Now we’re are going to analyze
the system which has block of mass m, spring with stiffness k and the damper
(dashpot) with coefficient of viscous damping c. The system is shown in next
figure.
 |
| Figure 1 – a) System with block
of mass m attached to the spring with stiffness k and the damper with
coefficient of viscous damping c b) Free body diagram showing external and
effective forces which act on the block of mass m. |
As you can see from the previous
figure we have derived the FBD diagram and showed which forces act on the body.
The external forces that act on the body are the spring force, damping force
and the gravity. The effective force is the inertial force which acts in opposite
direction to the spring force and the damping force. Now it’s time, using the
second Newton’s law to derive the differential equation of the analyzed system.
$$\begin{align}
& +\downarrow \sum{{{F}_{y}}}=m\ddot{x}, \\
& -k\left( x+\Delta \right)-c\dot{x}+mg=m\ddot{x}, \\
& mg=k\Delta , \\
& -kx-k\Delta -c\dot{x}+k\Delta =m\ddot{x}, \\
& m\ddot{x}+kx+c\dot{x}=0 \\
\end{align}$$
The general form of the differential equation for
the displacement of a particle in a one-degree-of freedom linear system where
viscous damping is present is
$$m\ddot{x}+c\dot{x}+kx=0$$
Let’s assume the solution of the previous system in
the following form:
$$x(t)=C{{e}^{ht}}$$
Now we need to derive the first and second
derivative of the assumed solution:
$$\begin{align}
& \dot{x}(t)=Ch{{e}^{ht}}, \\
& \ddot{x}(t)=C{{h}^{2}}{{e}^{ht}}. \\
\end{align}$$
Insert those derivatives into the
differential equation:
$$\begin{align}
& mC{{h}^{2}}{{e}^{ht}}+cCh{{e}^{ht}}+kC{{e}^{ht}}=0, \\
& m{{h}^{2}}+ch+k=0 \\
\end{align}$$
As you can see we have transformed the differential
equation into quadratic equation. So now we need to find the solution of the
quadratic equation. The solution can be written in the following form:
$${{h}_{1,2}}=\frac{-c\pm \sqrt{{{c}^{2}}-4mk}}{2m}=-\frac{c}{2m}\pm \sqrt{{{\left( \frac{c}{2m} \right)}^{2}}-\frac{k}{m}}$$
If we write the purposed solution as:
$$\begin{align}
& {{x}_{1}}(t)={{C}_{1}}{{e}^{{{h}_{1}}t}},{{x}_{2}}\left( t \right)={{C}_{2}}{{e}^{{{h}_{2}}t}}, \\
& x(t)={{x}_{1}}(t)+{{x}_{2}}(t), \\
& x(t)={{C}_{1}}{{e}^{{{h}_{1}}t}}+{{C}_{2}}{{e}^{{{h}_{2}}t}}. \\
\end{align}$$
Then we get:
$$x(t)={{C}_{1}}{{e}^{\left\{ -\frac{c}{2m}+\sqrt{{{\left( \frac{c}{2m} \right)}^{2}}-\frac{k}{m}} \right\}t}}+{{C}_{2}}{{e}^{\left\{ -\frac{c}{2m}-\sqrt{{{\left( \frac{c}{2m} \right)}^{2}}-\frac{k}{m}} \right\}t}}$$
The next step is to derive formulas for critical
damping and the damping ration. Let’s look at the discriminant of quadratic
equation.
$$\begin{align}
& {{\left( \frac{{{c}_{c}}}{2m} \right)}^{2}}-\frac{k}{m}=0, \\
& \frac{{{c}_{c}}}{2m}=\sqrt{\frac{k}{m}}, \\
& {{c}_{c}}=2m\sqrt{\frac{k}{m}}=2\sqrt{\frac{k{{m}^{2}}}{m}}=2\sqrt{km}, \\
& {{c}_{c}}=2m{{\omega }_{n}} \\
\end{align}$$
For every damped system the
damping ration is defined as the ration of the damping constant to the critical
damping constant:
$$\varsigma =\frac{c}{{{c}_{c}}}$$
Now we need to derive the damping
ratio and the first variable of determinant.
$$\frac{{{c}_{c}}}{2m}=\frac{c}{{{c}_{c}}}\frac{{{c}_{c}}}{2m}=\varsigma {{\omega }_{n}}$$
When we insert previously defined expressions into
the solution of the quadratic equation we will get:
$$\begin{align}
& {{h}_{1,2}}=-\frac{c}{2m}\pm \sqrt{{{\left( \frac{c}{2m} \right)}^{2}}-\frac{k}{m}} \\
& {{h}_{1,2}}=-\zeta {{\omega }_{n}}\pm \sqrt{{{\left( \zeta {{\omega }_{n}} \right)}^{2}}-{{\omega }_{n}}^{2}} \\
& {{h}_{1,2}}=\left( -\zeta \pm \sqrt{{{\zeta }^{2}}-1} \right){{\omega }_{n}} \\
\end{align}$$
Inserting the constant h into the
assumed solution of the differential equation we will get:
$$x(t)={{C}_{1}}{{e}^{\left( -\zeta +\sqrt{{{\zeta }^{2}}-1} \right){{\omega }_{n}}t}}+{{C}_{2}}{{e}^{\left( -\zeta -\sqrt{{{\zeta }^{2}}-1} \right){{\omega }_{n}}t}}$$
The nature of the roots h1 and h2
and the behavior of previous solution depends on the magnitude of damping. So
now we need to examine those cases.
FIRST CASE – The system is
underdamped and has these conditions:
$$\varsigma <1,c<{{c}_{c}}\text{ or }\frac{c}{2m}<\sqrt{k/m}$$
For these conditions the expression
under the square root is negative and we can write the solutions h1 and h2 as:
$$\begin{align}
& {{h}_{1}}=\left( -\zeta +i\sqrt{1-{{\zeta }^{2}}} \right){{\omega }_{n}}, \\
& {{h}_{2}}=\left( -\zeta -i\sqrt{1-{{\zeta }^{2}}} \right){{\omega }_{n}} \\
\end{align}$$
And the assumed solution can be
written in the following form:
$$\begin{align}
& x(t)={{C}_{1}}{{e}^{\left( -\zeta +i\sqrt{1-{{\zeta }^{2}}} \right){{\omega }_{n}}t}}+{{C}_{2}}{{e}^{\left( -\zeta -i\sqrt{1-{{\zeta }^{2}}} \right){{\omega }_{n}}t}} \\
& \text{ }={{e}^{-\zeta {{\omega }_{n}}t}}\left\{ {{C}_{1}}{{e}^{i\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t}}+{{C}_{2}}{{e}^{-i\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t}} \right\}, \\
& \text{ }={{e}^{-\zeta {{\omega }_{n}}t}}\left\{ \left( {{C}_{1}}+{{C}_{2}} \right)\cos \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t+i\left( {{C}_{1}}-{{C}_{2}} \right)\sin \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t \right\}, \\
& \text{ }={{e}^{-\zeta {{\omega }_{n}}t}}\left\{ {{{{C}'}}_{1}}\cos \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t+i{{{{C}'}}_{2}}\sin \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t \right\}, \\
& \text{ }={{X}_{0}}{{e}^{-\zeta {{\omega }_{n}}t}}\sin \left( \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t+{{\phi }_{0}} \right), \\
& \text{ }=X{{e}^{-\zeta {{\omega }_{n}}t}}\cos \left( \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t-\phi \right). \\
\end{align}$$
.
Now we need to determine the
arbitrary constants form the initial conditions.
For the initial conditions:
$$\begin{align}
& x(0)={{x}_{0}}, \\
& \dot{x}(0)={{{\dot{x}}}_{0}}, \\
\end{align}$$
we have:
$$\begin{align}
& x(t)={{e}^{-\zeta {{\omega }_{n}}t}}\left\{ {{{{C}'}}_{1}}\cos \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t+i{{{{C}'}}_{2}}\sin \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t \right\}, \\
& \dot{x}(t)=-\zeta {{\omega }_{n}}{{e}^{-\zeta {{\omega }_{n}}t}}\left\{ {{{{C}'}}_{1}}\cos \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t+i{{{{C}'}}_{2}}\sin \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t \right\}+ \\
& {{e}^{-\zeta {{\omega }_{n}}t}}\left\{ -{{{{C}'}}_{1}}\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}\sin \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t+i{{{{C}'}}_{2}}\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}\cos \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t \right\}, \\
& x(0)={{e}^{-\zeta {{\omega }_{n}}0}}\left\{ {{{{C}'}}_{1}}\cos \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}0+i{{{{C}'}}_{2}}\sin \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}0 \right\}={{x}_{0}}, \\
& {{{{C}'}}_{1}}\cos \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}0={{x}_{0}}, \\
& {{{{C}'}}_{1}}={{x}_{0}}, \\
& \dot{x}(0)=-\zeta {{\omega }_{n}}{{e}^{-\zeta {{\omega }_{n}}0}}\left\{ {{{{C}'}}_{1}}\cos \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}0+i{{{{C}'}}_{2}}\sin \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}0 \right\}+ \\
& {{e}^{-\zeta {{\omega }_{n}}0}}\left\{ -{{{{C}'}}_{1}}\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}\sin \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}0+i{{{{C}'}}_{2}}\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}\cos \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}0 \right\}={{{\dot{x}}}_{0}} \\
& -\zeta {{\omega }_{n}}{{{{C}'}}_{1}}+i{{{{C}'}}_{2}}\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}={{{\dot{x}}}_{0}}, \\
& -\zeta {{\omega }_{n}}{{x}_{0}}+i{{{{C}'}}_{2}}\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}={{{\dot{x}}}_{0}}, \\
& {{{{C}'}}_{2}}\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}={{{\dot{x}}}_{0}}+\zeta {{\omega }_{n}}{{x}_{0}} \\
& {{{{C}'}}_{2}}=\frac{{{{\dot{x}}}_{0}}+\zeta {{\omega }_{n}}{{x}_{0}}}{\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}} \\
\end{align}$$
.
And hence the solution becomes:
$$x(t)={{e}^{-\zeta {{\omega }_{n}}t}}\left\{ {{x}_{0}}\cos \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t+\frac{{{{\dot{x}}}_{0}}+\zeta {{\omega }_{n}}{{x}_{0}}}{\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}}\sin \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t \right\},$$
The other two constants X,ϕ can
be determine from:
$$\begin{align}
& X={{X}_{0}}=\sqrt{{{\left( {{{{C}'}}_{1}} \right)}^{2}}+{{\left( {{{{C}'}}_{2}} \right)}^{2}}} \\
& X=\sqrt{x_{0}^{2}+{{\left( \frac{{{{\dot{x}}}_{0}}+\zeta {{\omega }_{n}}{{x}_{0}}}{\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}} \right)}^{2}}}=\frac{\sqrt{x_{0}^{2}\omega _{n}^{2}+\dot{x}_{0}^{2}+2{{x}_{0}}{{{\dot{x}}}_{0}}\varsigma {{\omega }_{n}}}}{\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}}, \\
& {{\phi }_{0}}={{\tan }^{-1}}\left( \frac{{{{{C}'}}_{1}}}{{{{{C}'}}_{2}}} \right)={{\tan }^{-1}}\left( \frac{{{x}_{0}}{{\omega }_{n}}\sqrt{1-{{\zeta }^{2}}}}{{{{\dot{x}}}_{0}}+\varsigma {{\omega }_{n}}{{x}_{0}}} \right), \\
& \phi ={{\tan }^{-1}}\left( \frac{{{{{C}'}}_{2}}}{{{{{C}'}}_{1}}} \right)={{\tan }^{-1}}\left( \frac{{{{\dot{x}}}_{0}}+\varsigma {{\omega }_{n}}{{x}_{0}}}{{{x}_{0}}{{\omega }_{n}}\sqrt{1-{{\zeta }^{2}}}} \right), \\
\end{align}$$
x
.
The following expression:
$${{\omega }_{d}}=\sqrt{1-{{\varsigma }^{2}}}{{\omega }_{n}}$$
is called the frequency of damped
vibration. It can be seen that the frequency of damped vibration is always less
than the natural frequency. The decrease in the frequency of damped vibration
with increasing amount of damping is shown graphically in next figure.
 |
| Figure 2- Variation of ωd with
damping |
SECOND CASE – Critically damped
system which has the following conditions:
$$\zeta =1,c={{c}_{c}},\frac{c}{2m}=\sqrt{\frac{k}{m}}$$
In this case the two roots h1 and
h2 are equal:
$${{h}_{1}}={{h}_{2}}=-\frac{{{c}_{c}}}{2m}=-{{\omega }_{n}}$$
Because of the repeated roots,
the solution is reduced to:
$$x(t)=({{C}_{1}}+{{C}_{2}}t){{e}^{-{{\omega }_{n}}t}}$$
The application of the initial
conditions:
$$\begin{align}
& x(0)={{x}_{0}}, \\
& \dot{x}(0)={{{\dot{x}}}_{0}}, \\
\end{align}$$
for this case gives:
$$\begin{align}
& x(t)=({{C}_{1}}+{{C}_{2}}t){{e}^{-{{\omega }_{n}}t}} \\
& \dot{x}(t)=-{{\omega }_{n}}\left( {{C}_{1}}+{{C}_{2}}t \right){{e}^{-{{\omega }_{n}}t}}+{{C}_{2}}{{e}^{-{{\omega }_{n}}t}} \\
& x(0)=({{C}_{1}}+{{C}_{2}}0){{e}^{-{{\omega }_{n}}0}}={{x}_{0}} \\
& {{C}_{1}}={{x}_{0}}, \\
& \dot{x}(0)=-{{\omega }_{n}}\left( {{x}_{0}}+{{C}_{2}}0 \right){{e}^{-{{\omega }_{n}}0}}+{{C}_{2}}{{e}^{-{{\omega }_{n}}0}}={{{\dot{x}}}_{0}}, \\
& -{{\omega }_{n}}{{x}_{0}}+{{C}_{2}}={{{\dot{x}}}_{0}} \\
& {{C}_{2}}={{{\dot{x}}}_{0}}+{{\omega }_{n}}{{x}_{0}} \\
\end{align}$$
.
And the solution becomes:
$$x(t)=\left( {{x}_{0}}+\left( {{{\dot{x}}}_{0}}+{{\omega }_{n}}{{x}_{0}} \right) \right)t{{e}^{-{{\omega }_{n}}t}}$$
It can be seen that the motion
represented in previous equation is non-periodic.
 |
| Figure 3 –Critically damped
system |
THIRD CASE – Overdamped system:
$$\varsigma >1,c>{{c}_{c}},\frac{c}{2m}>\sqrt{\frac{k}{m}}$$
As
$$\sqrt{{{\varsigma }^{2}}-1}>0$$
then the roots h1 and h2 are real
and distinct and are given by:
$$\begin{align}
& {{h}_{1}}=\left( -\varsigma +\sqrt{{{\varsigma }^{2}}-1} \right){{\omega }_{n}}<0 -="" align="" class="MTDisplayEquation" div="" end="" h="" left="" n="" omega="" right="" sqrt="" varsigma="">
In this case the solution can be
expressed as:
$$x(t)={{C}_{1}}{{e}^{\left( -\varsigma +\sqrt{{{\varsigma }^{2}}-1} \right){{\omega }_{n}}t}}+{{C}_{2}}{{e}^{\left( -\varsigma -\sqrt{{{\varsigma }^{2}}-1} \right){{\omega }_{n}}t}}$$
As before, we need to apply the
initial condition in order to determine the constants of integration C1 and C2.
$$\begin{align}
& x(t)={{C}_{1}}{{e}^{\left( -\varsigma +\sqrt{{{\varsigma }^{2}}-1} \right){{\omega }_{n}}t}}+{{C}_{2}}{{e}^{\left( -\varsigma -\sqrt{{{\varsigma }^{2}}-1} \right){{\omega }_{n}}t}}, \\
& \dot{x}(t)={{C}_{1}}\left( -\varsigma +\sqrt{{{\varsigma }^{2}}-1} \right){{\omega }_{n}}{{e}^{\left( -\varsigma +\sqrt{{{\varsigma }^{2}}-1} \right){{\omega }_{n}}t}}+{{C}_{2}}\left( -\varsigma -\sqrt{{{\varsigma }^{2}}-1} \right){{\omega }_{n}}{{e}^{\left( -\varsigma -\sqrt{{{\varsigma }^{2}}-1} \right){{\omega }_{n}}t}}, \\
& x(0)={{C}_{1}}{{e}^{\left( -\varsigma +\sqrt{{{\varsigma }^{2}}-1} \right){{\omega }_{n}}0}}+{{C}_{2}}{{e}^{\left( -\varsigma -\sqrt{{{\varsigma }^{2}}-1} \right){{\omega }_{n}}0}}={{x}_{0}}, \\
& {{C}_{1}}+{{C}_{2}}={{x}_{0}}\Rightarrow {{C}_{1}}={{x}_{0}}-{{C}_{2}}, \\
& \dot{x}(0)={{C}_{1}}\left( -\varsigma +\sqrt{{{\varsigma }^{2}}-1} \right){{\omega }_{n}}{{e}^{\left( -\varsigma +\sqrt{{{\varsigma }^{2}}-1} \right){{\omega }_{n}}0}}+{{C}_{2}}\left( -\varsigma -\sqrt{{{\varsigma }^{2}}-1} \right){{\omega }_{n}}{{e}^{\left( -\varsigma -\sqrt{{{\varsigma }^{2}}-1} \right){{\omega }_{n}}0}}={{{\dot{x}}}_{0}}, \\
& {{C}_{1}}\left( -\varsigma +\sqrt{{{\varsigma }^{2}}-1} \right){{\omega }_{n}}+{{C}_{2}}\left( -\varsigma -\sqrt{{{\varsigma }^{2}}-1} \right){{\omega }_{n}}={{{\dot{x}}}_{0}}, \\
& \left( {{x}_{0}}-{{C}_{2}} \right)\left( -\varsigma +\sqrt{{{\varsigma }^{2}}-1} \right){{\omega }_{n}}+{{C}_{2}}\left( -\varsigma -\sqrt{{{\varsigma }^{2}}-1} \right){{\omega }_{n}}={{{\dot{x}}}_{0}}, \\
& -\varsigma {{\omega }_{n}}{{x}_{0}}+{{x}_{0}}{{\omega }_{n}}\sqrt{{{\varsigma }^{2}}-1}+\varsigma {{\omega }_{n}}{{C}_{2}}-{{C}_{2}}{{\omega }_{n}}\sqrt{{{\varsigma }^{2}}-1}-\varsigma {{\omega }_{n}}{{C}_{2}}-{{C}_{2}}{{\omega }_{n}}\sqrt{{{\varsigma }^{2}}-1}={{{\dot{x}}}_{0}}, \\
& -2{{C}_{2}}{{\omega }_{n}}\sqrt{{{\varsigma }^{2}}-1}={{{\dot{x}}}_{0}}-{{x}_{0}}{{\omega }_{n}}\sqrt{{{\varsigma }^{2}}-1}+\varsigma {{\omega }_{n}}{{x}_{0}}, \\
& 2{{C}_{2}}{{\omega }_{n}}\sqrt{{{\varsigma }^{2}}-1}=-{{{\dot{x}}}_{0}}+{{x}_{0}}{{\omega }_{n}}\sqrt{{{\varsigma }^{2}}-1}-\varsigma {{\omega }_{n}}{{x}_{0}}, \\
& {{C}_{2}}=-\frac{{{x}_{0}}{{\omega }_{n}}\left( \varsigma -\sqrt{{{\varsigma }^{2}}-1} \right)+{{{\dot{x}}}_{0}}}{2{{\omega }_{n}}\sqrt{{{\varsigma }^{2}}-1}}, \\
& {{C}_{1}}={{x}_{0}}+\frac{{{x}_{0}}{{\omega }_{n}}\left( \varsigma -\sqrt{{{\varsigma }^{2}}-1} \right)+{{{\dot{x}}}_{0}}}{2{{\omega }_{n}}\sqrt{{{\varsigma }^{2}}-1}}=\frac{2{{x}_{0}}{{\omega }_{n}}\sqrt{{{\varsigma }^{2}}-1}-2{{x}_{0}}{{\omega }_{n}}\sqrt{{{\varsigma }^{2}}-1}+{{x}_{0}}{{\omega }_{n}}\varsigma +{{{\dot{x}}}_{0}}}{2{{\omega }_{n}}\sqrt{{{\varsigma }^{2}}-1}}, \\
& {{C}_{1}}=\frac{{{x}_{0}}{{\omega }_{n}}\varsigma +{{{\dot{x}}}_{0}}}{2{{\omega }_{n}}\sqrt{{{\varsigma }^{2}}-1}} \\
\end{align}$$
.
Previous equation shows that the
motion is aperiodic regardless of the initial conditions imposed on the system.
Since h1 and h2 are both negative, the motion diminishes exponentially with
time.
