III Second Order DE 3.7 Mechanical Vibration 3.8 Forced Vibration · 2016. 6. 26. · §3.7 Spring-Mass System Undamped Vibration §3.8 Forced Vibration Objective Purpose of this

§3.7 Spring-Mass SystemUndamped Vibration

§3.8 Forced Vibration

III Second Order DE

§3.7 Mechanical Vibration


Satya Mandal, KU

Satya Mandal, KU III Second Order DE §3.7 Mechanical Vibration §3.8 Forced Vib



Objective

◮ Purpose of this section is to cite applications of LSODEsto physics and engineering.

◮ Many applications concern motion of a particle/body.This stems from the fact that accelaration is the secondderivative u”(t), where u is the distance travelled.Among them are

◮ Motion of a Falling body,◮ Motion of a pendulum,◮ Vibration of springs.

◮ There are other examples in Electrical engineering.




Spring-Mass System

The textbook considers a Spring-Mass System (see diagram inpage 193):

◮ A spring of length l is hanging vertically, from a point.

◮ A mass m is attached to the spring, which adds anelongation L, at rest.

◮ The downward direction is considered as positivedirection.

◮ At rest, there are two forces acting on it. First, theweight w = mg , and the force due to the spring action.

◮ By Hooke’s law, the force Fs due to the spring isproportional to the elongation L, acting upward. So,Fs = −kL, where k is a constant.




Continued

◮ Static State: At the equilibrium (static) state, only otherforce acting on it is the force due to gravity w = mg ,acting downward (positive direction). So, we have

mg − kL = 0 =⇒ k =mg

L(1)




They Dynamic State

One would like to study the motion of the body, when it isdisturbed by some external force.

◮ Let u = u(t) is the position of the body, measureddownward, from the position of equilibrium. FromNewton’s law

mu”(t) = f (t) (2)

where f (t) is the total force acting on the body.




Forces Acting on the body

◮ f (t) is the sum of all the forces acting on the body.

◮ A better question: For the purpose of modeling, whichforce we include and which one we neglect, as "noise"?

◮ Among the forces to consider are:◮ The force due to gravity = mg ,◮ The force due to spring action = −k(u(t) + L),◮ The force due to drag = −γu′(t),◮ Other forces applied, denoted by F (t).

◮ So, net force:

f (t) = mg − k(u(t) + L)− γu′(t) + F (t)




The Equation of the Spring-Motion

◮ By Newton’s Law (2):

mu”(t) = f (t) = mg − k(u(t) + L)− γu′(t) + F (t)

◮ Since mg − kL = 0, the Equation of the spring motion is:

mu”(t) + ku(t) + γu′(t) = F (t) (3)

◮ (3) seems to be a reasonable model. Depending onprecision needed, the model could be more complex.




Choice of UnitsSample I (Ex. 5): undamped, no external forceSample II (Ex. 6): undamped, no external force

Undamped Vibration

◮ For simplicity,◮ we will only consider undamped vibrations. So, γ = 0.◮ For now, assume there is no external force. So F (t) = 0.

◮ So, (3), with initial conditions is:

mu”(t) + ku(t) = 0u(0) = u0

u′(0) = u′0

(4)

This is a homogeneous equation (IVP).





Continued

◮ The CE of the DE (4) is mr 2 + k = 0.

◮ So, r = ±√

km

i = ω0i where ω0 =√

km

.

◮ So, the general solution of (4) is

u = A cosω0t + B sinω0t where ω0 =

√

k

m(5)

◮ A,B are determined by the initial conditions on (4).





Amplitude-frequency form of solution (5)

◮ The solution (5) can also be written as

u = R cos(ω0t − δ) (6)

This form is more widely used in physics and engineering.





Continued

◮ R is called the Amplitude. This is the maximumdisplacement from equilibrium.

◮ ω0 is called the frequency. In fact, by (1),

ω0 =

√

k

m=

√

g

L(7)

◮ Define Period

T =2π

ω0

= 2π

√

k

m= 2π

√

g

L(8)

◮ δ is called the phase.





Continued

◮ A,B and R , δ are related as follows:

{

u = A cosω0t + B sinω0t

u = R cos(ω0t − δ) = R cos δ cosω0t + R sin δ sinω0t

◮ So,

{

A = R cos δB = R sin δ

=⇒{

R =√

A2 + B2

tan δ = BA

(9)





Choice of Units

◮ Either use Foot-Pound-Second (FPS) system

◮ or use centimeter-gram-second (CGS) system.





Sample I (Ex. 5)

The Problem Statement:

◮ An object has weight w = 2 lb is suspended from a spring.

◮ It stretches the spring L = 6 inches or L = .5 feet.

◮ The mass is pulled additional 3 inches and released. Thismeans the initial condition is u(0) = 3, u′(0) = 0.

◮ Determine the position u(t) at time t.

◮ Find amplitude, frequency, period and phase.





Solution

◮ Use FPS units. So, g = 32 feet/s2.

◮ By (7), frequency ω0 =√

g

L=

√

32

.5= 8

◮ So, period T = 2πω0

= π4

◮ By (6) u = R cos(ω0t − δ)= R cos(8t − δ).

◮ The Derivative u′ = −8R sin(8t − δ).

◮ Initial conditions:{

u(0) = R cos δ = 3

12= 1

4

u′(0) = 8R sin δ = 0=⇒

{

R = 1

4

δ = 0





Continued

◮ So, amplitude R = 1

4and phase δ = 0.

◮ u(t) = R cos(8t − δ) = 1

4cos(8t)





Sample II (Ex. 6)


◮ An object has weight w = 100g is suspended from aspring.

◮ It stretches the spring L = 5 cm.

◮ The mass is set in motion from equilibrium position witha downward velocity 10 cm/s. This means the initialcondition is u(0) = 0, u′(0) = 10.


◮ Find amplitude, frequency, period and phase.

◮ When does it first return to the equilibrium position?





Solution

◮ Use CGS units. So, g = 980 cm/s2.

◮ By (7), frequency ω0 =√

g

L=

√

980

5= 14

◮ So, period T = 2πω0

= 2π14

= π7

◮ By (6) u = R cos(ω0t − δ)= R cos(14t − δ).

◮ The Derivative u′ = −14R sin(14t − δ).


u(0) = R cos δ = 0u′(0) = 14R sin δ = 10

=⇒{

δ = π2

R = 10

14= 5

7





Continued

◮ So, amplitude R = 5

7and phase δ = π

2.

◮ u(t) = R cos(14t − δ) = 5

7cos

(

14t − π2

)

= 5

7sin(14t)

◮ Finally, u(t) = 0 =⇒ 5

7sin(14t) = 0 =⇒ 14t = 0, π, . . .

So, first time it returns to equilibrium t = π14

sec.





§3.7 Assignments and Homework

◮ Read Example 2. [They do it using (5), while I used (5).

◮ Homework: §3.7 (page 203) See Homework Site!




Sample III (§3.8 Ex 5): Undamped, ForcedSample IV (§3.8 Ex 10, edited): Undamped, ForcedSample V (§3.8 Ex 18): Undamped, Forced


◮ For simplicity, continue to assume there is no damp.So,γ = 0.

◮ In §3.8, we consider Spring-mass system, with additionalexternal force F (t). Since γ = 0, the model (3) reducesto mu” + ku = F (t).

◮ Further assume, F (t) = F0 cosωt. Here F0, ω are positiveconstants, representing the amplitude and frequency ofthe force. So, the model of the motion is:

mu” + ku = F0 cosωt (10)





Undamped Forced Vibration

◮ By (5) and (6), the general solution, of the correspondinghomogeneous DE mu” + ku = 0, has two forms

uc = c1 cosω0t + c2 sinω0t = R cos(ω0t − δ) (11)

◮ Assume ω 6= ω0.

◮ By method of undetermined coefficients (§3.5); in deed,by direct checking a particular solution of (10) is:

U =F0

m(ω2

0− ω2)

cosωt (12)

I will show this in the white board.





The General Solution

◮ Therefore, a general solution u = uc + U of (10) is:

u = c1 cosω0t + c2 sinω0t +F0

m(ω2

0− ω2)

cosωt (13)

OR

u = R cos(ω0t − δ) +F0

m(ω2

0− ω2)

cosωt (14)

◮ We recall by 7, ω0 =√

km=

√

g

L.





Sample III (Ex. 5)


◮ An object has weight w = mg = 4 lb is suspended from aspring.

◮ It stretches the spring L = 1.5 inches or L = 1.512

feet.

◮ The mass is pulled another 1 foot from its equilibriumposition and released. This means the initial condition isu(0) = 1, u′(0) = 0.

◮ There is no damp

◮ the mass is acted on by an external force F (t) = 2 cos 3t.

◮ Formulate the initial value problem.






Solution

◮ Use equation (10) and FPS system.

◮ First, by k = mg

L= 4

1.5/12= 32. Also, mass m = w

g= 4

32.

◮ By (10) the equation of motion: mu” + ku = F (t) =⇒

4

32u” + 32u = 2 cos 3t =⇒ u” + 256u = 16 cos 3t.

◮ So, the initial value problem is

u” + 256u = 16 cos 3t

u(0) = 1

6

u′(0) = 0





Continued

◮ Use (14). The numbers, we need

◮ The frequency ω0 =√

km

=√

32

4/32 = 16.

◮ ω = 3, F0 = 16

◮ By (14) the general solution is:


m(ω2

0− ω2)

cosωt

= R cos(16t − δ) +16

4

32(162 − 32)

cos 3t

u = R cos(16t − δ) +128

247cos 3t





Continued

◮ To use the initial conditions, compute derivative:

u′ = −16R sin(16t − δ) +384

247sin 3t


u(0) = R cos δ + 128

247= 1

u′(0) = 16R sin δ = 0=⇒

{

δ = 0R = 119

247





Continued

◮ Therefore,

u = R cos(16t−δ)+128

247cos 3t = R cos(16t)+

128

247cos 3t





Sample IV (Ex. 10, edited)


◮ An object has weight w = 8 lb is suspended from a spring.

◮ It stretches the spring L = 2 feet.

◮ The mass is pulled another 8 inches from its equilibriumposition and released. This means the initial condition isu(0) = 8

12= 2

3, u′(0) = 0.

◮ There is no damp

◮ the mass is acted on by an external forceF (t) = 128 sin 8t.






Solution: Rework the model (10)

◮ In (10), we had F (t) = F0 cosωt. For this problem,F (t) = F0 sinωt.

◮ So, for this situation, the model changes to

mu” + ku = F0 sinωt (15)

◮ As in (10) and its particular solution (12), particularsolution of (15) is (assume ω 6= ω0):

U =F0

m(ω2

0− ω2)

sinωt (16)





The General Solution

◮ Therefore, similar to (13, 14), a general solution,u = uc + U of (15) is:

u = c1 cosω0t + c2 sinω0t +F0

m(ω2

0− ω2)

sinωt (17)

OR


m(ω2

0− ω2)

sinωt (18)

◮ We recall by 7, ω0 =√

km=

√

g

L.





Solution

◮ Back to solution Ex. 10. Use FPS system.

◮ First, by k = mg

L= 8

2= 4 and m = w/g = 8/32 = .25

◮ By (15) the equation of motion: mu” + ku = F (t) =⇒

.25u” + 4u = 128 sin 8t =⇒ u” + 16u = 512 sin 8t.

◮ So, the initial value problem is

u” + 16u = 512 sin 8t

u(0) = 2

3

u′(0) = 0





Continued

◮ Use (14). The numbers, we need

◮ The frequency ω0 =√

gL=

√

32

2= 4.

◮ F0 = 128, ω = 8



m(ω2

0− ω2)

sinωt

= R cos(4t − δ) +128

8(42 − 82)sin 8t

u = R cos(4t − δ)− 1

3sin 8t





Continued


u′ = −4R sin(4t − δ)− 8

3cos 8t


u(0) = R cos δ = 2

3

u′(0) = 4R sin δ − 8

3= 0

=⇒{

R cos δ = 2

3

R sin δ = 2

3

=⇒

{

tan δ = 1R2 = 4

9+ 4

9

=⇒{

δ = π4

R = 2√

2

3





Continued

◮ Therefore,

u = R cos(4t−δ)−1

3sin 8t =

2√

2

3cos

(

4t − π

4

)

−1

3sin 8t





Sample V (Ex. 18)


◮ Consider the undamped forced system, given by the initialvalue problem:

u” + u = 3 cosωt

u(0) = 0u′(0) = 0

Assume ω 6= 1.

◮ Find the solution u(t) at time t.

◮ Give Matlab graphs of u versus t for ω = 0.7, 0.8, 0.9





Solution

◮ Use (14). We have ω0 =√

km=

√

1

1= 1.

◮ F0 = 3



m(ω2

0− ω2)

cosωt

= R cos(t − δ) +3

(1 − ω2)cosωt





Continued


u′ = −R sin(t − δ) +3ω

(1 − ω2)sinωt

◮ Initial conditions:

{

u(0) = R cos δ + 3

(1−ω2)= 0

u′(0) = R sin δ = 0=⇒

{

δ = 0, π

R =∣

∣

∣

3

(1−ω2)

∣

∣

∣

{

δ = 0 if 1 > ω2

R =∣

∣

∣

3

(1−ω2)

∣

∣

∣

OR

{

δ = π if 1 < ω2

R =∣

∣

∣

3

(1−ω2)

∣

∣

∣





Continued

◮ Therefore, if 1 > ω2 then

u =

∣

∣

∣

∣

3

(1 − ω2)

∣

∣

∣

∣

cos t +3

(1 − ω2)cosωt

◮ If 1 < ω2 then

u =

∣

∣

∣

∣

3

(1 − ω2)

∣

∣

∣

∣

cos(t − δ) +3

(1 − ω2)cosωt





Three cases

◮ If ω = .7 then

u =3

(1 − .49)cos t +

3

(1 − .49)cos .7t

◮ If ω = .8 then

u =3

(1 − .64)cos t +

3

(1 − .64)cos .8t

◮ If ω = .9 then

u =3

(1 − .81)cos t +

3

(1 − .81)cos .9t





Simplification

◮ If ω = .7 then

u = 5.8824 cos t + 5.8824 cos .7t

◮ If ω = .8 then

u = 8.3333 cos t + 8.3333 cos .8t

◮ If ω = .9 then

u = 15.7895 cos t + 15.7895 cos .9t





Three graphs with ω = .7, .8, .9:

0 10 20 30 40 50 60 70 80 90 100−40

−30

−20

−10

0

10

20

30

40blue(ω=.7), green(ω=.8), red(ω=.9)





§3.8 Assignments and Homework

◮ Read Example 3, 4. [They use (5), while I used (5).]

◮ Homework: §3.8 See Homework Site!


Documents

III Second Order DE 3.7 Mechanical Vibration 3.8 Forced Vibration · 2016. 6. 26. · §3.7 Spring-Mass System Undamped Vibration §3.8 Forced Vibration Objective Purpose of this