spring mass couple.pdf

Section 4.9; Section 5.6

Free Mechanical Vibrations/Couple Mass-SpringSystem

June 30, 2009

Free Mechanical Vibrations/Couple Mass-Spring System

Today’s Session


Today’s Session

A Summary of This Session:


Today’s Session


(1) Free Mechanical Vibration (no forcing term).


Today’s Session


(1) Free Mechanical Vibration (no forcing term).(2) Coupled Mass-Spring systems


Today’s Session


(1) Free Mechanical Vibration (no forcing term).(2) Coupled Mass-Spring systems(3) Our first exposure to systems of differential equations


Today’s Session


(1) Free Mechanical Vibration (no forcing term).(2) Coupled Mass-Spring systems(3) Our first exposure to systems of differential equations(4) Eigenvalues and Eigenvectors


Free Mechanical Vibrations

Free mechanical vibration=no forcing function, so f (t) = 0. Weare dealing with

my ′′ + by ′ + ky = 0

where m = mass attached to a spring of stiffness k, subject tofriction (or damping) proportional to speed with damping constantb.


Free Mechanical Vibrations

Free mechanical vibration=no forcing function, so f (t) = 0. Weare dealing with

my ′′ + by ′ + ky = 0

where m = mass attached to a spring of stiffness k, subject tofriction (or damping) proportional to speed with damping constantb.

Four cases: (1) undamped free case; (2) underdamped case; (3)overdamped case; (4) critical damped case.


Case 1: Undamped mass-spring system: b = 0.

The equation is given by:

my ′′ + ky = 0




my ′′ + ky = 0

or

y ′′ +k

my = 0




my ′′ + ky = 0

or

y ′′ +k

my = 0

Let ω2 = k

m. The quantity ω =

√

k

mis called the angular

frequency (measured in radians per second)




my ′′ + ky = 0

or

y ′′ +k

my = 0

Let ω2 = k


√

k



Period: T = 2πω (measured in seconds)




my ′′ + ky = 0

or

y ′′ +k

my = 0

Let ω2 = k


√

k




Frequency: f = 1T

= ω2π (measured in Hertz=1/seconds=# cycles

per second)




my ′′ + ky = 0

or

y ′′ +k

my = 0

Let ω2 = k


√

k




Frequency: f = 1T

= ω2π (measured in Hertz=1/seconds=# cycles

per second)The solution is given by: y = C1 cos ωt +C2 sinωt = A sin(ωt +φ).


Undamped case, cont’d

Here A is called the amplitude, and φ is called the phase. They aregiven by

A =√

C 21 + C 2

2 φ = arctanC1

C2.




A =√

C 21 + C 2

2 φ = arctanC1

C2.

Example A 1/8 kg manss is attached to a spring with stiffnessk = 16N/m. The mass is displaed 0.5 m to the right of theequilibrium point and given an outward initial velocity of

√2 m/s.

(a) Neglecting damping, find a formula of the displacement as afunction of time. Display the values of ω, T , f , A and φ.




A =√

C 21 + C 2

2 φ = arctanC1

C2.

Example A 1/8 kg manss is attached to a spring with stiffnessk = 16N/m. The mass is displaed 0.5 m to the right of theequilibrium point and given an outward initial velocity of

√2 m/s.

(a) Neglecting damping, find a formula of the displacement as afunction of time. Display the values of ω, T , f , A and φ.

(b) How long after release does the mass pass first through theequilibrium position?



Answers:

(a) ω = 8√

2 rad/sec; f = 1.8 cycles/sec (or Hz); T = 0.555 sec.



Answers:

(a) ω = 8√


y(t) =1

2cos 8

√2t +

1

8sin 8

√2t



Answers:

(a) ω = 8√


y(t) =1

2cos 8

√2t +

1

8sin 8

√2t

A =√

(12 )2 + (1

8 )2 =√

178 and tanφ = 1/2

1/8 = 4 so φ = 1.326 rad.

y(t) =

√17

8sin(8

√2t + 1.326)



Answers:

(a) ω = 8√


y(t) =1

2cos 8

√2t +

1

8sin 8

√2t

A =√

(12 )2 + (1

8 )2 =√

178 and tanφ = 1/2

1/8 = 4 so φ = 1.326 rad.

y(t) =

√17

8sin(8

√2t + 1.326)

(b) y(t) = 0 means sin(8√

2t + φ) = 0 so 8√

2t + φ = kπ for k

integer. Solving gives: t = kπ−φω = 0.16 sec. Every 1/2 period (or

0.28 sec) the mass goes through the equilibrium point.


Overdamped case: b2 − 4k m > 0

The equation ismy ′′ + by ′ + ky = 0




The characteristic equation is given by

mr2 + br + k = 0





mr2 + br + k = 0

so

r = −b

2m±

√b2 − 4k m

2m

The solution is given by:

y(t) = C1er1t + C2e

r2t





mr2 + br + k = 0

so

r = −b

2m±

√b2 − 4k m

2m

The solution is given by:

y(t) = C1er1t + C2e

r2t

Note that both r1 and r2 are negative (why?), so as t → ∞,y(t) → 0.


Underdamped case: b2 − 4k m < 0

With ω =√

4k m−b2

2m , the roots of the characteristic equation are:

r = −b

2m± i ω

So the solution is given by:

y(t) = C1e− b

2mt cos ωt + C2e

− b

2mt sinωt.



With ω =√

4k m−b2


r = −b

2m± i ω


y(t) = C1e− b

2mt cos ωt + C2e

− b

2mt sinωt.

Over time y(t) dies in an oscillatory fashion. ω is a quasi-angularfrequency, T = 2π/ω is called the quasiperiod and f = 1/T iscalled the quasi-frequency.



With ω =√

4k m−b2


r = −b

2m± i ω


y(t) = C1e− b

2mt cos ωt + C2e

− b

2mt sinωt.

Over time y(t) dies in an oscillatory fashion. ω is a quasi-angularfrequency, T = 2π/ω is called the quasiperiod and f = 1/T iscalled the quasi-frequency.The solution can be put in the form

y(t) = A e−b

2mt sin(ωt + φ)

where (as before):

A =√

C 21 + C 2

2 φ = arctanC1

C2.


Critically damped case: b2 − 4k m = 0

In this case, the characteristic equation has a double rootr = − b

2m , So the solution is

y(t) = C1 e−b

2mt + C2 t e−

b

2mt





y(t) = C1 e−b

2mt + C2 t e−

b

2mt

Example: (related to webassign question and example p. 234)





y(t) = C1 e−b

2mt + C2 t e−

b

2mt

Example: (related to webassign question and example p. 234)Find the value of b for which

y ′′ + b y ′ + 25y = 0 y(0) = 1, y ′(0) = 0

is critically damped. Solve for y(t) in this case and sketch it.





y(t) = C1 e−b

2mt + C2 t e−

b

2mt

Example: (related to webassign question and example p. 234)Find the value of b for which

y ′′ + b y ′ + 25y = 0 y(0) = 1, y ′(0) = 0

is critically damped. Solve for y(t) in this case and sketch it.

Answer: b = 10; y(t) = (1 − t)e−5t


5.6: Coupled Mass-Spring System: Example 1, p. 308

See the graph and description of the couple mass-spring system onp. 308.



See the graph and description of the couple mass-spring system onp. 308.I will show the derivation of the following equations in class:




We wish to solve:

{

m1x′′ = −k1x + k2(y − x)

m2y′′ = −k2(y − x)




We wish to solve:

{

m1x′′ = −k1x + k2(y − x)

m2y′′ = −k2(y − x)

In our example: m1 = 2kg ; k1 = 4N/m; m2 = 1kg ; andk2 = 2N/m. So:

{ 2x ′′ = −6x + 2yy ′′ = 2x − 2y


Example 1, p. 308, cont’d

Let’s clean this up a little bit:

{

x ′′ = −3x + y

y ′′ = 2x − 2y




{

x ′′ = −3x + y

y ′′ = 2x − 2y

This is a linear system of differential equations of second order(which is homogeneous; why?) We can put it in the form:

{ (D2 + 3)x − y = 0−2x + (D2 + 2)y = 0




{

x ′′ = −3x + y

y ′′ = 2x − 2y


{ (D2 + 3)x − y = 0−2x + (D2 + 2)y = 0

Therefore(D2 + 2)(D2 + 3)x − (D2 + 2)y = 0




{

x ′′ = −3x + y

y ′′ = 2x − 2y


{ (D2 + 3)x − y = 0−2x + (D2 + 2)y = 0

Therefore(D2 + 2)(D2 + 3)x − (D2 + 2)y = 0

We note that: (D2 + 2)y = 2x .




{

x ′′ = −3x + y

y ′′ = 2x − 2y


{ (D2 + 3)x − y = 0−2x + (D2 + 2)y = 0

Therefore(D2 + 2)(D2 + 3)x − (D2 + 2)y = 0

We note that: (D2 + 2)y = 2x .So:

(D2 + 2)(D2 + 3)x − 2x = 0.




{

x ′′ = −3x + y

y ′′ = 2x − 2y


{ (D2 + 3)x − y = 0−2x + (D2 + 2)y = 0

Therefore(D2 + 2)(D2 + 3)x − (D2 + 2)y = 0

We note that: (D2 + 2)y = 2x .So:

(D2 + 2)(D2 + 3)x − 2x = 0.

We clean this up a little bit:

(D4 + 5D2 + 4)x = 0.



Then factor:(D2 + 1)(D2 + 4)x = 0



Then factor:(D2 + 1)(D2 + 4)x = 0

Therefore:

x(t) = C1 cos t + C2 sin t + C3 cos 2t + C4 sin 2t

and

y = (D2 + 3)x = 2C1 cos t + 2C2 sin t − C3 cos 2t − C4 sin 2t


Example 2

Use the method of the previous example (the annihilator method)to solve the first order (homogeneous) linear system:

{

x ′ = 3x + 4yy ′ = 4x + 3y


Example 2

Use the method of the previous example (the annihilator method)to solve the first order (homogeneous) linear system:

{

x ′ = 3x + 4yy ′ = 4x + 3y

Answer:

x(t) = C1e7t + C2e

−t

andy(t) = C1e

−t − C2e−t


spring mass couple.pdf

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