lecture 8 vibrations

7/29/2019 Lecture 8 Vibrations

1/27

Copyright Ren Barrientos Page 1

VIBRATIONS AND PERIODIC PHENOMENA

Lecture 8

Introduction

What do the following systems have in common?

(a) (b) (c) (d)

Clearly the first three are mechanical in nature. They are composed of springs, strings, and masses. The lastone, on the other hand, is an electrical system with no moving parts. What makes them members of the same

family is the mathematical equation that governs their behavior; all these systems obey a second order

differential equation of the form

; ,

which, as you shall see, often results in periodic motion. Periodic phenomena are central to much of scienceand manifest themselves in a myriad of ways: your heart beats in a relatively periodic way somewhere

between 60 and 80 beats per minute depending on your health and age. The Earth revolves around the sun once

approximately every 365.25 days1 and about its axis once every 24 hours. Even the tiny particles thatconstitute matter behave in periodic ways that give rise to their chemical and physical properties.

Structures, such as hanging bridges and buildings, oscillate under the influence of natures forces. Electrical

circuits exhibit periodic phenomena as well. In this and the next lecture, we study the mathematics ofvibrations in the context of simple electromechanical systems.

Some of the language used to describe periodic motion is borrowed from circular motionso we should birelfy

discuss it.

Uniform circular motion describes the motion of a point along a circle in such a way that its angular speed is

constant. We define the average angular speedas the ratio of the change in the central angle to the time ittakes for that change to take place: when this speed is remains constant for arbitrary periods of time , we call it uniform circular motion and it ischaracterized by this constant angular speed .

1 This is an average time. That is why we have the leap year every four years.

Earth-Sun: 365.25 days Earth: 24hours Human heart: 70 5 bpm

Front view

Side view

P

SimplePendulum

E

R

L

C

Electriccircuit

Spring-masss stem

Shockabsorber

F


2/27


A point on the rim also has a linear velocity which is defined as lim / where the vector is thepoints position vector relative to the circles center. This velocity is not constant because it is always changing

its direction. However, if the point moves uniformly the magnitude ||remains constant. Hence, uniformcircular motion also means that that a point on the rim moves with constant linear speed given by l i m where

is a small arc along the circle. Periodic motion is characterized by aperiod and a frequency, defined

by the time it takes to complete one revolution. Say it takes seconds to do it; then we call theperiod ofrotation and its basic unit is thesecond. Since this is the time it takes for one cycle, we also say that the periodis so many seconds per cycle.One cycle corresponds to a rotation by 2 radians. Thus, if it takes seconds to complete this cycle, the angularspeed is

This is an important relation which you should commit to memory.

The frequency of rotation refers to the number of cycles per second and its basic unit is the Hertz2.Therefore,

These are the fundamental relations that we use to describe uniform circular motion and which will be used todescribe any other type of periodic motion as well:

Variable Unit Symbol Important relation

Period second (sec) Frequency Hertz (Hz) 1

Angular speed rad/sec 2

2Example 1 A rotating disk completes 10 revolutions in 4 seconds. What are (a) its period, (b) its frequency,and (c) its angular speed?Solution

(a) Since it completes 10 revolutions in 4 seconds, its period is 4 sec/10 rev or . . Wecould also say that the period is 2/5 seconds per revolution, but it is understood that we are referringto one revolution when we use the term period.

(b) The frequency is 1/. Thus, . or2.5 cycles per second.(c) The angular speed is 2 . /.

We will also refer to the angular speed as angular frequency or circular frequency.

Spring-mass SystemsA spring-mass system consists of an block of mass attached to a secured spring (coil) and free to move eithervertically or horizontally, as shown below. The differential equation governing either system is the same soeither one serves the purpose of illustrating the effect restoring forces, which are responsible for forcing the

system back to equilibrium.

2 Named after the German physicist Heinrich Hertz 1857 1894

/


3/27


Vertical spring-mass system Horizontal spring-mass system

Below are identified some of the forces at play. Friction and drag always oppose motion and are not shown inthe figure.

The free-body force diagram on the left show that the force due to gravity does not come into play for the horizontal systembecause thenormal force andgravitational force cancel out. Therefore, the only forces that accelerates the block inthis case are the restoring forcedue to the spring, frictional and drag forces, and perhaps other external agents. On theother hand, the vertical block is accelerated by the combined effect of the force of gravity and the restoring force. The natureof the springs restoring force will be discussed below.

At first glance it is very non-intuitive that the same differential equation should govern both the vertical and thehorizontal systems. After all, gravity plays an active role in the vertical system whereas it is cancelled by the

normal force (see figure) exerted by the table on the block in the horizontal one. Nevertheless, as we willfind out, these two systems are mathematically identical.

In order to study the dynamics of this problem, we need to identify the nature of all forces at play.The Coil

Everyone is familiar with the length-preserving property of coils such as the ones used in shock absorbers forexample. We will use the term spring to refer to devices which can be both stretched and compressed and

which have the property of restoring themselves to their original length when disturbed away from it.

Robert Hooke3 studied the properties of springs and determined that, as long as they are not stretched too muchand are operating under normal conditions4, the force with which they pull or push back is proportional to the

amount by which its length has changed relative to its natural state.

If the springs natural length is l0 and it is stretched by an amount then the magnitude of the force withwhich it pulls back is given by

where k is a constant called Hookes constant, also called the spring constant, whose unit is the lb/ft in the

English system or Newton/meter in the SI system. In order to stress the fact that the force always acts to restore

the coil to its natural length we sometimes write The negative sign is used to indicate the restorative nature of, which always opposes the direction in whichthe coil is changed.

If 0 then the coil has been stretched and will act to compress it. If 0, then the coil has been3 Robert Hooke (1635-1703) was a British physicist contemporary of Sir Isaac Newton.4 Temperature, for example, has a significant effect on springs.


4/27


compressed and the force will act in opposition to restore it to its natural length.

We will assume that our springs and coils obeys Hookes Law.

Friction

Friction and drag are the other two important forces that act on moving objects. We make the simplifying

assumption that these resistive forces can be represented by a single vector such that where is the velocity vector and a positive constant. The negative sign tells us that always pointsopposite the direction of velocity.

Units

In the SI system, we use the kilogram (Kg) for mass, the meter (m) for distance, and the second (sec) for time.

Near the earths surface, objects are subjected to gravitys force which causes them to accelerate with an

acceleration of approximate magnitude 9.8m/sec2. sometimes we will use 10 instead of 9.8 in order tosimplify calculations in which precision is not essential.In the English system we use the slug for mass, the foot (ft) for distance, and the second (sec) for time. Theacceleration due to Earths gravitational force in these units is 32ft/sec2.

Do not confuse weight with mass: the weight of an object is the product . Thus,an object weighing120 lb has a mass of 3.75 slug

Having stated these concepts, we are ready to study the spring-mass system in the free, undamped case.

Free Undamped Motion

Attach a block of mass

to a spring5 with spring constant

, as shown below. Assume that friction and all

other dissipative forces can be neglected. Displace the object from equilibrium and set it in motion by releasing

it with some initial velocity [if the object is simply released, we take 0. Then the mass will begin tooscillate periodically to and fro with some frequency.We are interested in describing this periodic motion; that is, to determine the objects position function,

frequency of oscillation, and other information about its dynamics.

The force due to gravity and the normal contact force cancel out since the object has no acceleration inthe vertical direction. In the horizontal direction, Newtons second law tells us that The only force acting on the object is the restoring force exerted by the spring, hence:

5This is a spring that has both the capacity to be stretched and compressed, i.e., a coil. In practice we use two springs that have beenslightly stretched and attach them to opposite ends of the mass which is free to slide on an air track.

The figure shows the situation when the object is tothe right of its equilibrium position 0. We takethis to be the positive direction. is theunit vector inthe x-direction.

x

0


5/27


or

Equating components,

Therefore,

Equation (1) has characteristic polynomial / whose roots are / and /Thus, / /Using Eulers identity,

c os / s in /The units of/ are the radian per second the same as those of angular velocity. For that reason, we callthis quantity the angular (or circular) frequency and also denote it by : /Do not confuse this with the of angular velocity of circular motion. This describes the frequency ofoscillation of the mass in radians per second and may be converted to frequency in cycles per second via therelation /2It represents the number of times the block moved to and fro in one second.

Equation (2) is the standard solution of equation (1) and it is a perfectly reasonable answer - we expect themotion to be periodic. The constants and are obtained from the initial conditions which state where theobject is and what is its velocity when 0. Here are some possible initial conditions:1) The object is displaced to the right 0 and released: 0 , 0 0.2) The object is displaced to the left 0 and released: 0 , 0 0.3) The object is pulled to the right ( 0) and given an additional impulse to the right ( 0: 0 ,0 Of course there are other possibilities. For example, the block starts from equilibrium ( 0) and is givenimpulse one way or the other ( 0). The point is that the system has to be disturbed in order to beginoscillating.

1 2 3 4 5 6

-1.0

-0.5

0.5

1.0

0 1

2

0(right of equilibrium)

0(left of equilibrium)

0 0


6/27


Do not confuse the actual motion of the block with the graph of its position function. The actual motion is to and fro. The graph abovedescribes the position function . Also note that the object starts to the right of equilibrium: 0 0. Furthermore, since theinitial velocity is to the right, 0 0.

It is often convenient to express (2) in terms of a single trigonometric function:

Example 2 Express 3 s in2 4cos2 as a cosine function.Solution

Here

4and

3. Therefore,

4

3

5. The phase angle obeys the relationships

c os 45 s i n 35which means that is in quadrant I. In this quadrant, angle and its reference angle coincide.Therefore, any of the inverse trigonometric functions gives us the correct angle. Using the inverse sine

for example, s i n 0.644

Thus,

.Example 3 Express c os 3 2 s in3 as c o s .Solution

Here 1 and 2. Therefore, 1 2 5andc o s 15 s i n 25

The last set of equations uniquely determine the positive angle in the interval 0,2 and place it inquadrant

IV. We can use the inverse tangent function as follows:

The reference angle may be obtained from tan .Notice that we ignore the signs of and . Having obtained we canobtain the phase angle by subtracting from 2. Of course thisadjustment will be different if is in quadrants II orIII. In those caseswe subtract from or add it to , respectively.

c o s c os s i n

Trigonometric Result

The function

c os s in, can be expressed as

Where is a positive number given by andis a positive angle in the interval 0,2 suchthat

The numberC is called theamplitudeand thephase angle.

21

5


7/27


Thus, 2 t a n2 21.1071. The phase angle is approximately 5.18 radians and wemay write .The graphs below illustrate both versions of the function. Since we approximated the graphs are notexactly identical.

Expressing linear combinations of sines and cosines in terms of a single trigonometric function can be very

useful. For example, the amplitude of the wave in the previous example is not clear from its original equation

c os 3 2 s in3 However, it is easily identified in the equation 5cos35.18. Similarly, obtaining the first zero ofthe function identified by the red dot in the graphs above requires that we solvec o s 3 2 s i n 3 0It is easier to solve 5cos35.18 0which we do by setting 3 5 .1 8 2 12

or

2 110.366 where n is an integer. If we seek the fist zero on the interval 0, we must select values of for which2 110.36 is non-negative and that is when 2. Thus, the first zero (corresponding to 2 takes place at 310.366 0.156secExample 4 A 5 Kg block is attached to a spring whose spring constant is 20Nt/m and placed on an air track ofnegligible friction. After displacing it a distance of30 cm to the right of equilibrium it is released and allowedto oscillate freely. Assuming that air drag can also be neglected, determine (a) the frequency and period of

oscillation, (b) the position function, (c) the time at which the object first passes through the equilibrium

position, (d) the velocity function, and (e) the velocity at that time.

Solution(a) We are given 5, 2 0. Therefore, 20/5 /In order to find the frequency in Hz., we may use 2

2 22 10.318 Hz.

0.5 1.0 1.5 2.0 2.5 3.0

-2

-1

1

2

PlotSqrt5Cos35.18,,0,Pi

0.5 1.0 1.5 2.0 2.5 .0

-3

-2

-1

1

2

3

PlotCos32Sin3,,0,Pi


8/27


The period is given by 1 / : 3.14 sec(b) satisfies the equation

5 2 0 0 ; 0 0.3,0 0which has as general solution

where 2. Thus, c os 2 s in2 Applying the initial conditions in order to determine the arbitrary constants:0 0.3 cos 0 sin 0 0.3Therefore, . . To find , we will apply the second conditionwhich requires that we computethe first derivative: 2sin22cos2Thus, 0 0 2 s in0 2 c os 0 0. This equation implies that . Therefore,

0.3cos2(c) The object goes through equilibrium when 0. This occurs when 2 2 1/2 orwhen 2 14 Since 0, we insist that 0. Therefore, the first time the object goes through equilibrium iswhen 0. Thus, 4 .

(d) The velocity function is given by 0.6sin2(e) The velocity vector is given by

0.6sin2

Thus, 0.79 0.6sin1.580.6This means that the object is moving to the left with a speed of0.6 m/s.

1 2 3 4 5 6 7

-0.6

-0.4

-0.2

0.2

0.4

0.6

0.79


9/27


Mass Hanging Vertically

If the mass hangs vertically, the general analysis is as follows: set up the spring-mass system as shown below.

After the object attains its equilibrium position we disturb it, say by pulling it down and releasing it. We want

to describe the up-and-down motion that ensues by means of its position function as we did in thehorizontal case.

Take the positive x direction as indicated in the figure so that the force of gravity has a positive component.

Applying Newtons Second Law:

3When the block is located at , the springs restoring force is given by Thus, Since the object moves in one dimension (up and down), we may replace the previous equation with a scalar

equation But

[because when the object just hangs vertically, the spring force and the force due to gravity

cancel]. Therefore, from which we obtain the equation

This equation is exactly the same as that derived in the horizontal case.

Example 5 An block weighing 12 lbs. stretches a spring a distance of 6 inches. Once in equilibrium, theblock is given a push downward with an initial speed of

1.5feet/sec. The medium in which the object moves

offers no resistance. Determine (a) its position function, (b) how long before it returns to equilibrium for the

first time, and (c) the period, frequency and angular frequency.

SolutionA 6-inch initial displacement corresponds to 6/12 feet and a 12 lb object has mass of12/32 slug.The figure below summarizing the given information.

0 4

+xdirection

t= 0

0

t> 0

Fs


10/27


(a) Since mg = 12, slug. Thus,38 2 4 0is the equation of motion. The initial conditions are 0 0 and 0 1.5 [since the block startsat equilibrium, 0 0and is initially moving down, 0 1.5} Thus,3

8 2 4 0 ; 0 0 0 1.5The auxiliary equation is 38 2 4 0 6 4 0Its roots are 8 . Accordingly, and are solutions of the differential equation and its general solution is cIn trigonometric form, c os 8 s in8 Applying the initial conditions, 0 0

0 1 . 5 Thus, 0 c o s 8 316 sin8 316 sin8(b) The object goes through equilibrium when 0. Thus, we solve316 s i n 8 0This is true when 8 or

8

Since 0 we demand that 0. The first nonnegative value of for which this condition holds is 0, which corresponds to 0. At this time, the object is just set in motion by the impulse force.Thus, 1 corresponds to the next value of for which 0. Hence, the object goes throughequilibrium on its way up at. 8 0.393secc) The period of oscillation is given by 2 / . With 8 we obtain

mg = 12

6

1 2

24/

t= 0

v0 = +1.5 ft/secx(0) = 0


11/27


4 and 1 4That is, 0.785 sec, 1.27 Hz. and of course 8 rad/sec.

Example 6 A block weighing 4 lbs. is suspended from a spring whose spring constant is k= 8 lb/ft. It is pulleddown a distance of 6 inches from its equilibrium position and given an initial upward speed of 1 ft/sec.Assuming no frictional forces, find the position function and express it as a single sine function. At what times

will the object have maximum speed?

SolutionIf we had to make a graph of , what would it look like? Even before solving this problem weknow that it can be represented as a sine wave (or a cosine wave if one prefers). The initial conditions

tell us that 0 0 and 0 0 [since the object is pushed up, toward the negative direction].Hence, we can expect a curve like this one:

Let us proceed with the usual calculations.

In this case, most of the information has been given outright and we can write the differential equation

of motion: 18 8 0 ; 0 12 , 0 1The angular frequency is given by / 6 4 8. Hence, the general solution is c os 8 sin8Instead of applying the initial conditions to obtain the arbitrary constants as before, we wish to

illustrate another approach to obtaining a solution that involves a single trigonometric function (as wasmentioned before, this is often desirable). The existence and Uniqueness Theorem guarantees that ifwe find a solution which satisfies the initial conditions, we will have found the only solution. With

this in mind, we seek a solution of the form s i n8 ; 0We need to determine the values ofCand subject to the initial conditions. Once we find a solutionwhich satisfies the initial conditions, we found the only solution.

1 2 3 4 5 6

-1.0

-0.5

0.5

1.0

t= 0

mg=4 or slugs 6

1 ft/sec

8 lb/ftx(0) = 6/12 = 1/2

v0


12/27


Since 0 1 / 2 , s i n 1 / 2. The second initial condition implies that 8 c o s 1. Bydividing these two expressions we obtainsin8cos 0.5 t a n 4On the interval 0,2 this last equation is satisfied by two phase angles: one in quadrant II and theother in quadrant IV. We need to select the one that is compatible with the initial conditions, which

require that s i n 0 and c o s 0 [this is so because it is assumed that 0]. Hence the phaseangle must be in quadrant II. Therefore , using the reference angle approach,tan 41 1.3258 1 .3 2 5 8 1 .8 1 6We may obtain from either equation above. For example,

0.5sin1.816 0.5154Thus, the position function is 0.5154sin81.816 We can quickly verify that at least 0 0.5 is satisfied by calculating 0:

0 0.5154sin1.8160.49998

Given the rounding that took place, this is acceptable. The objects velocity function is given by 4.123cos81.816 Since |cos81.816 | 1, the objects maximum speed is 4.123ft/sec and it is attained preciselywhen |cos81.816 | 1 or when 81.816 where is an integer. Thus, solving for

1.8168 Since 0, the first value of for which this is true is 1 , which gives us 0.166 sec.Computing the position at this time,0.166 0.5154sin80.1661.816 0.0012Because of rounding, this value is not identically

0. However, had we used exact values it would be,

indicating that the object is going through the equilibrium position. Furthermore,0.166 4.123cos80.1661.816 4.123 ft/secmeaning that the object is moving up, as is expected, for its first pass through the origin.

Free Damped Motion

Let us assume now that there are frictional and drag forces which are proportional to the speed with which the

object moves, say , 0 . Then equation (3) becomes 5And the corresponding scalar equation in the variablex is

But again . Therefore, Hence, when we include a damping term, the equation is


13/27


Using the more traditional dot notation and setting 0 (which can always be done),

0 0

, 0

Example 7 Suppose the 12 lb. block in example 5 is now in a medium in which it experiences a drag force where 3. Determine the position function.Solution

Once again, 2 4 but now we have a drag force with 3. Using (6) we have38 3 24 0; 0 0 0 1.5The auxiliary equation is 38 3 2 4 0Simplifying,

8 6 4 0The roots are 4 43 and 4 43. Accordingly, the general solution intrigonometric form is cos43 s i n 43Applying the first initial condition: 0 0 1 1 0 0Thus, and the solution takes the form s in43we must compute the derivative in order to apply the second condition: 4 sin4343cos43

0 1 . 5 43 32Thus, Finally, the position function is: 38 s in43What is the period? The zeros of this function take place whensin43 0This equation is satisfied when 43 ,0 ,1 ,2 ,Solving fort,

, , , ,

A full cycle occurs when the block goes to maximum extension, then maximum compression, andfinally back to the starting point. The object starts at the origin ( 0 , moves down to its maxextension and returns to the origin when Then it goes to maximum compression and returnsto the origin again when .Thus, the period is . This makes sense: in example 5 there was no friction; therefore, the object took a shorter time tocomplete a period.

0 6


14/27


The period can also be obtained from the period formula 2 /:

Example 8 Suppose an 8 lb weight is suspended from a spring, stretching it 4 inches. After the mass comes torest, it is displaced a distance of 4 inches above equilibrium and released from rest. If the resistive force isequal in magnitude to two times the instantaneous speed, what is the differential equation of motion?

SolutionThe figures below summarize the given information

Using 24 lb/ft, 1/4 slug, 2 and the initial conditions 0 , 0 0 , theequation of motion is 14 2 2 4 0 , 0 1/3 0 0The roots of the characteristic polynomial are

2 4 4 14 242 14 2 4 2 412 2 2512 4 4 5 Thus, cos45sin450 1/3 c os 0 s in0 1 /3 /We need to use the second condition:

4 1

3cos45 sin45

13 45 sin45 45 cos45Evaluating at 0,0 41/3cos0 sin0 13 45 s in0 45cos00 /

The solution is 13 cos45 320 sin45

0.5 1.0 1.5 2.0 2.5 3.0

-0.10

-0.05

0.05

0.10

Tt

24

84 0

t= 0

00 4 /1 2 1 /3


15/27


Example 9 Find the velocity of the object in the previous example at the moment it first passes through itsequilibrium position.

SolutionSince the object started above equilibrium, it will be moving down when it first passes through it.

Thus, we need to find the time when this occurs and evaluate the derivative ofx at that time.

Setting 0:0.33cos450.15sin45 0Since 0, we can cancel it and write the equation:0.33cos450.15sin45

tan45 0.330.15 2.2Computing tan2.2 in order to solve fortgive us a we negative number so we must be carefulwith our computations. Let us graph the functions 0.33cos45 and 0.15sin45. Thefirst point of intersection on the inerval 0 is the desired value of:

The first point of intersection of these graphs is at approximately0.22 sec. Therefore, this is the time atwhich the objects passes through equilibrium for the first time; all we need to do is compute0.22: 40.33cos45 0.15sin45 0.3345sin45 0 .1 5 45cos45Thus, 0.22 0.22 .1.32cos450.220.6sin450.22

0.3345sin450.220.1545cos450.22

. /This cumbersome computation illustrates the usefulness of the corresponding expressions of the form c o s or s i n Fact: The function c os s in can also be expressed as the sine function ; 0where and is a positive angle in the interval 0,2 such that

s in c os Its derivation is straightforward and is left as an exercise.

Consider once again our previous example: 0.33cos450.15sin45We are interested in expressing the quantity in parenthesis as a single sine function, as described above. Here 0.33 and 0 . 1 5. Thus, 0.33 0.15.

0.2 0.4 0.6 0.8

0.3

0.2

0.1

0.1

0.2

0.3

t


16/27


The phase angle is the angle such that s i n 0.330.36 c os 0.150.36This angle must be in the first quadrant because both the sine and cosine are positive. Thus, we can use either

the sin1 or cos1 functions to compute it. Using the former we obtain

s i n 0.330.36 1.16 rad

Using these results, the position function may be expressed more compactly as 0.36sin 451.16..With the position function in this form, it is much easier to calculate the velocity of the mass as it goes through

equilibrium for the first time. Again we must solve 0:0.36sin45 1 .1 6 0 sin45 1 .1 6 0This means that 45 1 . 1 6 where n is an integer. However,

must chosen so that

0in the expression above:

1 . 1 645 When 0, we obtain a negative value fort. The first value ofn for which 0 is 1. With this valuewe obtain . sec which confirms our earlier finding.Exercise: Use the position function 0.36sin451.16 to find the velocity at 0 .2 2.

Other Models

Oscillatory motion is not confined to spring-mass systems. For example, we are familiar with the phenomenon

of objects partially immersed in water which oscillate up and down even when there are no waves. Take a

heavy cork and put it in a glass of water, push it down a bit and let go. Careful observation will reveal that itbobs up and down for a while, exhibiting a form of periodic motion.

Example 10 A cylindrical object of mass , radius and altitude is placed in a fluid of density . After itcomes to rest with its base submerged to a depth h, it is pushed down a distance of and released. Find adifferential equation that describes the objects subsequent motion.

SolutionThe figure below illustrates the forces that act on the object when it is in equilibrium. The small red

circle labeled P is used as a reference point which moves in exactly the same way as the cylinder.

Applying Newtons Second Law, 0

is the force due to atmospheric pressure and

is the force due to the pressure at depth h. In

this coordinate system,

Since 0.

Area =A+ y 0

P


17/27


is s the buoyant force. In physics we learn that this force is equal to the weight of fluiddisplaced. If that volume of fluid displaced is V, then its weight is :

because 0 we must put the negative sign to ensure that 0. Now we push thecylinder further and release it. The buoyant and gravitational forces become unbalanced and the

cylinder starts bobbing up and down. This is the motion we wish to describe relative to a coordinate

system set up such that 0 corresponds to the equilibrium position (see previous figure). At somelater time the situation is as follows:

The figure above shows an instant in which the cylinder is submerged below its equilibrium point.

Since it is now acted on by forces that no longer cancel out, it will accelerate. Observe that when thecylinder is below equilibrium the buoyant force will exceed and a net upward force will be exertedon it. We have Where is the unit vector in the positive y direction6. Again, the negative is there to accommodate ourchoice of coordinate system in which 0. Therefore, when is also negative points up.Applying Newtons Second Law,

But 0. Therefore, 0; 0 , 0 0Since we initially pushed the cylinder down, 0. The angular frequency is given by

This last equation is exactly the same as that of the spring-mass and pendulum equations except for

the physical parameters inside the radical. You should check the units of the radicand as an exercise.

Energy Methods

Energy methods capitalize on the principle of conservation of energy. Every mechanical system may be

described by an energy function which specifies at each instant in time the total mechanical energy of thesystem. The conservation of energy principle states that if there are no dissipative forces such as friction for

6The moral of this story is: choose your coordinates well. It would have made a lot more sense to pick the downward direction as thepositive direction in this example

0+ y


18/27


19/27


20/27


1018 sin 10 10cos 10Thus, 0 0 sin0 1 0 0 0. Therefore, . We have the solution 18 cos 10(b) The angular velocity and linear (tangential) velocity are related by

Differentiating and using 1, 1018 sin 10At the bottom of the swing, 0. Therefore, we set 0 in order to find the times at which theobject is at that location. 18 cos 1 0 0 10 2 1/2

or

2 1210 where 0. The first time the object is at the bottom of the swing corresponds to 0 . Thus, / 210 sec. At this time,/210 1018 sin10 210 1018

which is approximately 0.55 m/sec.(c) The angular acceleration is given by :

59 cos 10Example 12 A wheel radius and moment of inertia is and mass m is attached to a spring whose springconstant is k. The spring is secured to a wall and to the wheel which is free to rotate without slipping. If thewheels center is displaced to some initial position units from its from its equilibrium position, find thedifferential equation that satisfies and the systems natural frequency.

SolutionThe energy equation is given by where stands for the wheels center of mass (i.e. its center). Once we identify each of these, it is amatter of differentiating and deriving the differential equation.

At some time 0, the object is moving with velocity , the wheel is rotating with angular velocity and the spring is stretched by an amount , which may be negative is the spring is compressed.

,

0

,

0

12


21/27


Thus, the total energy at time is given by 12 12 12

where (this is the non-slip condition) and / . Notice that this is just andaccounting equation it accounts for all the energy of the system which must remain constant as

long as there are no frictional losses. Setting the derivative of E equal to

0gives us the desired

differential equation:

12 12 12 0 0Using / and // we may write a differential equation in terms of the variable :

0

Cancelling / and simplifying,

1

0

or

0The systems natural frequency is given by

Example 13 An object of mass m is attached to a string which wraps around a pulley of radius and momentof inertia is . The string is attached to a spring with spring constant kwhich is secured to a wall. The mass isdisplaced from its equilibrium position and released. Determine a differential equation that governs its motion.

Kinetic energy

Rotationalenergypotiential energy

a

Kinetic Energy Rotational energy Potential energy

0

,


22/27


SolutionLet us place the zero gravitational potential energy line at the level at which the object is in

equilibrium7. Since the spring is stretched by some amount once the object is attached, there is somepotential energy stored in the spring. If denotes the amount by which this spring is stretched, thenthat initial spring potential energy is given by

.The equilibrium state is shown in the figure on the left:

If the object is disturbed from this equilibrium position it will begin to oscillate. Let us denote that

displacement by

with the convention that values above equilibrium are positive.

Now consider a point in time after the object is disturbed and is, say, units above equilibrium (figureto the right). Then the systems energy is distributed as gravitational potential and kinetic energy ofthe mass, potential energy of the spring (this energy is 0 when the spring is un-stretched), androtational energy of the pulley. The sum total of these constitutes the total mechanical energy of the

system at time : 12 12 12 and if there are no frictional losses, is constant. Therefore, 0.Differentiating the energy equation,

1

2 1

2 1

2 0

We are on our way to a differential equation of motion. To avoid excessive clutter, we will use theprime to denote differentiation with respect to time: 0The relation between and is 1 Therefore, 1 1 0Simplifying this expression,

1

0or 1 0But . Therefore, Finally, we state the initial conditions: , .

7This means that when the object is below the equilibrium position, its potential energy is negative.

0

0

0


23/27


As you can see, the energy method is a very powerful tool. The previous two problems would have been extremely difficultto solve using the usual vectorial approach.

Exercise: The energy of spring-mass system in a non-dissipative is given by . Use it toderive the equation

0

It is interesting to summarize the systems we have studied in the following way:

Equation: 0 0 0 0Frequency: General Qualitative Analysis

Now let us consider the qualitative aspects of spring-mass systems governed by the equation , This differential equations characteristic polynomial is whose roots are

4

2 Therefore, depending on the sign of the discriminant 4 we will have one of three possiblesituations: two distinct real roots (D> 0) , two coincident real roots(D = 0), or a complex conjugate pair(D < 0).Accordingly, the solutions will involve exponential functions with real arguments in the first two cases and

exponential functions with complex arguments which will lead to trigonometric functions in the third case.

To simplify the analysis, denote 4 by (the Greek letterxi). Let us consider each case in turn.Case I: 4 0 [two real distinct roots]In this case, the position function is given by

Using LHopitlals Rule one can show that this function has a finite limit as . Systems for which 4 0 are called over-damped systems because the damping forces are very large in magnitude and impedemotion severely.

Exercise: Show that, regardless of the values ofc1 and c2, lim 0.CASE II: 4 0 [two real and equal roots]In this case, we have repeated roots and the position function is given by

spring

0+ y

0


24/27


It is easy to see that this function also has limit 0 as t. This is expected since the object is being subjectedto retarding forces. Systems for which 4 0 are called critically damped systems.CASE III: 4 0 [complex roots]The roots are complex conjugates given by8

42 ||2 The position function is given by

Using Eulers equation, Once again, it is easy to show that lim 0.Systems for which 4 0 are called under-dampedbecause the damping forces are not strong enoughto prevent oscillatory motion. Nevertheless, the motion will cease eventually, driven by the exponentially

decaying term /.

Under- damped profile

Example 14 The equation 2 4 0 ; 0 1 0 0 corresponds to a spring that has been pulleddown 1 unit below equilibrium and released (0 initial speed). Since 4 1 6 8 8 0 , we have anover-damped system whose position function is

Applying the initial conditions, 0 1 10 0 2 2 2 2 0Solving the system: 2 2 2 2 1 0Thus, 2 222 2 222

8Since 4 0 , || . Therefore, is positive.

Over- damped profile


25/27


Finally, The figure below illustrates this situation.

2 222 2 222 Example 15 The equation 2 4 2 0; 0 0 0 1 corresponds to a spring that starts atequilibrium and is pushed in the positive direction with an initial speed of 1 unit/sec. In this case is

4 1 6 1 6 0. Therefore, we have a critically-damped system. The characteristic polynomial has a

root

1of multiplicity

2. Thus, the solution is given by

The initial conditions require that 0 and 1. Hence, is the solution. The only time this object is at equilibrium is when 0. Subsequent to that, the blow

pushes it down and the object takes forever to get to its equilibrium state again!

Example 16 Consider an over-damped spring-mass system with no forcing function which has the followinginitial conditions: 0 0, 0 . In other words, the system starts at rest and is set in motion by a blowin one or the other direction. Show that the only time at which the mass is at the equilibrium position is when 0. That is, if 0 then 0.Solution

The solution for over-damped systems has the form

where 4.Applying the first condition: 0 0 0Therefore, and we may replace these constants with a single constant, which we denote by .Thus, and . Thus,

0.5 1.0 1.5 2.0

.05

.10

.15

.20

2 4 6 8

.05

.10

.15

.20

.25

.30

.35


26/27


Clearly 0 0. Will it ever be 0 again? Assume that there is a value of 0 such that

0Since

0and

0,

0That is or

1But we assumed 0. Therefore, this equation is satisfied only if 0 or 0. This is acontradiction because we assumed that the system is over-damped so that 0. Therefore, the masswill never pass though equilibrium again and its position function will have one of the following

profiles, depending on the initial velocity:

Exercise 1: In the previous example, show that the position function can be written as

whereA is a constant. Also, find lim .Exercise 2:Show that in the critically damped case, the mass can pass through equilibrium at most once,depending on the initial conditions.

DATA FOR THE SPRING EXPERIMENT PERFORMED IN CLASS

Extension (cm) MASS (gm)

0 0

1.4 50

2.0 70

2.8 100

4.5 150

2 4 6 8

-0.4

-0.2

0.2

0.4

0

20

40

60

80

100

120

140

160

0 2 4 6

m

a

s

s

extensionWe can find from the slope of the line. Sincewe are using mass, if the spring truly obeysHookes Law, then we should find that or . In other words, the variable (mass) should be a linear function of.

0 0 0 0

2 4 6 8

-0.4

-0.2

0.2

0.4


27/27

The slope of the line is / so by calculating it, we can obtain thevalue of. The computed value of the slopeis 0.03226 Kg/cm or3.226 Kg/m Thus, 3.226 . /The real test now is to measure the natural frequency of vibration and compare it to the observed frequency.

Using the empirical values of

with a mass of

0.50 kg in the formula

/we obtain the theoretical

value of frequency: 12 31.61/0.5. .The observed frequency9 is 108.03 . .which is in remarkable agreement.

9In order to measure the frequency, we time 10 oscillations. Thus, dividing this time by 10 allows us to obtain the observed frequency.Several trials of this should be done in order to find an average that is more representative of the true frequency. Three measurementswere made resulting in 8.1, 8.0, and 8.0 seconds Thus, 8.03 sec. My gratitude to the physics department for lending me theequipment and to Ms. Isabella Orozco for assisting me with time measurements.

lecture 8 vibrations

Documents