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Topic E: Oscillations

CONTENT1. Introduction

1.1 Overview

1.2 Degrees of freedom

1.3 Simple harmonic motion

2. Undamped free oscillation2.1 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 oscillation3.1 The equation of motion

3.2 General solution for different damping levels

4. Forced oscillation4.1 Mathematical expression of the problem

4.2 Static load

4.3 Undamped forced oscillation

4.4 Damped forced oscillation

OVERVIEW

● Oscillations are the time-dependent response of systems

to disturbances in the presence of restoring forces.

● Applications include:

– earthquakes;

– human-induced oscillations: bridges; grandstands;

– flow-induced oscillations;

– unbalanced rotating machinery.

● Systems have (one or more) natural frequencies; these

depend on:

– the strength of restoring forces (stiffness);

– the resistance to change of motion (inertia).

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OVERVIEW

● Oscillations may be:

– free: initial perturbation only;

– forced: continuous periodic forcing.

● Resonance occurs if the forcing frequency is close to a

natural frequency.

● Frictional damping removes energy:

– under-damped (oscillates, with diminishing amplitude);

– over-damped (never oscillates).

DEGREES OF FREEDOM

● The degrees of freedom are the small number of

parameters which are just sufficient to describe the

system configuration.

● They are usually displacements 𝑥 or angles θ.

mg

L

T

x

kx

Mass-spring Simple pendulum

Degree of freedom: 𝑥 Degree of freedom: θ

Restoring force from spring Restoring torque from gravity

DEGREES OF FREEDOM

● Some systems have several modes of oscillation; e.g.

stretching, twisting, deflecting. Each has its own

associated degree of freedom.

● The equations of motion are the differential equations

describing the time variation of these parameters.

● In this course we consider only single-degree-of-freedom

(SDOF) systems.

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SIMPLE HARMONIC MOTION (SHM)

When the forces resulting from a displacement are:

‒ opposite in direction

‒ proportional in size

to the displacement, there is a sinusoidal variation with

time. This is called simple harmonic motion (SHM).

Model: generalised mass-spring systemx

kx

(k is stiffness)

(m is inertia)Equation of motion:

SHM equation:

ω is the natural circular frequency

𝐹𝑥 = −𝑘𝑥

𝑚d2𝑥

dt2= −𝑘𝑥

d2𝑥

dt2= −ω2𝑥 ω2 =

𝑘

𝑚

PERIOD OF OSCILLATION

Solutions are sinusoidal with 2 arbitrary constants:

SHM equation:

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

Period of oscillation:

Frequency:

d2𝑥

dt2= −ω2𝑥

𝑥 = 𝐴 sin(ω𝑡 + ϕ)

𝑥 = 𝐶 sinω𝑡 + 𝐷 cosω𝑡

𝑇 =2𝜋

ω

𝑓 =1

𝑇=

ω

2𝜋

EXAMPLE

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

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AMPLITUDE AND PHASE

Alternative forms of solution:

SHM equation:

(𝐴 = amplitude; = phase)

These are easily interconverted by expanding sin(ω𝑡 + ϕ) and

equating coefficients:

d2𝑥

dt2= −ω2𝑥

𝑥 = 𝐴 sin(ω𝑡 + ϕ)

𝑥 = 𝐶 sinω𝑡 + 𝐷 cosω𝑡

𝐴 = 𝐶2 + 𝐷2

ϕ = tan−1( Τ𝐷 𝐶)

EXAMPLE

Write the following expressions in amplitude / phase-angle

form,

(a)

(b)

12cos 3𝑡 + 5 sin 3𝑡

4 cos 2𝑡 − 3 sin 2𝑡

𝐴 sin(ω𝑡 + ϕ)

BOUNDARY CONDITIONS

General solution:

SHM equation:

Constants 𝐶 and 𝐷 are fixed by boundary conditions

– usually, initial displacement 𝑥0 and velocity d𝑥

d𝑡 0.

d2𝑥

dt2= −ω2𝑥

𝑥 = 𝐶 sinω𝑡 + 𝐷 cosω𝑡

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SPECIAL CASES

1. Start from rest at displacement 𝐴:

2. Start from equilibrium position with initial velocity 𝑣0:

𝑥 = 𝐴 cosω𝑡

𝑥 = 𝐴 sinω𝑡 𝑣0 = 𝐴ω

EXAMPLE

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.

10 kg

100 N/m

60 N/m

x

VELOCITY AND ACCELERATION

Displacement:

Velocity:

since

Maximum velocity:

Maximum acceleration:

Maximum displacement:

𝑥 = 𝐴 sin(ω𝑡 + ϕ)

𝑣 = 𝐴ωcos(ω𝑡 + ϕ)

𝑣2 +ω2𝑥2 = 𝐴2ω2 cos2 θ + sin2 θ = 1

𝑣2 = ω2(𝐴2 − 𝑥2)

𝑥max = 𝐴

𝑣max = ω𝐴

𝑎max = ω2𝐴

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EXAMPLE

(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.

DISPLACEMENT FROM EQUILIBRIUM

m

k

x

m

mg

kx

Equilibrium:

Equation of motion:

Change variables: . Note that

SHM ... about the equilibrium position

𝑚𝑔 = 𝑘𝑥𝑒 ֜ 𝑥𝑒 =𝑚𝑔

𝑘

𝑚d2𝑥

d𝑡2= −𝑘𝑥 +𝑚𝑔

= −𝑘(𝑥 −𝑚𝑔

𝑘)

𝑋 = 𝑥 −𝑚𝑔

𝑘

d2𝑋

d𝑡2=d2𝑥

d𝑡2

d2𝑋

d𝑡2= −

𝑘

𝑚𝑋

EXAMPLE

A block of mass 16 kg is suspended vertically by two light

springs of stiffness 200 N m–1.

By writing 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

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EXAMPLE

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

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.

m

2kk

EXAMPLE

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 Nm−1 and 𝑘𝑅 = 150 Nm−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.

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APPROXIMATE SHM

For some systems the equation of motion is not precisely

SHM … but approximately so for small displacements.

This is particularly true of rotational displacements.

Approximations:

(θ in radians)

sin

1

sin θ ≈ θ

cos θ ≈ 1 (sometimes 1 − 12θ2)

COMPOUND PENDULUM

Simple pendulum:

– all mass concentrated at one point;

– can be treated by linear or rotational dynamics.

linear rotational

Compound pendulum:

– distributed mass;

– must be treated by rotational dynamics.L

Mg

A

G

𝐹 = 𝑚d2𝑥

d𝑡2𝑇 = 𝐼

d2θ

d𝑡2

COMPOUND PENDULUM

L

Mg

A

G

torque = moment of inertia × angular acceleration

Small-angle approximation:

Compare

SHM with

𝑇 = 𝐼d2θ

d𝑡2

−𝑀𝑔 × 𝐿 sin θ = 𝐼d2θ

d𝑡2

d2θ

d𝑡2= −

𝑀𝑔𝐿

𝐼sin θ

sin θ ≈ θ

d2θ

d𝑡2= −

𝑀𝑔𝐿

𝐼θ

d2θ

d𝑡2= −ω2θ ω =

𝑀𝑔𝐿

𝐼

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EXAMPLE

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

ROTATIONAL OSCILLATIONS DRIVEN BY

ELASTIC FORCES

axis r

r

Spring extension:

Force (magnitude):

Torque:

𝑥 = 𝑟θ

𝐹 = 𝑘𝑥 = 𝑘(𝑟θ)

𝑇 = −𝐹𝑟 = −𝑘𝑟2θ

EXAMPLE

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

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SHM EQUATION FROM ENERGY PRINCIPLES

For an ideal mass-spring system:

1

2𝑚𝑣2 +

1

2𝑘𝑥2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

d

d𝑡(1

2𝑚𝑣2 +

1

2𝑘𝑥2) = 0

𝑚𝑣d𝑣

d𝑡+ 𝑘𝑥

d𝑥

d𝑡= 0

𝑚d𝑥

d𝑡

d2𝑥

d𝑡2+ 𝑘𝑥

d𝑥

d𝑡= 0

𝑚d2𝑥

d𝑡2+ 𝑘𝑥 = 0

d2𝑥

d𝑡2+ω2𝑥 = 0 ω2 =

𝑘

𝑚

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

length 𝐿. 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 .

(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.

M

LL

DAMPED FREE OSCILLATION

A freely oscillating system subject to friction may be:

• underdamped: does oscillate, but with decaying

amplitude and reduced frequency;

• critically damped: oscillation just suppressed; fast

return to equilibrium;

• overdamped: no oscillation; slow return to equilibrium.

-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

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LINEAR DAMPING MODEL

Damping force proportional, and of opposite sign, to velocity:

Equation of motion:

Anticipate decaying oscillation:

(provided damping is small)

m

k

xc

𝐹𝑑 = −𝑐d𝑥

d𝑡

−𝑘𝑥 − 𝑐d𝑥

d𝑡= 𝑚

d2𝑥

d𝑡2

𝑚d2𝑥

d𝑡2+ 𝑐

d𝑥

d𝑡+ 𝑘𝑥 = 0

d2𝑥

d𝑡2+ (

𝑐

𝑚)d𝑥

d𝑡+ ω2𝑥 = 0

൯𝑥 = 𝐴e−λ𝑡sin(ω𝑑𝑡 + ϕ

ω2 =𝑘

𝑚

EXAMPLE

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.

20 kg

c = 40 N/(m/s)

k = 180 N/m

x

SOLUTION OF THE DAMPED EQUATION

Homogeneous, linear equation:

Look for exponential solutions:

auxiliary equation:

Roots:

damping ratio

General solution:

d2𝑥

d𝑡2+ (

𝑐

𝑚)d𝑥

d𝑡+ ω2𝑥 = 0

𝑥 = 𝐴e𝑝𝑡

𝑝2 +𝑐

𝑚𝑝 +ω2 = 0

𝑝 =−𝑐𝑚±

𝑐𝑚

2− 4ω2

2= ω −

𝑐

2𝑚ω±

𝑐

2𝑚ω

2

− 1

=𝑐

2𝑚ω𝑝 = −ω ±ω 2 − 1

𝑥 = 𝐴e𝑝1𝑡 + 𝐵e𝑝2𝑡

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SOLUTION OF THE DAMPED EQUATION

General solution:

Three types of solution:

1. two complex conjugate roots if < 1;

2. two distinct real roots if > 1;

3. two equal (real) roots if = 1.

Where roots are complex conjugates:

𝑥 = 𝐴e𝑝1𝑡 + 𝐵e𝑝2𝑡

𝑝 = −ω ±ω 2 − 1 =𝑐

2𝑚ω

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

DIFFERENT DAMPING LEVELS

Case 1. ζ < 1: under-damped system

Decaying oscillation with reduced frequency

Case 2. ζ > 1: over-damped system

No oscillation; slow decay to zero

Case 3. ζ = 1: critically-damped system

No oscillation; rapid return to zero

𝑥 = 𝐴e𝑝1𝑡 + 𝐵e𝑝2𝑡 𝑝 = −ω ±ω 2 − 1 =𝑐

2𝑚ω

൯𝑥 = 𝐴e−λ𝑡sin(ω𝑑𝑡 + ϕ

λ = ω =𝑐

2𝑚

𝑥 = 𝐴e−λ1𝑡 + 𝐵e−λ2𝑡

𝑥 = (𝐴 + 𝐵𝑡)e−ω𝑡

ω𝑑 = ω 1− 2

λ1,2 = ω( ∓ 1− 2)

DIFFERENT DAMPING LEVELS

-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

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EXAMPLE

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?

40 kg

c = 60 N s/m

k = 700 N/m

ω =𝑘

𝑚

=𝑐

2𝑚ω

ω𝑑 = ω 1− 2

FORCED OSCILLATIONSystems allowed to oscillate at their own natural frequency

undergo free oscillation.

Systems subject to externally-imposed periodic forcing

undergo forced oscillation.

Examples of forced oscillation:

‒ concert halls and stadiums;

‒ bridges (traffic- and wind-induced oscillations);

‒ earthquakes.

The system will oscillate at the forcing frequency…

… However, the natural frequency and the damping ratio

affect the amplitude (and the phase) of response. When

forcing and natural frequencies are close there will be a very

large-amplitude response: resonance.

MATHEMATICAL MODEL

m

k

xc

F sin t0

General solution: complementary function (which decays) +

particular integral

The long-term behaviour is that of the particular integral, and

is a sum of multiples of sin Ω𝑡 and cosΩ𝑡.

𝑚d2𝑥

d𝑡2= −𝑘𝑥 − 𝑐

d𝑥

d𝑡+ 𝐹0 sin Ω𝑡

d2𝑥

d𝑡2+𝑐

𝑚

d𝑥

d𝑡+ ω2𝑥 =

𝐹0𝑚sin Ω𝑡 ω2 =

𝑘

𝑚

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COMPARISON – STATIC LOAD

m

k

xc

F sin t0

Comparison: steady force of same magnitude (static load):

d2𝑥

d𝑡2+𝑐

𝑚

d𝑥

d𝑡+ω2𝑥 =

𝐹0𝑚sin Ω𝑡 ω2 =

𝑘

𝑚

ω2𝑥 =𝐹0𝑚

𝑥𝑠 =𝐹0𝑚𝜔2 =

𝐹0𝑘

UNDAMPED FORCED OSCILLATION

Trial solution:

mk

x

F sin t0

response displacement

under static loadmagnification factor

d2𝑥

d𝑡2+ ω2𝑥 =

𝐹0𝑚sinΩ𝑡 ω2 =

𝑘

𝑚

𝑥 = 𝐶 sinΩ𝑡

−𝐶Ω2 sinΩ𝑡 + ω2𝐶 sinΩ𝑡 =𝐹0𝑚sinΩ𝑡

𝐶(ω2 − Ω2) =𝐹0𝑚

𝐶 =𝐹0

𝑚(ω2 −Ω2)=

𝐹0𝑚ω2(1 − Ω2/ω2)

=𝑥𝑠

1 − Ω2/ω2

𝐶 = 𝑀𝑥𝑠 𝑥𝑠 =𝐹0𝑚ω2 =

𝐹0𝑘

𝑀 =1

1 − ΤΩ2 ω2

UNDAMPED FORCED OSCILLATION

mk

x

F sin t0

● The response of the system to external forcing depends on the ratio of the

forcing frequency Ω to the natural frequency ω.

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

natural frequency (Ω → ω).

● 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.

● 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.

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

𝑥 = 𝐶 sinΩ𝑡 𝐶 =𝑥𝑠

1 − Ω2/ω2

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EXAMPLE

Write down the general solution of the

following differential equation:

d2𝑥

d𝑡2+ 4

d𝑥

d𝑡+ 5𝑥 = 130sin 2𝑡

DAMPED FORCED OSCILLATION

Trial solution:

m

k

xc

F sin t0

displacement

under static load

undamped

natural frequency

damping

ratio

d2𝑥

d𝑡2+𝑐

𝑚

d𝑥

d𝑡+ ω2𝑥 =

𝐹0𝑚sinΩ𝑡 ω2 =

𝑘

𝑚

𝑥 = 𝐶 sinΩ𝑡 + 𝐷 cosΩ𝑡

𝐶 =1 − ΤΩ2 ω2

1 − ΤΩ2 ω2 2 + Τ2Ω ω 2 𝑥𝑠, 𝐷 =−2 ΤΩ ω

1 − ΤΩ2 ω2 2 + Τ2Ω ω 2 𝑥𝑠

𝑥𝑠 =𝐹0𝑚ω2 =

𝐹0𝑘

=𝑐

2𝑚ω𝜔 =𝑘

𝑚

DAMPED FORCED OSCILLATION

m

k

xc

F sin t0

● The system response depends on both the ratio of forcing to natural

frequencies (Ω/ω) and the damping ratio .

● Damping prevents complete blow-up (𝑀 → ∞) as Ω → ω.

● The imposed frequency at which maximum amplitude oscillations occur is

slightly less than the undamped natural frequency ω.

● The phase lag varies from 0 (as Ω → 0) to π (as Ω → ∞).

)𝑥 = 𝑀𝑥𝑠sin(Ω𝑡 − ϕ

𝑀 =1

1 − ΤΩ2 ω2 2 + Τ2Ω ω 2

tan ϕ =2 ΤΩ ω

1 − ΤΩ2 ω2

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EXAMPLE

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

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