Circular Reasoning: A Primer Of Periodic Phenomena

CIRCULAR REASONINGA Primer On Periodic Phenomena

by David Alan RogersUniversity of California, Santa Barbara

Contents

Chapter 1. Angles: Circles, Friendly Pulleys and Bicycle Mounted Foes. 41.1. The angle. 41.2. Applications of angles in simple dynamicical problems. 71.3. Interesting physical phenomena. 11Section 1 finale 14

Chapter 2. Periodic Phenomena: Symmetry, Phase and Women. 172.1. Some seemingly new concepts. 17

3

CHAPTER 1

Angles: Circles, Friendly Pulleys and BicycleMounted Foes.

1.1. The angle.

The notion of an angle relates a points positon on a circle to a section of saidcircles total circumference. The fundamental definition of the angle is that it is anumber relating the radius of a given circle to an arc length. The arc length is justa portion of a circles entire circumference. In the below figure, a section of circle ispicture along with the angle subtending it.

Figure 1.1: The arc length of a circle as related to its angle and radius.

The figure 1.1 represents the fundamental relation

r = s

where s is the arc length subtended by the angle and r is the radius of the circle.

1.1.1. Degrees as a unit of angle. Our most common experience with anglesis given in terms of degrees. Degrees are defined such that to return to a point on thecircle that you started at, you must traverse a total degree of 360o. The origin on thecircle is at 0o and we choose that point to be the beginning of the first quadrant onthe circle. In the figure above, the angle 0o is where the horizontal line of the sectorconnects the center of the circle with its boundary.

The circle can be divided into four distinct quadrants (notice we head COUNTERclockwise about the circle, less we choose to traverse negative angles). Because thetotal number of degrees encapsulated by the circle are 360, each quadrant shouldcontribute 360o/4 = 90o to the total. Therefore, the range of degrees specifying apoint inside each quadrant are given by;

4

1.1. THE ANGLE. 5

[0, 90o) QUADRANT I[90o, 180o) QUADRANT II[180o, 270o) QUADRANT III[270o, 360o) QUDADRANT IV

Notice that in effect, all we are doing to differentiate the quadrants is rotatingthe lower limit on the preceding quadrant by 90o. One also notices that because acomplete rotation of 360o returns oneself to the same point, we can identify (regardas the same) the angles 0o and 360o. In general, we have that for any angle , if werotated by an additional 360o we return to the same angle, in mathematical term

+360o =.But note that if is say, 30o, our equation is saying

30o+360o = 390o.In order for our mathematical view of rotations to be logically consistent, we

must say that 390o is an equivalent expression of the angle 300. In higher levelmathematics (namely Abstract Algebra), this notion is a special case of so calledequivalence classes.

EXAMPLE 1. Suppose we are given the angle 750o and wish to determine thesmallest angle which equivalently expresses it. We are motivated to do this, becauseattempting to place a point on a circle at an angle of 750o would entail rotating allaround the circle several times and then adding the remainder of degrees after totalrotations to designate our point. Lets not do that.

Solution: We first note that all angles can be expressed as a sum = n(360o)+where n is an integer (...,2,1, 0, 1, 2, ...) and is some angle between 0o and 360o(this is the smallest equivalent angle that we want to find). To determine n, we askourselves how many times can we fit 360 into 750? The quickest way to determinethis is to simply divide,

750360

2.083.n is clearly 2, as the remainder (0.83) after division is not an integer. Therefore,

=n(360o)= 750o2(360o)= 30o.Our general method to finding the smallest equivalent angle is then to divide by

360, discard the remainder and apply the above formula. The formula is not neces-sary to memorize if you understand the reasoning, which is the point of mathematics.

NOTE: On the following page is a figure of a circle and its four quadrants. Thequadrants are labeled counter clockwise, and as angles can be equated to a negativerepresentation in addition to 360o rotated analogues, these numbers are also labeled.A superfluous detail of this convention is that it evolved out of the right-handednessof human being, oddly enough. In a world dominated by lefties, the convention wouldbe reversed. One notices that if they curl the fingers of their right hand in the direc-tion of increasing angle, the thumb distinctly points out of the page. This seeminglyirrelevant fact is important in many fields of mathematics, where the orientationof an object requires rigorous conventions in order for calculations to remain self-consistent. The circle in the Euclidean plane is an example of an orientable manifoldwith the right-handed convention.

1.1. THE ANGLE. 6

Figure 1.2: The various quadrants of the unit circle and their ranges of angles.

1.1.2. Radians as a unit of angle. While degrees are an appropriate choicefor discussing the magnitude of angles in day to day speech, they lack features thatmake them powerful as a tool for calculation. As we discussed at the start of thissection, the defining feature of an angle is that it relates the radius of a circle to asection of its circumference (an arc length),

r = s.Long ago in antiquity, it was proposed that there existed a number (that we call

pi) which related the diameter of a circle to its circumference. That is, a numberwhich relates the total length around a circle to the length between two oppositepoints. One of the most numinous aspects of nature is not only that this number piexists, but has a never ending decimal representation. The number pi is possibly themost fundamental number in all of nature. In the language of modern mathematics,we define pi as understood by primitive geometers with the equation

pid =Cwhere C is the circumference of a circle and d is its diameter. The most advanced

ancient civilizations had good approximations to the value pi and approaches to theevaluation of pi are still an interest of many computer scientists.

Our notion of angle has a defining feature that if we traverse the total angle T(it is common in math and science to label certain objects with a subscript, the T isjust a label, you do not manipulate it). around the circle, we ought to have coveredthe total arc length or circumference of the circle,

rT =C =pid,the last form can be written in terms of the radius, d = 2r,

1.2. APPLICATIONS OF ANGLES IN SIMPLE DYNAMICICAL PROBLEMS. 7

rT = 2pir.Low and behold, what we have arived at is the most natural representation of

the total angle contained within a circle, this motivates our adoption of a unit of anglecalled radians. To denote the rest of the angles in the circle, we simply subdivide thetotal angle T = 2pi. Half of the total rotation is pi, a quarter is pi/2, etc. These areobviously logically equivalent to the rotations in units of degrees; pi corresponds to180o, pi/2 corresponds to 90o, etc. Notice that in the case of the unit circle, where r = 1by default, an angles measurement in radians is numerically equal to the length ofa corresponding arc.

One may easily convert from radians to degrees, via the correspondance of 2pirad and 360o. Suppose we are given an angle in radians, there are 2pi radians perevery 360o rotation, ergo

(3602pi

)radians=

(180pi

)radians=o.

The figure below labels points on a circle and those points corresponding anglesin both degrees and radians.

Figure 1.3: Angles in both units of radians and degrees along the unit circle.

1.2. Applications of angles in simple dynamicical problems.

In the world of natural science, a problem which does not vary in time is saidto be static. An example would be a farmer calculating the area of his rectangular


field, given the side lengths of the property line. However, the notion of angle allowsus a method to discuss rotational dynamics, problems where objects are rotating intime. Because the reader lacks knowledge of calculus, we are constrained to systemswhich are said to evolve uniformly. For now, we will only concern ourselves withsystems which undergo uniform motion.

1.2.1. Angular frequency and uniform circular motion. If an object is ro-tating uniformly (does not speed up or slow down, nor deviates from a circular trajec-tory) we know the length it travels in any interval of time t is numerically equivalentto some arclength of a circle. In this case, the rate at which the angle of the objectchanges is given by

= t

so that,

rt= r tt= r = s

gives us the arclength traversed in the time interval [0, t]. is called the angularfrequency.

Note that we are concerned with a time INTERVAL, we have simply chose thestarting time to be t= 0. In general, the time interval need not start at t= 0 but canstart at any value, t0 (remember, the subscript 0 is just a label). In this case, ourtime interval is [t0, t] and the time traversed in this interval becomes t = (t t0).The notation t is just another short hand, reminding us that our time interval didnot start at t0 = 0. With this small generalization, we get

rt= rtt= r

(t t0)(t t0)=s.

Why s and ? Well, we often discuss an objects position on a circle relative tothe origin at = 0, at this point s= 0. But just as we did with time, we could look atsystems where an object starts at an arclength of s0 from the origin. In which case,as the object rotates, the arclength it covers is actually the difference s s0 whereboth s and s0 are measured relative to the origin. We must also take into accountthat the positions along the arc of the circle correspond to different angles. Lets lookat an example to help our understanding.

EXAMPLE 2. An object is engaging in uniform circular motion. An observer no-tices the object is located at an angle of pi/4 radians when his timer reads 7 seconds.When the timer reads 9 seconds this object has arrived at pi/2 radians.

Solution: Lets analyze this system. One thing we notice we can determine withthe information we are given is the objects angular frequency, as we are given thatthe object is engaging in uniform motion along a circle and we see the motion occursin the intervals [pi/4, pi/2] radians and [7, 9] seconds. Ergo we take,

t0 = 7t= 90 =pi/4 =pi/2

and find, t= 2 seconds and = pi

2 pi

4= 2pipi

4= pi

4radians.


Hence,

= t

= pi8

radianssecond .

Now lets suppose that we somehow determine the radius of the circular trajec-tory (in the next example well discuss an interesting way to do this). r =p2. Letsfigure out the amount of arclength the object has covered. There are infact two ways,

rt=sholds one method,

s= 2p

2pi

8= pi

p2

4= pi

p2

2 (p2)2 =pi

2p

2while

rt= r(

t

)t= r =s

holds another,s=

p2(pi

4

)= pi

2p

2.

Thus far we have only discussed the angular frequency of a rotating object. In thereal world, the concept of speed is a much more familiar topic. Everyone knows thatspeed (denoted v) is just (distance)/(time), but does everyone know its relationship toangular frequency? You do.

rt=s = r= st

= v.

We just need to be slightly careful here, v is actually called the tangential speedbecause it is the speed of an object moving along a circular path. The speed of anobject you are most likely familiar with is defined for objects moving in a straightdirection and depends on its displacement per unit time. That is, the distancerelevant to the true linear speed is the length of the straight line connecting itsinitial and final positions. When one learns calculus, they see that the displacementper unit time treatment is a very poor description for curved trajectories, such ascircular motion. The speed of an object is a much more subtle concept and requiresthe notion of instantaneous small displacements.

Lets see more interesting examples of a system engaging in uniform circularmotion before we press on.

EXAMPLE 3. TJ is a dick. He is on his bike and has just stolen your trigonometrytextbook! Like the blustering child he is, he engages in uniform circular around youat angular frequency , while taunting you. Every given number of seconds, heyelps out the word loser! He does so consistently. You first ask for you textbookback nicely, he refuses. You then demand, he does not care. Finally after the tenthloser he is at an angle 1 on this circular trajectory and you say enough is enough,right as he is about to say loser once more, he is pelted in the head with a rockthrown by you at a speed v. How far are you from his unconscious body located at2?

Solution: To do this problem, lets first note that what were being asked to do isdetermine the radius of the circular trajectory, as you are the observer located at theorigin and at all points in the motion TJ is r distance away from you.

We are given the two angles, 2 and 1. At 1 you are fed up, and at 2 he hasJUST been hit in the head with your rock, which instantaneously drops him to the


ground. Because the rate at which he shouts his insult is constant, you know thatthe time it takes the rock to hit him t is the same as the amount of time it takeshim to travel radians. We however, are not given this time. It must be eliminatedalgebraically. Note,

=t.Because you are stationary, the rock must have traveled the wanted distance r uponimpact, that is

r = vt.Solving for t in the lower equation and substituting into the upper one,

= rv= r = v

= v

(21) ,and we now have a formula for r in terms of quantities we are given. Notice thishas all been done symbolically. If I were to specify real values for all these variables,one could simply plug them into the result. Most the time in problems where we aregiven values for certain variables, it is best to work out the problem algebraically(symbolically) before plugging the relevant numbers in; it is an extremely good habitand as this text progresses we will trend towards doing examples where variablesare not given specific values at all.

In the next example, we will look at how circles of different radius are related toeach other.

EXAMPLE 4. Two circles of differing radius are said to be concentric if they sharethe same center. Suppose we have such a system and the interior circle has a radiusr1 = 2/3 while the exterior circle has a radius of r2 = 3. We start a clock and pointsat = 0 on both begin moving at the same uniform angular frequency, . If the pointon the interior circle covers an arclength of s1 =

ppi meters in t seconds, at what

angle is point on the second circle and how much arc has it traversed?Solution: Because the points move at the same angular frequency, we have the

very important condition:1 =2 =.

This means that,

1 =2 = 1t

= 2t

,

or equivalently 1 =2, and since we know both start at an initial angle of 0,we might as well omit the deltas and call 1 and 2 the final angles after time t,

1 = 2.In other words, we need only find the angle 1 for the interior circle and it wil be

the same as the exterior one. Lets just use

r11 = s1 = 1 = 2 = s1r1= 3

ppi

2.

To determine the arc, we apply the same formula but on the exterior circle;

r22 = s2 = (3)3ppi

2= 9

2ppi.

Note that the radius being 3 times larger has made the covered arc 3 timeslarger. This is no coincidence, as

r22 = r21

1.3. INTERESTING PHYSICAL PHENOMENA. 11

for concentric circles and therefore,

s2 = r2(s1r1

)=(r2r1

)s1.

1.2.2. The similarity of circles. What we learned from the previous exampleis that the ratio of the radiuses of circles determines how the arclength is scaled(made larger or smaller). In the very important case that the interior concentriccircle is the unit circle where r1 = 1, we see that as the exterior circle becomes larger,the arc length is simply rescaled by the size of the radius r2,

s2 = r2s1.This is why the unit circle in a sense the most fundamental. Problems involvingcircles of varying size are simply rescaled versions of problems involving the unit cir-cle. This property is called the similarity of circles. When we arrive at trigonometricfunctions, we will see that a change of amplitude corresponds to a rescaling of theunit circle.

1.3. Interesting physical phenomena.

There are two common special cases of rotational dynamics that I will brieflycover here. These two examples are of great importance in applied physics and en-gineering, as such I will try to present them in a context which gives some relevanceto the definitions.

1.3.1. Rolling without slipping. Imagine a bicycle wheel rolling on a flat sur-face. Lets put paint on the wheel, so that as it rolls it leaves a trail. If the wheeldoes not slip or skid, we know that every speckle of paint in the straight line trailit leaves behind on the ground over a given time period came from one and only onepoint on the surface of the wheel. If the wheel DOES skid or slip, that means thatsome of the paint in the trail didnt actually come from points on the wheel but wasinstead smeared along the trail. The mathematical relation describing this conditionis,

s=xwhere x is change in position (distance of travel) of the center of the wheel

relative to some origin and s is the arclength a point on the wheel has covered whilerolling through an angle of . In otherwords,

r =xwhere is corresponding angle of rotation and r is the radius of the wheel.This relationship should of course be made dynamical, since rolling occurs in

time. x/t = vwheel is the speed of the center of the wheel and therefore the bike,since the all points on bike move at that same speed (otherwise the thing would befalling apart!), so we arrive at,

vwheel =xt

= rt

= vwheel = r.NOTE: Now, were not so much interested in physics in this text but I really

want to add a quick asside because it is perplexing but cute. You are maybe con-fused as to how the relation s=x suddenly appeared, my heuristic analysis is a bitmessy. It is infact a simplified version of a more complicated mathematical expla-nation which would require a knowledge of calculus/analysis. There is also a very


important and cute physical treatment that describes rolling without slipping. Thephysical explanation of rolling without slipping is that as the wheel moves along thesurface, the point on the wheel that meets the ground at any given time has zerovelocity. Pretty zen, right? If this point did NOT have zero velocity, it would bemoving relative to the point of contact and we would have slippage contributing tothe distance traveled. So how does the wheel rotate in the first place? The answeris you need the concept of acceleration, as while the edge has zero velocity at thispoint, its acceleration is non-zero. The condition that the velocity be zero at contactensures that the translational motion of the wheel (movement of its center relativeto the surface) is exactly the same as the arclength covered on the surface of thewheel. But enough of that.

Lets quicky a prototypical example of a rolling without slipping problem.

EXAMPLE 5. A ball with radius r = 3/2 is rolling uniformly with angular fre-quency = pi/p2 without slipping in a straight line across the surface of a table.Find the translational speed of the ball and use this to determine how much dis-tance the ball will roll in 2 seconds.

Solution: Very simply, vball = r= 3pi2p2 and therefore,

vballt=x=3pi

2p

2(2)= 3pip

2.

So, that wasnt very hard. Lets ask a more interesting question.Suppose the length of the table is l = 30 and the ball in the example above starts

out at its edge and begins to roll towards the other. How long does it take for the ballto fall onto the floor?

Solution: What this is really asking is that when the distance traversed is x= l,how much time has gone by?

vballt= rt=x= l.

Ergo,

t= lr

= 30(3/2)(pi/

p2)= 20

p2

pi.

Sometimes were interested in other quantities, say the total angle rolled through.Staying with the example above, what is the total angle the ball rotates through

before it falls off the table?Solution: We know that,

r= vball = r

t= xt

, = r =x

When the ball has rolled of the table, x= l, so

r = l = = 30(3/2)

= 20.

Problems of this kind that are still available to the reader are simply going to beeasy algebraic manipulations of the fundamental equation describing rolling withoutslipping.


1.3.2. Coupled pulleys. Lets now turn our attention to another type of phys-ical problem, namely coupled rotation. A system of pulleys is said to be coupled iftheir motions are dependent on one another. Examples of such systems in the realworld include the toothed pulley and belt as well as the sprocket and chain any kidwith a bicycle is familiar with.

Figure 1.4: Different types of coupled pulleys found in day to day life.

One feature we may desire in a system of pulleys coupled by a belt is that thebelt does not slip on the surface of the pulleys. Indeed, this is precisely why in thecase of a bicycle one uses a sprocket and chain to couple the pedal to the wheel. Thisso that energy is not lost in the form of the chain slipping as one pedals. To describethis condition mathematically, let us first note that when we rotate the first wheelby an angle, a point on this wheel sweeps an arclength of

r11 =s1.Now, since the belt does not slip and is tight, s1 is also the length of belt that hasbeen pulled off of the second wheel (causing it to rotate as well). That means,

s1 =s2or equivalent,

r11 = r22.

SECTION 1 FINALE 14

The moral of the story is that when one rotates the first wheel by an angle 1,the second wheel rotates by an angle

2 = r1r21.

Note that if r1 > r2 then the angle of the second wheel is larger and visa-versa.The reason for this comes from the similarity of circles, as we discussed in example4. Refer back to this.

Of course, we are also interested in the dynamical case, as when we ride our bikewe are concerned with how fast we are moving in relation to the angular frequenciesin the pulley system,

r11 = r22.

Section 1 finale

Lets now do a non-trivial exercise that utilizes the concepts of a wheel rollinguniformly without slipping, a system of coupled pulleys and the concept of similarcircles (recall example 4). If you can follow this problem, you have made tremendousprogress. If you can do this problem independently you are a God. As promised,we will begin to do problems where variables are not given any real values. Whenan example says a quantity is known, given, or measured, this means that youare free to include it in your final expression of the uknown or the thing you wishto find. Also be aware that whatever notation you use is arbitrary, you can nameyour variables absolutely anything that you want! Be wary that consistency is veryimportant to make your work understandable to yourself and others, however. Iattempt to choose what in my opinion is the most logically consistent notation.

EXAMPLE 6. Today you want to go and steal TJ the Dicks backpack. You are in-terested in knowing how fast you can physically ride your bicycle given your ownfitness level to ensure your escape. You ask an accomplice to clock the averageamount of time it takes you to pedal one full rotation (this is called the period ofrotation) when you pedal full force. You also measure the diameters of the backwheel sprocket, the pedal sprocket it is coupled to via the chain and the wheel itself.Your wheels have very good traction, so assume that you roll without any slippage.Our goal is simply to manipulate the equations we have developed in this section todo this fun problem.

Solution: We want to find your maximum speed, well simply call this v. Letsstart with what we know. We know the average amount of time it takes for you topedal a full rotation, we will call this time T (motivated by what we will soon callthe period). We also know the diameters ds,w and ds,p of the wheel sprocket andpedal sprocket respectively. In addition we have the diameter of the wheel, well justcall this dw. In addition, we have the very basic fact that a full rotation of a circle isrepresented by 2pi radians. Believe it or not, this is absolutely all that we need. Tobe absolutely clear,{

What we know: T, ds,w, ds,p, dw, 2piWhat we want to know: v

.

First off, we want to know at what angular frequency you can pedal, as that is goingto determine the rate at which the wheel sprocket (and hence the wheel) can rotate

SECTION 1 FINALE 15

via our condition of coupled pulleys. Thats easy enough

s,p = 2piT .

Now, those diameters are a little unsightly since we are accustomed to usingradii, but we know that all we need to do is divide them by 2 to be able to convertthem to radii and substitute them into the relevant equations. So, our condition ofcoupled pulleys simply becomes,

12ds,ps,p = 12ds,ws,w.

Well look at that, we dont even need the 1/2s anyways, they simply multiply out

ds,ps,p = ds,ws,w.Since we want s,w,

s,w =ds,pds,w

s,p.

Now, the wheel sprocket is rigidly attached to the wheel, so they arent coupledin the same way two wheels are with a chain. That simply means that when werotate the wheel sprocket, it forces the wheel to rotate by the same angle (they areconcentric circles with points on their edges rotating at the same angular frequency,a la example 4!). Hence,

s,w =w =ds,pds,w

s,p.

Now we know the angular frequency at which the wheel rotates. Well, we alsoknow the diameter of the wheel itself AND that it rolls without slipping, so thatcondition is,

12dww = v.

We know both dw and w so the problem is actually complete, but lets put thingsinto terms we know directly so we dont have to plug real values in at each step whenwe want to actually calculate something.

v= 12dww = 12dw

ds,pds,w

s,p = 12dwds,pds,w

2piT

and after canceling the 2s and putting it in a more pleasant looking form, wehave

v= piT

(dwds,pds,w

).

Notice how we have the speed in terms of all the things we originally said weknew! We did accomplished exactly what we wanted.

Now to show you the power of deriving your very own formulas, lets actuallyplug some stuff into it. Suppose your wheel diameter is 1/2 of a meter, your wheelsprocket diameter is 1/7 of a meter, your pedal sprocket diameter is 1/16 of a meterand you average a very respectable time of a quarter of a second to pedal a fullrotation. We then have,

v= pi(1/4 seconds)

((1/2 meters)(1/16 meters)

1/7 meters

)= 28pi

32meterssecond .

SECTION 1 FINALE 16

If you are interested in a number you can understand, plug this sucker into a calcu-lator:

v 2.74 ms .Then again, youre American, so you might want to see how fast you move in milesper hour. Lets convert using dimensional analysis, the conversion factor from metersto miles I simply looked up,(

2.74 meterssecond)( 1

1609.34milesmeter

)(3600

secondshour

)= 6.12mph.

So, either you or your bike is crumby. Fortunately we were just making up num-bers, if you measured the real numbers for yourself you could determine a realisticanswer! Of course, this all assumes youre in vacuum, but youre not reading this tolearn physics.

Conclusion. This concludes the first section of this text. The last problem hope-fully gave you an indication of the cool stuff you can do already and convinced youthat you have honed a quality understanding of the material covered. In the nextsection we will introduce a few concepts that somewhat generalize what we havediscussed here before moving on to a discussion of the notion of periodic functions, ofwhich all trigonometric functions are simply a particular case.

CHAPTER 2

Periodic Phenomena: Symmetry, Phase and Women.

Thus far we have only concerned ourself with uniform circular motion as phys-ical problems outside of this realm are beyond the scope of your education at themoment. Uniform circular motion is an example of periodic phenomena. Roughlyspeaking, a periodic system is any system where things recur after a period of time.Periodic systems need not only involve objects rotating through space but includeany phenomena where something recurs consistently after a fixed amount of time.Not to be lewd, but the menstrual cycle of a woman is roughly periodic. The choiceof the term period to describe a semi-predictably time of the month when a womanwill crave BLTs and threaten her significant other with a broken wine bottle is nocoincidence. To a rough approximation, this biological cycle is indeed a periodic phe-nomenon.

2.1. Some seemingly new concepts.

Thus far we have only concerned ourself with uniform circular motion as phys-ical problems outside of this realm are beyond the scope of your education at themoment. Uniform circular motion is an example of periodic phenomena. Roughlyspeaking, a periodic system is any system where things recur after a period of time.Periodic systems need not only involve objects rotating through space but includeany phenomena where something recurs consistently after a fixed amount of time.Not to be lewd, but the menstrual cycle of a woman is roughly periodic. The choiceof the term period to describe a semi-predictably time of the month when a womanwill crave BLTs and threaten her significant other with a broken wine bottle is nocoincidence. To a rough approximation, this biological cycle is indeed a periodic phe-nomenon.

The time it takes a given event to recur is called its period and we generallydenote it T. The period gives us a way to speak of the frequency of an event as well.Every time the given even occurs (cycles), a time of T has traversed. This frequency(usually denoted or f . The is the greek letter nu and not a v, yes that causesconfusion sometimes) is defined to be = 1/T so that if a time of say 3 periods haspassed (t= 3T), the event has occurred t= 1T (3T)= 3 times.

2.1.1. The relationships between frequency, angular frequency, cycleand period. The units of frequency are cycles per second, or inverse seconds (alsocalled hertz). Why inverse seconds and not (something)/seconds? Remember fre-quency is a measure of the number of cycles that occur in a given time interval. Thenotion of cycles has no units, it is an abstract concept, not a well-define physicalnotion like distance or time. The units of angular frequency are radians per second,so the difference between the two is quite subtle.

17

2.1. SOME SEEMINGLY NEW CONCEPTS. 18

Think of their relationship in terms of dimensional analysis. For every completecycle on the unit circle, you have traversed 2pi radians. Therefore, the conversionfactor for this relationship is

2piradians

cycle.

Suppose a particle on the unit circle returns to its initial position with a frequency of cycles per second, we can convert this into an angular frequency via this conversionfactor, (

cyclessecond

)(2pi

radianscycles

)= 2pi

(cyclessecond

radianscycles

)= radians

second.

From this, we can write down some useful relationships for solving periodic prob-lems.

= 2pi= 2piTT = 1

= 2pi

= 1T = 2piI would advise that one not remember all of this formulae of course, they are all

just rearrangements of the two important facts:{= 2pi= 1T

and if you ever find yourself unable to recall these, just think about it!

Circular Reasoning: A Primer Of Periodic Phenomena

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