1Lecture 3-SDOF UndampedEOM

MEEM 3700 1 of 16

MEEM 3700MEEM 3700Mechanical VibrationsMechanical Vibrations

Mohan D. Rao Chuck Van Karsen

Mechanical Engineering-Engineering MechanicsMichigan Technological University

Copyright 2003

Lecture 3-SDOF UndampedEOM

MEEM 3700 2 of 16

m

k

xm

k

xm

Given some initial conditions, Determine the resulting motion

Key Points: System is un-damped No external forces Only vertical motion

2Lecture 3-SDOF UndampedEOM

MEEM 3700 3 of 16

m

k

x

At rest, x = 0 (static equilibrium)

m x

mg

k

mg = k

Lecture 3-SDOF UndampedEOM

MEEM 3700 4 of 16

m

k

xm x

mg

kx+k

Maintains DynamicEquilibrium

Note: x is measured from the static equilibrium position.

3Lecture 3-SDOF UndampedEOM

MEEM 3700 5 of 16

m X

mg

+ kkxFree Body Diagram

Apply Newtons 2nd Law

= xmF &&

xm)kkx(mg &&=+ =+ xmFx &&

0=+ kxxm &&Equation of motion (EOM)

Lecture 3-SDOF UndampedEOM

MEEM 3700 6 of 16

0=+ kxxm &&Equation of motion2nd order Differential equationhomogeneous

constant coefficientslinear

Forms of solution:

st We will use this form

x(t)=X sin(t + )x(t)=X cos(t - )x(t)=Ce ---

4Lecture 3-SDOF UndampedEOM

MEEM 3700 7 of 16

0=+ kxxm &&Equation of motionstCe)t(x Assume =

st

st

Ces)t(xsCe)t(x

2==

&&&

0=+2 stst kCeCems0=+2 stCe)kms(

0=+2 kmssolutiontrivial-non afor

Lecture 3-SDOF UndampedEOM

MEEM 3700 8 of 16

0=+ kxxm &&Equation of motion

2ms + k = 0 1,2ks = jm

1 2s t s t1 2x(t) = C e + C e

k kj -jm m

1 2x(t) = C e + C et t

C1 and C2 are arbitrary constants to be determined from initial conditions

5Lecture 3-SDOF UndampedEOM

MEEM 3700 9 of 16

k kj - j m m

1 2x(t) = C e + C et t

( )je =cos jsin()Recall Eulers identity:1 2

k k k kx(t) = C cos + j sin + C cos - j sin m m m mt t t t

1 2 1 2k kx(t) = (C + C ) cos + j (C - C ) sin m mt t

k kx(t) = A cos + B sin m mt t

A & B are always real since C1 and C2 areComplex conjugates

Lecture 3-SDOF UndampedEOM

MEEM 3700 10 of 16

k kx(t) = A cos t + B sin tm m

0 1 2 3 4 5 6 7 8 9 10-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

time - seconds

amplitude X

case 1: imaginary roots

T

nk 2

= = = natural frequency (rad/sec)m T

6Lecture 3-SDOF UndampedEOM

MEEM 3700 11 of 16

n

1 cyclesf = = natural frequency , or HzT sec

( ) ( )n nx(t) = A cos t + B sin t

( ) ( )n nx(t) = A cos 2f t + B sin 2f t

Lecture 3-SDOF UndampedEOM

MEEM 3700 12 of 16

( ) ( )n nx(t) = A cos t + B sin tA and B are determined from the Initial Conditions

x(0)

x(0)&

( ) ( )x(0) = A cos 0 + B sin 0 01

x(0) = A

( ) ( )n n n nx(t) = - A sin t + B cos t&( ) ( )n nx(0) = - A* 0 + B 1&

n

x(0) = B&

0

7Lecture 3-SDOF UndampedEOM

MEEM 3700 13 of 16

( ) ( )n nn

x(0)x(t) = x(0) cos t + sin t&

( ) ( )n nx(t) = A cos t + B sin t

0 1 2 3 4 5 6 7 8 9 10-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

time - seconds

amplitu

de X

case 1: imaginary roots

X(0)

)(x 0&

Lecture 3-SDOF UndampedEOM

MEEM 3700 14 of 16

( )( ) cos nx t X t =

( ) ( )2

22 2 00n

xX A B x

= + = + & ( )

( )10

tan0 n

xx

=

&

AmplitudePhase

Different forms of the same solution:

8Lecture 3-SDOF UndampedEOM

MEEM 3700 15 of 16

( )( ) sin nx t X t = +

2

2 2 2 (0)(0)n

XX A B X = + = +

( )( )10

tan0

nxx

= &

AmplitudePhase

Different forms of the same solution:

Lecture 3-SDOF UndampedEOM

MEEM 3700 16 of 16

Equation of Motion:

Solution Form:

o tJ + k = 0&&

( ) ( )n n(t)=Acos t +Bsin t

Compare with Translatory SystemEquation of Motion:

Solution Form: ( ) ( )n nx(t)=Acos t +Bsin tm + k x = 0x&&

tn

k 2Here = = = natural frequency (rad/sec)J T

nk 2 = = m T