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:

$$\begin{align} & x(t)={{C}_{1}}{{e}^{\left( -\zeta +i\sqrt{1-{{\zeta }^{2}}} \right){{\omega }_{n}}t}}+{{C}_{2}}{{e}^{\left( -\zeta -i\sqrt{1-{{\zeta }^{2}}} \right){{\omega }_{n}}t}} \\ & \text{ }={{e}^{-\zeta {{\omega }_{n}}t}}\left\{ {{C}_{1}}{{e}^{i\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t}}+{{C}_{2}}{{e}^{-i\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t}} \right\}, \\ & \text{ }={{e}^{-\zeta {{\omega }_{n}}t}}\left\{ \left( {{C}_{1}}+{{C}_{2}} \right)\cos \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t+i\left( {{C}_{1}}-{{C}_{2}} \right)\sin \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t \right\}, \\ & \text{ }={{e}^{-\zeta {{\omega }_{n}}t}}\left\{ {{{{C}'}}_{1}}\cos \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t+i{{{{C}'}}_{2}}\sin \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t \right\}, \\ & \text{ }={{X}_{0}}{{e}^{-\zeta {{\omega }_{n}}t}}\sin \left( \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t+{{\phi }_{0}} \right), \\ & \text{ }=X{{e}^{-\zeta {{\omega }_{n}}t}}\cos \left( \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t-\phi \right). \\ \end{align}$$

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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:

$$\begin{align} & x(t)={{e}^{-\zeta {{\omega }_{n}}t}}\left\{ {{{{C}'}}_{1}}\cos \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t+i{{{{C}'}}_{2}}\sin \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t \right\}, \\ & \dot{x}(t)=-\zeta {{\omega }_{n}}{{e}^{-\zeta {{\omega }_{n}}t}}\left\{ {{{{C}'}}_{1}}\cos \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t+i{{{{C}'}}_{2}}\sin \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t \right\}+ \\ & {{e}^{-\zeta {{\omega }_{n}}t}}\left\{ -{{{{C}'}}_{1}}\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}\sin \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t+i{{{{C}'}}_{2}}\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}\cos \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}t \right\}, \\ & x(0)={{e}^{-\zeta {{\omega }_{n}}0}}\left\{ {{{{C}'}}_{1}}\cos \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}0+i{{{{C}'}}_{2}}\sin \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}0 \right\}={{x}_{0}}, \\ & {{{{C}'}}_{1}}\cos \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}0={{x}_{0}}, \\ & {{{{C}'}}_{1}}={{x}_{0}}, \\ & \dot{x}(0)=-\zeta {{\omega }_{n}}{{e}^{-\zeta {{\omega }_{n}}0}}\left\{ {{{{C}'}}_{1}}\cos \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}0+i{{{{C}'}}_{2}}\sin \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}0 \right\}+ \\ & {{e}^{-\zeta {{\omega }_{n}}0}}\left\{ -{{{{C}'}}_{1}}\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}\sin \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}0+i{{{{C}'}}_{2}}\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}\cos \sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}0 \right\}={{{\dot{x}}}_{0}} \\ & -\zeta {{\omega }_{n}}{{{{C}'}}_{1}}+i{{{{C}'}}_{2}}\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}={{{\dot{x}}}_{0}}, \\ & -\zeta {{\omega }_{n}}{{x}_{0}}+i{{{{C}'}}_{2}}\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}={{{\dot{x}}}_{0}}, \\ & {{{{C}'}}_{2}}\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}={{{\dot{x}}}_{0}}+\zeta {{\omega }_{n}}{{x}_{0}} \\ & {{{{C}'}}_{2}}=\frac{{{{\dot{x}}}_{0}}+\zeta {{\omega }_{n}}{{x}_{0}}}{\sqrt{1-{{\zeta }^{2}}}{{\omega }_{n}}} \\ \end{align}$$

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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:

$$\begin{align} & x(t)=({{C}_{1}}+{{C}_{2}}t){{e}^{-{{\omega }_{n}}t}} \\ & \dot{x}(t)=-{{\omega }_{n}}\left( {{C}_{1}}+{{C}_{2}}t \right){{e}^{-{{\omega }_{n}}t}}+{{C}_{2}}{{e}^{-{{\omega }_{n}}t}} \\ & x(0)=({{C}_{1}}+{{C}_{2}}0){{e}^{-{{\omega }_{n}}0}}={{x}_{0}} \\ & {{C}_{1}}={{x}_{0}}, \\ & \dot{x}(0)=-{{\omega }_{n}}\left( {{x}_{0}}+{{C}_{2}}0 \right){{e}^{-{{\omega }_{n}}0}}+{{C}_{2}}{{e}^{-{{\omega }_{n}}0}}={{{\dot{x}}}_{0}}, \\ & -{{\omega }_{n}}{{x}_{0}}+{{C}_{2}}={{{\dot{x}}}_{0}} \\ & {{C}_{2}}={{{\dot{x}}}_{0}}+{{\omega }_{n}}{{x}_{0}} \\ \end{align}$$

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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.