topic e: oscillations spring 2019 · correspond to masses and springs. a ... two fixed...

Mechanics Topic E (Oscillations) - 1 David Apsley

TOPIC E: OSCILLATIONS SPRING 2020

1. Introduction

1.1 Overview

1.2 Degrees of freedom

1.3 Simple harmonic motion

2. Undamped free oscillation

2.1 Generalised mass-spring system: simple harmonic motion

2.2 Natural frequency and period

2.3 Amplitude and phase

2.4 Velocity and acceleration

2.5 Displacement from equilibrium

2.6 Small-amplitude approximations

2.7 Derivation of the SHM equation from energy principles

3. Damped free oscillation

3.1 The equation of motion

3.2 General solution for different damping levels

4. Forced oscillation

4.1 Mathematical expression of the problem

4.2 Static load

4.3 Undamped forced oscillation

4.4 Damped forced oscillation


1. INTRODUCTION

1.1 Overview

Many important dynamical problems arise from the oscillation of systems responding to

applied disturbances in the presence of restoring forces. Examples include:

• response of a structure to earthquakes;

• human-induced structural oscillations (e.g. stands at sports venues; footbridges);

• flow-induced oscillations (e.g. chimneys, pipelines, power lines);

• vibration of unbalanced rotating machinery.

Most systems when displaced from a position of equilibrium have one or more natural

frequencies of oscillation which depend upon the strength of restoring forces (stiffness) and

resistance to change of motion (inertia). If the system is left to oscillate without further

influence from outside this is referred to as free oscillation.

On the other hand, the examples above are mainly concerned with forced oscillation; that is,

oscillations of a certain frequency are imposed on the system by external forces. If the

applied frequency is close to the natural frequency of the system then there is considerable

accumulation of energy and resonance occurs, with potentially catastrophic consequences.

In general, all systems are subject to some degree of frictional damping that removes energy.

A freely-oscillating system may be under-damped (oscillates, but with gradually diminishing

amplitude) or over-damped (so restricted that it never oscillates). When periodic forces act on

structures, damping is a crucial factor in reducing the amplitude of oscillation.

1.2 Degrees of Freedom Many systems have several modes of oscillation. For example, a long power line or

suspended bridge deck may “gallop” (bounce up and down) or it may twist. Often the

geometric configuration at any instant can be defined by a small number of parameters:

usually displacements 𝑥 or angles θ. Parameters which are just sufficient to describe the

geometric configuration of the system are called degrees of freedom. Here, for simplicity, we

shall restrict ourselves to single-degree-of-freedom (SDOF) systems. The equation of motion

then describes the variation of that parameter with time.

Examples of SDOF dynamical systems and their degree of freedom are:

• mass suspended by a spring (vertical displacement);

• pivoted body (angular displacement).

In both systems, oscillations occur about a position of static equilibrium in which applied

forces or moments are in balance.

1.3 Simple Harmonic Motion

For many systems the forces arising from a small displacement are opposite in direction and

proportional in size to the displacement. The equation of motion is particularly simple and

solutions take the form of a sinusoidal variation with time. This ubiquitous and important

type of oscillation is known as simple harmonic motion (SHM).


2. UNDAMPED FREE OSCILLATION The oscillatory motion of a system displaced from stable equilibrium and then allowed to

adjust in the absence of externally-imposed forces is termed free oscillation. If there are no

frictional forces the motion is called undamped free oscillation.

2.1 Generalised Mass-Spring System: Simple Harmonic Motion This is a general model for a linear free-oscillation problem. It doesn’t physically have to

correspond to masses and springs.

A generalised mass-spring system is one for which the resultant

force following displacement from equilibrium is a function of the

displacement and of opposite sign (i.e. a restoring force). For ideal

springs (or, in practice, for small displacements) the relationship is

linear:

𝐹𝑥 = −𝑘𝑥 (1)

𝑘 is called the stiffness or spring constant. It is measured in newtons per metre (N m–1). The

equation of motion is then

𝑚d2𝑥

d𝑡2= −𝑘𝑥 (2)

To solve (2) note that d2𝑥/d𝑡2 = −(𝑘/𝑚)𝑥 and that 𝑘/𝑚 is a positive constant. Thus, it is

common to write the equation as

Simple Harmonic Motion (SHM) Equation:

d2𝑥

d𝑡2= −ω2𝑥 or

d2𝑥

d𝑡2+ ω2𝑥 = 0 (3)

where

ω = √𝑘

𝑚 = √

stiffness

inertia (4)

To solve (3) look for solutions 𝑥(𝑡) whose second derivative is proportional to the original

function, but of opposite sign. The obvious candidates are sine and cosine functions.1 The

general solution (with two arbitrary constants) may be written in either of the forms:

𝑥 = 𝐴sin(ω𝑡 + ϕ) (5)

𝑥 = 𝐶 sin ω𝑡 + 𝐷 cos ω𝑡 (6)

Any system whose degree of freedom evolves sinusoidally at a single frequency is said to

undergo simple harmonic motion (SHM). ω is called the natural circular frequency and is

measured in radians per second (rad s–1).

1 In your mathematics classes, when you have covered complex numbers, you will encounter a third form of the

general solution involving complex exponentials: 𝑥 = 𝐴𝑒iω𝑡 + 𝐵𝑒−iω𝑡

x

kx


2.2 Natural Frequency and Period

One complete cycle is completed when ω𝑡 changes by 2π. Hence:

Period of

oscillation: 𝑇 =

2π

ω (7)

Frequency: 𝑓 =1

𝑇=

ω

2π in cycles per second or hertz (Hz) (8)

Important note.

It is common in theoretical work to refer to ω rather than 𝑓 as the natural frequency, because:

(a) it avoids factors of 2π in the solution; (b) it is ω rather than 𝑓 which appears in the

governing equation. Thus, one should be quite careful about precisely what is being referred

to as “frequency”; usually it will be obvious from the notation or units: ω in rad s–1, 𝑓 in

cycles s–1 (Hz).

The following example demonstrates that SHM can occur without any elastic forces.

Example 1. (Meriam and Kraige)

Two fixed counter-rotating pulleys a distance 0.4 m apart are driven at the same angular

speed ω0. A bar is placed across the pulleys as shown. The coefficient of friction between bar

and pulleys is μ = 0.2. Show that, provided the angular speed ω0 is sufficiently large, the bar

may undergo SHM and find the period of oscillation.

0.4 m

x

mg0 0


2.3 Amplitude and Phase

The general solution of the SHM equation (two arbitrary constants) may be written as one of:

𝑥 = 𝐴 sin(ω𝑡 + ϕ)

𝑥 = 𝐶 sin ω𝑡 + 𝐷 cos ω𝑡

The first is called the amplitude/phase-angle form (𝐴 = amplitude, ϕ = phase). Whichever

form is more convenient may be used. They are easily interconverted as follows.

Expand the amplitude/phase-angle form:

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

Compare with the second form and equate coefficients of sin ω𝑡 and cos ω𝑡 to obtain

𝐶 = 𝐴 cos ϕ , 𝐷 = 𝐴 sin ϕ

Eliminating ϕ and 𝐴 in turn gives:

amplitude √𝐶2 + 𝐷2 (9)

phase angle ϕ = tan−1(𝐷/𝐶) (10)

Note that there are two alternative values of ϕ (in opposite quadrants) with the same value of

tan ϕ. These must be distinguished by the individual signs of C and D.

Example 2.

Write the following expressions in amplitude/phase-angle form, 𝐴 sin(ω𝑡 + ϕ):

(a) 12 cos 3𝑡 + 5 sin 3𝑡

(b) 4 cos 2𝑡 − 3 sin 2𝑡

In the general solution the two free constants can be determined if initial boundary conditions

are given for displacement 𝑥0 and initial velocity (d𝑥/d𝑡)0. There are two special cases:

(1) Start from rest at displacement 𝐴:

𝑥 = 𝐴 cos ω𝑡

(2) Start from the equilibrium position with initial velocity 𝑣0:

𝑥 = 𝐴 sin ω𝑡 where 𝑣0 = 𝐴ω

-1

0

1

t

2p

cost

sint

p 3p/2p/2


Example 3.

For the system shown, find: (a) the equivalent single spring; (b) the natural circular frequency

ω; (c) the natural frequency of oscillation 𝑓; (d) the period of oscillation; (e) the maximum

speed of the cart if it is displaced 0.1 m from its position of equilibrium and then released.

2.4 Velocity and Acceleration

From the general solution in phase-angle form,

𝑥 = 𝐴 sin(ω𝑡 + ϕ)

Differentiation then gives for the velocity:

𝑣 = 𝐴ω cos(ω𝑡 + ϕ)

Squaring and adding, using cos2θ + sin2θ = 1:

𝑣2 + ω2𝑥2 = 𝐴2ω2

Hence, we can find the velocity at any given position in the cycle:

𝑣2 = ω2(𝐴2 − 𝑥2) (11)

Note also the maximum displacement, velocity and acceleration:

𝑥𝑚𝑎𝑥 = 𝐴 (12)

𝑣𝑚𝑎𝑥 = ω𝐴 (13)

𝑎𝑚𝑎𝑥 = ω2𝐴 (14)

Example 4. (Exam 2018)

(a) A boat is riding the waves in high seas, oscillating with simple harmonic motion in a

vertical line. The period of oscillation is 7 seconds, and the height of the boat varies

between 2 m and 8 m below a nearby pier. Find:

(i) the maximum speed,

(ii) the maximum acceleration,

of the boat during the oscillations.

(b) A box of mass 15 kg sits on the deck of the boat. By considering the forces on the box

and the acceleration that it is undergoing, find the normal contact force from the deck

on the box:

(i) at the top,

(ii) at the bottom,

(iii) in the middle of the oscillation.

10 kg

100 N/m

60 N/m

x


2.5 Displacement From Equilibrium

For many oscillating systems it is the displacement from a position of static equilibrium that

is important, rather than the absolute displacement.

The equilibrium position and the oscillation about it can be obtained simultaneously by:

(1) writing down the equation of motion for any convenient degree of freedom;

(2) identifying the point of equilibrium as the point where the acceleration is zero;

(3) rewriting the equation of motion in terms of the displacement from this point.

Consider a mass 𝑚 suspended by a spring. The equilibrium extension 𝑥𝑒

could easily be obtained by balancing weight and spring forces:

𝑚𝑔 = 𝑘𝑥𝑒 ⇒ 𝑥𝑒 =𝑚𝑔

𝑘

Alternatively, writing down the general equation of motion in terms of

the spring extension:

𝑚d2𝑥

d𝑡2= −𝑘𝑥 + 𝑚𝑔

= −𝑘(𝑥 −𝑚𝑔

𝑘)

From this, the position of equilibrium can be identified by setting acceleration d2𝑥/d𝑡2 = 0.

Change variables to

𝑋 = 𝑥 −𝑚𝑔

𝑘 = 𝑥 − 𝑥𝑒 = displacement from equilibrium

and note that, since 𝑥𝑒 is constant, d2𝑋/d𝑡2 = d2𝑥/d𝑡2. Then:

d2𝑋

d𝑡2= −

𝑘

𝑚𝑋

● A constant force (such as gravity) changes the position of equilibrium.

● A linear-restoring-force system undergoes SHM about the equilibrium position.

Example 5.

A block of mass 16 kg is suspended vertically by two light springs of stiffness 200 N m–1.

By writing down the equation of motion, find: (a) the equivalent single spring; (b) the

extension at equilibrium; (c) the period of oscillation about the point of equilibrium.

16 kg

k = 200 N/mk = 200 N/m

m

k

x

m

mg

kx


Example 6.

A 4 kg mass is suspended vertically by a string of elastic modulus λ = 480 N and unstretched

length 2 m. What is its extension in the equilibrium position? If it is pulled down from its

equilibrium position by a distance 0.2 m, will it undergo SHM?

Example 7.

A mass 𝑚 is hung in the loop of a light smooth cable whose two ends are fixed to a

horizontal support by springs of stiffness 𝑘 and 2𝑘 (see figure). Find the period of vertical

oscillations in terms of 𝑚 and 𝑘.

Oscillation about a point of equilibrium can also be observed in multi-spring systems where

components are already loaded at equilibrium, as in the following example.

Example 8.

A particle of mass 0.4 kg is confined to move along a smooth horizontal plane between two

points A and B a distance 2 m apart by two light springs, both of natural length 0.8 m. The

springs connecting the particle to A and B have stiffnesses 𝑘𝐿 = 50 N m–1 and 𝑘𝑅 =150 N m–1 respectively.

(a) Write down the equation of motion of the particle in terms of the distance 𝑥 from A.

(b) Find the position of equilibrium.

(c) Show that, if released from rest half way between the walls, the particle undergoes

simple harmonic motion and calculate:

(i) the period of oscillation;

(ii) the maximum speed and maximum acceleration.

m

2kk


2.6 Small-Amplitude Approximations

Many oscillatory systems do not undergo exact simple harmonic motion (where the restoring

force is proportional to a displacement), but their motion is approximately SHM provided that

the amplitude of oscillation is small. Such small-amplitude approximations are particularly

common for rotational motion about a fixed point (where the restoring torque is often

provided by gravity or elasticity) and rely on the approximations

sinθ ≈ θ (15)

cosθ ≈ 1 (sometimes, 1 −1

2θ2) (16)

when θ is measured in radians. These may be derived

formally by power-series expansions, but their essential

validity is easily seen geometrically (see right).

The following table shows that the approximation for sin θ is accurate to about 1% or better

for angles as large as 15˚. If a structural member were actually displaced by this much then

oscillation would be the least of your worries!

θ (degrees) 5 10 15 20

θ (radians) 0.0873 0.175 0.262 0.349

sin θ 0.0872 0.174 0.259 0.342

Important warning: duplication of notation

When considering rotational oscillations we will run into problems with the same symbols

being used for different quantities.

• 𝑇 is used for both torque and period of oscillation (and sometimes tension); in this

section we will use it to mean torque and write “period” out in full.

• ω is used for both natural circular frequency and angular velocity; while dealing with

oscillations we shall reserve it to mean the natural circular frequency and write θ̇ or

dθ/d𝑡 for the angular velocity.

2.6.1 Rotational Oscillations Driven by Gravity – Compound Pendulum

In a simple pendulum all the mass is concentrated at one point. This can be treated using

either the momentum or angular-momentum equations. In a compound pendulum the mass is

distributed; the system must be analysed by rotational dynamics.

In the absence of friction, two forces act on the body: the weight of the body

(which acts through the centre of gravity) and the reaction at the axis. Only

the former has any moment about the axis.

Let the distance from axis to centre of gravity be 𝐿 and let the angular

displacement of line AG from the vertical be θ.

The line of action of the weight 𝑀𝑔 lies at a distance 𝐿 sin θ from the axis of rotation, and

hence imparts a torque (moment of force) of magnitude 𝑇 = 𝑀𝑔 × 𝐿 sinθ in the opposite

L

Mg

A

G

sin

1


sense to θ. The rotational equation of motion is

torque = moment of inertia × angular acceleration

𝑇 = 𝐼d2θ

d𝑡2

Hence,

−𝑀𝑔 × 𝐿 sin θ = 𝐼d2θ

d𝑡2

For small oscillations, sin θ ≈ θ, so that

d2θ

d𝑡2= −

𝑀𝑔𝐿

𝐼θ

This is SHM with natural circular frequency

ω = √𝑀𝑔𝐿

𝐼

Example 9. (Exam 2017)

A uniform circular disk of mass 3 kg and radius 0.4 m is suspended from a horizontal axis

passing through a point on its circumference and perpendicular to the plane of the disk. A

small particle of mass 2 kg is attached to the other side of the disk at the opposite end of a

diameter.

(a) Find the moment of inertia of the combination about the given axis.

(b) Find the period of small oscillations about the axis.

Axis

3 kg

2 kg


2.6.2 Rotational Oscillations Driven By Elastic Forces

If the rotational displacement θ is small then the additional extension of a spring at distance 𝑟

from an axis is essentially the length of circular arc:

𝑥 = 𝑟θ

This gives a force of magnitude

𝐹 = 𝑘𝑥 = 𝑘(𝑟θ)

opposing the displacement and hence a torque of

magnitude 𝐹𝑟 acting in the opposite direction to the

rotational displacement:

𝑇 = −𝑘𝑟2θ

Example 10.

A uniform bar of mass 𝑀 and length 𝐿 is allowed to pivot about a horizontal axis though its

centre. It is attached to a level plane by two equal springs of stiffness 𝑘 at its ends as shown.

Find the natural circular frequency for small oscillations.

L

k k

axis r

r


2.7 Derivation of the SHM equation from Energy Principles

For a body of mass 𝑚, subject only to elastic forces2 with stiffness 𝑘, total energy is constant:

1

2𝑚𝑣2 +

1

2𝑘𝑥2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Differentiating with respect to time gives

d

d𝑡(1

2𝑚𝑣2 +

1

2𝑘𝑥2) = 0

Apply the chain rule to each term:

𝑚𝑣d𝑣

d𝑡 + 𝑘𝑥

d𝑥

d𝑡= 0

Replacing 𝑣 by d𝑥/d𝑡:

𝑚d𝑥

d𝑡

d2𝑥

d𝑡2 + 𝑘𝑥

d𝑥

d𝑡= 0

Dividing by d𝑥/d𝑡:

𝑚d2𝑥

d𝑡2+ 𝑘𝑥 = 0

Hence,

d2𝑥

d𝑡2+ ω2𝑥 = 0, ω2 =

𝑘

𝑚

Example 11.

A sign of mass 𝑀 hangs from a fixed support by two rigid rods of negligible mass and length

𝐿 (see below). The rods are freely pivoted at the points shown, so that the sign may swing in

a vertical plane without rotating, the rods making an angle θ with the vertical.

(a) Write exact expressions for the potential energy and kinetic energy of the sign in

terms of 𝑀, 𝐿, 𝑔, the displacement angle θ and its time derivative θ̇.

(b) If the sign is displaced an angle θ = π/3 radians and then released, find an expression

for its maximum speed.

(c) Find an expression for the total (i.e. kinetic + potential) energy using the small-angle

approximations sin θ ≈ θ, cos θ ≈ 1 − 1

2θ2.

(d) Show that, for small-amplitude oscillations, the assumption of constant total energy

leads to simple harmonic motion, and find its period.

2 Actually, far more general. For small displacements about a stable equilibrium point in an arbitrary potential:

𝑉 = 𝑉0 +𝑥−𝑥0

1!(

d𝑉

d𝑥)

0+

(𝑥−𝑥0)2

2!(

d2𝑉

d𝑥2)0

+ ⋯ = constant +1

2𝑘(𝑥 − 𝑥0)2 + ⋯

the last because, for stable equilibrium, (d𝑉 d𝑥⁄ )0 = 0, whilst (d2𝑉 d𝑥2⁄ )0 is a positive constant.

M

LL


3. DAMPED FREE OSCILLATION

All real dynamical systems are subject to friction, which opposes relative motion and

consumes mechanical energy. For a system undergoing free oscillation, we shall show that:

• moderate damping decaying amplitude and reduced frequency;

• large damping oscillation prevented.

The level at which oscillation is just suppressed is called critical damping.

3.1 The Equation of Motion

Friction always acts in a direction so as to oppose relative motion. For a frictional force that

depends on velocity, the damping force is often modelled as

𝐹𝑑 = −𝑐d𝑥

d𝑡 (17)

𝑐 is the viscous damping coefficient. If 𝑥 is a displacement then 𝑐 has units of N s m–1. In

practice, 𝑐 may vary with velocity, but a useful analysis may be conducted by assuming it is a

constant, in which case the damping is termed linear.

The equation of motion (“𝐹 = 𝑚𝑎”) for a damped mass-spring system is

−𝑘𝑥 − 𝑐d𝑥

d𝑡= 𝑚

d2𝑥

d𝑡2

or

𝑚d2𝑥

d𝑡2+ 𝑐

d𝑥

d𝑡+ 𝑘𝑥 = 0 (18)

Dividing by 𝑚, this can be written

d2𝑥

d𝑡2+ (

𝑐

𝑚)

d𝑥

d𝑡+ ω2𝑥 = 0 (19)

where ω = √𝑘/𝑚 is the natural frequency of the undamped system.

Example 12. (Exam, May 2016)

A carriage of mass 20 kg is attached to a wall by a spring of stiffness 180 N m–1. When

required, a hydraulic damper can be attached to provide a resistive force with magnitude

proportional to velocity; the constant of proportionality 𝑐 = 40 N/(m s–1).

(a) If the carriage is displaced, write down its equation of motion, in terms of the

displacement 𝑥, in the case when the hydraulic damper is in place.

(b) If there is no damping, find the period of oscillation.

(c) If the hydraulic damper is attached, find the period of oscillation and the fraction by

which the amplitude is reduced on each cycle.

Do this question from first principles, rather than the damping formulae that follow.

m

k

xc

20 kg

c = 40 N/(m/s)

k = 180 N/m

x


3.2 General Solution For Different Damping Levels

The undamped system (𝑐 = 0) has solutions of the form 𝐴 sin(ω𝑡 + ϕ). In the presence of

damping we expect the solution to decay in magnitude and, possibly, have a slightly different

frequency. Hence we might anticipate solutions of the form

𝑥 = 𝐴e−λ𝑡sin(ω𝑑𝑡 + ϕ) (20)

where 𝐴 and ϕ are arbitrary constants and λ and ω𝑑 are to be found. If you don’t like the

analysis which follows, you can try simply substituting this into equation (19) and (after quite

a lot of algebra) deriving the same results as below.

Equation (19) is a homogeneous, linear, second-order differential equation with constant

coefficients. Seek solutions of the form:

𝑥 = 𝐴e𝑝𝑡

where 𝑝 is a constant to be evaluated. Substituting in (19) we obtain the auxiliary equation

𝑝2 +𝑐

𝑚𝑝 + ω2 = 0

with roots

𝑝 =

−𝑐𝑚 ± √(

𝑐𝑚)2 − 4ω2

2 = ω [−

𝑐

2𝑚ω± √(

𝑐

2𝑚ω)2 − 1]

(21)

(For convenience) define

ζ =𝑐

2𝑚ω (damping ratio) (22)

Then

𝑝 = −ωζ ± ω√ζ2 − 1 (23)

As you know from your maths course, there are 3 possibilities, depending on the sign of the

quantity under the square root (here, ζ2 − 1) in equation (23):

1. two complex conjugate roots if ζ < 1;

2. two distinct negative real roots if ζ > 1;

3. two equal (negative, real) roots if ζ = 1.

For complex conjugate roots (𝑝 = 𝑝𝑟 ± i𝑝𝑖)the general solution can be written

𝑥 = 𝐴e𝑝𝑟𝑡+i𝑝𝑖𝑡 + 𝐵e𝑝𝑟𝑡−i𝑝𝑖𝑡 = e𝑝𝑟𝑡(𝐶 cos 𝑝𝑖𝑡 + 𝐷 sin 𝑝𝑖𝑡)

where 𝐴 and 𝐵 (or 𝐶 and 𝐷) are arbitrary constants. The last bracket also has an equivalent

amplitude/phase-angle form.


Case 1: ζ < 1 (𝑐 < 2𝑚ω): under-damped system

The general solution may be written

𝑥 = 𝐴e−λ𝑡sin(ω𝑑𝑡 + ϕ) (24)

where

ω𝑑 = ω√1 − ζ2 λ = ωζ =

𝑐

2𝑚 (25)

There is oscillation with reduced amplitude (decaying exponentially as e−λ𝑡) and reduced

frequency, ω𝑑.

The amplitude reduction factor over one cycle (ω𝑑𝑡 = 2π or 𝑡 = 2π/ω𝑑) is

exp {−2𝜋ζ

√1 − 2

} (26)

This may be used to determine the damping ratio ζ experimentally. Case 1 also includes the

special case of no damping (ζ = 0), in which case λ = 0 and ω𝑑 = ω.

Case 2: ζ > 1 (𝑐 > 2𝑚ω): over-damped system

The general solution is of the form

𝑥 = 𝐴e−λ1𝑡 + 𝐵e−λ2𝑡 (27)

where

λ1,2 = ω(ζ ∓ √1 − 2) (28)

There is no oscillation and, as both of the roots 𝑝 are negative (i.e. λ positive), the

displacement decays to zero, more slowly as the damping ratio ζ increases.

Case 3: ζ = 1 (𝑐 = 2𝑚ω): critically-damped system

The general solution has the form:

𝑥 = (𝐴 + 𝐵𝑡)e−ω𝑡 (29)

There is no oscillation and the amplitude decays rapidly to zero.


Summary

The damped mass-spring equation

𝑚

d2𝑥

d𝑡2+ 𝑐

d𝑥

d𝑡+ 𝑘𝑥 = 0

has solutions that depend on:

undamped natural frequency: ω = √𝑘

𝑚

damping ratio: ζ =𝑐

2𝑚ω

● If ζ < 1 the system is under-damped, oscillating with exponentially-decaying

amplitude and reduced frequency ω𝑑 = ω√1 − ζ2.

● If ζ > 1 the system is over-damped and no oscillation occurs.

● The fastest return to equilibrium occurs when the system is critically damped (ζ = 1).

Example.

𝑚 = 1 kg, 𝑘 = 64 N m–1

ω = 8 rad s–1

𝑥0 = 0.05 m, (d𝑥/d𝑡)0 = 0.

Cases in the graph:

𝑐 = 1.6 N s m–1 (ζ = 0.1)

𝑐 = 16 N s m–1 (ζ = 1)

𝑐 = 64 N s m–1 (ζ = 4)

Exercise. Use any other computer program or tool to compute and plot the solution for

various combinations of 𝑚, 𝑘, 𝑐.

Example 13.

Analyse the motion of the system shown. What is the damping ratio? Does it oscillate? If so,

what is the period? What value of 𝑐 would be required for the system to be critically damped?

-0.05

-0.04

-0.03

-0.02

-0.01

0.00

0.01

0.02

0.03

0.04

0.05

0.0 0.5 1.0 1.5 2.0 2.5 3.0

dis

pla

cem

en

t (m

)

time (s)

over-damped

criticallydamped

under-damped

40 kg

c = 60 N s/m

k = 700 N/m


4. FORCED OSCILLATION

Systems that oscillate about a position of equilibrium under restoring forces at their own

“preferred” or natural frequency are said to undergo free oscillation.

Systems that are perturbed by some externally-imposed periodic forcing are said to undergo

forced oscillation. Examples which may be encountered in civil engineering are:

• concert halls and stadiums;

• bridges (e.g. traffic- or wind-induced oscillations);

• earthquakes.

The natural frequency and damping ratio are important parameters for systems responding to

externally-imposed forces. Large-amplitude oscillations occur when the imposed frequency is

close to the natural frequency (the phenomenon of resonance). In general, the system’s free-

oscillation properties affects both amplitude and phase of response to external forcing.

4.1 Mathematical Expression of the Problem

The general form of the equation of motion with harmonic forcing is

𝑚d2𝑥

d𝑡2= −𝑘𝑥 − 𝑐

d𝑥

d𝑡+ 𝐹0 sin Ω𝑡 (30)

or

d2𝑥

d𝑡2+

𝑐

𝑚

d𝑥

d𝑡+ ω2𝑥 =

𝐹0

𝑚sin Ω𝑡 (31)

where the undamped natural frequency is given by

ω2 =𝑘

𝑚

Second-order ODEs of the form (31) are covered in your mathematics courses. The general

solution is the sum of a complementary function (containing two arbitrary constants and

obtained by setting the RHS of (31) to 0) and a particular integral (any particular solution of

the equation, obtained by a trial function based on the RHS). The complementary function is

a free-oscillation solution and, if there is any damping at all, will decay exponentially with

time. The large-time behaviour of the system, therefore, is determined by the particular

integral which, from the form of the RHS, should be a combination of sin Ω𝑡 and cos Ω𝑡.

Example 14. Write down the general solution of the following differential equation:

d2𝑥

d𝑡2+ 4

d𝑥

d𝑡+ 5𝑥 = 130 sin 2𝑡

m

k

xc

F sin t0


4.2 Static Load

A useful comparison is with the displacement under a steady load of the same amplitude 𝐹0,

rather than a variable load 𝐹0 sin Ω𝑡. The steady-state displacement 𝑥𝑠 is given by the

position of static equilibrium:

𝑘𝑥𝑠 = 𝐹0

whence

𝑥𝑠 =𝐹0

𝑘 =

𝐹0

𝑚ω2 (32)

4.3 Undamped Forced Oscillation

In the special case of no frictional damping, the equation

of motion is

d2𝑥

d𝑡2+ ω2𝑥 =

𝐹0

𝑚sin Ω𝑡

The form of the forcing function suggests a particular integral of the form 𝑥 = 𝐶 sin Ω𝑡. By

substituting this in the equation of motion to find 𝐶, one finds that

𝐶 =𝐹0

𝑚(ω2 − Ω2) =

𝑥𝑠

1 − Ω2/ω2 (33)

where, as above, 𝑥𝑠 = 𝐹0/𝑚ω2 is the displacement under a static load of the same amplitude.

Hence the magnitude of forced oscillations is 𝑀𝑥𝑠, where the amplitude ratio or

magnification factor 𝑀 is given by

𝑀 = |1

1 − Ω2/ω2| (34)

Key Points

(1) The response of the system to external forcing depends on the ratio of the forcing

frequency Ω to the natural frequency ω.

(2) There is resonance (𝑀 → ∞) if the forcing frequency approaches the natural

frequency (Ω → ω).

(3) If Ω < ω the oscillations are in phase with the forcing (𝐶 has the same sign as 𝐹0),

because the system can respond fast enough.

(4) If Ω > ω the oscillations are 180 out of phase with the forcing (𝐶 has the opposite

sign to 𝐹0) because the imposed oscillations are too fast for the system to follow.

(5) If Ω ≫ ω (very fast oscillations) the system will barely move (𝐶 → 0).

mk

x

F sin t0


4.4 Damped Forced Oscillation

The complete analysis of oscillations for a SDOF system involves forced motion and

damping. The equation of motion is

d2𝑥

d𝑡2+ (

𝑐

𝑚)

d𝑥

d𝑡+ ω2𝑥 =

𝐹0

𝑚sin Ω𝑡

Because of the d𝑥/d𝑡 term, a particular integral must contain both sin Ω𝑡 and cos Ω𝑡.

Substituting a trial solution of the form

𝑥 = 𝐶 sin Ω𝑡 + 𝐷 cos Ω𝑡

produces, after a lot of algebra, a solution (optional exercise) of the form

𝐶 =1 − Ω2/ω2

(1 − Ω2/ω2)2 + (2ζΩ/ω)2𝑥𝑠 , 𝐷 =

−2ζΩ/ω

(1 − Ω2/ω2)2 + (2ζΩ/ω)2𝑥𝑠 (35)

where, as in Section 3, the damping ratio is

ζ =𝑐

2𝑚ω

and the displacement under static load is

𝑥𝑠 = 𝐹0/𝑚ω2

This is most conveniently written in the amplitude/phase-angle form

𝑥 = 𝑀𝑥𝑠sin(Ω𝑡 − ϕ) (36)

where the amplitude ratio 𝑀 is given by

𝑀 =√𝐶2 + 𝐷2

𝑥𝑠 =

1

√(1 − Ω2/ω2)2 + (2ζΩ/ω)2 (37)

and the phase lag ϕ by

tanϕ = −𝐷

𝐶 =

2ζΩ/ω

1 − Ω2/ω2 (38)

Key Points

(1) The response of the system to forcing

depends on both the ratio of forcing to

natural frequencies (Ω/ω) and the damping

ratio ζ.

(2) Damping prevents the complete blow-up

(𝑀 → ∞) as Ω → ω.

(3) The imposed frequency at which the

maximum amplitude oscillations

(resonance) occur is slightly less than the

undamped natural frequency ω. In fact,


Ωmax = ω√1 − 2ζ2

(4) The phase lag varies from 0 (as Ω → 0) to 𝜋 (as Ω → ∞). However, the phase lag is

always π/2 when Ω = ω, irrespective of the level of damping.

Example 15. (Meriam and Kraige, modified)

The seismometer shown is attached to a structure which has a horizontal harmonic oscillation

at 3 Hz. The instrument has a mass m = 0.5 kg, a spring stiffness 𝑘 = 150 N m–1 and a

viscous damping coefficient 𝑐 = 3 N s m–1. If the maximum recorded value of 𝑥 in its

steady-state motion is 5 mm, determine the amplitude of the horizontal movement 𝑥𝐵 of the

structure.

x (t)B

mc

x

k

topic e: oscillations spring 2019 · correspond to masses and springs. a ... two fixed...

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