Forced Vibration in Second Order Systems
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## 1 Introduction to Forced Vibration in Second Order Systems[1]

Periodic excitation is a fact of life.

• No rotating system is perfectly balanced.
• Translational systems sit on vibratory platforms.

A free damped system under vibration cannot be controlled; it is a free system. Controlling a damped system under vibration requires the ability to inpart a force. This force is our control signal.

### 1.1 Harmonic Forcing in Second Order (Damped) Systems

Some periodic forces are harmonic. Some are not. Those forces which are not harmonic can be represented as a series of harmonic functions using Fourier analysis techniques. Since both harmonic and non-harmonic forcing functions can be represented with harmonic functions it makes sense to study the response of forced damped systems to a harmonic forcing function.

## 2 Forced Vibration is based on a Free Damped Vibration (Second Order Systems) Model

The equation of motion for a free damped vibration is (free second order system with viscous damping)

 $LaTeX: m\ddot{x}+c\dot{x}+kx=0$ Free Second Order System

where

m is the mass,
c is the damping coefficient,
k is the spring stiffness, and
x is the displacement

and equilibrium is defined as

x=0 at t=0

For forced vibration the equation for viscous damping becomes

 $LaTeX: m\ddot{x}+c\dot{x}+kx=F\mbox{ sin}\left( \nu t \right)$ Forced Second Order System

where

F is the amplitude of the forcing function,
ν is the frequency of the forcing function, and
t is time in seconds.

Below is more information for viscous damping, coulomb damping, and hysteretic damping.

### 2.1 Viscous Damping[2]

For full derivation see Viscous Damping.

Figure 1: Viscous Damping for single DOF system

Figure 1 shows the model for a single degree of freedom system with forcing function and viscous damping. The equation of motion is

 $LaTeX: m\ddot{x}+c\dot{x}+kx=F\mbox{ sin}\left(\nu t\right)$ Second Order System with Viscous Damping

The solution to $LaTeX: m\ddot{x}+c\dot{x}+kx = 0$ is presented in Viscous Damping. It also represents the complementary function. It demonstrates that the initial vibration will quickly dissipate. For sustained motion simple harmoinc motion at the frequency of excitation is assumed. That solution is $LaTeX: x=X\mbox{ sin}\left( \nu t - \phi \right)$ and leads to

 $LaTeX: \dot{x}=X\nu\mbox{ cos}\left( \nu t - \phi \right)=X \nu \mbox{ sin}\left( \nu t - \phi + \frac{1}{2}\pi \right)$

and

 $LaTeX: \ddot{x}=-X\nu^2 \mbox{ sin}\left(\nu t - \phi \right)=X\nu^2 \mbox{ sin}\left(\nu t - \phi + \pi \right)$

Making the correct substitutions the equation of motion for a second order system with viscous damping becomes

 $LaTeX: mX\nu^2 \mbox{ sin} \left(\nu t - \phi + \pi \right) + cX\nu \mbox{ sin}\left(\nu t - \phi + \frac{\pi}{2} \right) + kX \mbox{ sin} \left(\nu t - \phi \right) = F \mobx{ sin} \left( \nu t \right)$

Figure 2: Force vector diagram

The vector diagram in Figure 2 leads to

 $LaTeX: F^2 = \left(kX-mX\nu^2 \right)^2 + \left( cX\nu \right)^2$

or

 $LaTeX: X = \frac{F}{\sqrt{\left( \left(k-m\nu^2 \right)^2 + \left(c\nu \right)^2 \right)}}$

and

 $LaTeX: \mbox{tan} \phi = \frac{cX \nu}{\left( kX - mX \nu^2 \right)}$

At steady-state the equation for a second order system with viscous damping is

 $LaTeX: x=\frac{F}{\sqrt{\left( \left( k-m\nu^2 \right)^2 + \left( c \nu \right)^2 \right)}} \mbox{ sin} \left( \nu t - \phi \right)$

where

 $LaTeX: \phi = \mbox{tan}^{-1} \left( \frac{c\nu}{k-m\nu^2} \right)$

The equation of transient motion for a second order system with viscous damping is

 $LaTeX: x = Ae^{-\zeta \omega t} \mbox{ sin} \left(\omega \sqrt{\left(1-\zeta^2 \right)^{2} t} + \alpha \right)$

where

$LaTeX: \zeta = \frac{c}{2 \sqrt{k m} }$

The following definitions will make the transient equation more convenient

 $LaTeX: \omega=\sqrt{\frac{k}{m}}$ rad/s and $LaTeX: X_{s}=\frac{F}{k}$

Then the dynamic magnification factor, $LaTeX: X/X_{s}$, is

 $LaTeX: \frac{X}{X_{s}}=\frac{1}{\sqrt{\left \{ \left [ 1 - \left( \frac{\nu}{\omega} \right)^2 \right ]^2 + \left [2\zeta \frac{\nu}{\omega} \right ]^2 \right \} }}$ and $LaTeX: \phi = \mbox{tan}^{-1} \left( \frac{2\zeta \left(\frac{\nu}{\omega} \right)}{1-\left( \frac{\nu}{\omega} \right)^2} \right)$

where

$LaTeX: X_{s}$ is the static deflection of the system under a steady force $LaTeX: F$ and
$LaTeX: X$ is the dynamic amplitude

Mechanical vibration becomes more important as X/Xs gets larger and ν gets closer to ω. As ν gets closer to ω a small harmonic force can produce a large amplitude of vibration. This is known as resonance and occurs when the forcing frequency is equal to the natural frequency (i.e. ν/ω = 1). The maximum value of X/Xs can be determined by

 $LaTeX: \left( \frac{\nu}{\omega} \right)_{\left( X/X_s \right)_{max}} = \sqrt{1-2\zeta^2} \approx 1$ for smal ζ

and

 $LaTeX: \left( \frac{X}{X_s} \right)_{max} = \frac{1}{\left( 2\zeta \sqrt{\left( 1-\zeta^2 \right)} \right)}$

From the other Second Order Systems article

 $LaTeX: M_p = \left| H \left( j \omega_p \right) \right| = \begin{cases} 1, & \mbox{if }\zeta \ge \frac{1}{\sqrt{2}} \\ \frac{1}{2 \zeta \sqrt{1-\zeta^2}}, & \mbox{if }0 \le \zeta < \frac{1}{\sqrt{2}}\end{cases}$

Note the similarities to $LaTeX: \frac{X}{X_s}$.

#### 2.1.1 Q factor

For small values of ζ, $LaTeX: X/X_{s} \approx 1/2\zeta$. $LaTeX: 1/2\zeta$ is a measure of the damping in a system; this is known as the Q factor.

#### 2.1.2 Alternative solution for Second Order Systems with Viscous Damping

An alternative solution to the equation of motion for a second order system with Viscous Damping can be obtained by substituting $LaTeX: F\mbox{ sin} \left( \nu t \right)=\mbox{Im}\left( Fe^{j\nu t} \right)$. Then $LaTeX: m\ddot{x} + c\dot{x} + kx = Fe^{j\nu t}$ and an assumed solution of $LaTeX: x=Xe^{j\nu t}$ leads to

 $LaTeX: \left( k-m\nu^2 \right) X + jc\nu X = F$ or $LaTeX: X=\frac{F}{\left(k-m\nu^2 \right) +jc\nu}$

where

$LaTeX: j=\sqrt{-1}$

Therefore

 $LaTeX: X=\frac{F}{\sqrt{\left( \left( k-m\nu^2 \right)^2 + \left( c\nu \right)^2 \right)}}$

#### 2.1.3 Rotating Second Order Systems with Viscous Damping

Rotating systems with an unbalance produce excitation force proportional to the square of the excitation frequency. The necessary variables are

$LaTeX: m_r$ is the unbalanced mass,
$LaTeX: r$ is the effective radious, and
$LaTeX: \nu$ is the rotating angular speed.

Therefore the excitation force is

$LaTeX: m_{r}r\nu^2$

If applied to a system like Figure 1, the component of the force in the direction of motion is $LaTeX: m_{r}r\nu^2\mbox{ sin}\left( \nu t \right)$ and the amplitude of vibration is

 $LaTeX: X=\frac{\left( \frac{m_r}{m} \right) r \left( \frac{\nu}{\omega} \right)^2}{\sqrt{\left( \left( 1 - \left( \frac{\nu}{\omega} \right)^2 \right)^2 + \left( 2\zeta \nu \omega \right)^2 \right)}}$

The value of ν/ω for maximum X can be found by differentiating the equation above to get

 $LaTeX: \left( \frac{\nu}{\omega} \right)_{X_{max}} = \frac{1}{\sqrt{1-2 \zeta^2}}$

The peak response occurs when ν > ω. Also,

 $LaTeX: X_{max}=\left( \frac{m_r}{m} \right) \frac{r}{2\zeta \sqrt{1 - \zeta^2}}$

### 2.2 Coulomb Damping[3]

Figure 3: Coulomb Damping for single DOF system

For full derivation see Coulomb Damping.

Figure 3 shows the model for a single degree of freedom system with forcing function and coulomb damping. The equation of motion is

 $LaTeX: m\ddot{x}+kx \pm F_{d}=F\mbox{ sin}\left(\nu t\right)$ Second Order System with Coulomb Damping

Note that this equation is non-linear. This is due to the friction force always opposing the direction of motion.

The motion can be discontinuous if Fd is large compared to F. However, in most systems Fd is small. This fact allows us to create a linear approximation.

Express Fd in terms of cd - the viscous damping coefficient.

 $LaTeX: c_{d}=\frac{4F_{d}}{\pi \nu X}$

 $LaTeX: X=\frac{F}{\sqrt{\left [ \left( k-m\nu^2\right)^2 + \left(c_{d}\nu \right)^2 \right]}}$

 $LaTeX: X=\frac{F}{\sqrt{\left [ \left( k-m\nu^2\right)^2 + \left( \frac{4F_{d}}{\pi X} \right)^2 \right]}}$

 $LaTeX: \frac{X}{X_{s}}=\frac{\sqrt{1-\left( \frac{4F_{d}}{\pi F} \right)^2}}{1-\left( \frac{\nu}{\omega} \right)^2}$

where

$LaTeX: \frac{4F_{d}}{\pi F}<1$ or $LaTeX: F_{d}>\left( \frac{\pi}{4} \right) F$

At resonance the amplitude is not limited by Coulomb friction.

### 2.3 Hysteretic Damping[4]

Figure 4: Hysteretic Damping for single DOF system

For full derivation see Hysteretic Damping.

Figure 4 shows the model for a single degree of freedom system with forcing function and hysteretic damping. The equation of motion is

 $LaTeX: m\ddot{x}+k^{*}x=F\mbox{ sin}\left(\nu t\right)$ Second Order System with Hysteretic Damping

where

$LaTeX: k^{*}=k\left(1+j\eta\right)$
$LaTeX: x=\frac{F\mbox{ sin}\left(\nu t\right)}{\left(k-m\nu^2\right)+j\eta k}$

and

$LaTeX: \frac{X}{X_{s}}=\frac{1}{\sqrt{\left(\left [ 1-\left(\frac{\nu}{\omega}\right)^2\right ] ^2 + \eta^2\right)}}$

Note this result requires the $LaTeX: c=\frac{\eta k}{\nu}$ substitution.

#### 2.3.1 Q factor

If $LaTeX: c=\frac{\eta k}{\nu}$, at resonance $LaTeX: c={\nu}\sqrt{km}$, then $LaTeX: \eta=2\zeta=\frac{1}{Q}$ .

## 3 References

Leigh, J. R. 2004 Control Theory. ISBN 0863413390

### 3.1 Notes

1. Leigh, pg. 46
2. Leigh, pp. 46-52
3. Leigh, pp. 69-70
4. Leigh, pp. 70-71