In noise reducing engineering, the consequences of changes made to a system must be understood. Questions posed could be on the effects of changes to the mass, stiffness, or losses of the system and how these changes can influence the vibration of or noise radiation from some structures. Real constructions certainly have many or in fact infinite modes of vibration. However, to a certain extent, each mode can often be modeled as a simple vibratory system. The most simple vibratory system can be described by means of a rigid mass, mounted on a vertical mass less spring, which in turn is fastened to an infinitely stiff foundation. If the mass can only move in the vertical direction along the axis of the spring, the system has one degree of freedom (1-DOF). This is a vibratory system never actually encountered in practice. However, certain characteristics of systems with many degrees of freedom, or rather, continuous systems with an infinite degree of freedom, can be demonstrated by means of the very simple 1-DOF model. For this reason, the basic mass–spring system is used in this chapter to illustrate some of the basic concepts concerning free vibrations, transient, harmonic and other types of forced excitation. Kinetic and potential energies are discussed as well their dependence on the input power to the system and its losses.

1.1 A Simple Mass–Spring System

A simple mass–spring system is shown in Fig. 1.1. The mass is m and the spring constant \(k_0 \). The foundation to which the spring is coupled is completely stiff and unyielding. It is assumed that the spring is mass less and that the spring force follows the simple Hooke’s law. Thus, when the spring is compressed the distance x, the reacting force from the spring on the mass is equal to \(k_0 x\). The damping force due to losses in the system is denoted \(F_\mathrm{d} \). When the system is at rest, the damping force \(F_\mathrm{d} \) is equal to zero. The static load on the spring is mg where g is the acceleration due to gravity. Due to the static load, the spring is compressed the distance \(\Delta x\). The reacting force on the mass is \(k_0 \cdot \Delta x\). Thus

$$\begin{aligned} \Delta x \cdot k_0 = mg \end{aligned}$$
(1.1)
Fig. 1.1
figure 1

Simple mass–spring system

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Therefore, in principle if the static deflection is known, then the spring constant can be determined. However, for real systems, the static and dynamic stiffness are not necessarily equal. In particular, this is quite evident for various types of rubber springs. In addition, a real mount has a certain mass.

Figure 1.1 shows a simple mass–spring system excited by an external force F(t) and a damping force \(F_\mathrm{d}\). The equation of motion for this simple system is

$$\begin{aligned} m\ddot{x} + k_0 x + F_\mathrm{d} = F(t) \end{aligned}$$
(1.2)

The deviation of the mass from its equilibrium position is \(x = x(t)\). When the mass is at rest, then \(x = 0\). The damping force \(F_\mathrm{d} \) is determined by the losses in the system. Various processes can cause these losses. Some examples of often used simple theoretical models are:

  1. (i)

    Viscous damping;

  2. (ii)

    Structural or hysteretic damping;

  3. (iii)

    Frictional losses or Coulomb damping;

  4. (iv)

    Velocity squared damping.

The damping forces can be illustrated by assuming a simple harmonic displacement of the mass. The motion is given by \(x = x_0 \cdot \sin (\omega t)\). Here time is t and \(\omega = 2\pi f\) where \(\omega \) is the angular frequency and f the corresponding frequency. The damping forces for the four cases are:

  1. (i)

    Viscous damping

    $$\begin{aligned} F_\mathrm{d} = c\dot{x} = cx_0\omega \cdot \cos (\omega t) \end{aligned}$$
    (1.3)

    The damping force is proportional to the velocity \(\dot{x}\) of the mass, c is a constant. The energy dissipated per cycle, i.e., in a time interval \(t_0 \leqslant t \leqslant t_0 + T\), where \(\omega T = 2\pi \), depends linearly on \(\omega \), the angular frequency of oscillation. This type of damping occurs for small velocities for a surface sliding on a fluid film and for dashpots and hydraulic dampers.

  2. (ii)

    Structural damping

    $$\begin{aligned} F_\mathrm{d} = \frac{\alpha }{\pi \omega } \cdot \dot{x} = \frac{\alpha }{\pi } \cdot x_0 \cdot \cos (\omega t) \end{aligned}$$
    (1.4)

    The amplitude of the damping force is proportional to the amplitude of the displacement but independent of the frequency for harmonic oscillations. The energy dissipated per cycle of motion is frequency independent over a wide frequency range and proportional to the square of the amplitude of vibration. The losses in solids can often be described in this way. In Eq. (1.4) \(\alpha \) is a constant.

  3. (iii)

    Frictional damping

    $$\begin{aligned} F_\mathrm{d} = \pm F \end{aligned}$$
    (1.5)

    The frictional force has a constant magnitude. The plus or minus sign should be determined so that the frictional force is counteracting the motion of the mass. The frictional force can be due to sliding between dry surfaces.

  4. (iv)

    Velocity squared damping

    $$\begin{aligned} F_\mathrm{d} = \pm q \cdot \dot{x}^2 = \pm q \cdot \left( {x_0 \omega \cdot \cos (\omega t)} \right) ^2 \end{aligned}$$
    (1.6)

The damping force is proportional to the velocity squared and the sign should be chosen so that the force again is counteracting the motion of the mass. In Eq. (1.6) q is a constant. A body moving fairly rapidly in a fluid could cause this damping force.

Fig. 1.2
figure 2

Force-displacement curve for sinusoidal motion of a 1-DOF system with viscous damping

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The structural damping of Eq. (1.4) only applies for a harmonic displacement of the mass. A more general description of structural damping is presented in Chap. 3. Examples of energy dissipation due to viscous, structural, and frictional losses are given in Problems 1.1–1.3.

The force required for moving the mass of a simple 1-DOF system depends on the type damping in the spring. For maintaining a motion \(x = x_0 \cdot \sin (\omega t)\) of the mass, the force which must be applied to the mass is obtained from Eq. (1.2) as \(F = m\ddot{x} + k_0 x + F_\mathrm{d} \). Two cases are illustrated in Figs. 1.2 and 1.3. In the first example, Fig. 1.2, the damping force is viscous. An ellipse represents the force-displacement relationship for this case. The minor axis of the ellipse is proportional to the angular frequency \(\omega \) and the parameter c of Eq. (1.3). Structural damping gives the same type of force-displacement curve. However, for this case the minor axis of the ellipse is just proportional to the coefficient \(\alpha \), Eq. (1.4), and not to the angular frequency. In the second example, Fig. 1.3, the mass is exposed to frictional damping. The arrow in the diagram indicates how force and displacement vary as time increases. The area enclosed by one loop is equivalent to the energy required to perform one cycle of motion of the mass. For a system with structural damping, the energy dissipated per cycle is independent of the angular frequency. This is not the case when the losses are viscous as discussed in Problems 1.1 and 1.2.

Fig. 1.3
figure 3

Force-displacement curve for sinusoidal motion of a 1-DOF system with frictional damping

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Fig. 1.4
figure 4

Force-displacement curve for sinusoidal motion of a 1-DOF system with distorted structural or hysteretic damping

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Real structures subjected to vibrations tend to show a force-displacement behaviour shown in Fig. 1.4. The force-displacement curve follows a distorted hysteresis loop, which is not readily described mathematically or physically. However, in general, the energy dissipated per cycle rather than the exact force-displacement relationship is of primary importance for real vibratory systems. Therefore, for most practical purposes viscous or material or for that matter a frequency dependent damping can be assumed for the simple harmonic motion of a structure.

Frictional losses and velocity squared damping result in nonlinear equations when introduced in Eq. (1.2). Examples of nonlinear equations and their solutions are presented for example Refs. [1–3]. The Runge–Kutta method, used for numerical solutions in the time domain, is discussed in Refs. [1, 4, 5].

For linear problems, damping is often described as viscous or structural. In practice, this is not necessarily the case. However, if, for example the losses are small and almost structural, the parameter \(\alpha \) in Eq. (1.4) can be allowed to depend on frequency for harmonic or approximately harmonic motion. This type of damping model is discussed in subsequent chapters. Further and most importantly, if viscous or structural damping is introduced in Eq. (1.2), the resulting equation of motion is linear.

For some applications, like experimental modal analysis and finite element calculations, a certain form of damping is often assumed, see Chap. 10. A specific damping model might be necessary for mathematical or numerical reasons. However, the model could violate the physical characteristics of the damping mechanism.

1.2 Free Vibrations

Free vibrations of a system occur if for example the system at a certain instant is given a displacement from its equilibrium position or an initial velocity at that particular time. After this initial excitation, no external forces are applied to the system. The resulting motion of the system is due to free vibrations. For all natural systems, there is always some type of damping present. For such systems, the free vibrations die out after a certain length of time. The initial energy of the system is absorbed by losses.

For a simple mass–spring system (1-DOF) with viscous losses, the equation of motion for free vibrations, \(F(t) = 0\), given by Eqs. (1.2) and (1.3) as

$$\begin{aligned} m\ddot{x} + c\dot{x} + k_0 x = 0 \end{aligned}$$
(1.7)

The general boundary conditions or initial values are

$$\begin{aligned} x(0) = x_0, \quad \dot{x}(0) = v_0 \end{aligned}$$
(1.8)

The traditional approach to solving this equation is to assume a solution of the form

$$\begin{aligned} x(t) = A \cdot \mathrm{e}^{\lambda t} \end{aligned}$$
(1.9)

The eigenvalue \(\lambda \) is obtained by inserting this expression in Eq. (1.7). Consequently,

$$\begin{aligned} \lambda ^2m + \lambda c + k_0 = 0 \end{aligned}$$
(1.10)

It is convenient to define the following parameters:

$$\begin{aligned} \omega _0^2 = k_0 / m, \quad \beta = c / (2m) \end{aligned}$$
(1.11)

Using these parameters, the solution to Eq. (1.10) is

$$\begin{aligned} \lambda _{1,2} = - \beta \pm \sqrt{\beta ^2 - \omega _0^2 } \quad \text {or}\quad \lambda _{1,2} = - \beta \pm i\sqrt{\omega _0^2 - \beta ^2} \end{aligned}$$
(1.12)

with \(i = \sqrt{- 1} \). In general, there are two solutions to Eq. (1.7). The expression for the displacement x is therefore of the form

$$\begin{aligned} x(t) = A_1 \cdot \mathrm{e}^{\lambda _1 t} + A_2 \cdot \mathrm{e}^{\lambda _2 t} \end{aligned}$$
(1.13)

The parameters \(A_1 \) and \(A_2\) are determined from the initial condition (1.8).

The general result (1.13) can be written in a more tractable way depending on the magnitude of \(\omega _0\) as compared to \(\beta \).

For \(\omega _0>\beta \) the eigenvalues \(\lambda _{1,2}\) are complex. Introducing the real parameter \(\omega _\mathrm{r} \) as

$$\begin{aligned} \omega _\mathrm{r} = \sqrt{\omega _0^2 - \beta ^2} \end{aligned}$$
(1.14)

the general solution (1.13) is written

$$ x(t) = A_1 \cdot \mathrm{e}^{ - \beta t - i\omega _\mathrm{r} t} + A_2 \cdot \mathrm{e}^{ - \beta t + i\omega _\mathrm{r} t} $$

or

$$\begin{aligned} x(t) = \mathrm{e}^{ - \beta t}\left( {B_1 \cdot \sin \omega _\mathrm{r} t + B_2 \cdot \cos \omega _\mathrm{r} t} \right) \end{aligned}$$
(1.15)

where \(B_1 \) and \(B_2\) are some new parameters to be determined from the initial conditions. The equalities

$$ \sin \alpha = \left( {\mathrm{e}^{i\alpha } - \mathrm{e}^{-i\alpha }} \right) / 2i\quad \text {and}\quad \cos \alpha = \left( {\mathrm{e}^{i\alpha } + \mathrm{e}^{-i\alpha }} \right) / 2 $$

have been used to obtain Eq. (1.15). The initial conditions (1.8) in combination with Eq. (1.15) yield for \(t \geqslant 0\)

$$\begin{aligned} x(t) = \mathrm{e}^{ - \beta t}\left( {x_0 \cos \omega _\mathrm{r} t + \frac{v_0 + \beta x_0 }{\omega _\mathrm{r} }\sin \omega _\mathrm{r} t} \right) \end{aligned}$$
(1.16)

The motion x(t) of the mass is described by a damped oscillatory motion as shown in Fig. 1.5 for \(\omega _0 > \beta \). The angular frequency of the oscillations is given by the system parameters: stiffness, mass, and losses. The decay rate of the oscillations depends on the damping of the system.

Fig. 1.5
figure 5

Damped oscillatory motion of a 1-DOF system

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When the damping or \(\beta \) is increased, the angular frequency \(\omega _\mathrm{r}\) decreases and approaches zero as \(\beta \) approaches \(\omega _0 \). For the limiting case

$$ \lim \limits _{\beta \rightarrow \omega _0 } \cos \omega _\mathrm{r} t = 1\quad \text {and}\quad \lim \limits _{\beta \rightarrow \omega _0 } \frac{\sin \omega _\mathrm{r} t}{\omega _\mathrm{r} } = t $$

Thus, for the case \(\beta = \omega _0 \) the solution (1.16) is

$$\begin{aligned} x(t) = \mathrm{e}^{ - \beta t}[x_0 + t(v_0 + \beta x_0 )] \end{aligned}$$
(1.17)

The result is a critically damped motion for \(t \geqslant 0\) shown in Fig. 1.6.

Fig. 1.6
figure 6

Motion of a critically damped 1-DOF system

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For a heavily damped system \(\beta > \omega _0 \) . For this case, the eigenvalues (1.12) are written:

$$\begin{aligned} \lambda _{1,2} = - \beta \pm \gamma , \quad \gamma = \sqrt{\beta ^2 - \omega _0^2 } \end{aligned}$$
(1.18)

The parameter \(\gamma \) is real for \(\beta > \omega _0 \). The solution for this case is obtained from (1.16) by replacing \(\omega _\mathrm{r}\) by \(i\gamma \) since \(\omega _\mathrm{r} = \sqrt{\omega _0^2 - \beta ^2} = i\sqrt{\beta ^2 - \omega _0^2 } = i\gamma \). Thus according to definition

$$ \begin{array}{cc} \cos \omega _\mathrm{r} t = \left( {\mathrm{e}^{i\omega _\mathrm{r} t} + \mathrm{e}^{-i\omega _\mathrm{r} t}} \right) / 2 = \left( {\mathrm{e}^{ - \gamma t} + \mathrm{e}^{\gamma t}} \right) / 2, \\ \sin \omega _\mathrm{r} t = \left( {\mathrm{e}^{i\omega _\mathrm{r} t} - \mathrm{e}^{-i\omega _\mathrm{r} t}} \right) / 2i = \left( {\mathrm{e}^{ - \gamma t} - \mathrm{e}^{\gamma t}} \right) / 2i \end{array} $$

Thus for \(\beta > \omega _0 \) and \(t \geqslant 0\) the displacement x(t) is given by

$$\begin{aligned} x(t) = \frac{\mathrm{e}^{ - \beta t}}{2\gamma }\left\{ {\mathrm{e}^{\gamma t}[x_0 (\gamma + \beta ) + v_0 ] + \mathrm{e}^{ - \gamma t}[x_0 (\gamma - \beta ) - v_0 ]} \right\} \end{aligned}$$
(1.19)

The result is a nonoscillatory motion of the mass as shown in Fig. 1.7.

Mathematically, the solutions to the three cases \(\beta < \omega _0 \), \(\beta = \omega _0 \) and \(\beta > \omega _0 \) are basically the same. However, the physical interpretations of the results are most apparent from the expressions (1.16), (1.17) and (1.19).

Fig. 1.7
figure 7

Motion of a heavily damped 1-DOF system

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In most practical cases in vibro-acoustics, losses are small, i.e., \(\beta < \omega _0 \). In fact, as discussed later, typical losses for vibrating structures, which radiates sound in the audible frequency range, satisfy the inequality \(\beta \ll \omega _0 \). These structures are consequently lightly damped. For a lightly damped 1-DOF system performing an oscillatory motion, the displacement is given by (1.16) as

$$ x(t) = \mathrm{e}^{ - \beta t}\left( {x_0 \cos \omega _\mathrm{r} t + \frac{v_0 + \beta x_0 }{\omega _\mathrm{r} }\sin \omega _\mathrm{r} t} \right) $$

The system oscillates with the frequency \(f_\mathrm{r} = \omega _\mathrm{r} / (2\pi ) = \sqrt{\omega _0^2 - \beta ^2} / (2\pi )\). Whatever the initial conditions are, a freely lightly damped vibrating 1-DOF system oscillates with a natural, eigen or characteristic frequency \(f_\mathrm{r} \), which is a function of mass, stiffness and damping of the system.

The displacement, Eq. (1.16), can also be written as

$$\begin{aligned} x(t) = C \cdot \mathrm{e}^{ - \beta t} \cdot \sin (\omega _\mathrm{r} t + \varphi ) = C \cdot \mathrm{e}^{ - \beta t} \cdot \left( \sin \omega _\mathrm{r} t \cdot \cos \varphi + \cos \omega _\mathrm{r} t \cdot \sin \varphi \right) \end{aligned}$$
(1.20)

The amplitude C and the phase angle \(\varphi \) are obtained by comparing the two expressions defining x(t). Consequently, for equality at any time t, it follows that

$$ C\cos \varphi = (v_0 + \beta x_0 ) / \omega _\mathrm{r}, \quad C\sin \varphi = x_0 $$

and thus

$$\begin{aligned} \tan \varphi = x_0 \omega _\mathrm{r} / (v_0 + \beta x_0 ), \quad C = \sqrt{v_0^2 + \omega _0^2 x_0^2 + 2\beta x_0 v_0 } / \omega _\mathrm{r} \end{aligned}$$
(1.21)

Based on the expressions (1.20) and (1.21) the kinetic and potential energies can be derived. The potential energy stored in a mass less spring with the spring constant \(k_0\) is

The time average of the potential energy over a period T where \(T\omega _\mathrm{r} = 2\pi \) is

(1.22)

The subscript T denotes that the potential energy is averaged over one period. In general, for a lightly damped system radiating sound in the audible frequency range \(\beta T \ll 1\). This means that the exponential function in the integral (1.22) is approximately constant and equal to \(\exp ( - 2\beta t)\) in the time interval of integration. Considering this the integral (1.22) defining can be approximated by the expression

(1.23)

Thus, the time average over a period t to \(t+T\) of the potential energy is for small losses approximately decaying as \(\exp ( - 2\beta t)\).

The velocity \(\dot{x}(t)\) of the freely vibrating mass is from (1.20) obtained as

$$\begin{aligned} \dot{x}(t) = C\mathrm{e}^{ - \beta t}\left[ {\omega _\mathrm{r} \cos (\omega _\mathrm{r} t + \varphi ) - \beta \sin (\omega _\mathrm{r} t + \varphi )} \right] \end{aligned}$$
(1.24)

The time average of the kinetic energy over one period is

$$\begin{aligned} \bar{\mathcal{T}}_T = \frac{m}{2T}\int _t^{t + T} {\mathrm{d}t \cdot \dot{x}^2} \end{aligned}$$
(1.25)

Following the same procedure as above, the average kinetic energy \(\bar{\mathcal{T}}_T \) is for \(\beta T \ll 1\) and when averaged over a time period t to \(t+T\) given as

$$\begin{aligned} \bar{\mathcal{T}}_T\,\approx \,&\frac{mC^2\mathrm{e}^{ - 2\beta t}}{2T}\int _t^{t + T} \mathrm{d}t \cdot \{ \omega _\mathrm{r}^2 \cos ^2(\omega _\mathrm{r} t + \varphi )\nonumber \\&+ \beta ^2\sin ^2(\omega _\mathrm{r} t + \varphi )- \omega _\mathrm{r} \beta \sin [2(\omega _\mathrm{r} t + \varphi )] \}\nonumber \\ =\,&\frac{m\omega _0^2 C^2\mathrm{e}^{ - 2\beta t}}{4} \end{aligned}$$
(1.26)

The parameter \(\omega _0^2 \) is equal to \(k_0/m\). Thus, it follows from the Eqs. (1.23) and (1.26) that

(1.27)

The time averages of kinetic and potential energies for a freely vibrating and lightly damped 1-DOF system are approximately equal. Equality holds for nondamped systems, i.e., for \(\beta =0\).

The velocity squared is as function of time shown in Fig. 1.8. For small losses the time average of the velocity squared decays as \(\exp ( - 2\beta t)\). This result can be used to estimate through measurements the losses in the system. For viscous or structural damping, the time average of the kinetic energy or for that matter the total energy is decaying exponentially with time as \(\exp ( - 2\beta t)\) if \(\beta \ll \omega _0\).

Fig. 1.8
figure 8

Decay of the temporal average of the velocity squared of a lightly damped 1-DOF system

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For Coulomb or frictional damping it can be shown (see Problem 1.3) that the decay in amplitude per cycle is constant. Two straight lines can envelop the amplitude of the oscillating motion. The decay rate of the amplitude as a function of time is shown in Fig. 1.9. The mass is at rest at \(x = {\varDelta }\). At this position, the frictional force exceeds the spring force.

Fig. 1.9
figure 9

Decay of the amplitude of a 1-DOF system with frictional damping

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1.3 Transient Vibrations

When an external force is suddenly applied to a mechanical system at rest, transient vibrations are excited. Examples of transient excitation are a ship hit by a large wave, a car running on a bumpy road, an engine started up, and even the extreme case of an earthquake. For this type of excitation, it is important to estimate for the resulting motion the maximum deflection, velocity, or acceleration of the system. A severe transient excitation can cause great damage to a system. In its milder form it could, for example, be a resiliently mounted engine in a car or a ship vibrating so violently that damage is caused to the adjoining structures.

A simple case of transient excitation can be demonstrated by means of the simple mass–spring system. The mass is initially at rest when at time \(t = 0\) the system or mass is given an impulse I. The equation of motion for the mass can be written as:

$$\begin{aligned} m\ddot{x} + c\dot{x} + k_0 x = I \cdot \delta (t) \end{aligned}$$
(1.28)

The displacement and the impulse are positive in the same direction. Since the mass initially is at rest, it follows that

$$\begin{aligned} x(t) = 0~~\text {for}~~t \leqslant 0 \end{aligned}$$
(1.29)

In Eq. (1.28) I is the magnitude of the impulse and \(\delta (t)\) the Dirac pulse defined so that

$$\begin{aligned} \delta (t) = 0~~\text {for}~~t \ne 0,\quad \lim \limits _{\varepsilon \rightarrow 0} \int _{ - \varepsilon }^\varepsilon {\mathrm{d}t \cdot \delta (t) = 1} \end{aligned}$$
(1.30)

The expression \(I \cdot \delta (t)\) is defined as the force F(t) acting on the system since

$$ \int {F(t)\mathrm{d}t = \int {I\delta (t)\mathrm{d}t = I} } $$

For \(t > 0\) no external force is applied to the system resulting in free vibrations. If, according to the discussion in Sect. 1.2, \(\beta < \omega _0 \) the result for \(t> 0\) is given by

$$\begin{aligned} x(t) = \mathrm{e}^{ - \beta t}\frac{v_0 }{\omega _\mathrm{r} }\sin \omega _\mathrm{r} t \end{aligned}$$
(1.31)

This is obtained from Eq. (1.16) for \(x_0 = 0\). The unknown parameter in Eq. (1.31) is the initial velocity \(v_0 \) of the mass. However, this velocity is obtained from Eq. (1.28). This expression is integrated with respect time as

$$\begin{aligned} \int _{ - \varepsilon }^\varepsilon {\mathrm{d}t(} m\ddot{x} + c\dot{x} + k_0 x) = \int _{ - \varepsilon }^\varepsilon {\mathrm{d}t} I \cdot \delta (t) = I \end{aligned}$$
(1.32)

The result is

$$\begin{aligned} m\dot{x}(\varepsilon ) - m\dot{x}( - \varepsilon ) + cx(\varepsilon ) - cx( - \varepsilon ) + \int _{ - \varepsilon }^\varepsilon {\mathrm{d}t} k_0 x = I \end{aligned}$$
(1.33)

Initially, the mass is at rest. Thus \(x( - \varepsilon ) = \dot{x}( - \varepsilon ) = 0\). Since x is a continuous function, the integral in Eq. (1.33) approaches zero as \(\varepsilon \) goes to zero. Considering the initial condition \(x(0) = 0\), the limiting case as \(\varepsilon \rightarrow 0\) gives

$$\begin{aligned} m\dot{x}(0) = I \end{aligned}$$
(1.34)

This is the second initial condition. The first, \(x(0) = 0\), is given by (1.29). With \(\dot{x}(0) = v_0 = I / m\) the motion of the mass is according to Eqs. (1.31) and (1.34)

$$\begin{aligned} x(t) = \mathrm{e}^{ - \beta t}\frac{I}{m\omega _\mathrm{r} }\sin \omega _\mathrm{r} t \end{aligned}$$
(1.35)

where as before

$$ \beta = \frac{c}{2m}, \quad \omega _\mathrm{r} = \sqrt{\frac{k_0 }{m} - \left( {\frac{c}{2m}} \right) ^2} = \sqrt{\omega _0^2 - \beta ^2} $$

The response to a unit pulse at \(t = 0\), i.e., \(I = 1\) is defined as:

$$\begin{aligned} h(t) = \mathrm{e}^{ - \beta t}\frac{1}{m\omega _\mathrm{r} }\sin \omega _\mathrm{r} t \end{aligned}$$
(1.36)

This result can form the basis for the solution to the general problem when an external time-dependent force F(t) excites the mass.

The equation of motion for the general case is

$$\begin{aligned} m\ddot{x} + c\dot{x} + k_0 x = F(t) \end{aligned}$$
(1.37)

where

$$ F(t) = 0~~\text {for}~~t < 0 $$

The force acting on the mass is shown in Fig. 1.10 as function of time. At the time \(t = \xi \), the system is given an impulse \(\Delta I = \Delta \xi \cdot F(\xi )\). According to Eq. (1.36), the resulting deflection \(\Delta x\) caused by the impulse \(\Delta I\) is

$$ \Delta x = \Delta I \cdot h(t - \xi ) = \Delta \xi \cdot F(\xi ) \cdot h(t - \xi ) $$
Fig. 1.10
figure 10

Force as function of time and fractional impulse at \(t =\xi \)

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The total response at time t is the sum of all separate impulses up to that time. In integral form, the result to Eq. (1.37) is

$$\begin{aligned} x(t) = \int _0^t \mathrm{d}\xi \cdot F(\xi ) \cdot h(t - \xi ) \end{aligned}$$
(1.38)

Since \(F(\xi ) = 0\) for \(\xi < 0\), Eq. (1.37) can also be written as

$$ x(t) = \int _{ - \infty }^t \mathrm{d}\xi \cdot F(\xi ) \cdot h(t - \xi ) $$

Since \(h(t - \xi ) = 0\) for \(\xi \geqslant t\) it follows that the upper limit of the integral can be extended to plus infinity. The integral is the convolution of the input F(t) with the impulse response h(t) or

$$\begin{aligned} x(t) = \int _{ - \infty }^\infty \mathrm{d}\xi \cdot F(\xi ) \cdot h(t - \xi ) \end{aligned}$$
(1.39)

Symbolically this result is often written

$$ x(t) = F(t) * h(t) $$

This expression can be written again in an alternative way by a change of variables or by introducing the parameter \(\tau \) as \(\tau = t - \xi \). Thus

$$ x(t) = \int _{ - \infty }^\infty \mathrm{d}\tau \cdot F(t - \tau ) \cdot h(\tau ) $$

This result can be achieved in alternative way as suggested by Green and demonstrated in Problem 1.12. The starting point are the two equations

$$ m\ddot{h} + c\dot{h} + k_0 h = \delta (t - \tau ), \quad m\ddot{x} + c\dot{x} + k_0 x = F(t) $$

The solution to the first is given by Eq. (1.36). By multiplying the first equation by x and the second by h and integrating the resulting equations with respect to t, the displacement x(t) is obtained in integral form as shown in Problem 1.12. The procedure proposed by Green is also used in Sect. 6.2 to determine the response of a beam excited by a point force. Compare also Sect. 7.3.

For an ideal mass–spring system, the impulse excitation of the mass can basically be caused by a sudden motion of the foundation or by an external force acting on a mass mounted resiliently on a firm foundation as shown in Fig. 1.11.

Fig. 1.11
figure 11

Excitation of a mass due to a external force b motion of foundation

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The equations describing the motion of the mass in the two examples are

$$\begin{aligned}&\mathbf{Case }\,\, \mathbf{A } \qquad \qquad \qquad m\ddot{x} + c\dot{x} + k_0 x = F(t)\end{aligned}$$
(1.40)
$$\begin{aligned}&\mathbf{Case }\,\, \mathbf{B }\qquad \qquad m\ddot{x} + c(\dot{x} - \dot{y}) + k_0 (x - y) = 0 \end{aligned}$$
(1.41)

In the last expression, the substitution \(z = x - y\) can be made. This leads to the following expressions

$$\begin{aligned} m\ddot{z} + c\dot{z} + k_0 z = - m\ddot{y} = F_\mathrm{f} (t) \end{aligned}$$
(1.42)

The general solutions to Eqs. (1.40) and (1.41) are according to Eq. (1.38)

$$\begin{aligned}&\mathbf{Case }\,\, \mathbf{A } \qquad \qquad \qquad x(t) = \displaystyle \int _0^tF(\xi ) \cdot h(t - \xi ) \cdot \mathrm{d}\xi \nonumber \\&\mathbf{Case }\,\, \mathbf{B } \qquad \qquad x(t) = y(t) + \displaystyle \int _0^t \mathrm{d}\xi \cdot F_\mathrm{f} (\xi ) \cdot h(t - \xi ) \end{aligned}$$
(1.43)
$$\begin{aligned} = y(t) - \displaystyle \int _0^t \mathrm{d}\xi \cdot m\ddot{y}(\xi ) \cdot h(t - \xi ) \end{aligned}$$
(1.44)

The function h(t) is defined in Eq. (1.36). Two basic types of transient excitation—a step and pulse displacement of the foundation are discussed below.

In the first case, the foundation is given the displacement

$$\begin{aligned} y = A\left( {1 - \mathrm{e}^{ - \gamma t}} \right) \end{aligned}$$
(1.45)

Consequently as the parameter \(\gamma \) is increased, the gradient of the displacement curve increases as shown in Fig. 1.12. According to Eq. (1.44), the response x(t) of the mass due to the motion of the foundation is given by

$$ x(t) = A\left[ {1 - \exp ( - \gamma t) + m\gamma ^2\int _0^t h(t - \xi ) \cdot \exp ( - \gamma x) \cdot \mathrm{d}\xi } \right] $$

where

$$\begin{aligned} h(t - \xi ) = \frac{\mathrm{e}^{ - \beta (t - \xi )}}{m\omega _\mathrm{r} }\sin \omega _\mathrm{r} (t - \xi ) \end{aligned}$$
(1.46)

The integral is solved by means of partial integration. The result is:

$$ x(t) = A\left[ {1 - \exp ( - \gamma t) + \frac{\gamma ^2\Gamma (t)}{\omega _\mathrm{r} \left[ {(\beta - \gamma )^2 + \omega _\mathrm{r}^2 } \right] }} \right] $$
$$\begin{aligned} \Gamma (t) = \omega _\mathrm{r} \mathrm{e}^{ - \gamma t} - (\beta - \gamma )\mathrm{e}^{ - \beta t}\sin \omega _\mathrm{r} t - \omega _\mathrm{r} \mathrm{e}^{ - \beta t}\cos \omega _\mathrm{r} t \end{aligned}$$
(1.47)

From Eq. (1.47), the velocity and acceleration as well as the maximum deflection, velocity, and acceleration can be calculated.

Fig. 1.12
figure 12

Step displacement of foundation

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In Fig. 1.13, the response x is presented for two different values of \(\gamma \). The other parameters \(m,c,k_0 \) and A are constant. The figure shows that the maximum amplitude is increased when \(\gamma \) is increased, i.e., as the slope of the deflection curve for the foundation is increased. As time goes to infinity, the deflection of the mass approaches A, which is also the final deflection for the foundation. In Fig. 1.14, the deflection of the mass is shown for two different damping ratios. When the damping is increased by a factor 2.5, the amplitude of the mass is decreasing more rapidly as a function of time. Despite this, the maximum amplitude is just slightly reduced.

Fig. 1.13
figure 13

Response of a mass caused by a step displacement of foundation

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Fig. 1.14
figure 14

Response of a mass with respect to time for two damping ratios, \(\gamma = 40\)

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In the second example, the displacement y of the foundation is defined as

$$\begin{aligned} \begin{array}{ll} y(t) = A\sin \omega t~~ \text {for}~~ 0 \leqslant t \leqslant t_0\\ y(t) = 0~~ \text {for}~~ t < 0~~ \text {and}~~ t > t_0\end{array} \end{aligned}$$
(1.48)

where \(t_0 = \pi / \omega \). The motion of the mass is again described by Eq. (1.44) and is

$$\begin{aligned} x(t) = A\left[ {\sin \omega t + m\omega ^2\int _0^t {h(t - \xi )\sin \omega \xi \cdot \mathrm{d}\xi } } \right] \end{aligned}$$
(1.49)

where again h(t) is defined in Eq. (1.36). Note that the upper limit t for the integral cannot exceed \(t_0\) as y(t) is only defined in the time interval \(0 \leqslant t \leqslant t_0\). For \(0 \leqslant t \leqslant t_0\), the solution to Eq. (1.49) is

$$\begin{aligned} \begin{array}{ll} x(t) = A\left[ {\sin \omega t + \dfrac{\omega ^2}{\omega _\mathrm{r} }\left( {H_1 - H_2 } \right) } \right] \\ H_1 = \dfrac{\beta \cos \omega t + (\omega + \omega _\mathrm{r} )\sin \omega t - \beta \mathrm{e}^{ - \beta t}\cos \omega _\mathrm{r} t + (\omega + \omega _\mathrm{r} )\mathrm{e}^{ - \beta t}\sin \omega _\mathrm{r} t}{2\left[ {\left( {\omega + \omega _\mathrm{r} } \right) ^2 + \beta ^2} \right] }\\ H_2 = \dfrac{\beta \cos \omega t + (\omega - \omega _\mathrm{r} )\sin \omega t - \beta \mathrm{e}^{ - \beta t}\cos \omega _\mathrm{r} t - (\omega - \omega _\mathrm{r} )\mathrm{e}^{ - \beta t}\sin \omega _\mathrm{r} t}{2\left[ {\left( {\omega - \omega _\mathrm{r} } \right) ^2 + \beta ^2} \right] }\end{array} \end{aligned}$$
(1.50)

From these expressions, the deflection \(x_0 = x(t_0 )\) and velocity \(v_0 = \dot{x}(t_0 )\) can be calculated. For \(t > t_0 \), no external force is acting on the system thus the motion of the mass is determined by free vibrations. According to Eq. (1.16), the motion of the mass for \(t \geqslant t_0 \) is

$$\begin{aligned} x(t) = \mathrm{e}^{ - \beta (t - t_0 )}\left\{ {x_0 \cos \left[ {\omega _\mathrm{r} (t - t_0 )} \right] + \frac{v_0 + \beta x_0 }{\omega _\mathrm{r} }\sin \left[ {\omega _\mathrm{r} (t - t_0 )} \right] } \right\} \end{aligned}$$
(1.51)

In Fig. 1.15, the deflection of the foundation is shown as a function of time for three different values of \(\omega \), i.e., for three different pulse lengths \(t_0\).

Fig. 1.15
figure 15

Displacement y(t) of foundation

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Fig. 1.16
figure 16

Response of mass caused by displacement y(t) of foundation

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The corresponding response x(t) for the mass is shown in Fig. 1.16. It is evident that the response very much depends on the length of time, \(t_0 \), of the exciting pulse with respect to the time T for one cycle corresponding to free vibrations of the mass–spring system. The peak amplitude A of the pulse is constant in the example. For a short pulse, the acceleration of the foundation is initially large. This results in a large early displacement of the mass. The resulting amplitude is thus a function of the slope of the curve defining the displacement of the foundation as well as the length of the pulse. The importance of the pulse length can be illustrated by considering a 1-DOF system mounted on a fixed foundation and excited by an external force F(t) corresponding to Case A discussed above. For a resiliently mounted mass excited by a force with the time dependence of a half sine wave, the response is obtained from the Eq. (1.43) and by comparing the results (1.49) and (1.50). Thus, let a force F(t) excite the mass. The force is defined as

$$\begin{aligned} \begin{array}{ll} F(t) = F_0 \sin (\omega t)~~\text {for}~~0 \leqslant t \leqslant t_0\\ F(t) = 0~~ \text {for}~~ t < 0~~ \text {and}~~ t > t_0 \end{array} \end{aligned}$$
(1.52)

where \(t_0 = \pi / \omega \). The response x(t) of the 1-DOF system is for a fixed foundation, \(y = 0\), equal to

$$\begin{aligned} x(t) = F_0 \frac{H_1 (t) - H_2 (t)}{m\omega _\mathrm{r} }~~\text {for}~~ 0 \leqslant t \leqslant t_0 \end{aligned}$$
(1.53)

The functions \(H_1\) and \(H_2\) are defined in Eq. (1.50). For \(t > t_0 \), i.e., for free vibrations, the response is given by Eq. (1.51). The parameters \(x_0 \) and \(v_0 \) are as before the displacement and velocity at \(t = t_0 \). In Fig. 1.17, the response due to a force in the shape of a half sine wave is shown. The duration or pulse length of the half sine wave is set equal to T / 4, T / 2, and T as in the previous case. The force functions have the same time dependence as the displacements y(t) shown in Fig. 1.15. The resulting response of the mass due to this type of force excitation is illustrated in Fig. 1.17. The initial slopes of the response curves are different as compared to the previous case (motion of foundation). For force excitation of the mass, its initial velocity is equal to zero. The response of the mass again depends on the duration of the pulse exciting the system. The maximum displacement of the mass can be calculated from Eq. (1.53). In Fig. 1.18, the maximum amplitude of the response is shown as function of \(t_0 / T\) or the ratio between the pulse length and the time corresponding to one free oscillation of the system. The maximum amplitude increases for increasing pulse lengths up to \(t_0 \approx 0.8T\) for half sine wave excitation. For other types of excitation pulses, the maximum displacement is obtained for other ratios between \(t_0 \) and T. See Problem 1.6.

Fig. 1.17
figure 17

Response of mass excited by a force \(F(t) = y(t)\)

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Fig. 1.18
figure 18

Maximum amplitude as function of pulse length

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One final example is of interest in connection with forced excitation discussed in the next section. Assume that the mass at time \(t = 0\) is started to be excited by a harmonic force \(F(t) = F_0 \cdot \sin (\omega t)\). The resulting displacement of the mass for \(t \geqslant 0\) is obtained from Eq. (1.43). The calculations are somewhat laborious but still straightforward. The result is

$$\begin{aligned} x(t)\,=\,&\frac{F_0 [(\omega _0^2 - \omega ^2)\sin (\omega t) - 2\omega \beta \cos (\omega t)]}{m[(\omega _0^2 - \omega ^2)^2 + (2\beta \omega )^2]} \nonumber \\&- \mathrm{e}^{ - \beta t}\frac{F_0 [(\omega _0^2 - \omega ^2 - 2\beta ^2)\sin (\omega _\mathrm{r} t)(\omega /\omega _\mathrm{r}) - 2\omega \beta \cos (\omega _\mathrm{r} t)]}{m[(\omega _0^2 - \omega ^2)^2 + (2\beta \omega )^2]} \end{aligned}$$
(1.54)

The result indicates that as time goes to infinity or rather when \(\exp ( - \beta t) \ll 1\), the time dependence is predominately determined by the time dependence of the force. In other words, if a periodic force is applied to a damped mass–spring system, the motion of the mass shows the same periodicity as the force once the transients can be neglected. If for example \(\beta t \approx 2.5\), the energy of the transient motion is of the order 1/10 of the energy of the stationary motion. After twice this time, i.e., for \(\beta t \approx 5\) the ratio is 1/100. For this case, the transient vibrations can be neglected.

Clearly, the response of a dynamic system strongly depends on the rise time of the force exciting the system. The sharper the rise, the more violent is the response of the system. The maximum amplitude of the system depends primarily on the rise time rather than on the losses of the system. In noise control engineering, the rise time of any force acting on a noise radiating structure should be made as long as possible. The response of a system excited by an impulse depends not only on the rise time of the pulse but also on its duration.

For more complicated dynamic systems, transient solutions are most easily obtained by using Laplace transforms. See for example Ref. [5]. Transient vibrations of dynamic systems are discussed at length in for example Ref. [3].

1.4 Forced Harmonic Vibrations

When a harmonic force excites a damped dynamic system, the system tends to vibrate at the same frequency as that of the force. Initially, when the force is applied, the total motion is composed of transient as well as forced vibrations. The result is as shown in Eq. (1.54). However, there are always losses in a dynamic system. Consequently the free vibrations, which are proportional to \(\exp ( - \beta t)\) will die out as shown in the previous section. After a certain time, depending on the losses, the forced motion will dominate completely. Transient vibrations are therefore neglected whenever a system is excited by a stationary force.

Periodic forces are typically induced by rotating machinery. A periodic force could, for example, be caused by a rotor imbalance. For each turn of the rotor, the time history of the force is repeated.

If a force F(t) is repeated after the time or period T, i.e., if

$$\begin{aligned} F(t) = F(t + T) \end{aligned}$$
(1.55)

then the force is said to be periodic with the period T. An example of a periodic force F(t), period T,  is

$$ F(t) = F_0 \sin \omega t $$

where the angular frequency \(\omega = 2\pi / T\). This is a harmonic force, which is the simplest type of periodic force. The force is stationary since angular frequency and amplitude are constant.

The equation of motion for the simple mass–spring system with a harmonic force applied to the mass is

$$\begin{aligned} m\ddot{x} + c\dot{x} + k_0 x = F_0 \sin \omega t \end{aligned}$$
(1.56)

The solution describing the forced motion must be of the form

$$\begin{aligned} x(t)&= A_1 \sin \omega t + A_2 \cos \omega t = A_0 \sin (\omega t + \varphi )\nonumber \\&= A_0 \sin \omega t\cos \varphi + A_0 \cos \omega t\sin \varphi \end{aligned}$$
(1.57)

In both cases, there are two unknown quantities. In the first case, \(A_1\) and \(A_2 \) and in the second case \(A_0 \) and \(\varphi \). An identification of the coefficients in Eq. (1.57) yields

$$\begin{aligned} A_0 = \sqrt{A_1^2 + A_2^2 }, \quad \tan \varphi = A_2 / A_1 \end{aligned}$$
(1.58)

The general solution to Eq. (1.56) must also include the free vibration solution corresponding to the case when \(F(t) = 0\). However due to the losses in the system, this solution can be neglected as time goes to infinity. In fact, the transient response can in most practical cases be neglected after some minute after the force have been turned on as discussed in Sect. 1.3.

If the first expression in Eq. (1.57) is inserted in Eq. (1.56) the result is

$$\begin{aligned}{}[A_1 (k_0\,-\,m\omega ^2)\,-\,c\omega A_2 ]\sin \omega t\,+\,[A_2 (k_0 - m\omega ^2) + c\omega A_1 ]\cos \omega t = F_0 \sin \omega t \end{aligned}$$
(1.59)

For this equation to be valid at any time t, the expression inside the bracket in front of the cosine term must be zero. Thus,

$$ A_2 = - A_1 \frac{c\omega }{k_0 - m\omega ^2} = - A_1 \frac{2\beta \omega }{\omega _0^2 - \omega ^2} $$

where

$$ \beta = \frac{c}{2m},\quad \omega _0^2 = \frac{k_0 }{m}. $$

In addition, the expression in front of the sine term on the left hand side of the equation must equal \(F_0 \). The parameters \(A_1 ,A_2 ,A_0 \) and \(\varphi \) are consequently obtained as

$$\begin{aligned} \begin{array}{ll} A_1 = \dfrac{F_0 (\omega _0^2 - \omega ^2)}{m[(\omega _0^2 - \omega ^2)^2 + (2\beta \omega )^2]} \\ A_2 = - \dfrac{F_0 2\omega \beta }{m[(\omega _0^2 - \omega ^2)^2 + (2\beta \omega )^2]}\\ A_0 = \dfrac{F_0 }{m\left[ {(\omega _0^2 - \omega ^2)^2 + (2\beta \omega )^2} \right] ^{1 / 2}}\\ \tan \varphi = - \dfrac{2\beta \omega }{\omega _0^2 - \omega ^2}\end{array} \end{aligned}$$
(1.60)

The same result is obtained from Eq. (1.54) when t approaches infinity. The displacement or the amplitude \(A_0 \) in (1.60) has a maximum for the denominator having a minimum. For small losses, this maximum amplitude or resonance is found when the frequency of the force exciting the system nearly coincides with the natural frequency of the system or rather when \(\omega = \sqrt{\omega _0^2 - 2\beta ^2} \) whereas the natural angular frequency is \(\omega _\mathrm{r} = \sqrt{\omega _0^2 - \beta ^2} \). In general, the two expressions are almost equal since the losses are small or rather \(\beta \ll 1\).

According to the second expression in Eq. (1.57), there is a phase difference \(\varphi \) between force and amplitude. The magnitude of this phase angle depends on the losses and the relative difference between the angular frequencies for the force and the undamped resonance \(\omega _0 \) for the system. Thus for a force described by \(F(t) = F_0 \sin \omega t\) the displacement of the mass is \(x(t) = A_0 \sin (\omega t + \varphi )\) and its velocity \(\dot{x}(t) = \omega A_0 \cos (\omega t + \varphi ) = \omega A_0 \sin (\omega t + \varphi + \pi / 2)\). These quantities can be drawn in a phase diagram as shown in Fig. 1.19. The time dependence of the functions is obtained by the projections of the various vectors on the vertical axis in the diagram. If the force is of the form \(\cos \omega t\), velocity and displacement are equal to the projection of the corresponding vectors on the horizontal axis.

For the simple 1-DOF system excited by the force \(F_0 \sin \omega t\), the time average of the velocity squared is given by

$$\begin{aligned} \bar{v}^2&= \bar{\dot{x}}^2 \nonumber \\&= \frac{1}{T}\int _0^T A_0^2 \omega ^2\cos ^2(\omega t + \varphi )\mathrm{d}t \nonumber \\&= \frac{\omega ^2A_0^2 }{2} \nonumber \\&= \frac{\omega ^2F_0^2 }{2m^2\left[ {(\omega _0^2 - \omega ^2)^2 + (2\beta \omega )^2} \right] } \end{aligned}$$
(1.61)

Since x and thus \(\dot{x}\) are periodic, it is sufficient to take the average over only one period \(T = 2\pi / \omega \).

Fig. 1.19
figure 19

Phase difference between force, displacement, and velocity

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The time average of the kinetic energy for the 1-DOF system is

$$\begin{aligned} \bar{\mathcal{T}} = \frac{m\bar{v}^2}{2} = \frac{\omega ^2F_0^2 }{4m[(\omega _0^2 - \omega ^2)^2 + (2\beta \omega )^2]} \end{aligned}$$
(1.62)

In a similar way, the time average of the potential energy is

(1.63)

For vibrations induced by a simple harmonic force described by \(F_0 \sin \omega t\), the time averages of the potential and kinetic energies are not equal except when the angular frequency \(\omega \) of the force is equal to \(\omega _0 \). For free vibrations, equality holds when there are no losses, or in practice when the losses are small.

The average power \(\bar{\Pi }\) fed into the system during one period—or rather, the time average of the power input—is defined as

$$\begin{aligned} \bar{\Pi }&= \frac{1}{T}\int _0^T {\mathrm{d}t \cdot F(t)} \cdot v(t) \nonumber \\&= \frac{1}{T}\int _0^T {\mathrm{d}t \cdot F_0 \sin \omega t\left[ {A_1 \omega \cos \omega t - A_2 \omega \sin \omega t} \right] } \nonumber \\&= - \frac{\omega A_2 F_0 }{2} \nonumber \\&= \frac{\beta \omega ^2F_0^2 }{m[(\omega _0^2 - \omega ^2)^2 + (2\beta \omega )^2]} \end{aligned}$$
(1.64)

From Eqs. (1.61) and (1.64) it follows that

$$\begin{aligned} \bar{\Pi } = 2\bar{v}^2m\beta \quad \text {or}\quad \bar{v}^2 = \frac{\bar{\Pi }}{2m\beta } = \frac{\bar{\Pi }}{c} \end{aligned}$$
(1.65)

This important relationship means that the average velocity squared is inversely proportional to the losses in the system if the power input to the system is constant.

1.5 Fourier Series

The response of a simple 1-DOF system is readily determined whenever the system is excited by a harmonic force. The straight forward solutions discussed in the previous section can be extended to include any 1-DOF system excited by a periodic force. Any periodic function or force can be expanded in a Fourier series. Thus if \(F(t) = F(t + T)\), then the function F(t) can be written as

$$\begin{aligned} F(t) = \frac{a_0 }{2} + \sum \limits _{n = 1}^\infty {(a_n \cdot \cos \omega _n t + b_n \cdot \sin \omega _n t)} \end{aligned}$$
(1.66)

where \(\omega _n = 2\pi n / T\) for \(n=1,2,\ldots \). The coefficients \(a_n\) and \(b_n\) are derived as demonstrated in Problem 1.8. For \(n = 0, 1, 2,\ldots \), the coefficients \(a_n\) and \(b_n\) are

$$\begin{aligned} a_n = \frac{2}{T}\int _0^T {F(t)\cos \omega _n t\cdot \mathrm{d}t }, \quad b_n = \frac{2}{T}\int _0^T {F(t)\sin \omega _n t \cdot \mathrm{d}t} \end{aligned}$$
(1.67)

There are alternative ways of expanding the functon F(t) in the time interval \(0 \leqslant t \leqslant T\). This is demonstrated by first defining a new function \(G_1 (t)\) as:

$$ G_1 (t) = F(t)~~\text {for}~~0 \leqslant t \leqslant T,\quad G_1 (t) = - F( - t)~~\text {for}~~ - T \leqslant t < 0 $$

The function \(G_1 (t)\) is assumed to be periodic with the period \(T_0 = 2T\). A Fourier expansion of \(G_1 (t)\) in the time interval \( - T \leqslant t \leqslant T\) as given by Eq. (1.66) yields:

$$ G_1 (t) = \sum \limits _{n = 1}^\infty {b_n \cdot \sin (\omega _n t / 2)} $$

However, \(G_1 (t) = F(t)\) for \(0 \leqslant t \leqslant T\). Thus in this time interval F(t) is obtained as

$$\begin{aligned} F(t) = \sum \limits _{n = 1}^\infty {b_n \cdot \sin (\omega _n t / 2)} = \sum \limits _{n = 1}^\infty {b_n \cdot \sin (n\pi t / T)} \end{aligned}$$
(1.68)

The coefficients \(b_n\) are given by Eq. (1.67) as

$$\begin{aligned} b_n = \frac{2}{T}\int _0^T {\mathrm{d}t \cdot F(t)\sin (n\pi t / T)} \end{aligned}$$
(1.69)

Yet another possibility to expand F(t) in a series is obtained by defining a function \(G_2 (t)\) as

$$ G_2 (t) = F(t)~~\text {for}~~ 0 \leqslant t \leqslant T,\quad G_2 (t) = F( - t)~~\text {for}~~- T \leqslant t < 0 $$

A Fourier expansion of \(G_2 (t)\) in the time interval \( - T \leqslant t \leqslant T\) yields:

$$ G_2 (t) = \frac{a_0 }{2} + \sum \limits _{n = 1}^\infty {a_n \cdot \cos (\omega _n t / 2)} $$

where \(\omega _n = 2\pi n / T\) for \(n= 1,2,\ldots \). However, \(G_2 (t) = F(t)\) for \(0 \leqslant t \leqslant T\). Thus in this time interval F(t) is obtained as

$$\begin{aligned} F(t) = \frac{a_0 }{2} + \sum \limits _{n = 1}^\infty {a_n \cdot \cos (\omega _n t / 2)} = \frac{a_0 }{2} + \sum \limits _{n = 1}^\infty {a_n \cdot \cos (n\pi t / T)} \end{aligned}$$
(1.70)

The coefficients \(a_n\) are given by Eq. (1.67) as

$$\begin{aligned} a_n = \frac{2}{T}\int _0^T {F(t)\cos (n\pi t / T)\cdot \mathrm{d}t } \end{aligned}$$
(1.71)

In general, any of the three methods could be used to expand F(t) in a series. However, the rate of convergence could differ considerably between the methods.

The response of a mass, which is excited by a periodic force, can also be expanded in a Fourier series. If the force is periodic, then the forced motion is also periodic with the same period as the force. Consequently, the response x(t) of such a forced system can be written as

$$\begin{aligned} x(t)&= \frac{A_0 }{2} + \sum \limits _{n = 1}^\infty {(A_n \cdot \cos \omega _n t + B_n \cdot \sin \omega _n t)}\nonumber \\&= \frac{A_0 }{2} + \sum \limits _{n = 1}^\infty {C_n \cdot \cos (\omega _n t + \varphi _n )} \end{aligned}$$
(1.72)

Each amplitude \(A_n \) and \(B_n\) can be solved as described for the simple harmonic force. In a similar way, the total amplitude \(C_n \) and the phase angle \(\varphi _n\) are determined for each \(\omega _n\) or frequency component.

The time average of the velocity squared is as before given by

$$ \bar{v}^2 = \frac{1}{T}\int _0^T { \dot{x}^2\cdot \mathrm{d}t } $$

This can be expressed by means of the components \(A_n \) and \(B_n \) of the Fourier series as:

$$\begin{aligned} \bar{v}^2 = \frac{1}{2}\sum \limits _{n = 1}^\infty \omega _n^2 (A_n^2 + B_n^2 ) = \frac{1}{2}\sum \limits _{n = 1}^\infty \omega _n^2 C_n^2 \end{aligned}$$
(1.73)

This result is obtained since

$$ \frac{1}{T}\int _0^T {\mathrm{d}t \cdot \sin \omega _n \cdot \sin \omega _k = } \frac{1}{2}~~\text {for}~~n = k~~ \text {and equal to zero for}~~ n \ne k. $$

The time average of the velocity squared can be illustrated in the frequency domain as a line spectrum. The amplitude at \(\omega = \omega _n \) is \(\omega _n^2 (A_n^2 + B_n^2 ) / 2\). The frequency domain of a signal and its significance are discussed in Chap. 2.

1.6 Complex Notation

For harmonic excitation, is often convenient to assume that the force applied to the system can be expressed in the form \(F_0 \exp (i\omega t)\). Since a physical quantity must be real, it can be assumed that the real force is equal to

$$ F(t) = \mathrm{Re}\left( {F_0 \mathrm{e}^{i\omega t}} \right) = F_0 \cos \omega t. $$

If on the other hand, the force is given by a sine function then

$$ F(t) = \mathrm{Im}\left( {F_0 \mathrm{e}^{i\omega t}} \right) = F_0 \sin \omega t. $$

The expressions above Re and Im define the real and imaginary parts of the force, respectively. In the phase diagram, Fig. 1.19, the axes can now be identified with the real and imaginary parts of a force vector in the complex plane.

If now the force is given by \(F_0 \exp (i\omega t)\), then the resulting forced displacement of the mass in the 1-DOF system must be expressed in the form \(x = x_0 \exp (i\omega t)\). With these expressions for F and x inserted in the basic equation of motion \(m\ddot{x}\, +\, c\dot{x}\, +\, k_0 x = F\), the amplitude \(x_0\) is obtained as

$$\begin{aligned} x_0&= \frac{F_0 }{ - m\omega ^2 + i\omega c + k_0 } \nonumber \\&= \frac{F_0 }{m[(\omega _0^2 - \omega ^2) + 2i\omega \beta ]} \nonumber \\&= \frac{F_0 [(\omega _0^2 - \omega ^2) - 2i\omega \beta ]}{m[(\omega _0^2 - \omega ^2)^2 + (2\omega \beta )^2]} \end{aligned}$$
(1.74)

If the force is given by \(F(t) = F_0 \cos \omega t = \mathrm{Re}\left( {F_0 \mathrm{e}^{i\omega t}} \right) \), the solution is

$$\begin{aligned} x(t)&= \mathrm{Re}(x_0 \mathrm{e}^{i\omega t}) = \mathrm{Re}(x_0 \cos \omega t + ix_0 \sin \omega t)\nonumber \\&= \frac{F_0 [(\omega _0^2 - \omega ^2)\cos \omega t + 2\omega \beta \sin \omega t]}{m[(\omega _0^2 - \omega ^2)^2 + (2\omega \beta )^2]} \end{aligned}$$
(1.75)

For the case where the force is described by \(F(t) = F_0 \sin \omega t = \mathrm{Im}\left( {F_0 \mathrm{e}^{i\omega t}} \right) \), the solution is

$$\begin{aligned} x(t) = \mathrm{Im}(x_0 \mathrm{e}^{i\omega t}) = \frac{F_0 [(\omega _0^2 - \omega ^2)\sin \omega t - 2\omega \beta \cos \omega t]}{m[(\omega _0^2 - \omega ^2)^2 + (2\omega \beta )^2]} \end{aligned}$$
(1.76)

The last result is the same as that given by the Eqs. (1.57) and (1.60) and obtained in the traditional way. The general solution (1.74) can be expressed in an alternative way as

$$\begin{aligned} x = \left| {x_0 } \right| \cdot \exp [i(\omega t + \varphi )] \end{aligned}$$
(1.77)

where

$$ \left| {x_0 } \right| = \dfrac{F_0 }{m\left[ {(\omega _0^2 - \omega ^2)^2 + (2 \omega \beta )^2} \right] ^{1/2}}, \quad \tan \varphi = - \dfrac{2 \omega \beta }{\omega _0^2 - \omega ^2 } $$

So again if the force acting on the mass is described by \(F(t) = F_0 \sin \omega t\), then

$$ x = \mathrm{Im}\left\{ {\left| {x_0 } \right| \cdot \exp [i(\omega t + \varphi )]} \right\} = \left| {x_0 } \right| \cdot \sin (\omega t + \varphi ) $$

This expression is identical to the solution given by the Eqs. (1.57) and (1.60). Compare Fig. 1.19.

Physical quantities of interest for the description of the response of a mechanical system are velocity squared, kinetic energy, potential energy, and power input and the time averages of these quantities. These quantities are real and nonnegative. In order to consider this, a convention with respect to the complex notation must be adopted. The convention can be made that, when a complex number is to be squared, the complex number is multiplied by its complex conjugate. Further, the average value of the square of a quantity represented by the complex function \(x(t) = A\exp (i\omega t)\) is equal to one-half of the square of the magnitude A.

This convention can be compared to results obtained when just using sine and cosine functions. For example, if a function x(t) is written as \(x(t) = A\cos (\omega t + \varphi )\) then

$$ \bar{x}^2 = \frac{1}{T}\int _0^T {A^2\cos ^2(\omega t + \varphi )\cdot \mathrm{d}t = \frac{A^2}{2}}~~\text {since}~~\omega T = 2\pi $$

Alternatively, x(t) is expressed as

$$ x(t) = A \cdot \exp [i(\omega t + \varphi )] $$

According to the convention above, the average of x squared is defined as

$$\begin{aligned} \bar{x}^2 = \frac{xx^ * }{2} = \frac{1}{2}A\exp [i(\omega t + \varphi )]A^ * \exp [ - i(\omega t + \varphi )] = \frac{\left| A \right| ^2}{2} \end{aligned}$$
(1.78)

In a similar way,

$$\begin{aligned} \dot{x}(t) = i\omega A \cdot \exp [i(\omega t + \varphi )], \quad \bar{\dot{x}}^2 = \frac{\omega ^2\left| A \right| ^2}{2} \end{aligned}$$
(1.79)

The time average of the input power is also a nonnegative and real quantity. With \(F(t) = F_0 \exp (i\omega t)\) and \(x(t) = x_0 \exp (i\omega t)\) where \(x_0 \) is given in Eq. (1.74), then in accordance with the discussion above the average input power is defined as:

$$\begin{aligned} \bar{\Pi }&= \frac{1}{2}\mathrm{Re}(F \cdot \dot{x}^ * ) = \frac{1}{2}\mathrm{Re}[F_0 \mathrm{e}^{i\omega t} \cdot ( - i\omega )x_0 ^ * \mathrm{e}^{-i\omega t}] \nonumber \\&= \frac{\omega }{2}\mathrm{Im}(F_0 x_0 ) = \frac{\beta \omega ^2\left| {F_0 } \right| ^2}{m[(\omega _0^2 - \omega ^2)^2 + (2\beta \omega )^2]} \end{aligned}$$
(1.80)

This expression is the same as that derived in Eq. (1.64). It is, however, much more convenient to derive the required results by means of the complex notation than through solutions which are a combination of sine and cosine functions and which eventually have to be integrated with respect to time. In addition, the complex amplitudes \(F_0\) and \(x_0 \) have a very significant meaning which will be discussed in Chap. 2. Compare also Sect. 2.9.

The simple equation of motion for a linear damped 1-DOF system is for \(c = 2\beta m\) and \(F = F_0 \exp (i\omega t)\) and according to Eq. (1.56) given by

$$ m\ddot{x} + 2m\beta \dot{x} + k_0 x = F_0 \cdot \mathrm{e}^{i\omega t} $$

The parameters \(m,\beta \) and \(k_0\) are real quantities. If the solution \(x = x_0 \exp (i\omega t)\) is assumed the result is, neglecting the transient response and omitting the time-dependent term \(\exp (i\omega t)\), equal to

$$ x_0 ( - m\omega ^2 + 2i\omega m\beta + k_0 ) = F_0 $$

This can be rewritten as:

$$\begin{aligned} x_0 [ - m\omega ^2 + k_0 (1 + i\delta _0 )] = F_0 \end{aligned}$$
(1.81)

where

$$ \delta _0 = \frac{2\omega \beta }{\omega _0^2 } $$

Thus for harmonic excitation of a 1-DOF system, the equation of motion is, using complex notation

$$\begin{aligned} m\ddot{x} + kx = F \end{aligned}$$
(1.82)

where

$$ k = k_0 (1 + i\delta _0 ) $$

The spring constant k is thus complex. The real parameter \(\delta _0\) is equal to the ratio between the imaginary part of the spring constant and its real part. The parameter \(\delta _0 \) is quite simply referred to as the loss factor of the spring.

From Eq. (1.81), the response of the mass is also written

$$ x(t) = \frac{F_0 \mathrm{e}^{i\omega t}}{k_0 (1 + i\delta ) - m\omega ^2} = \frac{F_0 \mathrm{e}^{i\omega t}}{m(2\pi )^2[f_0^2 (1 + i\delta ) - f^2]} $$

For free vibrations, the same technique cannot be used directly. The result described in Eq. (1.81) is only valid as long as the displacement of the mass is harmonic. Free oscillations of a damped 1-DOF system are not harmonic, i.e., the velocity is not given by \(i\omega \) times the displacement. However, if the losses are very small, the free vibrations are approximately harmonic within a period.

For \(\beta \ll 1\), the free vibrations of a 1-DOF system are, according to Eq. (1.15), described by

$$ x(t) = \mathrm{e}^{ - \beta t} \cdot [C\cos \omega _\mathrm{r} t + D\sin \omega _\mathrm{r} t] $$

According to Eq. (1.81), \(\beta = \omega _0 \delta _0 / 2\) since \(\omega _\mathrm{r} \approx \omega _0 \) for \(\beta \ll 1\). The free oscillations of a lightly damped 1-DOF system can therefore be approximated by the expression

$$\begin{aligned} x(t) = \exp ( - \omega _0 \delta _0 t / 2) \cdot [C\cos \omega _\mathrm{r} t + D\sin \omega _\mathrm{r} t] \end{aligned}$$
(1.83)

The total energy or the time average of the kinetic or potential energy is thus decaying as

(1.84)

where is the energy at \(t = 0\). Compare Eqs. (1.23) and (1.26). For standard solids, the loss factor decreases as the frequency is increased. However, there are notable exceptions, for example rubber and some composite structures as discussed in Chap. 3.