mechanical vibrations in many mechanical systems: the motion is an oscillation with the position of...

Mechanical Vibrations

In many mechanical systems:The motion is an oscillation with the position of static equilibrium as the center


Examples: http://www.aw-bc.com/ide/idefiles/media/JavaTools/smpharos.html

– The motion of the flywheel in a watch– The suspension of an automobile– A solid structure like a bridge or a concrete weir vibrates due

to the action of wind, water, or earth tremors.


Spring with damping:– Consider a spring of length l– A bob and a damping mechanism (as so-called dashpot, which is just a cylinder filled with fluid, and a piston, much like a shock absorber) are attached to the other end. – The other end is fastened to a vertical hook in such a way thatthe whole system hangs motionless– Then the spring will lengthen to a Length of l+h in this static position, with h a real positive number.


– Let the total mass on the lower end of the spring be m.– When the bob is displaced from the position of static equilibrium and then

projected at a given initial speed in a vertical direction, we would expect the bob to execute an oscillatory motion.

Elastic force of the spring: when a force is applied to the lower end of the spring, one feels intuitively that the spring should bear some relation to the magnitude of the force.

The empirical Law (Hookw’s law due to Robert Hooke (1635-1703): The force necessary to extend a spring or an elastic string of natural length l to a length l+h is proportional to the extension h, provided that h<<l.– The proportionality constant in this law is called the spring constant or stiffness,

usually denote as k.– Modulus: product of the spring constant and the natural length

l k


A Mathematical Model– Variables

• Independent variable: t (time)• State variable: x(t): displacement of the bob from its equilibrium (Here the x-axis

is chosen vertically downward with the origin at the position of the bob when the system is in static equilibrium)

– Assumptions:• The mass of the spring is much smaller than the mass of the bob• The amplitude (i.e.,the maximal displacement from the center) of the oscillations is

so small that it is less than h, and the Hooke’s law is valid.• The magnitude of the resistance to the motion (due to air resistance and the

damping mechanism) is directly proportional to the speed at any time t, and always in the opposite direction to the velocity.


– Construct the model based on the assumption• We ignore the weight of the spring itself.• When the system is hanging in equilibrium, the acceleration is zero.

Hence the resultant force on the bob must also be zero by Newton’s second law, i.e.

• Total forces:– The weight m g which is always downward – The force –k(x+h) which is always upward since |x|<h– The resistance (or damping) –c x’ which is always in the opposite

direction to x, and where c is the proportionality constant– An (possible) external force f(t) directed downwards when its sign is

positive, and upwards when its sign is negative.

k h m g


• Newton’s second law

– Initial value problem (IVP)

• m>0, k>0 & c>=0 are parameters and given• The initial values are given• The equation is second linear ODE with constant coefficients !!!• For given initial data, f(t) and the parameters, there exists a uniqueness

solution of the IVP !!!!!

''( ) ( ) '( ) ( )m x t m g k x h c x t f t

''( ) '( ) ( ) ( ), 0,

(0) , '(0)

m x t c x t k x t f t t

x x

&


Solution of IVP:– Case 1: free motion, i.e. f(t)=0

• Overdamping: • Critical damping: • Underdamping:• No damping (simple harmonic oscillator): c=0

– Case 2: forced motion, i.e. • Undamed motion: c=0

– Resonance• Damped motion: 0c

( ) 0f t

2 4c k m2 4c k m2 4c k m

Solution structures of the ODE

For the homogeneous equation

– If y1(t) is a solution, then a y1(t) is also a solution for any constant a !

– If y1(t) and y2(t) are two nonzero solutions, then c1

y1(t) + c2 y2(t) is also a solution (superposition)– If y1(t) and y2(t) are two linearly independent

solutions, then the general solution is c1 y1(t) + c2 y2(t)

''( ) '( ) ( ) 0, 0,m x t c x t k x t t

Solution structure of the ODE

Solution structure of linear problem

– If y1(t) is a solution of the homogeneous equation, y2(t) is also a specific solution. Then y2(t)+a y1(t) is also a solution for any constant a !

– If y1(t) and y2(t) are two solutions, y1(t)-y2(t) is a solution of the homogeneous equation.

– If y1(t) and y2(t) are two linearly independent solutions of the homogeneous equation, y3(t) is a specific solution, then the

general solution is c1 y1(t) + c2 y2(t)+y3(t)

''( ) '( ) ( ) ( ), 0,m x t c x t k x t f t t


Free motion,i.e.f(t)=0: http://www.aw-bc.com/ide/idefiles/media/JavaTools/massprng.html

– Since the equation is linear with constant coefficients, try the solution in the form

– Plug into the equation, we get the characteristic equation

– The two roots are

''( ) '( ) ( ) 0, 0,

(0) , '(0)

m x t c x t k x t t

x x

( ) with a real or complex number.tx t e

2 0m c k 2 24 4

,2 2

c c k m c c k mp q

m m


– Overdamping:• p and q are two real distinct negative numbers• By superposition, we get the general solution

• To satisfy the initial values, we have

• Solve it, we get

2 4c k m

( ) p t q tx t A e B e

(0) , '(0)x A B x A p B q

,q p

A Bq p p q


• The analytical solution

• Properties– Long time behavior: x(t)0, when t infinity– Question: How many times would the bob pass through the

origin of the x-axis (or the point of static equilibrium)?

( ) ( )( )

p t q tq e p ex t

p q

( )p q pe

q


» Condition for the bob not passing the origin: the bob will tend to the origin when t goes to infinity.

» Condition for the bob passing the origin only one time: the bob would pass only once through the origin and then tend to the origin as t goes to infinity!!

– In the case of overdamping, there is no oscillation of the bob.http://www.aw-bc.com/ide/idefiles/media/JavaTools/massprng.html

2 21| | | | & ( )( ) ( ) 0

. . 3, 1, 1, 3, 1

p q p q m c km

e g m c k

1, . . 0 & 1 or 1& 0p

e gq


– Critical damping:• p=q=-c/2m are two same real negative numbers, we only

have one function which satisfies the equation

• To find another solution, we try and find another function

• By superposition, we get the general solution


2 4c k m

( ) ( ) p tx t A B t e

(0) , '(0)2

cx A x B A

m

( ) p tx t e

( ) p tx t t e




• Properties– Long time behavior: x(t)0, when t infinity– The bob passes through the origin exactly once when

– Otherwise,the bob merely tends to the origin from its initial position – In this case, there is no oscillation of the bob.http://www.aw-bc.com/ide/idefiles/media/JavaTools/massprng.html

,2

cA B

m

2( )2

c t

mc tx t t e

m

20 & 0, . . 2, 1, 1, 1, 2

2 2

c me g c k m

m m c


– Underdamping:• p and q are complex conjugate numbers and we write

• We have two complex solutions

• Since the ODE is homogeneous and linear, linear combination of the two solutions are still a solution, we get two real solutions

2 4c k m

24 -, , with 0

2 2 2

c c km cp i q i

m m m

2 2&c t c t

i t i tm me e e e

2 2cos( ) & sin( )c t c t

m me t e t


• By superposition, we get the general solution



2( ) cos( ) sin( )c t

mx t e A t B t

(0) , '(0)2

cx A x A B

m

2,

2

m cA B

m



• Introduce a phase angle


2

2 2 2

2

2( ) cos( ) sin( )

2

2 sin( 4 ) cos( 4 )

2 24

c t

m

c t

m

m cx t e t t

m

m c t tkm c km c e

m mkm c

22

2 2

2 ( 2 )sin , cos , ,

2 4

m c c mD

m D D m

2( ) cos( )c t

mx t D e t


• Properties: http://www.aw-bc.com/ide/idefiles/media/JavaTools/massprng.html

– Long time behavior: x(t)0, when t infinity– The bob passes through the origin infinite many times with each time a fix

period, the maximum amplitude in each period decreases when time increases!

– Examples: k=m=1, c=0.5 or 0.25


– No damping (simple harmonic motion): c=0• The analytical solution



0 0 0( ) cos( ) sin( )k

x t A t B tm

0(0) , '(0)x A x B

0

,m

A Bk



• Introduce a phase angle


• The bob oscillates periodically with amplitude D0 around the position of static equilibrium x=0. The period of the oscillation is defined as the time for a complete oscillation

http://www.aw-bc.com/ide/idefiles/media/JavaTools/massprng.html

2 20 0 0 0

0

( ) cos( ) sin( ) cos( )m

x t t t tk

22 2 2

0 0 0 20 0 0 0

sin , cos , ,m

DD D k

0 0 0( ) cos( )x t D t

0

22

mT

k


Forced motion: http://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.html

The external force is periodic:– E and a are positive constants, b is any constant

Two different cases:– Undamped motion:

• Resonance:

– Damped motion: c>0

''( ) '( ) ( ) ( ), 0,

(0) , '(0)

m x t c x t k x t f t t

x x

( ) cos( )f t E a t b

00k

cm

0a


Undamped motion:

– For the homogeneous ODE

– We know the two linearly independent solutions:

– We need find a specific solution and try the form

00,c a

''( ) ( ) cos( ), 0,

(0) , '(0)

m x t k x t E a t b t

x x

''( ) ( ) 0, 0,m x t k x t t

0 0 0cos( ), sin( ),k

t tm

cos( )A a t b


– Plug into the equation, we get

– This implies

– The general solution

– To satisfy the initial condition, we have

– Solve the above equations, we get A and B and the solution.

2 cos( ) cos( ) cos( ), for all m a A a t b k A a t b E a t b t

202 2 2

0

( ) ,( )

E Ek ma A E A a

k ma m a

0 0 2 20

( ) cos( ) sin( ) cos( )( )

Ex t A t B t a t b

m a

02 2 2 20 0

(0) cos , '(0) sin( ) ( )

E E ax A b x B b

m a m a


– The solution: http://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.html

• With• It is the sum of two simple harmonic motions but with different

periods, in general, it is NOT a periodic function!!

2 20 2 2

0

( ) cos( ) cos( )( )

Ex t A B t a t b

m a

2 2 2 2

cos , sin , ,A B

A B A B


Resonance:

– For the homogeneous ODE

– We know the two linearly independent solutions:

– We need find a specific solution and try the form

00,c a 20 0''( ) ( ) cos( ), 0,

(0) , '(0)

Ex t x t t b t

mx x

20''( ) ( ) 0, 0,x t x t t

0 0 0cos( ), sin( ),k

t tm

0sin( )A t t b


– Plug into the equation, we get

– This implies

– The general solution

– To satisfy the initial condition, we have

– Solve the above equations, we get A and B and the solution.

0 0 02 cos( ) cos( ), for all E

A t b t b tm

00

2 ,2

E EA A

m m

0 0 00

( ) cos( ) sin( ) sin( )2

E tx t A t B t t b

m

0(0) , '(0) cos2

Ex A x B b

m


– The solution: http://www.aw-bc.com/ide/idefiles/media/JavaTools/vibefdmp.html

• Resonance: This phenomena of unbounded oscillation when the natural frequency coincides with the frequency of the external force

0 0 020 0 0

sin( ) cos( ) ( )sin( ) sin( )

2 2

E b E tx t t t t b

m m


– Examples of resonance• When men are marching over a bridge• When waves hit a concrete damwall• When wind impacts on a structure like a bridge or a tower

– In actual fact, the structure would break up when the oscillations get too large, so that the oscillation never really becomes unbounded!!!

More examples: e.g. damped motion, c>0http://www.aw-bc.com/ide/idefiles/navigation/toolindexes/12.htm#12

Other examples

Electrical networks (Kirchhoff’s law)– q(t): charge onThe plates of the Capacitor C– L: Inductance of transformer– R: resistor– E: electromotor

Method for linear constant ODE

Consider:

Find two linearly independent solutions for

– Try to find the solution in the form

– Plug into the equation, we get the characteristic equation

– Write the two solutions based on the two roots

Find a specific solution of the original equation

''( ) '( ) ( ) ( ), 0,x t a x t b x t f t t

''( ) '( ) ( ) 0, 0,x t a x t b x t t

( ) with a real or complex number.tx t e

2 0a b

Methods for linear constant ODE

How to determine two functions f(t) and g(t) are linearly independent?Wronskian of f and g:

– Linearly independent: the Wronskina is everywhere nonzero– Linearly dependent: otherwise

Examples:– Linearly independent: f(t)=sin t, g(t)=cos t, W=1– Linearly dependent: f(t)=sin t, g(t)=3 sin t, W=0.

( ) ( )( ) ( , ) ( ) '( ) ( ) '( )

'( ) '( )

f t g tW t W f g f t g t g t f t

f t g t

mechanical vibrations in many mechanical systems: the motion is an oscillation with the position of...

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