Lecture 41 DOF: Free Undamped Vibrations

MCG 4308Mechanical Vibration Analysis

1 DOF, free undamped oscillator

•Although a simple model, it represents ALL 1DoF, linear, lumped parameter engineering systems with mass and stiffness

•See Lecture 1 example on equivalent mass and stiffness

•See table of equivalent masses/springs of common engineering elements

Mass

Mk

xDisplacement

Spring

Equation of Motion (Newton)

( )0

0

0

( ) ( )unstretched length

( ) ( )

my t k y t ll

my t ky t kl

= − −

=

⇒ + =

( ) ( )( ) ( ) 0

mx t kx tmx t kx t

= −⇒ + =

Mass

Mk

x

Spring

y

simpler

Which is best coordinate?x or y?

Energy Methods

An alternative way to determine the equation of motion

the natural frequency of a system (often without even having to find the equation of motion first!)

Useful if the forces or torques acting on the object or mechanical part are difficult to determine

Potential and Kinetic Energy

The potential energy of mechanical systems V is often stored in “springs” (remember that for a spring F=-kx)

212springV kx=

The kinetic energy of mechanical systems T is due to the motion of the “mass” in the system

212transT mx=

Mk

x0

Mass Spring

x=0

Equation of motion from conservation of Energy

( )

0

0

021

21 22

=+

=+⇒

=

+=+

kxxmx

kxxmx

kxxmdtdUT

dtd

time all for zero be cannot Since

)(

Mk

x

Mass Spring

x=0For a simple, conservative (i.e. no damper), mass spring system the energy must be conserved:

constant

or ( ) 0

T Vd T Vdt

+ =

+ =

Derivation of the solution

2

2

2

1 2

1 2

Substitute into 0

0

0

Natural Frequency

( ) and ( )

( )

n n

n n

t

t t

n

n

jt jt

jt jt

x Ce mx kx

m Ce kCe

m k

k k j jm m

km

x t C e x t C e

x t C e C e

λ

λ λ

ω ω

ω ω

λ

λ

λ ω

ω

−

−

= + =

+ =

+ =

= ± − = ± = ±

= ⇔

= =

= +

This approach will be used again for more complicated problems

Standard Forms:

1 2

1 2

( ) sin( )( ) sin cos

( ) n n

n

n nj t j t

x t A tx t A t A t

x t C e C eω ω

ω φω ω

−

= +

= +

= +

2

2

( ) ( ) 0

( ) ( ) 0

where

n

n

mx t kx t

x t x t

km

ω

ω

+ =

+ =

=

Standard forms for solution

Use these

Standard form for equation

What can we learn from the solution?

( ) sin( ) nx t A tω φ= +We can write the solution:

Differentiating:

2 2

2

( ) cos( ) = sin( )

Velocity lags displacement by Amplitude multiplied by

( ) sin( ) - ( )

sin( )Acceleration lags displacement by

Amplitude multiplie

n n

n n

n

n n n

n n

x t A tA t

x t A t x t

A t

ω ω φω ω φ π

πω

ω ω φ ω

ω ω φ ππ

= +

+ + /2

/2

= − + =

= + +

2d by nω

t

x(t)

Peak Values

max

max2

max

max or peak value of:displacement: velocity:

acceleration: n

n

x Ax A

x A

ω

ω

=

=

=

Remember, sin and cos always oscillate between +1 and -1. So, we can find the maximum values of displacement, velocity and acceleration. This will be useful later when using Raleigh’s Method.

Relationship between Displacement, Velocity,Acceleration

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-1

0

1

x

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-20

0

20

v

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-200

0

200

Time (sec)

a

A=1, ωn=12

Note how the relative magnitude increases for wn>1

Initial Conditions

0

0

(0) initial displacement(0) initial velocity

x xx v

= →

= →

If a system is vibrating then we must assume that something must have (in the past) transferred energy into to the system and caused it to move. For example the mass could have been:

•moved a distance x0 and then released at t=0 (i.e. given Potential energy) or

•given an initial velocity v0 (i.e. given Kinetic energy) or

•Some combination of the two above cases

General Initial Conditions for Free Vibrations:

Plugging in Initial Conditions

1 2

1 2 2

1 2 1

2 0

01

00

Solution Form: ( ) sin cos(0) sin(0) cos(0)(0) cos(0) sin(0)

(0)(0)

( ) sin cos

n n

n n n

n n

n nn

x t A t A tx A A Ax A A A

A x xvxA

vx t t x t

ω ω

ω ω ω

ω ω

ω ωω

= +

= + == − =

⇒ = =

= =

= +

Phase-shifted form

[ ] [ ]1 2

2

1

02

1 0

2 2 2 2 21 2

Other solution Form: ( ) sin( )( )= sin( ) cos sin( ) sin cos( )

cos sin

sintancos

arctan arctan (1) (Check quadrant!)

sin

n

n n n

n

x t A tx t A t A t A t

A A A A

AA

xAA v

A A A A

ω φ

ω φ φ ω φ ω

φ φ

φφφ

ωφ

φ

= +

+ = +

⇒ = =

⇒ = =

⇒ = =

+ = + 2 2

22 2 2 01 2 0 2

22 00 2

cos

( ) sin( ) with given in (1)

n

nn

A

vA A A x

vx t x t

φ

ω

ω φ φω

=

= + = +

= + +

A note on arctangents

Note that calculating arctangent from a calculator requires some attention. First, most machines work in radians.

The argument atan(-/+) is in a different quadrant then atan(+/-), and usual machine calculations will return an arctangent in between -π/2 and +π/2, reading only the atan(-) for both of the above two cases.

φ

φ+

+

-

_+

+

-

-

Initial Conditions2 2 20 0

1( ) sin( )n nn

x t x v tω ω φω

= + +

φ

01

n

v Aω =

0 2x A=2 2 2

0 01

nn

A x vωω

= +

t

x(t)

0x Slope here is v0

nωφ

Period

n

Tω

π2=

Amplitude A

Maximum Velocity

Anω

0 0

Eof M ( ) ( ) 0

IC: (0) (0)eq eqm x t k x t

x x x v

⇒ + =

= =

2

22 2 201 2 02

Most important things we want to know:

Natural Frequency

Amplitude of Motion

eqn

eq

n

km

vA A A x

ω

ω

⇒ =

= + = +

( )1 2

2 20 21 2 0 1 2

1

Response is ( ) sin cos sin

arctan

n n n

n

x t A t A t A t

v AA A x A A AA

ω ω ω φ

φω

= + = +

= = = + =

Summary of Simple Harmonic Motion

Determined by parameters of structure, k m

Determined by initial conditions and ωn

Example 1: Effect of fuel on frequency of an airplane wing

Model wing as transverse beam

Model fuel as tip mass

Ignore the mass of the wing and see how the frequency of the system changes as the fuel is used up

Example 1: Transverse beam stiffness

Strength of materials and experiments yield:

3

3

3

3

mEI

EIk

n =

=

ω

f

m

x

Mass of pod: 10 kg empty, 1000 kg fullI = 5.2x10-5 m4, E =6.9x109 N/m, l = 2 m

Hence the natural frequency changes by an order of magnitude while it empties out fuel.

ω full = 3EIm3 = 3(6.9 ×109 )(5.2 ×10−5 )

1000 ⋅ 23

= 11.6 rad/s = 1.8 Hz

ωempty =3EIm3 =

3(6.9 ×109 )(5.2 ×10−5 )10 ⋅23

= 115 rad/s =18.5 Hz

Energy Conservation in Oscillatory Motion

Potential energy versus time

Kinetic energy versus time

Note Kinetic Energy is max when Potential energy is minimum and vice versa. Total energy is conserved.

Conservation of Energy

constant

or ( ) 0

T Vd T Vdt

+ =

+ =

For a simple, conservative (i.e. no damper), mass spring system the energy must be conserved:

When displacement is maximum, potential energy is maximum but kinetic energy is minimum .

max min min maxcan usually 0set to 0

max max

so

T V T V

T V

=

+ = +

=

Can use these factsto derive equationof motion and findnatural frequency.

Raleigh’s Method for natural frequency

2 2max max

max max

2 2

2

1 1 ( )2 2

Since 1 1 ( )2 2

n

n

n n

V kA T m A

T V

kA m A

kk mm

ω

ω

ω ω

= =

=

=

⇒ = ⇒ =

•Assume solution x(t)= Asin(wt+φ) (sinusoid)

•Then max x(t) = A, max dx/dt=ωnA.

•Equate maximum potential (Vmax) and kinetic energies (Tmax) to calculate the natural frequency of the system.

Mk

x0x=0

Example:

Raleigh Method: shortcut

2 2

If we use concept of equivalent mass and equivalent stiffnessfor a 1 DOF system with *linear* spring elements, we can write total kinetic and potential energies as:

1 1and 2 2

Then Ral

eq eqT m x V k x= =

2

eigh method essentially says:

eqn

eq

km

ω =

Reminders

Polar moment of inertia is a measure of an object's ability to resist torsion – resistance to twist.

It is analogous to the area moment of inertia, which characterizes an object's ability to resist bending.

Must not be confused with the mass moment of inertia, which characterizes an object's angular acceleration due to a torque – resistance to being rotated.