Mechanical Vibrations

Husnain Inayat Hussain

List of Figures

2.1 A Simple Spring Mass System . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.2 Phase Angle φ and associated Triangle . . . . . . . . . . . . . . . . . . . . . . 82.3 Kinematics of an Undamped Spring Mass System . . . . . . . . . . . . . . . . 92.4 A Simple Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.5 A circular Disc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.6 A Compound Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.7 (a) Springs in Parallel. (b) Springs in Series. . . . . . . . . . . . . . . . . . . 192.8 A Network of Different Springs . . . . . . . . . . . . . . . . . . . . . . . . . . 192.9 A Torsional Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.1 A Vertical Spring Mass System with a Damper . . . . . . . . . . . . . . . . . 243.2 Examples of Exponentially Decaying Envelopes for Different Values of β . . . 28

3

List of Tables

2.1 Relationship between Magnitudes of Kinematics for Various Values of Angular Frequencies. 102.2 Natural Frequencies for Various Systems. . . . . . . . . . . . . . . . . . . . . 21

3.1 Eq. (3.9) Parameters and their Equivalence. . . . . . . . . . . . . . . . . . . . 26

4

Contents

1 Introduction 1

2 Simple Oscillatory Systems 2

2.1 Spring Mass System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.1.1 Observing Change in System’s State . . . . . . . . . . . . . . . . . . . 32.1.2 Equation of Motion for the Spring Mass System . . . . . . . . . . . . 42.1.3 Spring Mass System Solution . . . . . . . . . . . . . . . . . . . . . . . 4

2.2 Initial Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2.1 A Neat Trignometric Trick . . . . . . . . . . . . . . . . . . . . . . . . 6

2.3 Kinematics of an Undamped Spring Mass System . . . . . . . . . . . . . . . . 92.4 Simple Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.4.1 Linearization of the System . . . . . . . . . . . . . . . . . . . . . . . . 122.4.2 Mass Moment of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.5 Physical Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.5.1 Center of Percussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.5.2 Radius of Gyration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.6 Energy Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.7 Equivalent Spring Constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.8 Angular Frequency for General Cases . . . . . . . . . . . . . . . . . . . . . . . 202.9 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3 Damped Oscillations 22

3.1 Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.1.1 Pressure / Drag Damping . . . . . . . . . . . . . . . . . . . . . . . . . 233.1.2 Viscous Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.2 A Spring Mass System with Damping . . . . . . . . . . . . . . . . . . . . . . 243.2.1 Equation of Motion from Unstretched Position . . . . . . . . . . . . . 253.2.2 Equation of Motion from the Stretched Position . . . . . . . . . . . . 25

3.3 Damped Oscillations Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.3.1 Solution Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.3.2 Roots of Differential Equation . . . . . . . . . . . . . . . . . . . . . . . 28

3.4 The Great Battle: Damping vs. Oscillation . . . . . . . . . . . . . . . . . . . 28

5

Nations are born in the hearts

of poets, they prosper and die

in the hands of politicians.

Allama Mohammad Iqbal(1877-1938)

1Introduction

A periodic movement of an object around some position of rest (equilibrium) can be the sim-plest explanation of what vibrations are. In our everyday life we see and observe vibrationon a routine basis. In fact if we for just the time being, substitute the word periodic motionfor vibration, then we will be able to realize that our entire lives, the universe we live in, theobjects we use and the actions we take exhibit a certain form of periodicity in one way or theother.

Please consult the power point slides for this chapter.

1

With faith, discipline and self-

less devotion to duty, there

is nothing worthwhile that you

cannot achieve.

Mohammad Ali Jinnah(1876-1948)

2Simple Oscillatory Systems

In this chapter we will consider the most basic oscillatory systems and then go for trying tounderstand them physically first and then we will elaborate on different mathematical waysof understanding the physical behaviour.

2.1 Spring Mass System

A spring is defined as a mechanical element capable of storing energy. A mass is defined asa mechanical element capable of exhibiting inertia which is a measure of resistance of masswhen its state is being changed. By state we mean the state of rest or the state of motion.A system which is constructed out of these two elements as shown in Fig. 2.1 can be termedas a simple spring mass system. In the figure we have considered a spring with stiffness kand negligible mass, attached to a block of mass m and the entire assembly is placed on ahorizontal frictionless surface. x denotes the displacement of the spring from its rest position.The block’s current position (rest or equilibrium position) is taken as x = 0. If we now pull

m

k x

x = 0

Figure 2.1: A Simple Spring Mass System

the block and release it, it will start moving in opposite direction to the direction in whichpull was applied. This happens because the rigidity of the spring reacts against the initial pulland develops a resistive force which always acts in opposite direction to the applied stimulus.In the present case the applied stimulus is the action of pulling the spring and letting it go.Let us assume that the spring was pulled to some arbitrary location x along the x-axis. Underthe conditions that the pull is not too hard and it is gentle enough that the spring is not

2

2.1. SPRING MASS SYSTEM 3

permanently deformed then Hooke’s Law states that:

fs = −kx (2.1)

fs is the spring force whcih has a negative sign indicating the fact that spring force acts inopposite direction to the direction of push or pull. Once the spring is released we see thatit immediately starts moving in the negative x direction towards its rest position. However,once at the rest position, the spring continues to move and goes further back. It continuesto move back until it reaches the other extreme. At this instant it stops, but only for someunnoticeable instant. Now the spring force causes it to move forward toward the equilibriumposition in the positive x direction again. As it once again reaches the equilibrium, the trou-bled block of mass m’s misery does not stop. Rather it continues to move in the forwarddirection until it reaches the first point where it was released.

The question at this stage is why does the block not stop at its zero position. The block doesnot stop because the inertia of the block refrains it to stop arbitrarily. It should be clear bynow that behaviour of this system is governed by sprng force and inertia . Spring forcecauses the system to move in opposite direction of the initial stimulus while inertia causes thesystem to keep moving and not stop even if equilibrium postion is reached. If there are noother forces such as the force of friction then the block will continue to move forever! This ofcourse is an idealization just to make the mathematics digestible before more realistic complexcases. Real world systems come to halt eventually. However, regardless of this idealization,simple spring mass system presents an interesting and fundamental case which is worth ourtime and attention as the simple concepts developed in this study will help us explore morecomplex cases with relative ease. Under the assumptions of only a spring force acting on thesystem, if we want to know how the block moves with respect to time then we need Newton’s

Second Law of motion. It states that sum of all the force be them internal or external isequal to the product of the mass of the object and the acceleration with which it moves. Thismay be expressed mathematically as:

∑

~f = m~a (2.2)

Since we are dealing with one dimensional motion, we can forget about the vector notation.The only force present in the system is the spring force. Equating Eq. (2.1) and Eq. (2.2),one obtains the following:

− kx = ma (2.3)

2.1.1 Observing Change in System’s State

We want to investigate how the system changes its state. A system’s state may be definedby many parameters. In particular if one wants to study the change in the kinematic stateof the system then there are three parameters of interest:

position

velocity

acceleration

4 2.1. SPRING MASS SYSTEM

In the spring mass system considered above, we want to study the same parameters. Letus consider changes in this system’s position with respect to time. We set two measurementpoints xa and xb. The time it takes the block to move from position a to position b can beobtained by taking the time difference between these two points. This may be expressed as:

changepos =xb − xatb − ta

(2.4)

We may further refine the above equation to get:

changepos =∆x

∆t(2.5)

The numerator of Eq. (2.5) is the dispacement of the block and the denominator is the elapsedtime. However, one must understand that no matter how large ∆x is, the change will alwaysbe the average change in the displacement with respect to time. This change has a particularname and we know it is the velocity v of the block. This could be written as:

v =∆x

∆t(2.6)

But what if we want to observe the change in the position of the block at every instant oftime. If we want to observe parameters at each and every instant then what will be the valueof ∆t. In this case we apply a limiting process on Eq. (2.6):

vins = lim∆t→0

∆x

∆t=

dx

dt= x (2.7)

By following exactly the same logic we may write the instantaneous acceleration as:

ains = lim∆t→0

∆v

∆t=

dv

dt= x (2.8)

2.1.2 Equation of Motion for the Spring Mass System

Now we can rewrite Eq. (2.3) by following the discussion in § (2.1.1).

− kx = mx (2.9)

Rearranging the above equation we get:

mx(t) + kx(t) = 0 (2.10)

2.1.3 Spring Mass System Solution

By observing Eq. (2.10) we realize that in order to solve this equation we need a functionwhich when differentiated twice gives itself as the function upto some constant. We alsoknow that sinusoidal functions have such properties and therefore intuitively there must be asinusoidal function which solves the above equation. Let us make a guess at such a solutionand see if it works:

x(t) = A cos (ωt) (2.11)

2.2. INITIAL CONDITIONS 5

A is an arbitrary constant giving us the amplitude of the proposed solution. The ω we usedhere has to have rad/s as its units. This is so because the product ωt must produce radiansas it is the argument of a sinusoidal function. For simple harmonic motion we also know sucha quantity characterizes the angular frequency of the system. From chapter 1 we know thatfor simple harmonic motion ω is given as:

ω =2π

T(2.12)

T is the time period of a complete cycle and 2π is the span of one complete cycle. Now thatwe have developed some intuition about ω, we substitute Eq. (2.11) in the equation of motionof a simple mass spring system.

A(

−mω2 + k)

cos (ωt) = 0 (2.13)

Eq. (2.13) can be equal to zero in one of the three ways

−mω2 + k = 0

cos (ωt) = 0

A = 0 (2.14)

We know that cos (ωt) is an oscillatory function. It is zero only at specific points. To bemore precise this function is zero at π/2 and all other integer multiples of that. A is alsonot zero, in fact, it can not be zero, because if it is so then we do not have any solution tostart with. However, to satisfy Eq. (2.13), we need something which is zero at all instants oftime t. Thus, the only choice we have is to put the first equation in Eq. (2.14) equal to zero.Doing that yields:

−mω2 + k = 0

=⇒ ω2 =k

m

∴ ω = ±√

k

m(2.15)

2.2 Initial Conditions

Eq. (2.10) is a second order differential equation. Elementary calculus tells us that solving asecond order differential equation results in two constants of integration. Two constants ofintegration mean two unknown values. This requires two equations as we want to solve for twounknowns. The two required equations come from what are known as initial conditons.Initial conditions are the values of the state of the system at the instant when you startobserving it. For you this is your time t = 0. One such set of conditions could be the initialdisplacement and initial velocity and could be given as:

x(t) = xo|t=0

x(t) = vo|t=0

Now let us go back to Eq. (2.11) which represents the solution of the simple mass springsystem. We realize that there are is just one arbitrary constant to play around with. However,

6 2.2. INITIAL CONDITIONS

two roots in Eq. (2.15) suggest that there are two solutions to the original differential equationand there should be two corresponding arbitrary constants. Since we are dealing with lineardifferential equations we may combine the two solutions as one superimposed solution as isshown below:

x(t) = A1 cos(ω1t) +A2 cos(ω2t) (2.16)

Here ω1 refers to +√

k/m and ω2 denotes −√

k/m. If we define ω to be equal to√

k/m thenEq. (2.16) may be rewritten as:

x(t) = A1 cos(+ωt) +A2 cos(−ωt) (2.17)

Since cos(−θ) = cos(θ), Eq. (2.17) may be rewritten as

x(t) = A1 cos(ωt) +A2 cos(ωt)

x(t) = (A1 +A2) cos(ωt)

x(t) = A cos(ωt) (2.18)

All of this hardwork brought us back to the original proposed solution with only one arbitraryconstant. What if we propose some other solution and see if it gives us two arbitrary constants.So we propose the following solution:

x(t) = A sin(ωt) (2.19)

It turns out that Eq. (2.19) is also a solution to Eq. (2.10). Following the exact same procedureas with A cos(ωt) solution, we realize that we end up with only one arbitrary constant. A wayout of this situation is to again use linearity of the differential equation as a basis of addingthese two solutions to get two arbitrary constants. So the final solution is given as:

x(t) = A1 cos(ωt) +A2 sin(ωt) (2.20)

2.2.1 A Neat Trignometric Trick

In this section we will introduce a method of rewriting Eq. (2.18). This equation representsone way of writing the solution to Eq. (2.10). We may write the solution in other forms. Thisis important as it will facilitate us in incorporating initial conditions into the solution. Thisother form is obtained by considering the following:

A1 = A cos (φ)

A2 = A sin (φ) (2.21)

Replacing the values of A1 and A2 from Eq. (2.21) into Eq. (2.20), the latter may be rewrittenas:

x(t) = A cos(ωt) cos (φ) +A sin(ωt) sin (φ) (2.22)

To simplify Eq. (2.22) we use the following trignometric identity:

cos(a− b) = cos(a) cos(b) + sin(a) sin(b)

Identifying a as ωt and b as φ, Eq. (2.22) may finally be rewritten as:

x(t) = A cos (ωt− φ) (2.23)

2.2. INITIAL CONDITIONS 7

Where now the arbitrary constants are the amplitude A and the phase φ. DifferentiatingEq. (2.23) we get:

x(t) = −ωA sin (ωt− φ) (2.24)

We can now substitute the initial conditions to obtain the following:

x(0) = A cos (−φ) = xo

x(0) = −ωA sin (−φ) = vo (2.25)

Multiplying the first equation by ω, squaring it and adding it to the square of the secondequation in Eq. (2.25) results in the following:

ω2A2[

cos2(−φ) + sin2(−φ)]

= ω2x2o + v2o

The term in square brackets is equal to 1. Simplifying the above further, one can easily obtainthe following:

A =

√

ω2x2o + v2oω

(2.26)

To obtain the phase φ one could simply divide the two equations in Eq. (2.25) to get:

− sin (−φ)

cos (−φ)=

voωxo

tanφ =voωxo

φ = tan−1

(

voωxo

)

(2.27)

Therefore the complete solution of a simple mass spring system can be written as:

x(t) =

√

ω2x2o + v2oω

cos

[

ωt− tan−1

(

voωxo

)]

(2.28)

One must note that the above is a solution to an undamped system with initial conditions asmentioned in Eq. (2.25). We will study the damped cases later.

Exercise 1

Show that at time t = 0, x(t) is indeed equal to xo.

Solution

Substitute in Eq. (2.28) the time t = 0 to get:

x(0) =

√

ω2x2o + v2oω

cos

[

− tan−1

(

voωxo

)]

From elementary calculus we know that cos (−φ) = cosφ. We also know that:

φ = tan−1

(

voωxo

)

8 2.2. INITIAL CONDITIONS

φ

xoω

vo√ x o2 ω

2 +v o2

Figure 2.2: Phase Angle φ and associated Triangle

Using the above value of φ, one can write the following relation:

tanφ =voωxo

=perp

base

Using now the above equation and the Pythagoras theorem we can easily construct thetriangle shown in Fig. 2.2.

Using the triangle in Fig. 2.2, we can rewrite the solution as:

x(0) =

√

ω2x2o + v2oω

cos

[

tan−1

(

voωxo

)]

=⇒ x(0) =

√

ω2x2o + v2oω

cosφ

=⇒ x(0) =

√

ω2x2o + v2oω

ωxo√

ω2x2o + v2o

∴ x(0) = xo

Exercise 2

Show that at time t = 0, x(t) is indeed equal to vo.

Solution

Most of the work for this has already been done in Exercise 1. We will simply substitutet = 0 in the expression for the velocity to obtain:

x(0) = −ω

√

ω2x2o + v2oω

sin

[

− tan−1

(

voωxo

)]

=⇒ x(0) = ω

√

ω2x2o + v2oω

sinφ

=⇒ x(0) = ω

√

ω2x2o + v2oω

vo√

ω2x2o + v2o

∴ x(0) = vo

2.3. KINEMATICS OF AN UNDAMPED SPRING MASS SYSTEM 9

2.3 Kinematics of an Undamped Spring Mass System

Now that we know how to solve the differential equation of motion of an undamped springmass system such as that shown in Fig. 2.1, we will as well see how the kinematics of thesystem changes for a given set of initial conditions.

Let us assume that initial velocity vo is zero. The system is only given a displacement pullequal to xo. Following from Ex. 1 we can easily determine the arbitrary constants. These aregiven as:

A = xo

φ = 0

We can also differentiate with respect to time the expression of displacement, once for thevelocity and twice for the acceleration. Performing these steps we obtain the folowing:

x(t) = xo cosωt

x(t) = −ωxo sinωt

x(t) = −ω2xo cosωt

Having obtained these quantities, we could plot them to assess the inter-relationship among

t

f(t)x(t) = A cos (ωt) x(t) = −ωA sin (ωt) x(t) = −ω2A cos (ωt)

π2 π 3π

2 2π 5π2 3π 7π

2 4π

AωAω2A

−A−ωA

−ω2A

Figure 2.3: Kinematics of an Undamped Spring Mass System

displacement, velocity and acceleration for the initial conditions which state that x(0) = xoand x(0) = 0 and which have also been indicated above leading us to a phase value φ = 0. Inthe plot it is assumed that ω > 1. Though the plot is not drawn to any scale, the fact thatω > 1 is represented by arbitrarily increased magnitudes of velocity and accleration. Thisplot is presented with the help of Fig. 2.3. It is important to note the following:

1. Displacement and velocity are 90o out of phase.

10 2.4. SIMPLE PENDULUM

2. Displacement and acceleration are 180o out of phase.

3. If the maximum amplitude of displacement isA then the maximum amplitude of velocityis ω times higher.

4. If the maximum amplitude of displacement is A then the maximum amplitude of accel-eration is ω2 times higher.

In general for varying values of ω Table (2.1) presents relationship between magnitudes ofdisplacement, velocity and acceleration.

ω |x(t)| |x(t)| |x(t)| Relationship

ω < 1A ωA ω2A

|x| > |x| > |x|ω > 1 |x| < |x| < |x|ω = 1 |x| = |x| = |x|

Table 2.1: Relationship between Magnitudes of Kinematics for Various Values of AngularFrequencies.

2.4 Simple Pendulum

Fig. 2.4 represents a simple pendulum. A simple pendulum consists of a string which isattached to a support at one end and to a point mass m at the other. The string is takento be inextensible with negligible mass. At the equilibrium position as shown by the dottedmass object hanging right below the origin O, there is no motion and the forces, therefore,are ought to be equal.

O

m

T

fg = mg

θ

m

l

m

θ

T

mg

l sin θ

mg

h

P

Figure 2.4: A Simple Pendulum

The forces acting on this system can be investigated in two phases.

2.4. SIMPLE PENDULUM 11

1. Static Phase: At this stage the following could be written:

T − fg = 0

T = fg

Where fg is the weight of the pendulum and T is the tension in the string acting counterto the weight of pendulum.

2. Dynamic Phase: In dynamic phase the mass is then raised from its equilibrium posi-tion to a height h and released. We observe that the mass moves back to its equilibriumposition. Once it is at its equilibrium position, it does not stop and keeps moving untilit reaches the other extreme which is also shown by a dotted line and mass at an angleθ from the center line. If there are no resistive forces acting on the system to dampenthe motion, then the mass would continue to follow the dotted circular trajectory.

∑

i

F(x)i = −T sin θ

∑

j

F(y)j = T cos θ −mg

By writing the balance of forces in the x and y directions we are implying that motion is takingplace in both x and y directions. This is true but the motion as we observe it is actuallytaking place in a circular direction and we ask ourselves: Is it possile to use Newton’s

Laws for circular motion?. The answer is yes and here is how it is expressed. WritingNewton’s Second Law for circular motion, we may state that the sum of all the torquesis equal to the product of mass moment of inertia of the object multiplied by its angularacceleration. Mathematically this could be written as:

∑

k

τk = Jθ (2.29)

Here τ represents the net torque on the system, J is the mass moment of inertia of the pen-dulum (J is defined with respect to the point of rotation of the pendulum which is O in thepresent case) and θ is the angular acceleration of the pendulum, which is also the double timederivative of the angular coordinate θ. It is implied that the motion is not due to a resultanttorque or moment which is causing the pendulum to move.

The only moment in the system around the point of rotation O is that which is caused bythe weight of the pendulum. We can write this torque as the cross product of the vector~OP with the weight vector ~fg. This is given as:

−→τ =−−→OPx

−→fg (2.30)

Since the motion is taking place in the xy plane only and the only non-zero torque is actingalong z axis only, we can drop the vectorial notation. However, we still need a magnitude ofthe above expression as well as the sign that goes along with the expression. This expressioncan be obtained as follows:

|−→τ | =∣

∣

∣

−−→OPx

−→fg

∣

∣

∣

=⇒ |−→τ | =∣

∣

∣

−−→OP

∣

∣

∣

∣

∣

∣

−→fg

∣

∣

∣sin θ

∴ τ = lmg sin θ (2.31)

12 2.4. SIMPLE PENDULUM

∣

∣

∣

−−→OP

∣

∣

∣is nothing but the length of the pendulum, which is equal to l and

∣

∣

∣

−→fg

∣

∣

∣is the weight

of the pendulum mg. Also what we know is that this torque is a restoring torque and henceit always acts in a direction opposite to the applied stimulus. Taking this into account, wecould state that the sign of the above torque is coordinate dependent. Not only that but alsothat it will always act in a direction opposite to the direction of motion and therefore it willalways be accompanied by a negative sign. Substituting in Eq. (2.29) we obtain:

−mgl sin θ = Jθ

∴ Jθ +mgl sin θ = 0 (2.32)

This is the equation of motion of a simple pendulum.

2.4.1 Linearization of the System

Eq. (2.32) is a non linear differential equation in θ. Although it is possible to solve it using nu-merical or some other complex methods, we still want to solve it using the standard methodsfor linear equations that we have learnt so far. Therefore, we need to linearize it. In order to dothat, we first want to investigate the Taylor’s Series expansion of sine and cosine functions.

A Taylor’s series for a function is its approximations using a summation of infinite numberof terms. This approximation is performed at a single point. If this single point happens tobe zero, then this series becomes Maclaurin’s Series. Without going into further details,we will simply state the approximation for the sine and cosine functions:

sin θ =∞∑

n=0

(−1)n

(2n+ 1)!θ2n+1 = θ − θ3

3!+

θ5

5!− . . .

cos θ =∞∑

n=0

(−1)n

(2n)!θ2n = 1− θ2

2!+

θ4

4!− . . .

If θ is small then we can approximate the above functions as:

sin θ ≈ θ

cos θ ≈ 1

If we now use these approximations in Eq. (2.32), we will get the following:

Jθ +mglθ = 0

=⇒ θ +mgl

Jθ = 0 (2.33)

Comparing it with the equation of motion of a spring mass system x+ω2x = 0, we can easilyrealize that:

ω2 =mgl

J

ω =

√

mgl

J(2.34)

2.4. SIMPLE PENDULUM 13

Where ω is the angular frequency of the system measured in rad/s. Since ω = 2π/T , therefore,the time period of a simple pendulum is given as:

T = 2π

√

J

mgl(2.35)

2.4.2 Mass Moment of Inertia

Eq. (2.34) and Eq. (2.35) give us the natural angular frequency and time period of a simplependulum in terms of I which is the mass moment of inertia of the system.

ri mi

x

y

z

Figure 2.5: A circular Disc

Mass Moment of Inertia of an object is a measure of its resistance against circular motion.It plays the same role as is played by mass of an object in rectilinear motion. To obtain ageneral expression of this quantity we consider a solid disc as shown in Fig. 2.5. In Fig. 2.5is shown a portion of the disc and we are interested in finding out the kinetic energy of thissmaller portion. If KEi denotes this energy then we can express it as follows:

KEi =1

2miv

2i (2.36)

And therefore the total energy can be obtained by summing the individual energies as givenbelow:

KE =1

2

n∑

i=1

miv2i

KE =1

2

∑

m1v21 +m2v

22 +m3v

23 + . . .+mnv

2n

KE =1

2mv2 (2.37)

14 2.5. PHYSICAL PENDULUM

Now let us reuse Eq. (2.36) and replace the rectilinear velocity with the angular velocity.Therefore, rewriting it in angular terms gives us:

KEi =1

2mir

2i ω

2 (2.38)

Note that we have used the simple relationship that vi = riω. Also note that since we areusing a solid disc, ω is the same for all mass particles. Now the total kinetic energy of thesystem can once again be obtained by summing up the individual energies over all elementarymass particles. This gives:

KE =1

2ω2

∑

i

mir2i (2.39)

Since the energy of the moving disc is a physical quantity and it must bear the same numericalvalue whether we calculate it using rectilinear terms or angular terms, therefore the result ofEq. (2.37) and Eq. (2.39) must be the same. Also we know that the angular counterpart ofthe rectilinear velocity v is ω, thus the angular counterpart of mass m of the disc must bethe quantity under the summation in Eq. (2.39). This quantity is known as mass moment

of inertia and it is given as:

J =∑

i

mir2i (2.40)

In other words the role that mass plays in rectilinear motion is the same as performed bymass moment of inertia in angular motion.

Mass moment of inertia for various solids is available in mechanics textbooks and onlineresources. This text will not go into further details for calculation of the same. However, thesystem at hand, namely a simple pendulum has a mass moment of inertia equal to ml2.Using this value of J in Eq. (2.35), the angular frequency ω is given as:

ω =

√

mgl

ml2

ω =

√

g

l(2.41)

2.5 Physical Pendulum

A physical pendulum also known as a compound pendulum is an arbitrarily shaped rigidbody which can swing around a pivot point . A pivot can be any supporting point, e.g., anail onto which an irregularly shaped plate is hung. This nail performs the job of a pivot asthe plate can swing around it while the irregularly shaped plate itself becomes the compoundpendulum. In fact the entire system when studied under isolation could be termed as a com-pound pendulum.

Such a pendulum is defined with the help of Fig. 2.6. Although, admittedly, this is quitenicely shaped; however, the model that we will develop will be applicable to any irregularshape also. O is the pivot point, C is the center of mass and CP is the center of percussion

which will be defined later and l is the length of the pendulum.

2.5. PHYSICAL PENDULUM 15

fg fg

θ

θr

r sin θ

fg

RR

lcpl h

O

C

C

CP

CP

Figure 2.6: A Compound Pendulum

So what happens physically is that the object such as that shown in Fig. 2.6 is initially inequilibrium. At this stage we can write the following:

fg = R (2.42)

Here fg denotes the weight of the object acting at the center of mass C, whereas, R is thereaction force which develops at the support / pivot point O. The object is then raised bya height h = r(1 − cos θ) and let go. It starts oscillating about the pivot point O. We areinterested in finding the angular frequency of this compound pendulum which will then definethe vibration characteristics of this object. From § (2.4) we realize that the only momentacting about the pivot point in this entire system is a restoring torque which is due to theweight of the object fg = mg acting at the center of mass C. The magnitude of this torqueis given by:

τ = mgr sin θ (2.43)

Using Newton’s law of balance of moments of torque we could write the following:

Jθ = −mgr sin θ

Jθ +mgr sin θ = 0

θ +mgr

Jsin θ = 0

θ +mgr

Jθ = 0 (2.44)

Where the − sign in the first line of Eq. (2.44) indicates that the RHS is a restoring torque.Also one immediately realizes that since we assume θ to be small therefore, we may approxi-mate sin θ ≈ θ. From the last line in Eq. (2.44) we can deduce the natural angular frequency

16 2.5. PHYSICAL PENDULUM

of the system to be:

ω =

√

mgr

J(2.45)

2.5.1 Center of Percussion

We want to find out the length of a simple pendulum whose point mass is equal to the massof the compound pendulum such as that shown in Fig. 2.5 and whose frequency is also thesame as that of this compound pendulum. The length which satisfies this criteria is called thelength of percussion and when this length is projected on the compound pendulum the thepoint where it intersects the center line is called the center of percussion . Mathematicallywe may obtain it as follows:

ω(s) = ω(c)

√

g

lcp=

√

mgr

J (c)

g

lcp=

mgr

J (c)

lcp =J (c)

mr(2.46)

Where J (c) is the mass moment of inertia of the compound pendulum, m is the mass andr is the distance from the pivot point to the center of mass C. The superscripts (s) and (c)respectively denote the simple and compound pendulums.

The significance of center of percussion is that it is believed that it defines a sweet spot

for racquets, bats and swords where an impact results in maximum velocity imparted tothe striking object such as a ball in case of a cricket or a baseball bat. For a sword that wouldmean that the required force to cut through will be small. However, having said that, to thebest of author’s knowledge, this belief is under active research consideration and some resultssuggest that this theory of sweet spot is probably not right.

2.5.2 Radius of Gyration

Since the compound pendulum is undergoing circular vibrational motion then the measure ofresistance offered by this system is its mass moment of inertia.

Consider a circular hoop, the mass m of which is the same as the mass m of the compoundpendulum. If we try to set this hoop into circular motion then it will offer some resistance.As with the compound pendulum the measure of this resistance is its mass moment of inertia.Now we are looking for a circular hoop which offers the same resistance as does the compoundpendulum. When we find such a hoop we term its radius as radius of gyration . This isobtained mathematically as:

J (h) = J (c)

mr2gy = J (c)

rgy =

√

J (c)

m(2.47)

2.6. ENERGY METHOD 17

From Eq. (2.46) we deduce that the mass moment of inertia of a compount pendulum can bewritten as:

J (c) = mrlcp (2.48)

Substituting this value in Eq. (2.47), radius of gyration associated with a compound pendulumcan be given as:

rgy =√

rlcp (2.49)

The significance of radius of gyration is that it gives us an effective radius such that if themass of the object is known than its mass moment of inertia can be easily calculated usingJ = mr2gy.

2.6 Energy Method

For conservative systems the equation of motion of any oscillatory system can be obtained bydifferentiating its total energy with respect to time. This can be given mathematically as:

d

dtTE(t) = 0 ∀t (2.50)

The RHS of Eq. (2.50) is zero because the method is used on conservative systems. Con-

servative systems are defined as systems which do not lose their energy during the courseof motion. This is, of course, ideal because no system runs forever and eventually all theenergy in the system is exhausted. But for the undamped cases, especially the onese whereone is only interested in the natural frequencies of the system this method is used. In general,for complex cases where it is difficult to identify forces acting on the objects or for higherdimensional cases, equation of motion cannot be easily obtained. In such a situation energymethod is used.

However, we will use a very simple case to elaborate the method. We will use a simplependulum as shown in Fig. 2.4. Let TE, KE and PE respectively denote the total energy,kinetic energy and potential energy of the system. Then at any instant of time t the totalenergy can be expressed in terms of its kinetic energy and its potential energy. This can beexpressed mathematically as:

TE(t) = KE(t) + PE(t) ∀t (2.51)

Substituting Eq. (2.51) in Eq. (2.50), we obtain the following:

d

dt[KE(t) + PE(t)] = 0 (2.52)

Now the only thing left is to figure out the expressions for the potential and kinetic energiesof a simple pendulum. These are respectively given as:

PE = mgh

KE =1

2J(

θ)2

(2.53)

18 2.7. EQUIVALENT SPRING CONSTANT

Referring to Fig. 2.4 we realize that the height h by which the pendulum is raised is given by:

h = l (1− cos θ) (2.54)

Substituting Eq. (2.54) and Eq. (2.53) in Eq. (2.52) and simplifying, we get the following:

d

dt

[

1

2J(

θ)2

+mgl (1− cos θ)

]

= 0

(

Jθ +mgl sin θ)

θ = 0

Jθ +mgl sin θ = 0

θ +mgl

Jsin θ = 0

θ +mgl

Jθ = 0 (2.55)

The method described here is a simplification of the general Lagrangian Energy Method

which is used for complex systems where the application of Newtons’s Laws to obtain theequation of motion becomes difficult.

2.7 Equivalent Spring Constant

Sometimes a spring with desired rigidity is not available in which case springs can be combinedin one of the following two ways:

1. in parallel

2. in series

When the springs are joined in parallel the equivalent spring constant is given by

keq = k1 + k2

and when the springs are connected in series the equivalent stiffness is given by:

keq =k1k2

k1 + k2

The way springs can be combined is explained with the help of Fig. 2.7. In Fig. 2.7 (a) twosprings of unequal stiffness, namely, k1 and k2 are combined in parallel while in the part (b),two springs having rigidities k3 and k4 are combined in series.

2.7. EQUIVALENT SPRING CONSTANT 19

k1

k2

keq = k1 + k2

k3 k4

keq =k3k4

k3 + k4

(a)

(b)

Figure 2.7: (a) Springs in Parallel. (b) Springs in Series.

Exercise 3

Calculate equivalent stiffness for the network of springs shown in Fig. 2.8.

m

kakb

kc

k1

k2

k3

k4 k5

Figure 2.8: A Network of Different Springs

20 2.8. ANGULAR FREQUENCY FOR GENERAL CASES

Solution

The best way to solve Ex. 3 is to isolate subsystems of springs which can be combined usingrules given in § (2.7). Therefore, three regions are identified and the equivalent stiffness valuesare termed respectively as ka, kb and kc. The values of these equivalent spring constants aregiven as:

ka =k1k2

k1 + k2kb = ka + k3

kc = k4 + k5

The equivalent spring constant for the top half of the spring mass system is kb and the thatfor the lower part of the system is kc At first glance one might think that the final springconstant would be the result of combining kb and kc in series, however, this is not true andthese two springs are connecting the mass at the same potential level (or same height if youwill). Therefore, the upper and lower systems in fact are in parallel to each other with respectto the mass m and thus the final spring constant will be given as:

keq = kb + kc

2.8 Angular Frequency for General Cases

We now have developped the idea that in finding out the natural frequency of a system, weeither need the restoring force or the restoring torque. In case where forces are difficult tofind, we resort to energy method (energy method was dealt with here at a very basic level).Having figured out the above we always find, atleast for the simple systems that we havestudied so far, that the natural frequency is of the following form:

ω =

√

Restoring Force Coefficient

Measure of Inertia

Following this basic idea, we may investigate the natural frequency of the torsional pendulumas shown in Fig. 2.9. The disc is given an initial rotation θ about the z axis. Because of thetorsional rigidity κ the disc rod twists back by a restoring torque whose magnitude is givenby:

τ = κθ

Where κ is given in terms of G the shear modulus of the rod, Jp the polar moment of inertiaof the rod and l the length of the rod as shown in Fig. 2.9. The fact that the resistance tomotion is offered by the mass moment of inertia of the circular disk which has a mass m andradius r, the angular frequency of the system can be given as:

ω =

√

κ

J (d)(2.56)

And the mass moment of inertia for the disc in Fig. 2.9 is given by:

J (d) =1

2mr2

2.9. CONCLUSION 21

m, r

l κ =GJpl

θ

x

y

z

Figure 2.9: A Torsional Pendulum

We have figured out the natural frequency of the torsional pendulum without explicitly solvingfor the equation of motion. This is left as an exercise and the reader may confirm this. Basedon this simplification, we can now generalize the method for finding angular frequencies forsimple systems studied so far. The natural frequencies for various systems are given with thehelp of Table (2.2). The superscripts (s), (c) and (d)respectively denote simple pendulum,

System Restoring Force Inertia Angular Frequency

Spring Mass −kx m

√

km

Pendulum −mglθ J (s)

√

mgl

J (s)

Compound Pendulum −mgrθ J (c)√mgr

J (c)

Torsional Disc −κθ J (d)√

κJ (d)

Table 2.2: Natural Frequencies for Various Systems.

compound pendulum and torsional disc.

2.9 Conclusion

In this chapter we studied very simple oscillatory systems and obtained very fundamentalcharacteristics such as the natural frequencies of vibrational systems and time periods. Weused Newtons’s Laws for balance of forces and torques to obtain the equation of motion. Wethen solved those equations to obtain the kinematics of those systems. We then introducedenergy method for finding equation of motion for systems with complex forces. The treatmentpresented was quite basic. Then equivalent spring constants were introduced and finally themethod to find out the angular frequencies was generalized.

Personal beauty is a greater

recommendation than any let-

ter of introduction.

Aristotle (384 BC-322 BC)

3Damped Oscillations

In this chapter, we will build upon the concepts developed so far and gain insight into realphysical systems. A real oscillatory system is one which will eventually die out because itwill have used up all its energy. The question is where has this energy come from and moreimportantly how does the system use up this energy. Energy may come from pulling the blockof mass in case of simple spring mass system and in case of a simple pendulum it might comefrom raising the bob of pendulum and letting it go. It is not the only method of impartingenergy, in fact, nstead of pulling the mass, if it is just given a push in any direction alongthe axis of motion, this will also be considered as imparting energy to the system. In thefirst case, we are giving the system a displacement and in the second case we are giving thesystem a velocity. Note that the system can be given both a displacement and a velocity atthe same time. This is how energy is imparted to the system. But why do these systems dieout eventually? The reason for these systems to eventually stop is that all the energy thatwas imparted to the system initially is ultimately dissipated as heat energy and that energydisseminates in the surrounding atmosphere. In vibration this loss of energy by the systemis attributed to a physical process which is called damping .

3.1 Damping

Damping is an inherent characteristic of any oscillatory system. This is to say that whenevera system such as a spring mass system is excited by any of the methods above, we will seethat after some time the system will stop. Although we did not specifically try to stop itby applying opposing forces for example, but even then the system comes to a halt. All realsystems behave in this manner. This means that the ideal oscillatory systems that we studiedin Chapter 2 need refinement. Sometimes the time it takes to completely exhaust all theenergy is very long and it takes more than just a couple of minutes for the oscillation to stop.In some other instants the damping is so high that we can not possibly notice the responseof the system and to us the system appears un moved whereas we did not realize that thesystem did move, but the dynamics of the system were such that it died out in fraction of a

22

3.1. DAMPING 23

second (e.g. a system which oscillates at 100 cycles per second and it requires only 50 cyclesto die out will come to a halt in only half a second).

Now we want to incorporate damping in our system as an oppsosing force which makes thesystem come to a stop. We already know one opposing force and that is the spring force whosemagnitude is kx. However, we also know that this term does not help the system to dissipateenergy, rather, this force makes the system vibrate forever. So, the force we are looking forneeds to be different from this force. Now spring force is proportional to the diplacement ofthe mass from the equilibrium position. So, we need to think in terms of velocity, the nextquantity after displacement. The truth is that one kind of damping forces is proportional tothe velocity of vibration of the oscillating system. There are others which for instance dependon the friction of the surface on which the system moves. However, we will take up thosemodels later and will only concentrate on velocity dependent damping forces. Let us firstproduce the mathematical form of the damping force and then elaborate. This form is givenwith the help of Eq. (3.1).

fd = −c1x− c2x2 (3.1)

3.1.1 Pressure / Drag Damping

When the oscillating object is moving fast, then the opposition in notion is more severe interms of the pressure or drag which is proportional to the square of the velocity as shownin Eq. (3.1). To understand what it feels to be opposed by pressure, take your hand out froma fast moving car and feel the wind at your hand. Please do it for a short while and withutter caution, because if you are careless, you might lose your arm. Or, think of a falling dropof rain. The drop is moving at a high velocity. The wind pressure opposing the fall will beproportional to the square of the velocity. The term which becomes important in this case isthe second term on the RHS of Eq. (3.1), and the damping force can be given as:

fd = −c2x2 (3.2)

3.1.2 Viscous Damping

When the oscillating object is moving slow, then the term that takes precedence is the firstterm on the RHS of Eq. (3.1), which represents the viscosity related damping and is pro-portional to the velocity of the moving object. To appreciate this force, think of an iron bobslowly drowning in a pool of oil. The upward force felt by the bob will come from the viscosityof the oil and this force can be given as:

fd = −c1x (3.3)

For all practical purposes and to keep the mathematics simple, we will consider Eq. (3.3) tomodel damping in real systems. Since we will only be using one form of damping, therefore,we will use the notation fd = −bx for viscous damping force with b acting as the viscousdamping coefficient.

24 3.2. A SPRING MASS SYSTEM WITH DAMPING

3.2 A Spring Mass System with Damping

The system in Fig. 3.1 will now be considered to model damping into oscillating systems. Amass is vertically hung from a spring and a damper as shown in Fig. 3.1. Before attaching themass m to the system, the spring and the damper are loosely hanging from the fixed support.The unstretched coordinate of the system is denoted by x which is the vertical coordinate.It is zero in this unstretched position (x = 0). This position is indicated in Fig. 3.1. In thisstate the system is in static equilibrium. Applying Newton’s 2nd Law on this system yields:

∑

x

Fx = 0

The mass m is then connected to the system. The force of gravity pulls the mass toward theearth which results in a deflection δ of the mass. The system is still in static equilibrium,therefore, the sum of the forces acting on the system is again zero. However, the old coordinatex is no longer zero. Its value now is x = δ. So, we define a new coordinate x whose value iszero at x = δ as shown in the figure. Looking at Fig. 3.1 one can immediately recognize the

x = 0

x = δx = 0 m

k

fg = mg

fs = −kxfd = −bxb

Figure 3.1: A Vertical Spring Mass System with a Damper

equivalence between the two coordinates as shown with Eq. (3.4).

x = x− δ (3.4)

There now are two coordinates x and x with the former representing the position of the massfrom the unstretched spring position and the latter is the coordinate of the mass from thestretched postion. In the stretched position when the mass is deflected by δ, the balance offorces in the vertical direction gives:

mg = kδ (3.5)

3.2. A SPRING MASS SYSTEM WITH DAMPING 25

3.2.1 Equation of Motion from Unstretched Position

The mass in Fig. 3.1 is pulled down and let go. It starts moving away from the directionof the pull and towards the equilibrium position x = δ. As of now it is not known whetherdue to damping the system will come to a complete stop without crossing the equilibriumposition or the system will oscillate a little before coming to a halt. The behaviour of the massunder the action of the pull will be discussed later. For now the equation of motion is obtained.

It is now clear that the coordinate at the unstretched position is given by x. From thispositon, writing the balance of forces in the vertical direction produces the following:

mx = −bx− kx+mg

mx+ bx+ kx = mg (3.6)

The non zero RHS of Eq. (3.6) implies that when the mass finally cones to rest, its positionwill be x = δ and not x = 0.

3.2.2 Equation of Motion from the Stretched Position

Now the coordinate at the stretched position namely x is considered. Instead of rewriting thebalance of forces, the second equation in Eq. (3.6) is utilized in conjunction with the value ofx in terms of x. From Eq. (3.4) it can be seen that x can be rewritten in terms of x:

x = x+ δ (3.7)

Substituting this value of x in Eq. (3.6) results in the following:

mx+ bx+ kx = mg

m¨x+ b ˙x+ k (x+ δ) = mg

m¨x+ b ˙x+ kx+ kδ = mg

m¨x+ b ˙x+ kx = 0 (3.8)

Eq. (3.8) has been simplified by considering that mg = kδ and also the fact that the first andthe second time derivatives of δ are zero and therefore, with mx and bx, the deflection δ doesnot appear.

With two coordinates it will be a hassle to keep track of them when both the coordinatesrepresent the motion of the same object. A more preferable approach is to keep the samecoordinate x for both the cases and only concentrate on the concept. The concept is explainedagain as follows:

When the equation of motion is derived from the unstretched spring position, the RHShas the gravitational force mg.

mx+ bx+ kx = mg

When the equation of motion is derived from the stretched spring position, the RHSdoes not have the gravitational force term.

mx+ bx+ kx = 0

26 3.3. DAMPED OSCILLATIONS SOLUTION

3.3 Damped Oscillations Solution

Now the distinction between the unstretched and stretched postions is clear, the latter(Eq. (3.8)) which is also the simpler one can be divided by m and rewritten to produce thefollowing:

x+ γx+ ωo2x = 0 (3.9)

It should be clear that using the stretched version of the equation, the force of gravity termmghas already been canceled out. γ is defined as the normalized damping coefficient whose valueis given as b/m. ωo

2 is the square of the natural frequency given by k/m. It is worth notingthat the dimensions of γ and ωo are the same. Table (3.1) highlights important parametersin Eq. (3.9) and their interpretations.

Description Symbol Equals

Normalized Damping Coefficient γ b/m

Natural Frequency ωo

√

k/m

Table 3.1: Eq. (3.9) Parameters and their Equivalence.

There are many ways of solving Eq. (3.9). One of those consists of considering the solutionfunction x(t) to be complex valued. However, since the system lives in the physical world,any transformation that is done to convert the real world displacement coordinate x into acomplex numbered coordinate has to be reverted back. A simple way of expressing such atransformation is given as shown below:

x(t) = ℜ(z(t)) (3.10)

z(t) is a complex valued displacement coordinate. Eq. (3.9) can therefore be rewritten as:

z + γz + ωo2z = 0 (3.11)

Next the complex valued solution z(t) is further expressed as:

z(t) = Aeiat (3.12)

Substituting Eq. (3.12) into Eq. (3.11) results in the following:

(

−a2 + iaγ + ωo2)

Aeiat = 0 (3.13)

The only non trivial solution is obtained by putting the terms in brackets on the LHS equal tozero. However, care must be taken in separating the real and imaginary parts as the quantityin the bracket is complex valued. Doing just the same gives rise to:

−a2 + ωo2 = 0

aγ = 0 (3.14)

It is impossible to solve Eq. (3.16) because the set of equations given by Eq. (3.16) suggeststwo values for the arbitrary constant a and a cannot have two contradictory values at the

3.3. DAMPED OSCILLATIONS SOLUTION 27

same time. There is only one way out of this confusion. That is, a has to be complex itselfand that shall resolve this problem. In other words a is considered to be:

a = α+ iβ (3.15)

Rewriting the bracketed expression in Eq. (3.13) in terms of Eq. (3.15), the following isachieved:

− (α+ iβ)2 + i (α+ iβ) γ + ωo2 = 0

−α2 + β2 − i2αβ + iγα− βγ + ωo2 = 0 (3.16)

It is apparent that Eq. (3.16) has complex numbered parameters. In order to solve thisequation, the real and imaginary parts have to be written separately. Writing these partsseparately means both parts would identically equal to zero. These parts are given below:

i (2αβ − γα) = 0

−α2 + β2 − βγ + ωo2 = 0 (3.17)

Solving Eq. (3.17) is pretty straight forward. Doing that results in the following:

β =γ

2(3.18)

α2 = ωo2 +

γ2

4− γ2

2

α = ±√

ωo2 − γ2

4(3.19)

α has two roots, one with plus sign and the other with minus sign. The desired function zwill, therefore, also have two possible solutions, one according to the plus root and the otheraccording to the minus root. The solution to z is thus given as:

z(t) = A1ei(α1+iβ)t +A2e

i(α2+iβ)t (3.20)

The above solution may be written in simplified form as:

z(t) = A1e−βteiα1t +A2e

−βteiα2t (3.21)

3.3.1 Solution Analysis

At no stage should the reader ever forget that z(t) is a complex valued solution to Eq. (3.8).The presence of complex exponential also affirms this. In addition to this, there is no re-striction on the two arbitrary constants A1 and A2 to be real valued. These two constantscould also be complex valued. However, oscillatory systems under study live in real worldand this fact leads to the question: Is a complex valued solution acceptable? The answer isobviously no and a simple solution to this inconvinience is to only consider the real part ofthe solution while discarding the imaginary part. This could mathematically be representedby reproducing Eq. (3.10) as shown below:

x(t) = ℜ (z(t)) (3.22)

28 3.4. THE GREAT BATTLE: DAMPING VS. OSCILLATION

3.3.2 Roots of Differential Equation

The complex roots of the differential equation Eq. (3.11) in terms of the constants α and βare given in Eq. (3.19) and Eq. (3.18) respectively. In general the behaviour of the systemdepends on the values of these constants. In particular the value of α decides if the systembehaves like an oscillatory system or not. The value of β determines the sharpness of theexponentially decaying envelope. Examples of such envelopes are given in Fig. 3.2. Observingthe figure makes evident that larger the value of β the quicker the curve goes to zero.

t

f(t) = e−βt

β = 0.5β = 1β = 3

Figure 3.2: Examples of Exponentially Decaying Envelopes for Different Values of β

The other part of complex roots appears in the form of α. This constant has two possiblevalues, one postive and the other negative as shown in Eq. (3.19). But there really is more toα than meets the eye. In fact this constant may also assume values based on the discriminantpresent under the square root. The value of this discriminant decides the way in which thesystem would oscillate. This is dealt with in details in the next section.

3.4 The Great Battle: Damping vs. Oscillation