137330987 mechanical vibrations (1)

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8 Zachary S Tseng B-3 - 1

chanical Vibrations mass m is suspended at the end of a spring, its weight st retches the spring

a length L to reach a static state (the equil ibrium posit

system).

u(t ) denote the

lacement, as a funct

e, of the mass relats

ilibrium position. Reca

textbook’s conv

hat

wn ward is posit

n, u > 0 mean

spring is

stretched be

its

equilibriu m

l e ngt h, wh

< 0 m

tha t th

spri n g i

c ompr e s se d. The

m as

t hen set in

motion (b

one of semeans).

Ó 2008 Zachary S Tse

B-3 - 2

The

equation s govern a mas

spring syst

At

equ ilibri

(by Hooke

Law

)



= kL

ile in motion:

² + g u¢ + k u = F (t )

s is a second order linear differential equation with constant coefficients.

sually comes with two initial conditions: u(t

u

d u¢(t

u¢

mmary of terms:

u(t ) = displacement of the mass relative to its equilibrium position.

m = mass (m > 0)

g = damping constant ( g ³ 0)

k = spring (Hooke’s) constant (k > 0)

g = gravitational constant

elongation of the spring caused by the weight

F (t ) = Externally applied forcing function, if any u(t

nitial displacement of the mass

u¢(t

nitial velocity of the mass


damped Free Vibration ( g = 0, F (t ) = 0)

simplest mechanical vibration equation occurs when g = 0, F (t ) = 0.

s is the undamped free vibration. The motion equation is

m u ² + k u = 0.

characteristic equation is m

= 0. Its solutions are



±

general solution is then

) = C

s w

C

w

ere

alled the natural frequency of the system. It is the

uency at which the system tends to oscillate in the absence of any

mping. A motion of this type is called sim pl e har monic motion .

ment : Just like everywhere else in calculus, the angle is measured in

ans, and the (angular) frequency is given in radians per second. The

uency is not given in hertz (which measures the number of cycles or

olutions per second). Instead, their relation is: 2 p radians/sec = 1 hertz.

(natural) period of the oscillation is given by

conds).


get a clearer picture of how this solution behaves, we can simplify it with

identities and rewrite it as

u(t ) = R cos (w



d ).

displacement is oscillating steadily with constant amplitude of

llation

+

angle d is the phase or phase an g l e of displacement. It measures how

ch u(t ) lags (when d > 0), or leads (when d < 0) relative to cos(w

ch has a peak at t = 0. The phase angle satisfies the relation

re explicitly, it is calculated by:

if C

,



+

if C

,

if C

and C

,

−

if C

and C

,

The angle is undefined if C

C

.


example of simple harmonic motion:



Graph of u(t ) = cos(t ) − sin(t ) Amplitude:

Phase angle: d = − p/4


mped Free Vibration ( g > 0, F (t ) = 0)

en damping is present (as it realistically always is) the motion equation

he unforced mass-spring system becomes

m u ² + g u¢ + k u = 0.



ere m , g , k are all positive constants. The characteristic equation is m

+ k = 0. Its solution(s) will be either negative real numbers, or complex

mbers with negative real parts. The displacement u(t ) behaves differently

ending on the size of g relative to m and k . There are three possible

ses of behaviors based on the possible types of root(s) of the

racteristic polynomial.

e I. Two distinct (negative) real roots en g

mk , there are two distinct real roots, both are negative. The

lacement is in the form

+

mass-spring system with such type displacement function is called

r d am ped . Note that the system does not oscillate; it has no periodic

mponents in the solution. In fact, depending on the initial conditions the

s of an overdamped mass-spring system might or might not cross over

quilibrium position. But it could cross the equilibrium position at most

e.




ures: Displacement of an Overdamped system

Graph of u(t ) = e

e



Graph of u(t ) = − e

2e


e II. One repeated (negative) real root en g

mk , there is one (repeated) real root. It is negative:

displacement

is

in

the

form



) = C e

C

ystem exhibits this behavior is called criticall y d am ped . That is, the

mping coefficient g is just lar ge enough to prevent oscillation. As can be

n, this system does not oscillate, either. Just like the overdamped case,

mass could cross its equilibrium position at most one time.

ment : The value g

mk ®

alled critical d am pin g . It

he threshold level below which damping would be too small to prevent

system from oscillating.


ures: Displacement of a Critically Damped system



Graph of u(t ) = e

e2



Graph of u(t ) = e

e2


e III. Two complex conjugate roots en g

mk , there are two complex conjugate roots, where thei r common

part, l , is always negative. The displacement is in the form

) = C

s m



C

m

ystem exhibits this behavior is called und er d am ped . The name means

the damping is small compares to m and k , and as a result vibrations will

ur. The system oscillates (note the sinusoidal components in the

ution). The displacement function can be rewritten as

u(t ) = R e

(m

d ).

formulas for R and d are the same as in the previous (undamped free

ation) section. The displacement function is oscillating, but the

plitude of oscillation, R e

decaying exponentially. For all particular

utions (except the zero solution that corresponds to the initial conditions

0, u¢( t

0), the mass crosses its equilibrium position infinitely

n.



Damped oscillation: u(t ) = e

(2t )


displacement of an underdamped mass-spring system is a quasi-period ic

ction

(that

is,

it

show

s

periodic-like

motion,

but

it

is

not

truly

periodic

ause its amplitude is ever decreasing so it does not exactly repeat itself).



oscillating at quasi-frequency, which is m radians per second. (It’s just

frequency of the sinusoidal components of the displacement.) The peak-

eak time of the oscillation is the quasi-period :

conds).

ddition to cause the amplitude to gradually decay to zero, damping has

ther, more subtle, effect on the oscillating motion: It immediately

reases the quasi-frequency and, therefore, lengthens the quasi-period

mpare to the natural frequency and natural period of an undamped

em). The lar ger the damping constant g , the smaller quasi-frequency and

longer the quasi-period become. Eventually, at the critical damping

shold, when

e quasi-frequency vanishes and the

lacement becomes aperiodic (becoming instead a critically damped

em).

e that in all 3 cases of damped free vibration, the displacement function

ds to zero as t ® ¥ . This behavior makes perfect sense from a

servation of ener gy point-of-view: while the sy stem is in motion, the

mping wastes away whatever ener gy the system has started out with, but

e is no forcing function to supply the system with additional ener gy.

nsequently, eventually the motion comes to a halt.


m pl e: A mass of 1 k g stretches a spring 0.1 m. The system has a

mping constant of g = 14. At t = 0, the mass is pulled down 2 m and

ased with an upward velocity of 3.5 m/s. Find the displacement function.

at are the system’s quasi-frequency and quasi-period?

m = 1, g = 14, L = 0.1;

= 9.8 = kL = 0.1 k ® 98 = k .

The motion equation is u² + 14 u¢ + 98 u = 0, and

the initial conditions are u(0) = 2, u¢(0) = −3.5.



The roots of characteristic polynomial are r = −7 ± 7i:

u(t ) = C

7

C

7

Therefore, the quasi-frequency is 7 (rad/sec) and the quasi-period is

conds).

Apply the initial condition and we get C

, and C

/2. Hence

u(t ) = 2e

7

1.5e

7


mmary: the Effects of Damping on an Unforced Mass-Spring System nsider a mass-spring system under going free vibra tion (i.e. without a



ing function) described by the equation:

m u ² + g u¢ + k u = 0, m > 0, k > 0.

behavior of the system is determined by the magnitude of the damping

fficient g relative to m and k .

Undamped system (when g = 0)

Displacement: u(t ) = C

w

C

w

illation: Yes, periodic (at natural frequency

Notes: Steady oscillation with constant amplitude

+

Underdamped system (when 0 < g

mk )

placement: u(t ) = C

m

C



m

illation: Yes, quasi-periodic (at quasi-frequency m)

Notes: Exponentially-decaying oscillation

Critically Damped system (when g

mk )

placement: u(t ) = C

C

illation: No

Overdamped system (when g

mk )

placement:

+

illation:

No




ritically Damped Oscillation placement: u(t )= C

C

s crosses equilibrium at most once.

Mechanical Vibrations, F (t ) = 0

nderdamped System oscillates with amplitude decreasing

exponentially overtime,

Displacement: u(t )= C

mt + C

t ,

Oscillation quasi periodic: T /m

verdamped No Oscillation,

Displacement: u(t )= C

Mass crosses equilibrium at most once.



damped Displacement: u(t )= C

w

C

w

al frequency: w

, Steady oscillation with constant amplitude

4mk 0= 4mk

< 4mk




rced Vibrations damped Forced Vibration ( g = 0, F (t ) ¹ 0)

w let us introduce a nonzero forcing function into the mass-spring system.

keep things simple, let damping coefficient g = 0. The motion equation is

² + k u = F (t ).

articular, we are most interested in the cases where F (t ) is a periodic

ction. Without the losses of generality, let us assume that the forcing

ction is some multiple of cosine:

m u² + k u = F

wt .

s is a nonhomogeneous linear equation with the complementary solution

u

= C

w

C



w

form of the particular solution that the displacement function will have

ends on the value of the forcing function’s frequency, w.

e I. When w ¹ w

¹ w

n the form of the particular solution corresponding to the

ing function is

Y = A cos wt + B sin wt .

ving for A and B using the method of Undetermined Coefficients, we find

ω

refore, the general solution of the displacement function is



ω

cos


interesting instance of such a forced vibration occurs when the initial

ditions are u(0) = 0, and u¢(0) = 0. Applying the initial conditions to the

eral solution and we get

−

and C

.

s,



)

ω

in, a clearer picture of the behavior of this solution can be obtained by

riting it, using the identity:

sin( A) sin( B) = [cos( A − B) − cos( A + B)] / 2.

displacement becomes



behavior exhibited by this function is that the higher-frequency, of

w) / 2, sine curve sees its amplitude of oscillation modified by its

er-frequency, of (w

w) / 2, counterpart.

s type of behavior, where an oscillating motion’s own amplitude shows

odic variation, is called a beat .




example of beat:

Graph of u(t ) = 5 sin(1.8t ) sin(4.8t )


e II. When w = w

he periodic forcing function has the same frequency as the natural

uency, that is w = w

en the form of the particular solution becomes



Y = A t cos w

B t sin w

the method of Undetermined Coefficients we can find that

A = 0, and

ω

general solution is, therefore,

ω

+



.

first two terms in the solution, as seen previously, could be combined to

ome a cosine term u(t ) = R cos (w

d ), of steady oscillation. The third

m, however, is a sinusoidal wave whose amplitude increases

portionally with elapsed time. This phenomenon is called resonance.

Resonance: graph of u(t ) = t sin(t )


hnically, true resonance only occurs if all of the conditions below are

sfied:

There is no damping: g = 0,



A periodic forcing function is present, and

The frequency of the forcing function exactly matches the

ural frequency of the mass-spring system.

wever, similar behaviors, of unexpectedly lar ge amplitude of oscillation

to a fairly low-strength forcing function occur when damping is present

is very small, and/or when the frequency of forcing function is very e to the natural frequency of the system.


rcises B-3.1:

4 Solve the following initial value problems, and determine the natural

uency, amplitude and phase angle of each solution.

u² + u = 0, u(0) = 5, u¢(0) = −5.

u² + 25u = 0, u(0) = −2, u¢(0) = 3

.

u² + 100u = 0, u(0) = 3, u¢(0) = 0.

4u² + u = 0, u(0) = −5, u¢(0) = −5.

10 Solve the following initial value problems. For each problem,

rmine whether the system is under-, over-, or critically damped.

u² + 6u¢ + 9u = 0, u(0) = 1, u¢(0) = 1.

u² + 4u¢ + 3u = 0, u(0) = 0, u¢(0) = −4.

u² + 6u¢ + 10u = 0, u(0) = −2, u¢(0) = 9.

u² + 2u¢ + 17u = 0, u(0) = 6, u¢(0) = −2.

4u² + 9u¢ + 2u = 0, u(0) = 3, u¢(0) = 1.

3u² + 24u¢ + 48u = 0, u(0) = −5, u¢(0) = 6.

Consider a mass-spring system described by the equation

+ 3u¢ + k u = 0. Give the value(s) of k for which the system is under-,

r-, and critically damped.

Consider

a

mass-spring

system

described

by

the

equation

+ g



36u = 0. Give the value(s) of g for which the system is under-,

r-, and critically damped.

One of the equations below describes a mass-spring system under going

nance. Identify the equation, and find its general solution.

u² + 9u = 2cos 9t (ii.) u² + 4u¢ + 4u = 3sin 2t

) 4u² + 16u = 7cos 2t

8

Zachary

S

Tseng

B-3

-

21

Find the value(s) of k , such that the mass-spring system described by

h of the equations below is under going resonance .

8u² + k u = 5sin 6t (b) 3u² + k u = − p cos t

wers B-3.1:

u = 5cos t − 5sin t , w

, R = 2

, d = − p / 4

u = −2cos 5t +

5t , w

, R = 4, d = 2 p / 3

u = 3cos 10t , w

0, R = 3, d = 0

u = −5cos t / 2 − 10sin t / 2, w

/ 2, R =

d = p + tan

u = e

t e

itically damped

u = 2e

e



verdamped

u = −2e

t + 3e

, underdamped

u = 6e

4t + e

4t ,

underdamped

u = 4e

verdamped

u = −5e

4t e

itically damped

Overdamped if 0 < k < 9 / 8, critically damped if k = 9 / 8, underdamped

> 9 / 8.

Underdamped if 0 < g < 24, critically damped if g = 24, overdamped if

24. When g = 0, the system is undamped (rather than underdamped).

(iii),

2

k k

137330987 mechanical vibrations (1)

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