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1.5 Damped Simple Oscillator

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Now we will consider free (natural) response of a

simple oscillator in the presence of energy

dissipation (damping).

Assume viscous damping, and consider the

oscillator shown in Figure 1.14. The free-body

diagram of the mass is shown separately.

We use the following notation:

vn ¼ undamped natural frequency

vd ¼ damped natural frequency

vr ¼ resonant frequency

v ¼ frequency of excitation

The concept of resonant frequency will be

addressed in Chapter 2.

Usually, the viscous damping constant of a single

DoF is denoted by b (but, sometimes c is used

instead of b; particularly for multi-DoF systems).

Apply Newton’s Second Law. From the free-body diagram in Figure 1.14, we have the equation of

motion mx€ ¼ 2kx 2 bx_

mx€ þ bx_ þ kx ¼ 0 ð1:46Þ

x€ þ 2zvnx_ þv2

nx ¼ 0 ð1:47Þ

FIGURE 1.14 A damped simple oscillator and its freebody

diagram.

1-16 Vibration and Shock Handbook

This is a free (or unforced, or homogeneous) equation of motion. Its solution is the free (natural) response

of the system and is also known as the homogeneous solution. Note that vn ¼

ffiffiffiffiffi

k=m p ; which is the natural

frequency when there is no damping, and

2zvn ¼

m ð1:48Þ

Hence,

z ¼

ffiffiffiffi

z ¼

bffiffiffiffi

km p ð1:49Þ

Note that z is called the damping ratio. The formal definition of, and the rationale for, this terminology

will be discussed later.

Assume an exponential solution:

x ¼ C elt ð1:50Þ

This is justified by the fact that linear systems have exponential or oscillatory (i.e., complex exponential)

free responses. A more detailed justification will be provided later.

Substitute Equation 1.50 into Equation 1.47 to obtain

½l2 þ 2zvnl þ v2

n􀀉C elt ¼ 0

Note that C elt is not zero in general. It follows that, when l satisfies the equation

l2 þ 2zvnl þ v2

n ¼ 0 ð1:51Þ

then Equation 1.50 will represent a solution of Equation 1.47.

Equation 1.51 is called the characteristic equation of the system. This equation depends on the natural

dynamics of the system, not on forcing excitation or initial conditions.

The solution of Equation 1.51 gives the two roots:

l ¼ 2zvn ^

ffiffiffiffiffiffiffiffi

z2 2 1

vn ¼ l1 and l2 ð1:52Þ

These are called eigenvalues or poles of the system.

When l1 – l2; the general solution is

x ¼ C1 el1 þ C2 el2 t ð1:53Þ

The two unknown constants C1 and C2 are related to the integration constants and can be determined by

two initial conditions which should be known.

If l1 ¼ l2 ¼ l; we have the case of repeated roots. In this case, the general solution (Equation 1.53)

does not hold because C1 and C2 would no longer be independent constants to be determined by two

initial conditions. The repetition of the roots suggests that one term of the homogenous solution should

have the multiplier t (a result of the double integration of zero). Then the general solution is

x ¼ C1 elt þ C2t elt ð1:54Þ

We can identify three categories of damping level, as discussed below, and the nature of the response will

depend on the particular category of damping.

Time-Domain Analysis 1-17

1.5.1 Case 1: Underdamped Motion (z < 1)

In this case it follows from Equation 1.52 that the roots of the characteristic equation are

l ¼ 2zvn ^ j

ffiffiffiffiffiffiffiffi

1 2 z2

vn ¼ 2zvn ^ jvd ¼ l1 and l2 ð1:55Þ

where the damped natural frequency is given by

vd ¼

ffiffiffiffiffiffiffiffi

1 2 z2

vn ð1:56Þ

Note that l1 and l2 are complex conjugates. The response (Equation 1.53) in this case may be

expressed as

x ¼ e2zvn t

C1 ejvd t þ C2 e2jvd t

ð1:57Þ

The term within the square brackets of Equation 1.57 has to be real, because it represents the time

response of a real physical system. It follows that C1 and C2 have to be complex conjugates.

Note:

ejvd t ¼ cos vdt þ j sin vdt

e2jvd t ¼ cos vdt 2 j sin vdt

So, an alternative form of the general solution would be

x ¼ e2zvn t ½A1 cos vdt þ A2 sin vdt􀀉 ð1:58Þ

Here, A1 and A2 are the two unknown constants. By equating the coefficients it can be shown that

A1 ¼ C1 þ C2 and A2 ¼ jðC1 2 C2Þ ð1:59Þ

Hence,

C1 ¼

2 ðA1 2 jA2Þ and C2 ¼

2 ðA1 þ jA2Þ ð1:60Þ

1.5.1.1 Initial Conditions

Let xð0Þ ¼ x0; x_ð0Þ ¼ v0 as before. Then,

x0 ¼ A1 and v0 ¼ 2zvnA1 þ vdA2 ð1:61Þ

A2 ¼

vd þ

zvnx0

vd ð1:62Þ

Yet, another form of the solution would be

x ¼ A e2zvn t sinðvdt þ fÞ ð1:63Þ

Here, A and f are the unknown constants with

A ¼

ffiffiffiffiffiffiffiffiffiffi

A21

þ A22

and sin f ¼

ffiffiffiAffiffi1ffiffiffiffiffi

A21

þ A22

p ð1:64Þ

Also,

cos f ¼

ffiffiffiAffiffi2ffiffiffiffiffi

A21

þ A22

p and tan f ¼

A2 ð1:65Þ

Note that the response x ! 0 as t ! 1: This means the system is asymptotically stable.

1-18 Vibration and Shock Handbook

1.5.2 Logarithmic Decrement Method

The damping ratio z may be experimentally determined from the free response by the logarithmic

decrement method. To illustrate this approach, note from Equation 1.63 that the period of damped

oscillations is

T ¼

vd ð1:66Þ

Also, from Equation 1.63, we have

xðtÞ

xðt þ nTÞ ¼

A e2zvn t sinðvdt þ fÞ

A e2zvn ðtþnTÞ sin½vdðt þ nTÞ þ f􀀉

But, sin½vdðt þ nTÞ þ f􀀉 ¼ sinðvdt þ f þ 2npÞ ¼ sinðvd þ fÞ:

Hence,

xðtÞ

xðt þ nTÞ ¼

e2zvn t

e2zvn ðtþnTÞ ¼ ezvn nT ð1:67Þ

Take the natural logarithm of Equation 1.67, the logarithmic decrement,

zvnnT ¼ ln

xðtÞ

xðt þ nTÞ

􀀒 􀀓

However,

vnT ¼ vn

vd ¼

vn2p ffiffiffiffiffiffiffiffi

1 2 z2

vn ¼

2p ffiffiffiffiffiffiffiffi

1 2 z2

Hence, with xðtÞ=xðt þ nTÞ ¼ r we have the logarithmic decrement.

2pnz ffiffiffiffiffiffiffiffi

1 2 z2

p ¼ ln r

Note that ð1=nÞln r is the “per-cycle” logarithmic decrement and ð1=2pnÞln r is the “per-radian”

logarithmic decrement. The latter is

z ffiffiffiffiffiffiffiffi

1 2 z2

p ¼

2pn

ln r ¼ a ð1:68Þ

Then, we have

z ¼

ffiffiffiffiffiffiffiffiffiffi

1 þ a2

ð1:69Þ

This is the basis of the logarithmic decrement method of measuring damping. Start by measuring a point

xðtÞ and another point xðt þ nTÞ at n cycles later. For high accuracy, pick the peak points of the response

curve for the measurement of xðtÞ and xðt þ nTÞ: From Equation 1.68 it is clear that, for small damping,

z ¼ a ¼ per-radian logarithmic decrement.

1.5.3 Case 2: Overdamped Motion (z > 1)

In this case, roots l1 and l2 of the characteristic equation (Equation 1.51) are real. Specifically, we have

l1 ¼ 2zvn þ

ffiffiffiffiffiffiffiffi

z2 2 1

vn , 0 ð1:70Þ

l2 ¼ 2zvn 2

ffiffiffiffiffiffiffiffi

z2 2 1

vn , 0 ð1:71Þ

Time-Domain Analysis 1-19

and the response equation (Equation 1.53) is nonoscillatory. Also, it should be clear from Equation 1.70

and Equation 1.71 that both l1 and l2 are negative. Hence, x ! 0 as t ! 1: This means the system is

asymptotically stable.

From the initial conditions

xð0Þ ¼ x0; x_ð0Þ ¼ v0

we obtain

x0 ¼ C1 þ C2 ðiÞ

and

v0 ¼ l1C1 þ l2C2 ðiiÞ

Multiply the first initial condition (Equation i) by l1 :

l1x0 ¼ l1C1 þ l1C2 ðiiiÞ

Subtract Equation iii from Equation ii:

v0 2 l1x0 ¼ C2ðl2 2 l1Þ

We obtain

C2 ¼

v0 2 l1x0

l2 2 l1 ð1:72Þ

Similarly, multiply the first initial condition (Equation i) by l2 and subtract from Equation ii. We obtain

v0 2 l2x0 ¼ C1ðl1 2 l2Þ

Hence,

C1 ¼

v0 2 l2x0

l1 2 l2 ð1:73Þ

1.5.4 Case 3: Critically Damped Motion (z 5 1)

Here, we have repeated roots, given by

l1 ¼ l2 ¼ 2vn ð1:74Þ

The response for this case is given by (see Equation 1.54)

x ¼ C1 e2vn t þ C2t e2vn t ð1:75Þ

Since the term e2vn t goes to zero faster than t goes to infinity, we have t e2vn t ! 0 as t ! 1: Hence, the

system is asymptotically stable.

Now use the initial conditions xð0Þ ¼ x0; x_ð0Þ ¼ v0: We obtain

x0 ¼ C1

v0 ¼ 2vnC1 þ C2

Hence,

C1 ¼ x0 ð1:76Þ

C2 ¼ v0 þ vnx0 ð1:77Þ

Note: When z ¼ 1 we have the critically damped response because below this value the response is

oscillatory (underdamped) and above this value, the response is nonoscillatory (overdamped). It follows

1-20 Vibration and Shock Handbook

that we may define the damping ratio as

z ¼ damping ratio ¼

damping constant

damping constant for critically damped conditions ð1:78Þ

1.5.5 Justification for the Trial Solution

In the present analysis, the trial solution (Equation 1.50) has been used for the response of a linear system

having constant parameter values. A justification for this is provided now.

1.5.5.1 First-Order System

Consider a first-order (homogeneous, no forcing input) linear system given by

2 l

􀀏 􀀐

x ¼ x_ 2lx ¼ 0 ð1:79Þ

This equation can be written as

x ¼ l dt

Integrate:

ln x ¼ lt þ ln C

Here, ln C is the constant of integration. Hence,

x ¼ C elt ð1:80Þ

This is then the general form of the free response of a first-order system. It incorporates one constant of

integration, and hence will need one initial condition.

1.5.5.2 Second-Order System

We can write the equation of a general second-order (homogeneous, unforced) system in the operational

form:

2 l1

􀀏 􀀐

2 l2

􀀏 􀀐

x ¼ 0 ð1:81Þ

By reasoning as before, the general solution would be of the form x ¼ C1 el1 t þ C2 el2 t : Here, C1 and C2

are the constants of integration, which are determined using two initial conditions.

1.5.5.3 Repeated Roots

The case of repeated roots deserves a separate treatment. First, consider

d2x

dt2 ¼ 0 ð1:82Þ

Integrate twice:

dt ¼ C; x ¼ Ct þ D ð1:83Þ

Note the term with t in this case. Hence, a suitable trial solution for the system

2 l

􀀏 􀀐

2 l

􀀏 􀀐

x ¼ 0 ð1:84Þ

would be x ¼ C1 elt þ C2t elt :

The main results for the free (natural) response of a damped oscillator are given in Box 1.2.

Time-Domain Analysis 1-21

Box 1.2

FREE (NATURAL) RESPONSE OF A DAMPED

SINGLE OSCILLATOR

System Equation: mx€ þ bx_ þ kx ¼ 0 or x€ þ 2zvnx_ þv2

nx ¼ 0

Undamped natural frequency vn ¼

ffiffiffiffiffi

k=m p

Damping ratio z ¼ b=2

ffiffiffiffi

km p

Characteristic Equation: l2 þ 2zvnl þ v2

n ¼ 0

Roots (Eigenvalues or Poles): l1 and l2 ¼ 2zvn ^

ffiffiffiffiffiffiffiffi

z2 2 1

Response:

x ¼ C1 el1 t þ C2 el2 t for unequal roots ðl1 – l2Þ

x ¼ ðC1 þ C2tÞelt for equal roots ðl1 ¼ l2 ¼ lÞ

Initial Conditions: xð0Þ ¼ x0 and x_ð0Þ ¼ v0

Case 1: Underdamped (z < 1)

Poles are complex conjugates: 2zvn ^ jvd

Damped natural frequency vd ¼

ffiffiffiffiffiffiffiffi

1 2 z2

x ¼ e2zvn t ½C1 ejvd t þ C2 e2jvd t􀀉 ¼ e2zvn t ½A1 cos vdt þ A2 sin vdt􀀉 ¼ A e2zvn t sinðvdt þ fÞ

A1 ¼ C1 þ C2 and A2 ¼ jðC1 2 C2Þ

C1 ¼ 1

2 ðA1 2 jA2Þ and C2 ¼ 1

2 ðA1 þ jA2Þ

A ¼

ffiffiffiffiffiffiffiffiffiffi

A21

þ A22

and tan f ¼

Initial conditions give:

A1 ¼ x0 and A2 ¼

v0 þ zvnx0

Logarithmic Decrement per Radian:

a ¼

2pn

ln r ¼

z ffiffiffiffiffiffiffiffi

1 2 z2

where r ¼ xðtÞ=½xðt þ nTÞ􀀉 ¼ decay ratio over n complete cycles.

For small z: z ø a

Case 2: Overdamped (z > 1)

Poles are real and negative: l1; l2 ¼ 2zvn ^

ffiffiffiffiffiffiffiffi

z2 2 1

x ¼ C1 el1 t þ C2 el2 t

C1 ¼

v0 2 l2x0

l1 2 l2

and C2 ¼

v0 2 l1x0

l2 2 l1

Case 3: Critically Damped (z 5 1)

Two identical poles: l1 ¼ l2 ¼ l ¼ 2vn

x ¼ ðC1 þ C2tÞe2vn t with C1 ¼ x0 and C2 ¼ ___________v0 þ vnx0

1-22 Vibration and Shock Handbook

1.5.6 Stability and Speed of Response

The free response of a dynamic system, particularly a vibrating system, can provide valuable information

concerning the natural characteristics of the system. The free (unforced) excitation may be obtained, for

example, by giving an initial-condition excitation to the system and then allowing it to respond freely.

Two important characteristics which can be determined in this manner are:

1. Stability

2. Speed of response

The stability of a system implies that the response will not grow without bounds when the excitation

force itself is finite. This is known as bounded-input bounded-output (BIBO) stability. In particular, if

the free response eventually decays to zero, in the absence of a forcing input, the system is said to be

asymptotically stable. We have seen that a damped simple oscillator is asymptotically stable, but an

undamped oscillator, while being stable in a general (BIBO) sense, is not asymptotically stable.

The speed of response of a system indicates how fast the system responds to an excitation force. It is

also a measure of how fast the free response (1) rises or falls if the system is oscillatory; or (2) decays, if

the system is nonoscillatory. Hence, the two characteristics, stability and speed of response, are not

completely independent. In particular, for nonoscillatory (overdamped) systems these two properties are

very closely related. It is clear then that stability and speed of response are important considerations in

the analysis, design, and control of vibrating systems.

The level of stability of a linear dynamic system depends on the real parts of the eigenvalues (or poles),

which are the roots of the characteristic equation. Specifically, if all the roots have real parts that

are negative, then the system is stable. Also, the more negative the real part of a pole, the faster the decay

of the free response component corresponding to that pole. The inverse of the negative real part is the

time constant. The smaller the time constant, the faster the decay of the corresponding free response, and

hence, the higher the level of stability associated with that pole. We can summarize these observations

as follows:

Level of stability. Depends on decay rate of free response (and hence on time constants or real parts

of poles)

Speed of response. Depends on natural frequency and damping for oscillatory systems and decay rate

for nonoscillatory systems

Time constant. Determines stability and decay rate of free response (and speed of response in

nonoscillatory systems)

Now let us consider the specific case of a damped simple oscillator given by Equation 1.47.

Case 1 (z < 1)

The free response is given by x ¼ A e2zvn t sinðvdt þ fÞ

Time constant t ¼

zvn ð1:85Þ

The system is asymptotically stable. The larger zvn; the more stable the system. Also, the speed of

response increases with both vd and zvn:

Case 2 (z > 1)

The response is nonoscillatory, and is given by

x ¼ A1 el1 t ðdecays slowerÞ þ A2 el2 t ðdecays fasterÞ

where l1 ¼ 2zvn þ

ffiffiffiffiffiffiffiffi

z2 2 1

vn and l2 ¼ 2zvn 2

ffiffiffiffiffiffiffiffi

z2 2 1

vn:

Time-Domain Analysis 1-23

This system has two time constants:

t1 ¼

ll1l and t2 ¼

ll2l ð1:86Þ

Note that t1 is the dominant (slower) time constant. The system is also asymptotically stable. The larger

the ll1l the faster and more stable the system.

Consider an underdamped system and an overdamped system with damping ratios zu and zo;

respectively. We can show that the underdamped system is more stable than the overdamped system if

and only if

zo 2

ffiffiffiffiffiffiffiffi

o 2 1

, zu ð1:87aÞ

or equivalently,

zo .

u þ 1

2zu ð1:87bÞ

where zo . 1 . zu . 0 by definition.

Proof

To be more stable, we should have the underdamped pole located farther away than the dominant

overdamped pole from the imaginary axis of the pole plane; thus

zuvn . zovn 2

ffiffiffiffiffiffiffiffi

o 2 1

Hence,

zu . zo 2

ffiffiffiffiffiffiffiffi

o 2 1

Now, bring the square-root term to the left-hand side and square it:

o 2 1 . ðzo 2 zuÞ2 ¼ z2

o 2 2zozu þ z2

Hence,

2zozu . z2

u þ 1

zo .

u þ 1

2zu

This completes the proof. A

To explain this result further, consider an undamped ðz ¼ 0Þ simple oscillator of natural frequency vn:

Its poles are at ^jvn (on the imaginary axis of the pole plane). Now let us add damping and increase z

from 0 to 1. Then the complex conjugates poles 2zvn ^ jvd will move away from the imaginary axis as

z increases (because zvn increases) and hence the level of stability will increase. When z reaches the value

1 (critical damping) we obtain two identical and real poles at 2vn: When z is increased beyond 1, the

poles will be real and unequal, with one pole having a magnitude smaller than vn and the other having a

magnitude larger than vn: The former (closer to the “origin” of zero) is the dominant pole, and will

determine both stability and the speed of response of the overdamped system. It follows that, as z

increases beyond 1, the two poles will branch out from the location 2vn; one moving towards the origin

(becoming less stable) and the other moving away from the origin. It is now clear that as z is increased

beyond the point of critical damping, the system becomes less stable. Specifically, for a given value of

zu , 1; there is a value of zo . 1; governed by Equation 1.87, above which the overdamped system is less

stable and slower than the underdamped system.

1-24 Vibration and Shock Handbook

Example 1.4

Consider the simple oscillator shown in Figure 1.14, with parameters m ¼ 4 kg; k ¼ 1:6 £ 103 N=m; and

the two cases of damping:

1. b ¼ 80 N=m=sec

2. b ¼ 320 N=m=sec

We will study the nature of the free response in each case.

The undamped natural frequency of the system is

vn ¼

ffiffiffiffi

ffiffiffiffiffiffiffiffiffiffiffiffiffi

1:6 £ 103

rad=sec ¼ 20:0 rad=sec

Case 1

2zvn ¼

or 2z £ 20 ¼

Then,

zu ¼ 0:5

The system is underdamped in this case.

Case 2

2z £ 20 ¼

320

Then,

zo ¼ 2:0

The system is overdamped in this case.

Case 3

The characteristic equation is

l2 þ 2 £ 0:5 £ 20l þ 202 ¼ 0

l2 þ 20l þ 202 ¼ 0

The roots (eigenvalues or poles) are

l ¼ 210 ^ j

ffiffiffiffiffiffiffiffiffiffiffiffi

202 2 102

¼ 210 ^ j10

ffiffi

3 p

The free (no force) response is given by

x ¼ A e210t sin

􀀍

ffiffi

3 p t þ f

􀀎

The amplitude A and the phase angle f can be determined using initial conditions.

Time constant t ¼

10 ¼ 0:1 sec

Case 4

The characteristic equation is

l2 þ 2 £ 2 £ 20l þ 202 ¼ 0

Time-Domain Analysis 1-25

l2 þ 80l þ 202 ¼ 0

The roots are

l ¼ 240 ^

ffiffiffiffiffiffiffiffiffiffiffiffi

402 2 202

¼ 240 ^ 20

ffiffi

3 p

¼ 25:36; 274:64

The free response is given by

x ¼ C1 e25:36t þ C2 e274:64t

The constants C1 and C2 can be determined using

initial conditions. The second term on the righthand

side goes to zero much faster than the first

term, as shown in Figure 1.15. Hence, the first term

will dominate and will determine the dominant

time constant, level of stability, and speed of

response. Specifically, the response may be

approximated as

x ø C1 e25:36t

Hence,

Time constant t ¼

5:36 ¼ 0:19 sec

This value is double that of Case 1. Consequently, it is clear that the underdamped system (Case 1)

decays faster than the overdamped system (Case 2). In fact, according to Equation 1.87b, with zu ¼ 0:5

we have

zu .

0:52 þ 1

2 £ 0:5 ¼ 1:25

Hence, an overdamped system of damping ratio greater than 1.25 will be less stable than the

underdamped system of damping ratio 0.5.

Table 1.1 summarizes some natural characteristics of a damped simple oscillator under three different

levels of damping. The nature of the natural response for these three cases is sketched in Figure 1.16.

In general, the natural response of a system is governed by its eigenvalues (or poles), which are the roots

of the characteristic equation. The poles may be marked on a complex plane (s-plane), with the

horizontal axis representing the real part and the vertical axis representing the imaginary part. The nature

of the free response depending on the pole location of the system is shown in Figure 1.17.

Response

e−74.64t

e−5.36t

FIGURE 1.15 The free (homogeneous) response

components of an overdamped system.

TABLE 1.1 Natural Characteristics of a Damped Oscillator

Damping Ratio

z , 1 z . 1 z ¼ 1

Level of damping Underdamped Overdamped Critically damped

Oscillatory response Yes No No

Stability Asymptotically stable

(less stable than z ¼ 1

case but not necessarily

less stable than the

overdamped case)

Asymptotically stable

(less stable than the

critically damped case)

Asymptotically stable

(most stable)

Speed of response Better than overdamped Lower than critical Good

Time constant 1=ðzvnÞ 1=ðzvn ^

ffiffiffiffiffiffiffiffi

z2 2 1

vn Þ 1=vn

1-26 Vibration and Shock Handbook

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