Introduction

1

Examples

Examples

Equation of motion

Newton's 2nd law of motion

The rate of change of

momentum of a mass is equal to the force acting

on it.

Newton's 2nd law of motion

Energy conservation

0

constant

UTdt

d

UT

Potential energy

2

2

1kxU

Kinetic energy 2.

2

1xmT

0..

kxxm

0....

..2

2

kxxmxmkxtF

xmdt

txdm

dt

txdm

dt

dtF

Vertical system

0..

; ..

kxxm

kmgmgxkxm stst

Solution to the equation of motion:

0

or

0 0)(

:yields (1)eqation intosolution thesubstitute

solution. of roots theare andconstant is B where)(

: type theofsolution thehasmay (1)equation osolution t

frequency natural theasknown theis / where

0

as, written becan (1)Equation equation. aldifferentiordinary

order 2nd homogenuos ait (1), 0

22

22

2

n

ststn

st

n

n

s

eBes

SBetx

mk

xx

kxxm

(2)

Then

)( and Put

sin)(cos)()(

,sincos

: thatNote

BB)(

:as written be nowcan (1)equation osolution t The

.

212211

2121

21

tsinωAtcosωAx(t) n2n1

BBjABBA

tBBjtBBtx

Therefore

tjte

eetx

jsei

nn

nntj

tjtj

n

n

nn

motion. theof applying

from obtained becan ) and (or and constants The :

.frequancy and amplitudean of

motion harmonic a of consists system theof vibrationfree hesay that tcan we

angle. phase theis tan

andmotion of amplitude theis Where

(3)

:form in the written be alsocan (1)equation osolution t The

21

n

1

21

22

21

nditionsinitial co

AAANote

A

A

A

AAA

φ)tAcos(ωx(t) n

. any timeat ion consideratunder system

theofnt displaceme of valueobtain the toused becan (4)Equation

tan and

:become angle phase theand amplitude The

)........(4 sincos)(

:becomessolution general then the and

:yields (3)or (2)solution general theinto conditions

thesegSustitutin . )( and )( ; 0at Let

:conditions Inatial

1

2

20

0

0

.

201

0

..

0

t

XA

ttXtx

XAXA

XtxXtxt

nn

n

n0

0

.

n

0

.

n

0

.

ωXX

ωX

φ

ωX

A

Graphical representation x(t ) = A cos(ωnt – ϕ )

Torsional vibration

rad/sec.

as obtained becan frequency natural which thefrom and

motion ofequation theis 0

0

:law 2 Newtons Applying

ineria. ofmoment polar theis 32

where

as; calculated isshaft theof

stiffness that theNote shown. as ineria ofmoment of

disk a and K stiffness torsionalofshaft a has systemA

n

nd

4

t

D

t

tDDt

D

pp

t

D

J

K

KJJK

JM

dI

L

GIK

J

Natural Frequency (ωn) :

It is a system property. It depends, mainly, on the stiffness and the mass of vibrating system.

It has the units rad./sec, or cycles/sec. (Hz)

It is related to the natural period of oscillation (τn) such that, τn = 2π/ωn

and ωn = 2 π fn where fn is the natural frequency in Hz.

Example 2.1 solution:

Initial assumptions:

1. the water tank is a point mass

2. the column has a uniform cross section

3. the mass of the column is negligible

Example 2.1 solution:

a. Calculation of natural frequency:

1. Stiffness: , But:

So:

2. Natural frequency :

3

3

l

EIk 44444 3475.24.23

6464mddI io

mNxxx

k / 812,28990

3475.2103033

9

srxm

kn /ad 9829.0

103

812,2895

Example 2.1 solution:

b. Finding the response:

1. x(t ) = A cos (ωn t - ϕ )

So, x(t ) = 0.3 cos (0.9829 t )

mXX

XA on

o

o 3.0

. 2

2

0

0tan

.

tan 11

nono

o

XX

X

Example 2.1 solution:

c. Finding the max. velocity:

Finding the max. acceleration :

0.9829)*/(2then t

/20.9829t when maximum is )(

9829.0sin9829.03.0.

tx

ttx

smx /2949.09829.03.0.

max

9829.0/then t

9829.0 when maximum is ..

9829.0cos9829.03.0..

2

ttx

ttx

22max /2898.09829.03.0

..smx

Example 2.2 :

mm

kk

kk

m

m

eqneq

21

21eq

eqeq

2121

2211

21

k isfrequency natural theand 0,xkx

thenismotion ofequation the,4

1k

:becomes system theof stiffness equivalent for x the

ngsubstitutiby then x,k wsuch that k stiffness and

mass of system spring mass a toequivalent is system The

k

2w

k

2w2or x 2x2x x

thatNote g. w where xk 2w and xk 2w

: thatshow

pulleys theof diagramsbody free The ly.respective xand xare

2 and 1 pulleys of centres theofnt displaceme Let the

Example 2.2 : solution

Simple pendulum

Governing equation:

Assume θ is very small

Natural frequency (ωn)

.. OOO JJM

0sin..

mglJO

sin

0..

mglJO

mgl

J

J

mgl O

On 2

Solution

n

oon

o

AAt

At

.

,.

0.

0

22

1

ttt nn

o

no sin

.

cos

tAtAt nn sincos 21

0sin..

WdJO

0..

WdJO

OOn J

mgd

J

Wd

Example 2.5: Q2.45

Draw the free-body diagram and derive the equation of motion using Newton s second law of motion for each of the systems shown in Fig