Vibration Under General Forcing Conditions

Vibrations in mechanical systems occur when they are subjected to various external forces. These forces could be constant, time-dependent, or random, and depending on their nature, the system's response may vary significantly. Understanding how a system responds to different forcing conditions is crucial in designing systems that can endure real-world dynamic loads, maintain structural integrity, and reduce fatigue over time.

Introduction to Forced Vibration

A forced vibration occurs when a time-varying external force is applied to a mechanical system. Unlike free vibration, where the system oscillates due to its initial displacement or velocity, forced vibration is driven by an external source of energy. The behavior of the system under forced vibration is determined by several factors, including the nature of the force, damping, and the system's natural frequency.

In a general sense, a mechanical system subjected to external forces follows Newton’s second law of motion:

Equation of motion:
\begin{equation} m \ddot{x} + c \dot{x} + kx = F(t) \end{equation}

Where:

• \( m \) is the mass of the system,
• \( c \) is the damping coefficient,
• \( k \) is the stiffness of the system,
• \( x \) is the displacement, and
• \( F(t) \) is the external force acting on the system.

Types of Forcing Conditions

The external force acting on a system can take several forms. These are commonly categorized into deterministic and non-deterministic (random) forces:

1. Harmonic Forces

Harmonic forces are sinusoidal in nature and can be represented as:
\begin{equation} F(t) = F_0 \cos(\omega t) \end{equation}
Where \( F_0 \) is the amplitude and \( \omega \) is the angular frequency of the force.

This type of forcing commonly occurs in rotating machinery and vibrating components subjected to periodic motion. Harmonic forces play a significant role in resonant conditions when the driving frequency of the force matches the natural frequency of the system, potentially causing large oscillations.

2. Impulsive Forces

Impulsive forces are forces applied over a short period. Mathematically, these are represented by a Dirac delta function or as a very short-duration force of high magnitude:
\( F(t) = P \delta(t) \)

Impulsive forces cause an immediate reaction from the system, such as when a hammer strikes a metal plate. These forces are essential in analyzing shock responses and are often used in impact testing.

3. Step Forces

A step force is a constant force applied at \( t = 0 \) and sustained over time. It can be represented as:
\( F(t) = F_0 H(t) \)
Where \( H(t) \) is the Heaviside step function.

Step forces are commonly used in static loading scenarios where a load is suddenly applied and remains constant. The system response under step forcing is often analyzed to determine the system’s transient and steady-state behavior.

4. Random Forces

In real-world applications, external forces are often random or non-deterministic. Random forces are characterized by their statistical properties, such as their mean value, variance, and power spectral density.

Vibrations induced by random forces are commonly encountered in systems subjected to environmental loads, such as wind, earthquakes, and ocean waves. Analyzing these forces involves stochastic methods, and the system’s response is usually described in probabilistic terms.

System Response to Forced Vibrations

The response of a vibrating system to external forces can be classified into two components:

1. Transient Response

The transient response occurs immediately after the force is applied. It reflects the system's behavior before reaching steady state and usually depends on the initial conditions of the system (displacement and velocity). The transient response generally dies out over time, depending on the damping characteristics.

2. Steady-State Response

After the transient phase, the system enters a steady-state phase where the response oscillates with the same frequency as the external force. For harmonic forcing conditions, the steady-state response is in the form:
\( x(t) = X \cos(\omega t - \phi) \)

Here, \( X \) is the amplitude of the steady-state response, and \( \phi \) is the phase difference between the force and the displacement. The system’s amplitude and phase are influenced by the driving frequency, natural frequency, damping, and force amplitude.

Damping Effects on Forced Vibration

Damping plays a critical role in a system’s response to external forces. In real-world systems, damping is present due to energy dissipation through friction, air resistance, material properties, etc. The effect of damping can be categorized into three regimes:

1. Underdamped System

In an underdamped system, the damping is relatively small, and the system oscillates before gradually coming to rest. For forced vibrations, underdamped systems can exhibit large amplitudes near resonance conditions.

2. Critically Damped System

A critically damped system returns to equilibrium as quickly as possible without oscillating. In mechanical design, this is often the desired behavior for systems that need to respond rapidly without overshoot.

3. Overdamped System

In an overdamped system, the damping is so large that the system returns to equilibrium very slowly. There are no oscillations, and the system is sluggish in its response.

Resonance in Forced Vibration

Resonance occurs when the forcing frequency matches the natural frequency of the system. In this condition, even a small external force can cause significant amplitude in the system’s response. Resonance can lead to catastrophic failures in structures and machinery if not properly managed.

Example: The Tacoma Narrows Bridge collapse in 1940 is one of the most well-known examples of resonance. The bridge collapsed due to wind forces that caused oscillations matching the natural frequency of the bridge structure.

To prevent resonance, systems are typically designed so that their natural frequencies are far from the anticipated forcing frequencies. Additionally, damping can be introduced to mitigate the effects of resonance by reducing the amplitude of vibrations.

Practical Applications of Forced Vibrations

Understanding forced vibrations is crucial in various engineering fields:

• Automotive Engineering: Vehicle suspension systems are designed to minimize forced vibrations from road surfaces and improve comfort and safety.
• Aerospace Engineering: Aircraft wings and fuselage are subject to random and harmonic forces from airflow, requiring careful vibration analysis to ensure structural integrity.
• Civil Engineering: Buildings and bridges must withstand random forces from wind and seismic activity, requiring analysis of vibrations and resonance conditions.
• Mechanical Engineering: Rotating machinery is subjected to periodic forcing from imbalanced components, necessitating harmonic analysis to reduce wear and tear.

Conclusion

Vibration under general forcing conditions is a fundamental topic in mechanical engineering, with broad applications in structural design, automotive systems, aerospace, and civil infrastructure. The analysis of different types of forces and their impact on mechanical systems helps engineers create more reliable and resilient designs, preventing failures caused by excessive vibrations. By understanding the nature of the external forces, the system’s response, and damping effects, engineers can optimize system performance and enhance safety in dynamic environments.

Free vibrations of one-degree-of-freedom systems with viscous damping

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.