§3.4: Mechanical Vibrations A Mass-Spring Systemspot.pcc.edu/~kkidoguc/m256/m256_03.4.pdf3.4: Mechanical Vibrations Free Damped Motion –Eigenvalue Computation 30 October 2019 11

130 October 2019 Kidoguchi, Kenneth

2

2 0d x dxm c kxdtdt

⇒ + + =2

2

R SF F F

d x dxm c kxdtdt

= +

= − −

, ^ 0RdxF cv c c cdt

= − = − ∈ >

Assuming a frictional force proportional to the speed of the mass:

where c is called the damping constant.If there are no additional external forces, FE = 0, then:

, ^ 0SF kx k k= − ∈ >

km

Equilibrium Position @ x = 0

x > 0x < 0For an ideal Hooke's Law spring, the force exerted on the mass, m, by the spring is:

where k is called the spring constant.

§3.4: Mechanical VibrationsA Mass-Spring System

30 October 2019 Kidoguchi, Kenneth2

22 20 02

c d x c dxx x x xm m dtdt

+ +ω ≡ + +ω

N.B.: A dot over the variable, “dot notation” , means we are considering a time dependent dynamic variable and differentiating with respect to time.

20

km

⇐ ω =2

202 0d x c dx x

m dtdt+ +ω =

2

2 0d x c dx k xm dt mdt

+ + =

2

2 0d x dxm c kxdtdt

+ + = km


x > 0x < 0

§3.4: Mechanical VibrationsA Mass-Spring System

30 October 2019 3 Kidoguchi, Kenneth

L

-mg

0θ >0θ <

Let s be the distance travelled by the mass on a circular arc and assume a damping force proportional to v = ds/dt, we have:

Consider a simple pendulum with a mass m attached to a massless rod of length L.

R GF F F= +

§3.4: Mechanical VibrationsA Simple Pendulum


L

-mgss = Lθ

sin( )mg− θ

cos( )mg− θ

0θ >0θ <

θ

2

2 sin( ) 0d s dsm c mgdt dt

+ + θ =



L

-mgss = Lθ

sin( )mg− θ

cos( )mg− θ

0θ >0θ <

θ0 0 2g LT

L gω = ⇒ = π

2

2

2202

sin( ) 0

0

d c d gdt m dt L

d c ddt m dt

θ θ+ + θ =

θ θ+ +ω θ =


830 October 2019 Kidoguchi, Kenneth

2 2

0

0

0

0

amplitude

angular frequency /phase angle arctan( / )period 2 /frequency 1/ /(2 )time lag /

C A B

k mB A

Tf T

= +

ω =

α =

= π ω

ν = = = ω π

δ = α ω

Where:

( ) ( )( )0 0 00

cos cos cosC t C t C t α

= ω −α = ω − = ω −δ ω

km


x > 0x < 0

( ) ( )0 0( ) cos sinx t A t B t= ω + ω

With general solution:

2202 0d x x

dt+ω =

If the system is undamped, c = 0, so:

§3.4: Mechanical VibrationsSimple Harmonic Motion – Free Undamped Motion

Kidoguchi, Kenneth30 October 2019 9

phase trajectory

akaphase

portrait

1sin2 2 2

t π π = − −

t-domain

0

amplitude 1angular frequency / 2phase angle / 4period 4frequency 1/ 4time lag 1/ 2

C

Tf

=ω = πα = π=

ν = =δ =

( )( )0 0( ) ( ) sinv t x t C t= = − ω ω −δ

1cos2 2

t π = −

( )( )0( ) cosx t C t= ω −δUndamped system response description:

3.4: Mechanical VibrationsSimple Harmonic Motion – Free Undamped Motion


2202 2

c crm m

= − ± −ω

which requires: Eigenvalues

Characteristic polynomial

2 20 0rtcr r e

m + +ω =

Upon substitution:

or: 2 20 0cr r

m+ +ω =

2

( )

( )

( )

rt

rt

rt

x t e

x t re

x t r e

=

=

=

Assume a solution of the form:

20 00, where / natural (angular) frequencycx x x k m

m+ +ω = ω = = Given:

§3.4: Mechanical VibrationsFree Damped Motion – Eigenvalue Computation


1 21 2( )

( )

r t r tx t a e a edxv tdt

= +

=

phase trajectory

akaphase

portrait

t-domain

/55 t te e− −= +

The system is said to be overdampedwith general solution:

22

1 0

22

2 0

2 2

2 2

c crm m

c crm m

= − + −ω

= − − −ω

220If , eigenvalues are real and distinct.

2cm

> ω

§3.4: Mechanical VibrationsDistinct Real Eigenvalues – Overdamped System


2202 2

c crm m

= − + −ω

220If , eigenvalues are complex.

2cm

< ω

§3.4: Mechanical VibrationsComplex Eigenvalues – Underdamped System


2

1 1 00

, where 12 2c cr im m

= − + ω ω =ω − ω

220

220

0

2

00

2 2

12 2

12 2

c crm m

c cm m

c cim m

= − + −ω

= − + −ω − ω

= − + ω − ω

220If , eigenvalues are complex.

2cm

< ω



phase trajectory

akaphase

portrait

t-domain

( ) dxv t xdt

= =

/2 1cos2 2

te t− π = −

( ) ( )

( ) ( )( )

2 21 1 2 1

21 1 2 1

( ) cos sin

cos sin

c ct tm m

c tm

x t a e t a e t

e a t a t

− −

−

= ω + ω

= ω + ω

The system is said to be underdampedwith general solution:

2

1 00

where 12

cm

ω =ω − ω

12cr im

= − + ω

220Hence, if ,

2cm

< ω



1 2( )

( )

rt rtx t a e a t edxv t xdt

= +

= =

The system is said to be critically damped with general solution:

220If , we have one distinct eigenvalue,

2 2c cr -m m

= ω =

§3.4: Mechanical VibrationsRepeated Eigenvalues – Critically Damped System


) 2 0, (0) 2, (0) (0) 0d x x x x x v+ + = = = =

) 0, (0) 2, (0) (0) 0c x x x x v+ = = = =

) 4 13 0, (0) 6, (0) (0) 0b x x x x x v+ + = = = =

) 7 10 0, (0) 2, (0) (0) 13a x x x x x v+ + = = = = −

• find x(t), the solution to the IVP and dx/dt and• sketch graphs of x(t) and dx/dt in the t-domain and in the phase plane.

For each of the following initial value problems (IVPs):

§3.4: Mechanical VibrationsHarmonic Oscillator – Example IVPs


) 7 10 0, (0) 2, (0) (0) 13a x x x x x v+ + = = = = −

§3.4: Mechanical VibrationsHarmonic Oscillator– Example IVPs



) 4 13 0, (0) 6, (0) (0) 0b x x x x x v+ + = = = =






) 0, (0) 2, (0) (0) 0c x x x x v+ = = = =





) 2 0, (0) 2, (0) (0) 0d x x x x x v+ + = = = =




Category Eigenvalues Parameters Decay Phase Trajectory

Undamped imaginary c = 0 no decay ellipse

Underdamped complex c2 – 4mk < 0 e-ct/(2m) spiral to origin

Overdamped real c2 – 4mk > 0 e-ct/(2m) decay to origin

Critically damped repeated c2 – 4mk = 0 e-rt decay to origin

km


x > 0x < 0

where: m > 0, k > 0, and c > 0 are real, constants.

0m x c x k x+ + =

The damped harmonic oscillator model:

§3.4: Mechanical Vibrations Mechanical Vibrations – Response Summary


0m x c x k x+ + = The damped harmonic oscillator model:

§3.4: Mechanical Vibrations Mechanical Vibrations – Response Summary Revisited

§3.4: Mechanical Vibrations A Mass-Spring Systemspot.pcc.edu/~kkidoguc/m256/m256_03.4.pdf3.4: Mechanical Vibrations Free Damped Motion –Eigenvalue Computation 30 October 2019 11

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