Physics 71 Notes_ Finals

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Physics 71 Final Exam Reviewer

Diandrew Lexter L. DyThis is a very condensed reviewer for the Finals. This is

essentially just writing all the needed equations, with someconcepts. Exercises are suggested from our previous prob-lem sets, quizzes, and long exmas. This summary in nature

can be used as a guide for things you have to learn for theexam. If you cannot understand it, then reading Young andFreedman would be helpful.

1 Ch 1: Units, Physical Quantities and

Vectors

The first chapter deals with preliminary topics for physics.This is not included in the exam, but the skills discussedhere are prevalent all throughout the remaining chapters ofYoung and Freedman, as well as Physics in general. Let sbriefly discuss the important skills:

1.1 What is Physics?

Such an important question that has various interpretations.In the words of my arrogrant physicist friend, Physics iseverything!. I however think Physics is Math in disguise,while Mathematicians think they are Philosophers, Philoso-phers want to be Psychologists, Psychologists goal is toexplain Biology, Biologists envy Chemists, and Chemistrywants to be Physics (so circle!). Now what about the En-gineers? Engineers want to be in Science, but they are toobusy getting rich. Architects think they are gods.

1.2 Units

Units are important for experimentalist and engineers be-cause they want to base the physical variables by somestandard measurement. These standard scales however arebased on convention, the most popular and useful of whichareS.I. units.

The S.I. units used for Physics are as follows:

Variable Units

length meter [m]time seconds [s]mass kilogram [kg]

Force Newton [N]Energy Joule [J]Power Watt [W]

Pressure Pascal [Pa]

The fundamental units are the first three: length, timeand mass. The rest of the variables has units that are ex-pressible in terms of the fundamental units. The reasonthey introduce these units is probably convenience.

Familiarization of units for the variables has importantconsequences in any Physics exam. First, you can guess theformula for the variable by looking at the units. For exam-ple, power is in Watts, which is Js . So you can probably

guess that:

Power =energy

time , (1.1)

which is the correct expression for average power. You canalso guess the force and speed relation of average power.However, this method will not work when there are unitlessconstants involved. For example, the escape speed should

be in ms, and must involve G, with units [G] = m3

kgs2.

Noting that [G mass/distance] =m2

s2

, you can probably

guess that:

vesc=

Gm

r . (1.2)

This formula however is off by a factor of

2, hence is wrong.You can still probably get away by guessing the formulabecause the exam is multiple choice, thus you can probablyguess the factor.

You can also guess the correct answer in multiple choice

exams by the correct units. If the question requires work(pun intended), your answer should be in Joules, and youmay discard the choices with different units. Remember,guessing is an art and a science, and is not just completelyrandom.

An amazing example of guessing, and my favorite prob-lem so far, is in your third exam, the Launch problem (some-where at the end). The problem states that you launch aprojectile at 4

kms

from the ground, and you are asked of

its speed at a height 20km off the ground. The solutionhere is through energy conservation, then find speed after a(slightly) long algebra and calculator exercise (that is why it

is someewhere at the end). However, you could have guessedthe answer by simply looking at the choices. Looking at thechoices, you can see that only one option has speed lesserthan the initial speed. And intuitively, the speed shoulddecrease at the top.

So in short: guessing works sometimes, and can alsobe used as a check to your answers. Why am I babblingabout this? Maybe because I realized that guessing is animportant skill, and must be taught! Sorry, I should havediscussed this before, never too late I hope.

One of the advantages of SI units are the SI prefixes thatare all based on powers of 10. In Physics 71, the important

prefixes are:Prefix power syntax (in meter)

kilo 103 kmcenti 102 cm

The rest of the prefixes will be useful later in yourPhysics career. Hence conversion of units is a breeze! Thetime conversion however obeys the following rules:

time in seconds

1 minute [min] 60 [s]1 hour [hr] 60 min =3600 [s]1 year [yr] 525600 moments!

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And also for days, months and other time quantities. Thispeculiarity of time is that it isnt really S.I. because of thebase 10 wasnt converted. Maybe its too difficult to changethe times. Theoretical physicists dont care however, assometimes they will just measure time in meters, hence theSI base 10 property still works!

1.3 Scalars and Vectors

Scalars are quantities without direction, while vectors havedirection. Several vectors can be added using the parallelo-gram rule.

Vectors can be broken down into components. We usu-ally use Cartesian components, where the basis (unit vec-tors) are perpendicular to each other. Hence we can expressany vector in three space dimensions as:

V= Vxi + Vyj + Vzk (1.3)where

i,

j,

k are the basis unit vectors, while Vx, Vy, Vz are

the components of V, which are scalar numbers (can be

negative!). The magnitude of the vector can be solved usingthe Pythagorean theorem:

V = |V| =

(Vx)2 + (Vy)2 + (Vz)2. (1.4)

The magnitudeVis never negative! The direction can alsobe solved. For a vector F = Fxi+ Fyj, the usual angle issolved as:

= ArctanFyFx

(1.5)

For vectors, important operations include: inner prod-

uct (e.g. W =F s) or cross product (e.g. =r F). Fa-miliarization of these operations are important in the nextchapters.

That ends Chapter 1!

2 Ch 2: Motion along a straight line

2.1 Dictionary

Consider an object moving in a straight line. The motionof the object hence can be described by its position at alltimes, x(t). This position can be any real number (can benegative!). From x(t), we can define displacement, definedas:

displacementt1 to t2 =x(t2) x(t1) = x. (2.1)

Hence displacement is dependent only at the endpoints ofthe position, and the sign implies the direction (positivemeans its going to more positive position). This is differentthan our notion of distance, where we add magnitude of dis-placements even with a change of direction. Rule of thumb:Going in a full circle: zero displacement, distance > 0.

We next define velocity vx as:

velocity: vx =dx

dt. (2.2)

This quantity is a vector (vx is actually the component,

direction isi) with the sign implying direction (can be neg-ative). We define speed v (no subscript) as:

speed: v= |vx| . (2.3)

Now we can also add instantaneous as an adjective tothe velocity and speed because it is evaluated at one time.Because derivatives are hard to imagine, we can define theaverage velocity and speed from time t1 to t2 as:

ave. velocity: vx,ave,(t1 to t2) =x(t2) x(t1)

t2 t1 =xt

.

ave. speed: vave = |vx,ave| . (2.4)

Note that in the limit that the interval t1 to t2 becomesinfinitesimal t2 t1 = t, then the definition of averagevelocity becomes velocity!

We finally define acceleration as:

acceleration: ax = dv

dt =

d2x

dt2. (2.5)

and the average counterpart:

ave. acceleration: ax,ave,(t1 to t2) =v(t2) v(t1)

t2 t1 =v

t.

(2.6)

These formulas are important and must be memorized!Trivia: da

dt is a jerk.

Examples

You can find several examples in Young (see the syllabus).In the first exam, understand Kinematics, Hindi Tamad andStopping Distance.

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2.2 Plottingx t, vx t, ax t diagramsTo picture the properties of moving objects, we can intro-duce graphs. These are the 2D graphs that we did with thehorizontal axis as time, and the vertical axis can either bethe position (x), velocity vx and acceleration ax.

Important concepts include that the derivative of a func-tion is related to the slope of the graph. For example, the

velocity att1 is the slope of the x t graph at t1.

Examples

You can find several examples in Young (see the syllabus).In the first exam, understand X vs T, Plot I and Plot II.Also read about Problem Set 1, where most of the numbersare about diagrams.

2.3 Motion with constant acceleration

Let us first consider when vx is a constant. Then we obtain(by integration) the position at time t1 as:

x(t1) =x(t0) + vx(t1 t0) (2.7)wheret0is the initial time, hencex(t0) is the initial position.This is hence relatively easy to memorize, but also easy toderive by integration.

Next, we consider the case where ax is a constant (inde-pendent of time, hence is not a jerk!). We then obtain theequations of motion at time t1:

x1= x0+ vx0(t1 t0) + 12ax(t1 t0)2 (2.8)

vx1= vx0+ ax(t1 t0) (2.9)

(vx1)2 = (vx0)

2 + 2ax(x1

x0) (2.10)

where we introduce the notation: x0 = x(t0) as the initialposition and vx0= vx(t0) as the initial velocity, while x1=x(t1), vx1= vx(t1).

Note that these equations should be familiar to you. Iusually derive these equations by integration, but you canalso memorize! In practice, I can survive constant accelera-tion problems with these three equations, and I just deriveother things by algebra. I hate memorizing! Another possi-bly important relation is the average formula:

x1 x0= vx1+ vx02

(t1 t0), (2.11)

which can be derived from the previous three/For freely falling bodies, we can approximate the motion

as with acceleration constant ay =g. Note that we referthey-axis because it is associated with going up! Hence, wejust replaceax andx in the above equations as:

y1= y0+ vy0(t1 t0) 12g(t1 t0)2 (2.12)

vy1= vy0 g(t1 t0) (2.13)

(vy1)2 = (vy0)

2 2g(y1 y0) (2.14)Note: These equations should be natural to you! Likebreathing...

Examples

We have done lots of examples for this (too many to men-tion). See PS1,2,3 and Q2,3 for example. In your exam,Stopping Distance, Skid marks, Catch, In free fall are im-portant examples.

That ends Ch 2!

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3 Ch 3: Motion in 2 or 3D

We now extend our discussion of the previous chapter tohigher dimensions.

3.1 Dictionary

Again we introduce the position, but this time in 3D as:

r= xi + yj + zk. (3.1)From hereon, if you want to consider 2D, then dont includethe third component, or set z = 0. We can also definedisplacement as:

displacementt1 to t2 =r(t2) r(t1) = r. (3.2)

We next define velocity v as:

velocity: v=dr

dt =

dx

dti + dy

dtj + dz

dtk. (3.3)

Hence we can define the components vx, vy, vz of velocity.We next define speed v as:

speed: v= |v| =

(vx)2 + (vy)2 + (vz)2. (3.4)

We can also define the average velocity and speed from timet1 to t2 as:

ave. velocity: vave,(t1 to t2)= r(t2) r(t1)

t2 t1 .ave. speed: vave = |vx,ave| . (3.5)

We finally define acceleration as:

acceleration: a=dv

dt =

d2r

dt2. (3.6)

and the average counterpart:

ave. acceleration: aave,(t1 to t2)= v(t2) v(t1)

t2 t1 . (3.7)

These definitions are important and must be memorized!

Examples

See young, Problem sets, in Quiz:Acceleration, Aave, Save

3.2 Projectile Motion

A projectile is defined as any body given an initial velocityand its trajectory is purely governed by gravity and possiblyair resistance. The rule of thumb here is: solve horizontaland vertical components separately, the horizontal motionhas constant velocity while the vertical motion is constantacceleration (g). The equations are therefore:

x(t1) =x(t0) + vx(t1 t0) (3.8)

y1= y0+ vy0(t1

t0)

12

g(t1

t0)

2 (3.9)

vy1= vy0 g(t1 t0) (3.10)

(vy1)2 = (vy0)

2 2g(y1 y0), (3.11)

where the x refers to the horizontal position, whiley is thevertical position.

From experience, the usual problems usually start withthe object given an initial speed at a certain angle. Fromthese initial conditions, you should be able to deduce theinitial velocity componentsv0x, v0y. The projectile then fol-lows a trajectory, and several properties of the projectilecan be asked. For example, you may be asked the timeof to reach maximum height, the time of flight, the range,the maximum height and the speed at certain points amongothers. You may be also asked from several scenarios, whichcan include on a flat ground or from a certain height. Youshould be able to solve them quickly. Here are my tips:

1. Determine the time immediately! Time is preciousfor these problems, and is usually the first thing to solve forthese problems.

2. A sketch usually helps, especially with complicatedproblems.

3. Practice! And practice quickly..

Examples: You have to recall how to solve several prob-lems about projectile. Know how to compute for maximumheight, the time of flight, the range, the maximum heightand the speed at certain points. I emphasize that you shouldnever memorize these things, except for the equations ofmotion stated above. The reason is that, your memory canbe brutally misleading, especially the range and maximumheight. Focus your memorization powers elsewhere in the

exam.

3.3 Motion in a circle

We now discuss motion in a circle. When an object is mov-ing in a circle at constant speed v , then its net accelerationis always directed towards the center of the circle with mag-nitude:

arad = v2

r (3.12)

wherer is the circles radius. Now if the object is increasing

speed along the circular path, then there must be a tan-gential accelerationatan along the direction of motion, suchthat the net acceleration is the vector sum arad+ atan. Ifthe motion is decreasing speed, then the tangential acceler-ation must be opposite to the movement. These conceptscan be easier understood through the succeeding figure.

Examples

There are several problems that we did involving motion ina circle. Among which are in PS4-3, Q2-6, LE1-CircularRoad, Toy Car. You may also be asked about the period ofthe motion, which we can get through T = 2r/v.

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Figure 1: Object moving in a circle with speed increasing.Notice the direction of the acceleration component vectors.

3.4 Relative motion

The objects velocity is always relative to the one who ismeasuring it. In Amsterdam high way for example, theyhave a speed limit of 120kph (as opposed to CommonwealthAvenues 60kph limit). So even though the speed limit isfast, there is a tendency for the cars to have the same speedof about 120kph. So if you are in a car, then it seems thatthe car in front of you is not moving at all. However, with

respect to the person standing along the road, the cars aremoving very fast. Thus, when speed is mentioned, theremust always be a relative to next to it. In our discussionof speed so far, it is assumed that the observer is relative.We now generalize to those moving observers.

We introduce the notation:

vA|B = velocity of A relative to the observer B. (3.13)

So in our car example, we have vcar1|car2 = 0 whilevcar1|person = 120kph. We can now compare relative speedof objectAfor two observersB, Cby the important relation:

vA|B =vA|C+ vC|B. (3.14)

Notice the placement of the indices, which is a useful wayof memorizing this equation.

Example

The raindrop problem in a moving car is my favorite exam-ple for this problem. Also, the boat in a moving stream isa common problem LE1-River.


4 Newtons Laws of Motion

Isaac Newton tried to describe the motion of objects bydescribing a new physical variable: Force! Let us first defineforce as:

The Force is an energy field created by all living things. It

surrounds us and penetrates us. It binds the galaxy together.

Life creates it, makes it grow. Its energy surrounds us and

binds us. Luminous beings are we, not this crude matter.You must feel the Force around you; between you, me, the

tree, the rock, everywhere. Yes, even between the land and

the ship. A Jedis strength flows from the Force. But beware

of the dark side. Anger, fear, aggression; the dark side of

the Force are they. Easily they flow, quick to join you in a

fight. If once you start down the dark path, forever will it

dominate your destiny, consume you it will, as it did Obi-

Wans apprentice!

Seriously though, force is the push and pull of any twoobjects. In effect it defines how two objects interact andcauses the other to move. We can naively give two types of

foces depending if theres physical contact (contact forces)and none (non-contact forces). Contact forces include forcesdue to pushing, or tensile forces. Non-contact forces includegravity, which does seem like theres no physical contact.

Newton decided to define clearly forces through the New-tons laws of motion through:

Law 1: Inertial reference frame: a= 0.

Law 2: Fnet= manet.

Law 3: Action-reaction: Fa on b = Fb on aHere, we introduce the notation thatFa on b is the force

of object a on another object b. The Newtons laws of mo-tion are only valid for inertial reference frames. Hence it

should be obvious to you that a rotating reference frame isnot inertial.

For the second law, we emphasize the net subscript! Thismeans if there are several forces acting on an object, thenet force (or the sum of all the forces) will be related tothe acceleration of the object. The net force is actuallynot a physical force, unless of course theres only one forceacting on the object. However, a common mistake of theseproblems is confusing between forces and the net force.

The third law, action-reaction, must be given in order tobe consistent with the conservation of momentum that willbe define.

4.1 Weight, Normal force, Tension

We now define weight of the object as the force of earth onthe object. For this section, we assume that the accelerationdue to gravity isg, hence the weight is W =mg directeddownwards (by 2nd law).

Weight :W = mg(downwards) (4.1)

From the above definition of weight, then a book on topof a table must have a weight towards the ground. However,the book is not accelerating at all towards the ground. This

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wherek is the coefficient of kinetic friction, and fn is thenormal force. The direction of the frictional force is oppositeto the motion.

For the static friction fs, the relation is:

fs sfn (5.3)

where s is the coefficient of static friction. Notice the in-equality, such that the static friction depends on the amountof force you push the object. To see this, imagine pushing abox on a rough surface. If you push with a small force, thenthe box wont move because the static friction opposes yourforce equally. Increasing your force would also increase thestatic friction, until it reachessfn. After that, the box willmove, with a constant kinetic friction. It is important thatyou understand these differences.

Examples

You must be ready with solving frictional problems. You

must understand the concepts, as well as draw FBD likeyour LE1-FBD. Memorize the key definitions of frictionalforces.

5.3 Banked curves

Banked curves are a cute application of frictional forces onroad surfaces. The result is:

Figure 3: Banked curves and FBD.

tan = v2

gR (5.4)

which you should be able to derive from the FBD.

That ends Ch 5!

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Diandrew Lexter L. Dy

6 Ch 6: Work and Kinetic Energy

6.1 Work

Work to me is essentially a fancy word for effort. Precisely

though, and in Physics context, it is defined as:

W =

r2r1

F(r) dr (6.1)

where it shows that work is just a line integral, calculusspeaking. In one dimension, this is just:

W =

x2x1

F(x)dx (6.2)

For those who are not really good at integration, what thismeans is that work is the area under the curve of an F

x

graph. From this, you shouuld be able to compute for thework given an F t graph! Examples include: LE2 Jess1and Jess2, plus the Quiz4-9. Be careful that by area underthe graph, there can be negative area also. Understandingis the key!

When force is a constant, such that F(r) = F, then wecan express work (W) as:

Work: W =F s (6.3)

where s is the objects displacement. Note that work can

be negative, zero and positive, depending on the directionofF and s. If they are along the same direction, then workis positive, and if in opposite then it is negative. If they areperpendicular, then work is zero.

The work concept has been used several examples, in-cluding LE2 Work I, II and III, Motor Work among others.Read through Quiz 3 or PS4 for example to familiarize howto solve work problems.

6.2 Kinetic Energy

Kinetic energy (K) is a measure of how energetic an objectis, in the layman sense. It is defined through an objectsmass and speed as:

Kinetic Energy: K=1

2

mv2 (6.4)

Note that this definition will always be consistent indepen-dent on the force of interaction.

6.3 Work-Kinetic Energy Theorem

Work-Kinetic energy theorem is a staple and sometimesoverused theorem for several problems encountered inPhysics 71. It states that:

Wnet = K=1

2mv2f

1

2mv2i (6.5)

Emphasis is on the net work done on the object, and notjust due to one force.

Examples include: LE2 Jess II or PS 4-4, Q4-2. YOushould understand when to use this theorem, which is quiteuseful and maybe a bit overused.

6.4 Power

Power is the rate of work. A requirement for power is ofcourse knowledge! Google Ernie Baron + knowledge +power! Power is defined as:

Power: P =dW

dt

(6.6)

When force is a constant, we can express this as:

Power, constant force: P =F v (6.7)

In many instances where the force is a constant, you canuse the above formula and hence integration is not needed.Sometimes though the speed is not given but the averagespeed. For this, you may use average power as:

Power, average: Pav =F vav (6.8)

where vav is the average velocity defined earlier and you

should memorize by now.Examples: Several examples include PS4-4, and several

Young examples. Please keep in mind that these definitionsmust be memorized because some takes too much time toderive.

That ends Ch 6!!

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7 Potential Energy and Energy conserva-

tion

7.1 Potential Energy

The potential energy U(x) is the energy stored from theforce. It is related by the force as:

Fx =

dU(x)

dx (7.1)

where we just consider one dimension for simplicity. Hencethe force is negative to the slope of the U x graph.

From this expression, we can find the position where itsin equilibrium, or when force is zero. This is when the slopeis zero. We can however classify these equilibrium pointsthrough its concavity. These are:

1. Stable equilibrium: U-x graph is concaved up!d2U/dx2 >0.

2. Unstable equilibrium: U-x graph is concaved down!d2U/dx2


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8.3 Center of Mass

The center of mass of any object is the location where theweighted average of the mass for a system of particles. Theweighted average formula is just:

rCM =

i rimi

i mi(8.10)

wheremi are the masses of individual particles. The centerof mass is not necessarily the geometric center of an object.However, for highly symmetric objects that are homoge-nous (uniform density), then these two different conceptsare identical (e.g. homogenous sphere, cube).

The center of mass doesnt have to be at the object,e.g.a homogenous donuts center of mass is at the hole in themiddle.

The center of mass for highly symmetric shapes can beestimated easily. You can also calculate it by the center ofmass formula above, but it can be difficult.

Familiarize yourselves with this concept through exam-ples such as: LE2 Smiley.

That ends Ch 8!

9 Ch 9: Rotation of Rigid Bodies

We now discuss rotation. This is different than the previouschapters, where the motion is translational. For these cases,we shall employ angular variables.

9.1 Dictionary of Angular variables

By angular variables, it means the movement of an object is

along a certain angle, instead of being translational (alongx-axis for example). We define the position of an object as

s= r (9.1)

wheres is the arclength of a circle of radius r over an angle [in radians]. The displacement s can be thought of as achange of position in the same way we have:

x x0= r( 0) (9.2)

We therefore can obtain an analogy for these two types ofvariables: translational

rotational

translational rotationalposition: x angular position:

velocity: vx = dxdt angular velocity: = ddt

acceleration: ax = dvdt angular acceleration: = ddt

mass: m Moment of Inertia: IKtan =

12mv

2 Krot = 12I2

force: F= dpdt torque: = dLdt =r Fmomentum: p angular momentum: L= r p

Note that these variables will be discussed later in thischapter and the next.

9.2 Constant and rotation

Similar to the result for constantv, we obtain the equationsof motion for constant as:

(t1) =(t0) + x(t1 t0) (9.3)

wheret0is the initial time, hence(t0) is the initial position.Note that the resemblance to the translational analog shouldbe obvious to you.

Next, we consider the case where is constant of whichwe have the equations of motion at time t1:

1= 0+ 0(t1 t0) + 12(t1 t0)2 (9.4)

1= 0+ (t1 t0) (9.5)

(1)2 = (0)

2 + 2(1 0) (9.6)

where we introduce the notation: 0 = (t0) as the ini-tial position and 0 = (t0) as the initial velocity, while1 = (t1), 1 = (t1). Again, the resemblance to the

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translational should be obvious to you. These equationsshould be easy to derive once you know the translationalcase: just change the translational variables to the rota-tional variables.

Finally, we stress that from the analogous relation be-fore: s = r that the velocity and angular velocity arerelated by:

v=r

arad =r. (9.7)

Here, we emphasize arad because it is possible to have anatan depending on how is changing.

A trick in solving these types of problem is units. Whenyou are asked about angular speed, then your units shouldbe in

rads

, while if its speed, it should be

ms

. For that

reason, in solving these types of problems, you must alsosolve for the units. This is helpful because sometimes youforget to multiply r , and that you will notice your answersis missing a meter in the units. So my tip here is units!

Examples

Familiarize yourselves by looking at Young, or LE2 Sushiproblems.

9.3 Rotational Kinetic energy

There is also the rotational counterpart for kinetic energy,defined as:

Krot = 1

2I2 (9.8)

whereI is the moment of inertia, which is different depend-ing on the geometry of the rotating rigid object. You candetermine the moment of inertia through integration, butyou can also refer to a table:

Figure 4: Moment of inertia for several geometries.

You have to know how to calculate the moment of inertiagiven the formulas above.

The total kinetic energy of an object is the sum of itsrotational and translational i.e. Ktot =Ktrans+Krot. For

an object that is just rotating on an axis, the total kineticenergy is just the rotational. If it is also moving translation-ally,e.g.the center of mass is moving linearly at a distance,then you have to add the translational kinetic energy.

9.4 Parallel Axis Theorem

Sometimes you are only explicitly given the formula of the

moment of inertia at a certain po . However you can solvefor the moment of inertia about a parallel axis through:

Ip = Icm+ M d2 (9.9)

whereMis the mass of the rigid object, and dis the distanceof the parallel axis to that of the center of mass. You shouldtry using parallel axis on several geometries, like solving theparallel axis at the end of a rod, or a sphere. Try reviewingmost of PS5. This is an integral part!

9.5 Conservation of Energy

Conservation of energy also holds for these systems. Youmust be careful though of accounting all of the kinetic en-ergy. A common mistake is just considering the transla-tional or rotational kinetic energy to the energy, which leadsto a mistake in the calculation. Please be careful about this!

Examples: Try to solve PS5 again, also LE2: Axis, Rankthe Cylinders and Disks and Rolling down, Ball 1 and Ball2 problems are useful concepts that you have to know!

That ends Ch 9!

10 Ch 10: Dynamics of Rotational Motion!

This is just a continuation of Ch 9 for rotation.10.1 Torque

Torque is the rotational analogue of force. Recall that F =ma, of which we obtain the analogue:

torque: =I =r F (10.1)Here, has direction along the axis of rotation. In thesecond definition = r F, r is the moment arm, whichis the position vector from the axis of rotation to the pointwhere the force F is acted upon.

The direction of the torque is obtained by the right handrule. The cross product can be computed either of the threerelations:

=rFsin = rF =rF (10.2)

In the first equality, is the angle between r and F vec-tors. You must know how to find the angle! This approachcan be difficult if you dont know how to find the angle.

The second and third approach are equivalent, that isfind the component of any of r or F that is perpendicu-lar to the other and just multiply them. This approach issometimes easier and faster, and I highly recommend you to

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study it! To find the direction, you just use the right-handrule.

Examples for torque problems can be found in Young orLE3 Torques. Static equilibrium problems in Chapter 11also uses torque a lot!

10.2 Rolling without slipping

A circular object (cylinder or sphere) rolling in a surface caneither slip (i need this!) or can be rolling without slipping(or in tagalog.. puyat! like what I am doing right now,thank you very much! :))

When the circular object of radius r is rolling withoutslipping, then the center of mass obeys:

scm =r

vcm =racm= r (10.3)

The total kinetic energy is then Ktot = Ktrans + Krot!You have to include the translational as well!

When an object is rolling without slipping, there shouldbe friction involved. This is a static friction, not kineticfriction! And since it is a static friction, the work doneby this friction is zero (displacement is zero!). Hence themechanical energy will not diminish and the object will rollforever (even though theres friction!) Try to understandthese concepts, or at least keep telling it to yourselves untilthese become true.

When an object is rolling with slipping, friction cancause the object to slow down until friction is enough suchthat the rolling without slipping condition will hold. If thereare no friction, then the object will slip forever (which i wishi could do.. but im not yet suicidal though.. its really late..

sorry.. almost done!).

10.3 Dynamics of rotational motion

In the same way that FBD + Newtons second law are usedin the translational counterpart, so will it work in the rota-tional, but note that the Newtons second law counterparthas the form:

net = Inet (10.4)

Knowing this, you should be able to solve the yo-yo prob-lem, objects rolling down an incline problem, pulley prob-lems (LE2 Pulley 1,2). These concepts to be honest aredifficult to understand. However, with practice comes famil-iarity. Familiarity breeds correctness. It may also lead tocontempt, but you dont have much choice do you?? Thatstoo sad. I dont like rotation also.. nakakahilo e.

10.4 Work and Power in Rotation

Again, we can calculate work in rotation analogous to thetranslational part. We do this through:

W =

2

1

zd (10.5)

where z is the torques only component along the axis ofrotation. There is also a work-kinetic energy counterpart:

W = Krot = Krot,2 Krot,1 (10.6)

which is very useful (LE2-Merry2).

We can also define rotational power for constant torqueanalogously as:

P = (10.7)

which resembles P =F v.10.5 Angular momentum

Angular momentum is defined similarly as:

L= r p= I (10.8)

Notice that it is similar to torque in the sense that the mo-ment arm r is multiplied by its translational counterpart.

The Newtons second law in rotation can be expressedas:

net= dL

dt (10.9)

which should resemble something familiar. Thus when thenet external torque of a system is zero, thenL is a constant,and hence conserved.

Solving through conservation of angular momentumshould be similar to that of momentum, but with an addedcross product. In many cases, r and p are perpendicular,and that you just multiply in magnitude. You must be care-ful in the direction though (use right-hand rule), as theremight be a sign error otherwise. You may also use I defi-nition, which is useful for rotating objects.

This is probably the most difficult thing to comprehendin the exam, yet very easy to calculate algebraically. Tryto read more about Young, and study LE2 stick. Theseproblems are also important, and must be understood!

That ends Chapter 5

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Diandrew Lexter L. DyThis is a very condensed reviewer for the third Long

exam. This is essentially just writing all the needed equa-tions, with some concepts. Exercises are suggested from ourprevious problem sets and Quizzes. This summary in nature

can be used as a guide for things you have to learn for theexam. If you cannot understand it, then reading the textwould be helpful.

11 Ch 12: Static Equilibrium and Elastic-

ity

This is the first Chapter for the Third long.

11.1 Static Equilibrium

For this section, you have to know what is static equilibrium.Basically, static equilibrium are situations when objects arenot moving at all. Not even rotating, or moving at aconstant speed. The two conditions of static quilibrium are:

Fnet= 0; (11.1)

net= 0 [at all points] (11.2)

To differentiate this from equilibrium depends on howyou define equilibrium. Young didnt elaborate, hence westate that an object in equilibrium when it is moving trans-lationally at constant speed. Dont worry about this pos-sible confusion, because I made sure that your exam, youonly have to worry about static equilibrium, which has aprecise definition.

11.2 Center of GravityA related topic is the center of gravity. This is related tothe center of mass, but this time its the gravity that isbeing averaged over. More precisely, given a collection ofnparticles, then the center of gravitys location is calculatedusing:

r=

ni=1 riWini=1 Wi

(11.3)

where for object i, ri is its position and Wi is its weight,which is justmigi. When the acceleration due to gravity,gi,is constant for all particles, then the position of the center

of gravity reduces to:

r=

ni=1 rimini=1 mi

(11.4)

which is just the center of mass location. Hence the center ofgravity corresponds to the center of mass wheng is constantfor all particles. This is true when the objects are near thesurface of the earth, where the g = 9.81

ms2

is roughly

constant.

We can also find the center of gravity by noting thatthere is no net torque when the object is balanced at thecenter of gravity. This is what I usually use because I usuallyforget the average formula above.

Examples for Static Eq and CoG

You have to familiarize yourselves with some possible prob-lems that you can encounter for static equilibrium. Typesof problems for this section include:

1. See-saw problems

These are important problems because its pretty basic

and unconvoluted. Examples include (PS6-1, PS6-2, PS6-3,PS6-5, PS6-6, PS6-13, PS6-16). You really have to knowhow to solve these problems before the exam, and whilethese problems might be repetitive in nature, they can beuseful.

Note that for many problems above, you can surviveusing only the condition net= 0. For other problems, youmay have to useFnet = 0. These problems include Q6-3, PS6-14, PS 6-15. You dont have to solve each problem though,just know how to solve them, and solve them quickly.

2. Hanging beam problems

These problems are usually a beam attached to a hingeon a wall and supported by a string to keep it in static equi-librium. Problem examples include PS 6-11, PS 6-15, Q6-1.These questions most of the time requires trigo, because thestring is usually at an angle. Determining the angle is attimes the tricky part for these types of problem.

A skill involved for these problems is finding the per-pendicular component of the tension, so that you can justmultiply it with the level armr, hence the calculation is wayeasier. This skill is important if you wish to solve the prob-lems quickly, and hence gives you enough time to review theanswer.

3. Ladder problems These problems are quite convu-luted in nature, because it requires you to solve a system of

equations. Problem examples include: PS6-7 to 10, PS6-19.

11.3 Elasticity

Objects are not rigid realistically, and they are at timesstretched, squeezed or twisted in the presence of a force.For an object given a force F perpendicular to its crosssectional area A, we can define tensile stress as:

Tensile stress = F

A (11.5)

and we can also define tensile stress as:

Tensile strain = l

l0(11.6)

where l0 is the original length of the object, and l isthe amount of stretching and compression. We next defineYoungs modulus Y as:

Y =Tensile stress

Tensile strain. (11.7)

In many objects, the Youngs modulus are relatively con-stant. We can similarly define bulk stress, bulk strain andbulk modulus when the force is all around the object, and

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which we obtain the relations:

Bulk stress = F

A ; Bulk strain =

V

V0

Bulk modulus = Bulk stress

Bulk strain. (11.8)

And similarly for shear stress:

Shear stress = F

A ; Shear strain =

l

l0

Shear modulus = Shear stress

Shear strain. (11.9)

In many objects, we can treat the modulus as constant.This is okay for small l. For larger l however, Y is notconstant, and the behavior of stress vs strain is not linear.It has the shape as shown in Figure below

Figure 5: Stress vs strain of a metal

You have to understand this figure. Point a is the pointwhere the strain is still proportional to the strain. Theregion between a and b is a bit curved, but will not deformonce released. Beyondb, plastic deformation follows, wherethere is permanent deformation once the object is relievedof stress. Finally, at point d, the object breaks.

Examples

Examples of this part involves mainly plugging the defini-tion of stress or strain, or manipulating the equations tofind the needed variable. Examples include PS7-1 to 5 andQ6-4. While some problems may require you to use previousknowledge or FBD like the elevator problem PS7-4.

These examples are basic in nature, and does not re-quire much derivations, but you have to at least know thedefinitions by heart.

That ends Chapter 11

12 Chapter 12: Gravity

We now review gravity. In our level, its classical, startingwith the implications of Newtons law of Gravitation.

12.1 Newtons Law of Gravitation

Given two objects with masses m1 and m2, the magnitudeof the force ofm1 on m2 is given by:

F =Gm1m2

r2 (12.1)

where r is the distance between the two objects and G =6.67 1011N m2/kg2 is the gravitational constant. Thisforce is an attractive force, such that the direction of theforce on m2 is towards m1. Also, note that the Newtonsthird law holds, because the reactive force has the same formas that above, that is:

Fm1onm2 =Fm2onm1 = Gm1m2

r2

(12.2)

and the direction is opposite with each other (because itsattractive).

Examples

Several examples include the pretty straightforward: PS7-8 or PS7-16. When there are many objects involved, thensuperposition is needed: Q6-7.

12.2 Acceleration due to gravity, mass andweight

When you consider one of the object as that of the earth,then we can calculate the force of earth on any object.

The weight of the object is essentially the force by earth,and hence decreases as the object gets farther from earth.

The acceleration at the surface of the earth on any massm1 can be computed by equating to m1g the force due toearth, hencem1 cancels and we obtain:

g= Gme

r2e(12.3)

whereme = 5.97 1024 [kg] is the mass of earth, and re =6.38

106 [m] is the average radius of the earth.

In any planet of massmp and radiusrp, the accelerationof gravity has a similar form:

gp =Gmp

r2p. (12.4)

Examples

Examples include comparison of the surface gravity orweight at several heights: (PS7-6, PS7-22) and that of sev-eral planets (PS7-7, PS7-20). These problems are very basicand fundamental in nature, so it should be obvious that thisis one of the many focus in any exam.

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13 Ch 14: Fluid Mechanics

Fluids is a nice chapter!

13.1 Density

Density is defined as:

= mass

Volume

= m

V

. (13.1)

It is an intrinsic property: it is independent of the size ofan object.

Examples

The calculation of density usually involves calculation of thevolume. Now volume of regular objects like the sphere orcube should be familiar to you, and thus formulas will notbe given on the exam. Problems include being careless (likeforgetting the formula, or not taking the cube of a factor),or panicking because there is no time. Examples includePS8-10 to 11

13.2 Pressure in a fluid

Pressure by a liquid is defined as:

p= F

A (13.2)

where A is the cross sectional area of the fluid, and F isthe force perpendicular to the area, similar to our definitionof stress. For liquids, pressure varies with depth through:

p= p0+ gh (13.3)

whereh is the depth. We can selectp0 to be exposed to the

earths atmosphere (e.g. sea-level) such that p0 = patm =101325[Pa] = 1 [atm].

The pressurepabove is also called absolute pressure. Wealso define gauge pressure as p patm. When p0 = patm,then the gauge pressure (pg) is:

pg =gh (13.4)

Gauge pressure is hence easier to calculate. But be care-ful. The gauge doesnt include the atmospheric pressure, sodont include it in your calculations.

Examples

The examples include PS8-12,13. This concept is important,and you have to study how to solve these problems.

13.3 Buoyant force

When an object is floating or submerged in a fluid, thenthere is buoyant force brought about by the depth variationof the pressure. The buoyant force has the form

F= gVdispj. (13.5)Here Vdisp means the volume displaced by water, which isthe volume of the object that is submerged in water. hereis thedensity of the fluid, not the object.

Examples

Familiarize yourselves with this concept through several ex-amples: PS8-13 to 15, Q7-1, Q7-3, Q7-5. You must knowhow to solve these problems before exam!This is important.

13.4 Continuity equation

For incompressible fluids, the volume flow rate:

dV

dt =Av = constant (13.6)

whereAis the cross sectional area and v is the fluid speed,This implies that:

A1v1= A2v2 (13.7)

This is the continuity equation.

Examples

Examples include: PS 8-18, Q7-2, Q7-5, Q7-6, the showerhead problem done in class. Familiarize yourselves withthese examples as this will be important!

13.5 Bernoulli equation

The Bernoulli equation states that:

P+ gh +1

2v2 = constant (13.8)

While this is a long equation, you can think of this as thefluid counterpart for energy conservation, with the mass be-ing the density. This has important concepts like as speed

of fluid increases, the pressure decreases.

Examples

Examples include: PS 8-17, Q7-2, Q7-5. Another importantconcept!

This ends Chapter 14!

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14 Ch 13: Oscillation

Oscillations are by definition objects with back and forthmotion.

14.1 Simple Harmonic Motion (SHM)

Simple harmonic motion has the property that you can ex-press your equations of motion as:

d2x(t)

dt2 = 2x(t) (14.1)

where is called the angular frequency. The solution forthis equation is:

x(t) =A cos(t + ) (14.2)

whereis some phase angle that is determined at the initialposition of the object. We can select this to be zero, whichimplies thatx is initially at its amplitude:x(0) =A.

The plot of x(t) is sinusoidal. From the graph, you

should be able to find the period and the amplitude. Theangular frequency can then be computed byT = 2/. SeeExample: PS9-1

The objects speed and acceleration can then be solvedstraightforwardly by differentiating the position with time.Hence:

v(t) = dx

dt = A sin(t + ) (14.3)

a(t) = d2x

dt2 = A2 cos(t + ) (14.4)

We therefore see that a = 2x at all times. Also,we note that when the position is maximum in magnitude(and hence also acceleration), then the speed is a minimum.And also vice versa. This is because when cos is max (inmagnitude), then sin is minimum (in magnitude).

The energy of SHM can be computed by noting thatwhen the kinetic energy is zero, thex= A, henceE= U=12

kA2, wherek = m2. This energy remains constant at alltimes. The kinetic energy is maximum when U=x = 0, orwhen the system is at equilibrium.

Shadow

The shadow of an object moving in uniform circular motionexhibits SHM. We can see this from the diagram below:

The shadow then is the x-component of the position ofthe object. Hence x = r cos . Noting that the angularspeed s = v/r is a constant, then we can express =st + 0. Hence the shadow has the solution:

x(t) =r cos(st + 0). (14.5)

From the solution to SHM (14.2), we see that = 0 and = s and the amplitude is r. Hence the shadow exhibitsSHM.

Figure 6: Shadow of an object in uniform circular motionexhibits SHM

Spring-mass system

Spring-mass system exhibits SHM, with 2 = km , where kis the springs constant and m is the objects mass. Theperiod is hence:

T =2

= 2

m

k (14.6)

You must memorize these formulas, or derive it from FBD.The latter is however not advisable for the exam because ittakes too much time.

The spring-mass system has the same period if it is mov-ing horizontally or vertically.

Examples include: PS8-3, PS8-4.

Simple pendulum

For simple pendulum of small oscillations, we have2 = gl ,where l is the length of the spring. The period is hence:

T = 2

= 2

l

g (14.7)

You must memorize these formulas!!Very important! I thinkam more familiar with the period, because it is used veryoften. Then from that, I just calculate the frequency fromT = 2/.

Examples include: PS8-6, PS8-8. You really must un-derstand this part. Otherwise, you dont have much hopefor this part of the exam.

Physical pendulum

For physical pendulums, or for rigid bodies of small oscilla-tions, we have 2 = mgdI , where m is the mass of the rigidobject,d is the distance of the axis of rotation to the centerof mass, andIis the moment of inertia about the axis. The

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period is hence:

T =2

= 2

I

mgd (14.8)

It doesnt have to be said, but I will say it anyway.. this isimportant!

Examples include: PS8-9, PS8-12. These examples arereally important, so please understand them!

Damped Oscillation

This part is more conceptual. For SHM, the amplitude doesnot decrease with time, which is not the case when frictionis present. When the amplitude decays in time, then thisis what we call damped oscillation. For several models, thefrictional force has the formbv such that the equations ofmotion has the form:

d2x

dt

2 =

2x

b

dx

dt

(14.9)

Note that b is the damping parameter. When b is large,then the systen is more damped, or high in friction. Thesolution to this equation is:

x(t) =Aeb/2m cos(t + ) (14.10)

where has the form:

=

k

m b

2

4m2. (14.11)

You dont have to memorize these equations for the thirdexam, because its a bit convoluted I must say. But you

have to understand the following cases:Caseb = 0: then it is SHM.

Caseb 2m, then is imaginary. This is called over-damping.

Also, you have to understand the plots for these cases asshown below.

Figure 7: Oscillation for different damping

What this means is that underdamping oscillates butwith amplitude decaying to zero. The critical damping how-ever does not oscillate and the amplitude immediately goes

to zero. For overdamping, the rate of decay is slower thanthat of critical damping, but it also doesnt oscillate butjust decays.

Forced Oscillation

Oscillations can also be driven, that is you can simultane-

ously give a force on the object together with the restoringforce. An interesting scenario occurs when the force is alsosinusoidal, that is of the form F = F0cos(Dt), where Dis the driving angular frequency. When this happens, thenwe obtain an amplitude plot with D as:

Figure 8: Amplitude vs driving frequency for differentdamping parameters

We therefore see that asD approach the natural angu-lar frequency (angular frequency without damping or driv-ing force, or that solved using SHM), then the amplitudepeaks! This peaking of the amplitude is called resonance,

hence the resonance frequency is also the natural frequency.Note that: frequency is f and angular frequency is .

This is important! Do not be confused about this. They arerelated by = 2f, such that T = 1/f.


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15 Ch 15: Mechanical Waves

Mechanical waves are waves that travel in a medium. Theyhave the form

f(x vt); wave propagating to the positive xf(x + vt); wave propagating to the negative x (15.1)

For sinusoidal mechanical waves, it has the form:

y(x, t) =A cos(kx t + ) orA cos(kx + t + ) (15.2)

depending on the propagation of the wave. We call stillas the angular frequency, while k is the wave number. Wedefine = 2/k as the wavelength of the wave, which issimilar to the period but with the space instead of time. Thespeed of propagation of the wave is defined as v= /k = f .Let us list the important formulas:

k=

2

= 2f

T = 1

f =

2

v=

k =f . (15.3)

These are equations that you have to learn by heart!

When given a plot of y vs x and y vs t, you should be ableto determine easily the amplitude, the period and the wave-length. After this, you should be able to compute f , k, , v.Do this quickly!Remember, if it takes you a minute to findany of the variables, then you are too slow, and hence needsmore practice.

Examples: See PS10-1, PS10-3 for example. Do this

quickly! Solve each in under two minutes, and you shouldbe fine. Remember, practice makes cliches!!

15.1 Types of Mechanical Waves

Types of Mechanical waves: Transverse waves and Longitu-dinal waves.

Transverse waves are those where the particles in themedium move in a transverse direction (perpendicular tothe direction of propagation). Hence,

y(x, t) =A cos(kx t + ) (15.4)

is a transverse wave, because the particles of the medium ismoving in the y-direction.

Longitudinal waves are however, when the movement ison the same direction of the direction of propagation.

Note about transverse waves:

The propagation speed is the speed of the movement ofthe pulse. The medium did not move in the propagationdirection, but just in the y direction. But there is a seemingmotion of the pulse, because the crest seems to move.

We can also define transverse speed as the speed of theparticles of the medium. This is just vy = dy/dt. This isnot the same as the propagation speed!

15.2 Transverse wave on a string

When a string with tension is given a transvers speed vy,then there is a speed of the transverse wave, or propagationspeed, v given by:

v= FT

(15.5)

where = mass/length = m/L is the mass density of thestring. HereFT is the tension of the string.

The average power due to the string is given by:

Pav = 1

2

FT

2A2 (15.6)

where is the angular frequency of the wave, and A is itsamplitude. There is also the maximum powerPmax= 2Pav.Dont bother too much with Pmax, just Pav.

The loudness of the sound of the string is described by

the intensity, defined by:

I= Pav4r2

(15.7)

where r is the distance of the listener to the source of thesound. These formulas are important and must be memo-rized, because it will be too involved to derive them.

Examples

Several examples that require some or all of these formu-

las include PS10-4,5 PS 10-14,15. You have to know theseproblems, because they are important!

15.3 Standing waves

When two waves collide, then they superpose, or you justadd them! There are some cases though that after two wavesare added, the resulting wave doesnt seem to propagate,that is a crest of the wave doesnt move. These cases arecalled standing waves. Two waves that have the same fre-quency and wave number, and same amplitude but movingin opposite propagation direction can be a standing wave.

We define the node of a standing wave as the points

which does not move. The antinode are the minimum andmaximum values of the wave, which

Figure 9: Node and antinode

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Note that you must know how to draw a wave with acertain number of nodes or antinodes. For example: drawa string of length L with 10 nodes. You should also beable to show that the frequency of the same stringwith 10 antinodes is higher than that of 10 nodes,or 5 antinodes or 5 nodes.

15.4 Harmonic frequencies of a stringFor a string with both ends fixed, you can obtain severalnatural frequencies as shown in the diagram below.

Figure 10: Harmonic frequencies

The fundamental wavelength is the largest wavelengththat you can achieve. This happens when 1 = 2L. Withv= f , then we obtain the fundamental frequency as:

f1= v

2L (15.8)

For other harmonics, we can calculate the wavelength andthe frequencies as:

n =

1

n =

2L

n ; fn = nf1=

nv

2L ; n= 1, 2, 3... (15.9)

The fundamental frequency has the highest amplitude,and is usually heard most prominently (I A2).

You have to memorize these formulas, or just draw thelargest wavelength you can achieve when both ends arefixed. However, it might be easier to derive them.

Examples

Familiarize yourselves with some important examples aboutstrings. For example: PS 10-2, PS 10-6, PS 10-9, PS 10-13.


16 Ch 16.8: Doppler Effect

When a source emitting a frequency is moving, then thefrequency heard by the listener will be different than thefrequency from the source. This is Doppler effect.

These are the important concepts you have to under-

stand for Doppler effect: When the source and/or listeneris approaching each other, then the frequency heard by thelistener will increase. If its going away, then its the opposite.That is:

fL> fs: approaching

fL< fs: moving away (16.1)

Next.. you have to remember this factor:

fL= v vLv vs fs (16.2)

where v is the speed of sound in air. Note that we can

choose either plus or minus in both numerator and denom-inator, depending on whether the direction of movement ofthe listener and the source. We just have to remember thatif it is approaching, then the frequency of the listener has toincrease, and when moving away, it has to decrease. Thenwe adjust the signs accordingly.

For example, suppose that the listener is going to the xdirection with speedvLand the source is going to the +xdi-rection with speedvs. Then both movement should cause anincrease in fL because both are approaching. Hence to in-crease, the numerator should add, and denominator shouldbe negative. Hence the equation is:

fL= v+ vLv vs fs (16.3)

Now suppose that the listener is going to the x directionwith speed vL and the source is also going to the +xdirec-tion with speed vs. Then the listener is going away to thesource, hence causing a decrease of the frequency, so thatwe take the minus sign.Hence the equation is:

fL= v vLv vs fs (16.4)

Try it for other cases just to familiarize yourselves.While these equations look complicated, it is actually notas convuluted as it seems. You just have to know that:

fL> fs: approaching

fL< fs: moving away

fL=v vLv vs fs (16.5)

Examples

Examples include: PS10-10 and 10-11. Try to read moreexamples in the book for practice. This is very important!

That ends Ch 16.8!

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