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1.2 Undamped Oscillator

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Consider the mechanical system that is schematically

shown in Figure 1.1. The inputs (or

excitation) applied to the system are represented

by the force f ðtÞ: The outputs (or response) of

the system are represented by the displacement y:

The system boundary (real or imaginary)

demarcates the region of interest in the analysis.

What is outside the system boundary is the

environment in which the system operates. An

analytical model of the system may be given by

one or more equations relating the outputs to

the inputs. If the rates of changes of the response

(outputs) are not negligible, the system is a dynamic system. In this case the analytical model, in the

time domain, becomes one or more differential equations rather than algebraic equations. System

parameters (e.g., mass, stiffness, damping constant) are represented in the model, and their values

should be known in order to determine the response of the system to a particular excitation. State

variables are a minimum set of variables, which completely represent the dynamic state of a system at

any given time t: These variables are not unique (more than one choice of a valid set of state variables

is possible). For a simple oscillator (a single-degree-of-freedom (DoF) mass – spring – damper system)

an appropriate set of state variables would be the displacement y and the velocity y_: An alternative set

would be y_ and the spring force.

In the present section, we will first show that many types of oscillatory systems can be represented by

the equation of an undamped simple oscillator. In particular, mechanical, electrical, and fluid systems will

be considered. The conservation of energy is a straightforward approach for deriving the equations of

motion for undamped oscillatory systems (which fall into the class of conservative systems). The

equations of motion for mechanical systems may be derived using the free-body diagram approach, with

the direct application of Newton’s Second Law. An alternative and rather convenient approach is the use of

Lagrange’s equations. The natural (free) response of an undamped simple oscillator is a simple harmonic

motion. This is a periodic, sinusoidal motion.

1.2.1 Energy Storage Elements

Mass (inertia) and spring are the two basic energy storage elements in mechanical systems. The concept

of state variables may be introduced as well through these elements.

1.2.1.1 Inertia (m)

Consider an inertia element of lumped mass m; excited by force f ; as shown in Figure 1.2.

The resulting velocity is v:

System

Boundary

Environment

System

Outputs

System (Response)

Inputs

(Excitation)

f(t)

Dynamic System

State Variables (y,y)

Parameters (m,k,b)

FIGURE 1.1 A mechanical dynamic system.

1-2 Vibration and Shock Handbook

Newton’s Second Law gives

dt ¼ f ð1:1Þ

Kinetic energy stored in the mass element is equal

to the work done by the force f on the mass. Hence,

Energy E ¼

f dx ¼

dt ¼

fv dt ¼

ð m dv

v dt

¼m

v dv

Kinetic energy KE ¼

mv2 ð1:2Þ

Note: v is an appropriate state variable for a mass element because it can completely represent the energy

of the element.

Integrate Equation 1.1:

vðtÞ ¼ vð02Þ þ

ðt

f dt ð1:3Þ

Hence, with t ¼ 0þ; we have

vð0þÞ ¼ vð02Þ þ

ð0þ

f dt ð1:4Þ

Since the integral of a finite quantity over an almost zero time interval is zero, these results tell us that a

finite force will not cause an instantaneous change in velocity in an inertia element. In particular, for a

mass element subjected to finite force, since the integral on the right-hand side of Equation 1.4 is zero,

we have

vð0þÞ ¼ vð02Þ ð1:5Þ

1.2.1.2 Spring (k)

Consider a massless spring element of lumped

stiffness k; as shown in Figure 1.3. One end of the

spring is fixed and the other end is free. A force f is

applied at the free end, which results in a

displacement (extension) x in the spring.

Hooke’s Law gives

f ¼ kx or

dt ¼ kv ð1:6Þ

Elastic potential energy stored in the spring is equal to the work done by the force on the spring. Hence,

Energy E ¼

f dx ¼

kx dx ¼ 1

2 kx2 ¼

dt ¼

fv dt ¼

dt ¼

f df ¼

f 2

Elastic potential energy PE ¼

kx2 ¼

f 2

k ð1:7Þ

Note: f and x are both appropriate state variables for a spring, because both can completely represent the

energy in the spring.

FIGURE 1.2 A mass element.

f = kx

FIGURE 1.3 A spring element.

Time-Domain Analysis 1-3

Integrate Equation 1.6:

f ðtÞ ¼ f ð02Þ þ

ðt

v dt ð1:8Þ

Set t ¼ 0þ: We have

f ð0þÞ ¼ f ð02Þ þ

ð0þ

v dt ð1:9Þ

From these results, it follows that at finite velocities there cannot be an instantaneous change in the force

of a spring. In particular, from Equation 1.9 we see that when the velocities of a spring are finite:

􀀄

0þ

􀀅

¼ f ð02Þ ð1:10Þ

Also, it follows that

􀀄

0þ

􀀅

¼ xð02Þ ð1:11Þ

1.2.1.3 Gravitation Potential Energy

The work done in raising an object against the

gravitational pull is stored as gravitational potential

energy of the object. Consider a lumped mass

m; as shown in Figure 1.4, which is raised to a

height y from some reference level.

The work done gives

Energy E ¼

f dy ¼

mg dy

Hence,

Gravitational potential energy PE ¼ mgy ð1:12Þ

1.2.2 The Method of Conservation of Energy

There is no energy dissipation in undamped systems which contain energy storage elements only. In

other words, energy is conserved in these systems, which are known as conservative systems. For

mechanical systems, conservation of energy gives

KE þ PE ¼ const: ð1:13Þ

These systems tend to be oscillatory in their natural motion, as noted before. Also, analogies exist with

other types of systems (e.g., fluid and electrical systems). Consider the six systems sketched in Figure 1.5.

1.2.2.1 System 1 (Translatory)

Figure 1.5(a) shows a translatory mechanical system (an undamped oscillator) which has just one degree

of freedom x: This may represent a simplified model of a rail car that is impacting against a snubber. The

conservation of energy (Equation 1.13) gives

mx_2 þ

kx2 ¼ const: ð1:14Þ

f = mg

FIGURE 1.4 A mass element subjected to gravity.

1-4 Vibration and Shock Handbook

Here, m is the mass and k is the spring stiffness. Differentiate Equation 1.14 with respect to time t: We

obtain

mx_x€ þ kxx_ ¼ 0

Since generally x_ – 0 at all t; we can cancel it out. Hence, we obtain the equation of motion:

x€ þ

x ¼ 0 ð1:15Þ

1.2.2.2 System 2 (Rotatory)

Figure 1.5(b) shows a rotational system with the single DoF u: It may represent a simplified model of a

motor drive system. As before, the conservation energy gives

J_u 2 þ

Ku 2 ¼ const: ð1:16Þ

In this equation, J is the moment of inertia of the rotational element and K is the torsional stiffness of the

shaft. Then, by differentiating Equation 1.16 with respect to t and canceling u_; we obtain the equation of

motion:

u€þ

u ¼ 0 ð1:17Þ

1.2.2.3 System 3 (Flexural)

Figure 1.5(c) is a lateral bending (flexural) system, which is a simplified model of a building structure.

Again, a single DoF x is assumed. Conservation of energy gives

mx_2 þ

kx2 ¼ const: ð1:18Þ

J θ

k1 k=k1+k2 k2

Mass density = r

(a)

(b)

(c)

(d)

(e)

(f)

+ −

FIGURE 1.5 Six examples of single-degree-of-freedom oscillatory systems: (a) translatory; (b) rotatory; (c) flexural;

(d) pendulous; (e) liquid slosh; (f) electrical.

Time-Domain Analysis 1-5

Here, m is the lumped mass at the free end of the support and k is the lateral bending stiffness of the

support structure. Then, as before, the equation of motion becomes

x€ þ

x ¼ 0 ð1:19Þ

1.2.2.4 System 4 (Pendulous)

Figure 1.5(d) shows a simple pendulum. It may represent a swinging-type building demolisher or a ski

lift, and has a single-DoF u. We have

KE ¼

mðlu_Þ2

Gravitational PE ¼ Eref 2 mgl cos u

Here, m is the pendulum mass, l is the pendulum length, and g is the acceleration due to gravity. Hence,

conservation of energy gives

ml2u_ 2 2 mgl cos u ¼ const: ð1:20Þ

Differentiate with respect to t after canceling the common ml:

lu_u€þ g sin u u_ ¼ 0

Since, u_ – 0 at all t; we have the equation of motion:

u€þ

sin u ¼ 0 ð1:21Þ

This system is nonlinear, in view of the term sin u:

For small u; sin u is approximately equal to u: Hence, the linearized equation of motion is

u€þ

u ¼ 0 ð1:22Þ

1.2.2.5 System 5 (Liquid Slosh)

Consider a liquid column system shown in Figure 1.5(e). It may represent two liquid tanks linked by a

pipeline. The system parameters are: area of cross section of each column ¼ A; mass density of

liquid ¼ r; length of liquid mass ¼ l:

We have

KE ¼

2 ðrlAÞ_y 2

Gravitational PE ¼ rAðh þ yÞ

2 ðh þ yÞ þ rAðh 2 yÞ

2 ðh 2 yÞ

Hence, conservation of energy gives

rlAy_ 2 þ 1

2 rAgðh þ yÞ2 þ 1

2 rAgðh 2 yÞ2 ¼ const: ð1:23Þ

Differentiate:

ly_y€ þ gðh þ yÞy_ 2 gðh 2 yÞy_ ¼ 0

But, we have

y_ – 0 for all t

Hence,

y€ þ gðh þ yÞ 2 gðh 2 yÞ ¼ 0

1-6 Vibration and Shock Handbook

y€ þ

y ¼ 0 ð1:24Þ

1.2.2.6 System 6 (Electrical)

Figure 1.5(f) shows an electrical circuit with a single capacitor and a single inductor. Again, conservation

of energy may be used to derive the equation of motion.

Voltage balance gives

vL þ vC ¼ 0 ð1:25Þ

where vL and vC are voltages across the inductor and the capacitor, respectively.

Constitutive equation for the inductor is

dt ¼ vL ð1:26Þ

Constitutive equation for the capacitor is

dvC

dt ¼ i ð1:27Þ

Hence, by differentiating Equation 1.26, substituting Equation 1.25, and using Equation 1.27, we

obtain

d2i

dt2 ¼

dvL

dt ¼ 2

dvC

dt ¼ 2

d2i

dt2 þ i ¼ 0 ð1:28Þ

Now consider the energy conservation approach for this electrical circuit, which will give the same

result. Note that power is given by the product vi:

1.2.2.7 Capacitor

Electrostatic energy E ¼

vi dt ¼

dt ¼ C

v dv ¼

Cv2

2 ð1:29Þ

Here, v denotes vC: Also,

v ¼

i dt ð1:30Þ

Since the current i is finite for a practical circuit, we have

Ð0þ

02 i dt ¼ 0:

Hence, in general, the voltage across a capacitor cannot change instantaneously. In particular,

vð0þÞ ¼ vð02Þ ð1:31Þ

1.2.2.8 Inductor

Electromagnetic energy E ¼

vi dt ¼

i dt ¼ L

i di ¼

Li2

Time-Domain Analysis 1-7

Here, v denotes vL: Also,

i ¼

v dt ð1:32Þ

Since v is finite in a practical circuit, we have

Ð0þ

02 v dt ¼ 0:

Hence, in general, the current through an inductor cannot change instantaneously. In particular,

ið0þÞ ¼ ið02Þ ð1:33Þ

Since the circuit in Figure 1.5(f) does not have a resistor, there is no energy dissipation. As a result, energy

conservation gives

Cv2

2 þ

Li2

2 ¼ const: ð1:34Þ

Differentiate Equation 1.34 with respect to t:

dt þ Li

dt ¼ 0

Substitute the capacitor constitutive equation 1.27:

iv þ Li

dt ¼ 0

Since i – 0 in general, we can cancel it. Now, by differentiating Equation 1.27, we have

di=dt ¼ Cðd2v=dt2Þ: Substituting this in the above equation, we obtain

d2v

dt2 þ v ¼ 0 ð1:35Þ

Similarly, we obtain

d2i

dt2 þ i ¼ 0 ð1:36Þ

1.2.3 Free Response

The equation of free motion (i.e., without an excitation force) of the six linear systems considered above

(Figure 1.5) is of the same general form:

x€ þv2

nx ¼ 0 ð1:37Þ

This is the equation of an undamped, simple oscillator. The parameter vn is the undamped natural

frequency of the system. For a mechanical system of mass m and stiffness k; we have

vn ¼

ffiffiffiffi

ð1:38Þ

To determine the time response x of this system, we use the trial solution:

x ¼ A sinðvnt þ fÞ ð1:39Þ

in which A and f are unknown constants, to be determined by the initial conditions (for x and x_Þ; say,

xð0Þ ¼ x0; x_ð0Þ ¼ v0 ð1:40Þ

Substitute the trial solution into Equation 1.37. We obtain

ð2Av2

n þ Av2

nÞsinðvnt þ fÞ ¼ 0

1-8 Vibration and Shock Handbook

This equation is identically satisfied for all t:

Hence, the general solution of Equation 1.37 is

indeed Equation 1.39, which is periodic and

sinusoidal.

This response is sketched in Figure 1.6. Note

that this sinusoidal oscillatory motion has a

frequency of oscillation of v (radians/sec). Hence,

a system that provides this type of natural motion

is called a simple oscillator. In other words, the

response exactly repeats itself in time periods of T

or a cyclic frequency f ¼ 1=T (Hz). The frequency

v is in fact the angular frequency given by v ¼ 2p f : Also, the response has an amplitude A; which is the

peak value of the sinusoidal response. Now, suppose that we shift this response curve to the right through

f =v: Consider the resulting curve to be the reference signal (with signal value ¼ 0 at t ¼ 0; and

increasing). It should be clear that the response shown in Figure 1.6 leads the reference signal by a time

period of f=v: This may be verified from the fact that the value of the reference signal at time t is the same

as that of the signal in Figure 1.6 at time t 2 f=v: Hence, f is termed the phase angle of the response, and

it is a phase lead.

The left-hand-side portion of Figure 1.6 is the phasor representation of a sinusoidal response.

In this representation, an arm of length A rotates in the counterclockwise direction at angular speed

v: This is the phasor. The arm starts at an angular position f from the horizontal axis, at time

t ¼ 0: The projection of the arm onto the vertical ðxÞ axis is the time response. In this manner,

the phasor representation can conveniently indicate the amplitude, frequency, phase angle, and the

actual time response (at any time t) of a sinusoidal motion. A repetitive (periodic) motion of the

type 1.39 is called a simple harmonic motion, meaning it is a pure sinusoidal oscillation at a single

frequency.

Next, we will show that the amplitude A and the phase angle f both depend on the initial conditions.

Substitute the initial conditions (Equation 1.40) into Equation 1.39 and its time derivative. We obtain

x0 ¼ A sin f ð1:41Þ

v0 ¼ Avn cos f ð1:42Þ

Now divide Equation 1.41 by Equation 1.42, and also use the fact that sin2f þ cos2f ¼ 1: We obtain

tan f ¼ vn

􀀏 􀀐2

Avn

􀀏 􀀐2

¼ 1

Hence,

Amplitude A ¼

ffiffiffiffiffiffiffiffiffiffiffi

0 þ

ð1:43Þ

Phase f ¼ tan21 vnx0

v0 ð1:44Þ

Example 1.1

A simple model for a tracked gantry conveyor system in a factory is shown in Figure 1.7.

The carriage of mass ðmÞ moves on a frictionless track. The pulley is supported on frictionless bearings,

and its axis of rotation is fixed. Its moment of inertia about this axis is J: The motion of the carriage is

restrained by a spring of stiffness k1; as shown. The belt segment that drives the carriage runs over the

2π − f

π − φ

− φ Time t

Response x

FIGURE 1.6 Free response of an undamped simple

oscillator.

Time-Domain Analysis 1-9

pulley without slippage, and is attached at the

other end to a fixed spring of stiffness k2: The

displacement of the mass is denoted by x; and the

corresponding rotation of the pulley is denoted by

u: When x ¼ 0 (and u ¼ 0) the springs k1 and k2

have an extension of x10 and x20, respectively, from

their unstretched (free) configurations. Assume

that the springs will remain in tension throughout

the motion of the system.

Solution Using Newton’s Second Law

A free-body diagram for the system is shown in

Figure 1.8.

Hooke’s Law for the spring elements:

F1 ¼ k1ðx10 þ xÞ ðiÞ

F2 ¼ k2ðx20 2 xÞ ðiiÞ

Newton’s Second Law for the inertia elements:

mx€ ¼ F 2 F1 ðiiiÞ

Ju€ ¼ rF2 2 rF ðivÞ

Compatibility:

x ¼ ru ðvÞ

Straightforward elimination of F1; F2; F; and u in Equation i to Equation v, using algebra, gives

m þ

􀀏 􀀐

x€ þ ðk1 þ k2Þx ¼ k2x20 2 k1x10 ðviÞ

It follows that

Equivalent mass meq ¼ m þ

Equivalent stiffness keq ¼ k1 þ k2

Solution Using Conservation of Energy

Kinetic energy T ¼

mx_2 þ

Ju_ 2

Potential energy ðelasticÞ V ¼

k1ðx10 þ xÞ2 þ

k2ðx20 2 xÞ2

Total energy in the system:

E ¼ T þ V ¼

mx_ 2 þ

Ju_ 2 þ

k1ðx10 þ xÞ2 þ

k2ðx20 2 xÞ2 ¼ const:

Differentiate with respect to time:

mx_x€ þ Ju_u€þ k1ðx10 þ xÞx_ 2 k2ðx20 2 xÞx_ ¼ 0

Substitute the compatibility relation, x_ ¼ ru_; to obtain

mx_x€ þ

r2 x_x€ þ k1ðx10 þ xÞx_ 2 k2ðx20 2 xÞx_ ¼ 0

x r

Frictionless

Rollers

No-Slip Belt

And Pulleys

Frictionless

Bearings

FIGURE 1.7 A tracked conveyor system.

F2 F2

FIGURE 1.8 A free-body diagram for the conveyor

system.

1-10 Vibration and Shock Handbook

Eliminate the common velocity variable x_ (which cannot be zero for all t). We obtain

m þ

􀀏 􀀐

x€ þ ðk1 þ k2Þx ¼ k2x20 2 k1x10

which is the same result as before.

Solution Using Lagrange’s Equations

Lagrangian L ¼ T 2 V ¼

mx_2 þ

Ju_ 2 2

k1ðx10 þ xÞ2 2

k2ðx20 2 xÞ2

Now, substituting for u in terms of x (Equation v), we obtain

L ¼

m þ

􀀏 􀀐

x_2 2

k1ðx10 þ xÞ2 2

k2ðx20 2 xÞ2

Use Lagrange’s equation (see Box 1.1):

›L

›q_i

›L

›qi ¼ Qi for i ¼ 1; 2; …; n

where the generalized coordinate qi ¼ x and the corresponding generalized force Qi ¼ 0 because there

are no nonconservative and external forces. We obtain

›L

›x_ ¼ m þ

􀀏 􀀐

›L

›x ¼ 2k1ðx10 þ xÞ þ k2ðx20 2 xÞ

From this, we obtain the equation of motion:

m þ

􀀏 􀀐

x€ þ ðk1 þ k2Þx ¼ k2x20 2 k1x10

which is identical to what we obtained before.

Natural Frequency

From the equivalent translational system, the natural frequency (undamped) of the system is obtained as

vn ¼

ffiffiffiffiffiffiffiffiffi

keq=meq

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ðk1 þ k2Þ= m þ

s 􀀏 􀀐

ðviiÞ

Substitute for x and its derivatives into Equation vi using the compatibility condition (Equation v). We

obtain the equivalent rotational system:

ðr2m þ JÞ u€þ r2ðk1 þ k2Þu ¼ rk2x20 2 rk1x10

The equivalent moment of inertia:

Jeq ¼ r2m þ J

The equivalent torsional stiffness:

Keq ¼ r2ðk1 þ k2Þ

Therefore, the corresponding natural frequency is

vn ¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

􀀄

k1 þ k2

􀀅

􀀄

r2m þ J

q 􀀅

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ðk1 þ k2Þ= m þ

s 􀀏 􀀐

ðviiiÞ

This result is identical to the previous result (vii). This is to be expected, as the system has not changed

(only the response variable was changed).

Time-Domain Analysis 1-11

Common methods of developing equations of motion for mechanical systems are summarized in

Box 1.1.

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