams we obtained the following equation of motion for this system:

* showing forces producing acceleration of mass m, not static equilibrium

We now find a function x(t) that is the general solution to this equation. If you are not familiar with solving second-order homogeneous linear differential equations using the characteristic equation method refer to the maths tutorial.

In the tutorial on free undamped vibrations we found that k/m = ω_{n}^{2}
where ω_{n} is the natural frequency of the undamped system

Roots r_{1} and r_{2} of this equation from the
quadratic formula are:

We now manipulate this expression as follows to provide a
dimensionless factor ζ (zeta) called the ** damping ratio**.

(dimensions of the parameters of ζ are: c =
M/T ω_{n} = 1/T m = M)

From the maths tutorial we know the
general solution to the equation of motion using roots r_{1 }
and r_{2} of the characteristic equation is:

We can examine the general forms of this solution for different
values of ζ by considering coefficients r_{1} and r_{2}
of exponent* t.*

- In equation (3) √(ζ
^{2}- 1) is positive meaning that solutions are real numbers and can be obtained directly from equation (3). - √(ζ
^{2}- 1) < ζ thus [- ζ ± √(ζ^{2}- 1)] must be negative. Since ω_{n}is always positive, both r_{1}and r_{2}must be negative. Therefore the motion must have a decreasing exponential characteristic.

The following example illustrates a damped spring, mass and
dashpot system where ζ > 1 with initial conditions for
x(t) = x_{0 }and dx(t)/dt = v_{0} at time t=0.

To find expressions for constants C1 and C2 in equation (3) we let:

At time t = 0: x_{0} = C_{1} + C_{2}
and v_{0} = dx/dt = aC_{1} + bC_{2 }
(because e^{0} = 1)

Solving for C_{1} and C_{2} gives:

Figure 2 below shows a plot of x(t) against time t for a system with the following parameters:

m = 1.0 kg k = 10 N/m c = 12
Ns/m x_{0} = 0.1 m v_{0}
= 1.0 m/s which give ω_{n}
= 3.16 rad/s and ζ = 1.90

The motion characterised by
ζ >1 decreases exponentially with time and is defined as **
aperiodic.** It is often described as

In equation (3) if ζ = 1 then [- ζω_{n}
± ω_{n}√(ζ ^{2} - 1)] = - ω_{n}

From the maths tutorial we know the general solution to this equation is:

To find constants C_{1} and C_{2} use initial
conditions x_{0} and v_{0} at time t = 0

Figure 3 below shows a plot for x(t) against time t for the system with ζ = 1 using the values for m, k and initial conditions in the previous example. In this case the damping coefficient c = 6.32 Ns/m

The motion characterised by ζ = 1 is also aperiodic and has
the shortest time for decay to the limit x(t) = 0. It is called the **
critically damped** condition.

Consider roots r_{1}, r_{2} of the characteristic
equation for motion of the system when ζ < 1

If ζ < 1 then (ζ ^{2} - 1) is negative and the roots are
complex numbers. From the maths tutorial
we know the roots are:

r_{1} = α + i β and r_{2} = α -
i β where α = -ω_{n}
ζ and β = ω_{n}√(1
- ζ ^{2} )

because √(ζ ^{2} - 1) = √(-1)(1 - ζ ^{2}
) = i.√(1 - ζ ^{2} )

To find constants C_{1} and C_{2} in equation (5) use initial
conditions x_{0} and v_{0} at time t = 0

From equation (5) x(0) = x_{0 }= C_{1}

which for v_{0} = dx/dt at time t = 0 gives:

Figure 4 below shows a plot of x(t) against time t for the system with ζ < 1 using the values for m, k and initial conditions in the previous examples In this case the damping ratio ζ = 0.1 and the damping coefficient c = 0.63 Ns/m.

When ζ < 1 damping produces a decaying oscillatory motion, often described as *
light damping*. As it is the
most important vibratory behaviour in engineering applications we
examine the characteristics in more detail.

From the maths tutorial we know this equation can be expressed as:

From above α = -ω_{n}
ζ and β = ω_{n}√(1
- ζ ^{2} )

Where A = **√**((e^{αt}C_{1})^{2}
+(e^{αt}C_{2})^{2}) and φ = tan^{-1}(e^{αt}C_{2})
/ e^{αt}C_{1}

Equation (6) can be interpreted as damped oscillatory motion in two parts.

**cos(ω**_{n}**(√1 -
ζ ** ^{2}** ).t - φ)**
expresses harmonic motion at frequency ω_{n}√(1
- ζ ^{2} ). We call this the **damped natural frequency****
ω**_{d} which differs
from the natural frequency of the free vibrating system by
the factor √(1
- ζ ^{2} ). In most practical applications ζ << 1 and
ω_{d} ≈ ω_{n} . In the example above
ζ = 0.1 and √(1
- ζ ^{2} ) = 0.995. This deviation is impossible to
detect on the plot in Figure 4.

We obtained the plot in Figure 5 by assigning a known value to the damping ratio ζ. It is useful to have an experimental method to find ζ by measuring and plotting the displacement of the system mass against time. The following method involves measuring successive peaks in the decaying motion.

From Figure 5 the period of each cycle of the decaying harmonic motion is T. The ratio of the peak displacements between n cycles can be expressed as;

Constant A and the cosine terms cancel giving:

Typically ζ << 1 thus √(1 - ζ ^{2}) ≈ 1 giving:

We can estimate ζ for the system in the example above from the
plot in Figure 5 using peaks x_{1} and x_{3} for
which n = 2

x_{1} ≈ 0.30 m x_{3} ≈
0.08 m

Compare with the value assigned for the example ζ = 0.1

This has been quite a meaty tutorial but if you take the maths slowly step by step to their conclusions interpreting the outcomes is rewarding.

In the next tutorial we consider forced vibration without damping

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