1Lecture 5-Energy MethodMEEM 3700 1

MEEM 3700MEEM 3700Mechanical VibrationsMechanical Vibrations

Mohan Rao Chuck Van Karsen

Mechanical Engineering-Engineering MechanicsMichigan Technological University

Copyright 2003

Lecture 5-Energy MethodMEEM 3700 2

M

K

X

Single Degree of Freedom Free VibrationSingle Degree of Freedom Free Vibration--Energy Energy MethodMethod

M

K

XM

K

Given an initial condition, Determine the resulting motion

2Lecture 5-Energy MethodMEEM 3700 3

m

K

X

Energy Method: Conservation of Energy

m X

K.E. +P.E. = constant

T + U = constant

21T= mX2

&

Total Kinetic energy in the system

Lecture 5-Energy MethodMEEM 3700 4

Energy Method: Conservation of Energy

KU = Total potential energy in the system

U = potential energy in the spring +change in potential energy due to elevation

21U=mgx+ Kx -mgx2

X

3Lecture 5-Energy MethodMEEM 3700 5

Energy Method: Conservation of Energy

T + U = constant

Therefore the rate of change of system energy must be zero

0=+dt

)UT(d

2 21 1T+U= mx + Kx2 2

&

d(T+U) 2 2= mxx+ Kxx=0dt 2 2

&&& &&&

Lecture 5-Energy MethodMEEM 3700 6

0mx Kx+ =&&Equation of motion

Energy Method: Conservation of Energy

0=22+2

2=+ XKXXXmdt

)UT(d &&&&

tsin)(xtcos)(x)t(x nn

n 0+0= &

Solution to equation of motion (as before)

n

eqn eq

KmKor m

=

=eq = equivalent values

4Lecture 5-Energy MethodMEEM 3700 7

The classic pendulum problemThe classic pendulum problem--Energy MethodEnergy Method

m

L Kinetic Energy

2

21= L

dt)(sindmT

Determine the equation of motionand the natural frequency

for small angles sin

( )221= LmT &

Case 1: Energy method

Lecture 5-Energy MethodMEEM 3700 8

m

L

Potential energy U, is proportional tothe change in elevation of the mass

)cosLL(mgU =cosL L

5Lecture 5-Energy MethodMEEM 3700 9

m

L

constant UT =+( ) )cosLL(mgLmUT +21=+ 2&

0=+dt

)UT(d

0=+22 2 &&&& sinmgLLm

0=+ gL && Equation of motion

Lg

n =

Lecture 5-Energy MethodMEEM 3700 10

The classic pendulum problemThe classic pendulum problem--NewtonNewtons Methods MethodCase 2: Newtons method

m

L

= xmF && = &&JM

6Lecture 5-Energy MethodMEEM 3700 11

Free body diagram

L

mmg

T

mgcos

mgsin

mg

Lecture 5-Energy MethodMEEM 3700 12

Free body diagram

mg

T

mgsin

= &&JM= &&JLsinmg0=+2 mgLmL &&

0=+ gL &&L

gn =

7Lecture 5-Energy MethodMEEM 3700 13

m

Connecting barMass M, length L

Pendulum problem: mass of bar is consideredPendulum problem: mass of bar is considered

o Kinetic Energy

2 2 2

2 22

2 2

1 12 2

2 12 212 3

bar bob o

o cg

T T T J mL

L ML LJ J M M

MT m L

= + = + = + = +

= +

& &

&

Lecture 5-Energy MethodMEEM 3700 14

m

Connecting barMass M, length L

Pendulum problem: mass of bar is consideredPendulum problem: mass of bar is considered

Potential Energy

( cos ) ( cos )2 2

bar bobU U UL LU Mg mg L L

= += +

8Lecture 5-Energy MethodMEEM 3700 15

Equation of Motion( )

2 2

2

2

0

1 (1 cos ) (1 cos ) 02 3 2

sin 03 2

for small angles sin

03 2

d T Udtd M Lm L Mg mgLdt

M Mm L m gL

M Mm L m gL

+ = + + + =

+ + + = =

+ + + =

&

&&

&&

2

3

n

Mm g

Mm L

+ = +

Lecture 5-Energy MethodMEEM 3700 16

Compound Pendulum problemCompound Pendulum problem

o

pivotd

cg

0oJ mgd + =&&

0 0M J = &&0sinmg d J = &&