Splash Physics

Physics Problem SolvingRavi Charan, Anu Sinha, David Xiao MIT ESP Splash 11/19/2010

All problems are taken from David Morins textbook Introduction to Classical Mechanics.David Morin is a professor at Harvard and his website can be found at:

www.physics.harvard.edu/people/facpages/morin.html

For materials from this course, including a digital copy of this handout, the LATEXcode, thetemplate, help with solutions, or any other inquiries, contact Ravi Charan ([email protected]),David Xiao ([email protected]), and/or Anu Sinha ([email protected]).

As a preface to all of this: please dont be intimidated. If you find this way too hard anddont think youd ever have to do this, dont worry, MIT students who are not physics majors (ormechanical engineering) are never expected to know a lot of this material (you do need to knowthe calculus and almost all majors require differential equations however). If you find this materialeasy, we strongly suggest you do self study, do physics olympiad, and come to MIT and be beastat physics (but feel free to major in whatever you want)

The textbook is a problem solving based approach to mechanics and covers Newtons laws,oscillation, energy, the Lagrangian method, central forces, angular momentum, accelerating framesof reference, and an intro to relativity, and is highly recommended for further study for those witha strong mathematics background.

This handout is a brief selection of just a few interesting problems and topic, designed toconvince you that physics is really interesting and has a lot to learn that your high school physicsteacher may not ever mention exists. We almost certainly wont make it through this entire handouttoday, but well try to get through the first four sections. The first is a quick introduction to a coolidea, the second is by far the most difficult problem, so well cover it here. The third is a way tocalculate some wacky things, and if we can get to the fourth, well do a demo of the result, whichis moderately surprising. The fifth is a fun puzzle. The remainder of the sections are yours to lookat if you are interested, including three more technical sections (the last two being largely aboutmath). They should be approachable by yourself.

If you wish to do self study and are interested in using Morin, we highly recommend it. You willcertainly need differential calculus. All of the differential equations are explained in the textbook,much as in the second section. Morin strenuously avoids linear algebra, so you wont need any ofit if you dont want it, but it certainly helps. Youll want to be good with integrals, and finallyvectors are extremely important in physics, though we dont use them much here, so cross productsare nice. The appendices in Morin contain a lot of information about the multi variable calculus.In general though, Morin will introduce most of the math that youll need, and being forced to usea lot of the math is a good way to make yourself learn that math really well. If youre interested inself study, we are more than happy to provide a syllabus and good exercises. We recommend youalso be prepared to persevere through some pretty dense readings though.

Finally, we strongly recommend you try problems before reading the solutions (here, and ingeneral). Especially this first one. Its pretty easy:

1 of 18

Physics Problem Solving Ravi Charan, Anu Sinha, David Xiao

1 Unit Analysis

Problem 1. (Unit Analysis) Suppose you are given two quantities: a, with units of kg, b, withunits of m/s2, and c, with units of m. Find an expression in a, b, and c (not necessarily all needappear however) with units of m/s

Solution Clearly the only possible combination is something of the form kbc for k a scalar

constant. Note that a cannot appear in the equation, because nothing would be able to cancel itsunits of kg. Likewise, we can deduce that b and c must appear in this combination.

Obviously this is as uninteresting as a problem can possibly get (and were more than happy toprove ourselves wrong if you want us to)

Problem 2. (Unit Analysis) A satellite of mass m travels in a circular orbit at a height r abovethe center of the earth. Find an expression for its speed

Solution From an intuitive standpoint, the only thing the answer should depend on is m, r,and g. It wouldnt make sense (at reasonable velocities) for the answer to depend on c, the speedof light, or R, the gas constant. In fact, we can even guess that it wont depend on m, but that isa less reasonable assumption that non-dependence on c or R, and we wont need it.

But then the units of these three quantities are exactly in the above problem. So the answermust be k

rg. In fact, k = 1, but even without knowing that, this is still a lot of information to

gain just from unit analysis.

The limerick below is courtesy David Morin, and illustrates the extensive humor availablethroughout his textbook (and its actually funny) and on his website. We strongly recommend theextensive litany of chicken-crossing-the-road jokes.

Always check your units!

Your units are wrong! cried the teacher.Your church weighs six joules - what a feature!And the people insideAre four hours wide,And eight gauss away from the preacher!

One of our goals today is going to be to convince you that, while physics is a lot of math, thereis more than math to physics. Hopefully unit analysis is a good start. Next were going to showsomething really cool

2 The Quantum Pencil

Problem 3. (3.13a Balancing a Pencil) Idealize a pencil as a mass m at the end of massless rodof length l. Consider a pencil standing upright on its tip. Assume it makes a small angle 0 > 0with the vertical, and has initial angular speed 0 > 0. Let its angle from the vertical as a functionof time be (t). While is small, what is (t) explicitly?

MIT ESP Splash 11/19/2010 2 of 18


Solution At the risk of throwing you in headfirst a bit too fast, this problem is going to re-quire a little bit of basic calculus, some differential equations, some knowledge of torque, and useof the most important mathematical concept in physics: The Small Angle Approximation

Torque is the force applied (mg) times the distance from the pivot to the point of application,l, times the sin of the angle between the force vector and the pivot/radial vector. since the mass isat the end only, it is not l/2. The torque on the pencil is

= mgl sin (1)

The moment of inertia of the pencil isI = ml2 (2)

(it would be ml2/3 if we modeled the pencil as a continuous mass). Also, we know that:

= I (3)

Where is the torque, I is the moment of inertia, and is the angular acceleration, which is thederivative of the angular velocity , which is the derivative of the angle, :

=d

dt=d2

dt2(4)

(If that equation is too easy for you, then the vector form is ~ = I~, where I is a 33 matrix calledan inertia tensor, and and need not point in the same direction. If you know cross productsand linear algebra, you should find time to read about them). Also be aware that often physicistswill write = = where the dots represent time derivatives (one or two, I dont think theybother with three), for convenience. Anyways, this application gives us that:

d2

dt2= =

I=mgl sin

ml2= g sin /l (g/l) (5)

Note the use in the last (approximate) equality of the small angle approximation, sin = . As arule, whenever a problem says small angle, you can make this approximation. However, if youare serious about physics, its worth understanding when and why this is actually valid. See thesection on the binomial theorem for a justification. In the meantime, just do this.

This is a good time to perform the two most important sanity checks in physics:

1. Are the units correct? is unitless, so has units of 1/s2. (g/l) also has units 1/s2. Phew.

2. Is the sign correct? Well, as increases, we expect it to fall faster and faster, so yes the signis correct.

Please note: If the sign is wrong, dont just add a minus sign (like your physics teacher mighttell you to do). Go figure out where you missed the minus sign earlier. Did two things pointin opposite directions? Imagine if the force was applied so it caused to decrease. Then should be negative instead.

At this point we are going to throw some differential equations at you. Most physics youll bedoing through early college (except for weird things like modeling turbulent drag flow) is linear-ish



and produces linear differential equations. Well take a break and show how to solve the two mostcommon types of differential equations:

d2

dt2= +2 (t) = Aet +Bet (6)

Produces exponential functions. A and B are constants depending on initial conditions. At t,if you expect the function not to go to infinity, then you should have A = 0.

d2

dt2= 2 (t) = A sin(t+ ) (7)

For A and a constant, determined by the initial conditions. We can also write this as (t) =A sin(t) + B cos(t). Believe it or not, this equation is the same as the one before it, except becomes imaginary. Solving linear differential equations is important. It requires some knowledgeof calculus. Taylor series will help, and so will linear algebra. It is probably the most intuitive ofall of subjects mentioned so far to pick up.

Back to our original problem. We can apply (6) to (5) with =g/l to get

(t) = Aetg/l +Bet

g/l so

d

dt= (t) = A

g

letg/l B

g

letg/l (8)

But we have, at t = 0 that the exponential terms vanish, and that (0) = 0, (0) = 0

A+B = 0 and AB = 0l

g(9)

So solving for gives our master equation (isnt it ugly?)

(t) =1

2

(0 + 0

l

g

)(etg/l)

+1

2

(0 0

l

g

)(etg/l)

(10)

We can discuss how the various relations between 0 and 0 lead to different behaviors as t.The physical correspondence is letting the pencil swing all the way under the table (say its rotationon a bar) and back up again. If 0 is great enough, then the first term takes over, and the pencilkeeps swinging in circles. If 0 is smaller, then this shouldnt happen. Somehow our equationsdont allow for this... weird.

This is because of the Small Angle Approximation. Once gets large, sin is nowhere near. whoops. So this only holds for small .

Phew. That is ugly. And totally useless. But being able to do algebra, especially by hand, isreally important.

The skill to do math on a pageHas declined to the point of outrage.Equations quadraticaAre solved on Mathmatica,And on birthdays we dont know our age.



So, now for the really cool result, with all that stuff out of the way:

Problem 4. (3.13b The Heisenberg Uncertainty Principle) Due to Heisenbergs uncertainly prin-ciple, it is impossible to balance a pencil perfectly: intuitively, the pencil is only balanced if0 = 0 = 0. But then we know the position and momentum of the pencil exactly! Heisenbergsuncertainty principle reads

xp ~ = 1.05 1034JsInterpret this to mean

(l0)(ml0) ~ (11)Find the minimum time for to become reasonably large, say = 1 (in radians), given the restrainton 0 and 0. At this point it should be clear that the pencil has largely fallen (a little of 30

fromhorizontal), but sin(1) = .84 1 so our small angle approximation hasnt completely broken downyet. Hint: read the first couple lines of the solution

Solution I promise you that I read this from a solution too. Ill explain why its valid now,and if you pursue physics, you should eventually be able to make similar claims after a little bit ofthinking, and solve ridiculously hard looking problems.

So looking at equation (10), we can see that solving this for when = 1 is going to be... difficult.However, unlike in math, we dont need a completely exact solution. Since we want to maximizethe falling time, we want to make 0 and 0 as small as possible (we cans ee this intuitively orfrom the equation). Thus the constants A and B will be on the order of

~ aka really tiny. So

by the time that etg/l gets large enough that is not as super tiny as A, et

g/l will be even

super-tinier. In fact, itll make so little of a difference that we can just ignore it completely.So throwing out the last term we get

(t) 12

(0 + 0

l

g

)(etg/l)

(12)

Now, we want to find the 0 and 0 that maximize the time needed to fall. Then since theexponential term will not change, we want to minimize its coefficient. From (11), we get

0 ~ml20

(13)

If the coefficient of the exponential is c, then we have

c 12

(0 +

(~ml2

g

l

)1

0

)(14)

Calculus or AM-GM (we use this here so we can show you the calculation. In general be preparedto use a lot of calculus) gives (equality when the two terms are equal)

c 12

(0 +

(~ml2

g

l

)1

0

)

~ml2

g

l(15)

Finally, set = 1



ln

((ml2

~

)2l

g

)1/4= t

g

l t = 1

4

l

glnm2l3g

~2(16)

Put m = .010kg = 10g, l = .1m = 10cm, g = 10m/s2, h = 1.06 1034Js gives t 3.5sSo you cant make a pencil balance for more than about 3.5 seconds.

This concludes the really hard problem.Okay that was pretty ugly... but the next problem is also really cool, and involves very little

math.

3 Unusual Moments of Inertia

We recall two facts about moments of inertia that well need, and a third one that we wont. Pleasenote that all objects are assumed to be pancake objects in two dimensions, and axes of rotationare perpendicular to the plane of the object unless otherwise stated (otherwise you need inertiatensors, as mentioned earlier)

1. The parallel axis theorem: Let an object have mass M and moment of inertia ICM aboutsome the center of mass. Then if R is the distance from some point z to the center of massCM , the moment of inertia about z is

Iz = MR2 + ICM (17)

2. Scaling of moments of inertia: Suppose we scale an object A by doubling all of its dimensionsby 2 (including its mass) to get B. Then the moment of inertia of B about its center of massis 8 times the moment of inertia of A about its center of mass. This holds for other scalingfactors.

We can see this because

I =

r2 dm (18)

And we are doubling every r as well as m, so a factor of 8 comes out of the integral

3. The perpendicular axis theorem: For a pancake object in two dimensions, lying on the x yplane, its moment of inertia about the z axis Iz, is the sum of its moment of inertias aboutthe two horizontal axes: Iz = Ix + Iy. The use of this theorem is very rare. Proof:

Ix =

(y2 + z2) dm Iy =

(x2 + z2) dm Iz =

(x2 + y2) dm (19)

But z = 0 on the pancake object, so Iz = Ix + Iy.

Problem 5. Find the moment of inertia of a (uniform density) bar about its center, withoutusing any integrals, and using only the above facts

Solution Our strategy is to calculate I = I2 for a bar of length 2L in terms of I = I1 for abar of length L, and solve for I1.



Imagine cutting a bar of length 2L in half. Then the moment of inertia of the bar of length 2Labout its center I2 is just twice the moment of inertia of a bar of length L, about its end, which wewill denote I 1. So

I2 = 2I1 (20)

Now we use the parallel axis theorem to relate the moment of inertia through the end of a rodof length L, I 1, to the moment of inertia through the center of the rod, I1

I2 = 2I1 = 2

(M

(L

2

)2+ I1

)(21)

But since I2 = 8I1,

8I1 =ML2

2+ 2I1 I1 1

12ML2 (22)

Which is in fact the same answer obtained through the integral.

Well thats a neat trick. But it also has some pretty weird applications

Problem 6. (8.8a) Find the moment of inertia about its center of the Cantor set: take a bar oflength L, and remove the middle third, then the middle third of the remaining pieces, and so on,ad infinitum. Let the resulting object have mass M .

Solution Well, let the moment of inertia of the Cantor set be I. The Cantor set has two halves,each of which are one third the size of the original cantor set, and have half the mass, but are ge-

ometrically identical. Then their moment of inertia about their center should beI

18. The distance

of each halves center from the original center is1

3. So we have

I = 2

((M

2

)(L

3

)2+

1

18I

)=ML2

9+

1

9I I = ML

2

8(23)

Note this is larger than for a rod, because the mass is distributed farther away.

Note that this solution was done slightly differently than the first one: originally we had twoobjects, one of length 2L, one of length L. Here we took two objects, one of length L, one of lengthL/3 (the alternative would have been 3L and L). Feel free to do whichever makes the most senseto you. You should get the same answer. (There are fewer fractions in the first method, but itrequires slightly more thinking).

Problem 7. (8.8b,c) Find the moment of inertia about the center of mass of a Sierpinski triangleand a Sierpinski carpet each having mass M and side lengths L

Solution The solution is left as an exercise; however the answers are1

9ML2 and

3

16ML2

In general any self similar object (a square, and equilateral triangle, but not a circle) can haveits moment of inertia calculated this way. More precisely, you must be able to make the object upof a bunch of smaller copies of itself.

Well, those are probably the two coolest problems. But the remainder have some interestingresults as well. The next problem has a physically realizable solution which is pretty interesting toobserve. Also, the solution illustrates how useful it can be to keep track of minus signs early on,when you dont know which direction a force is applied



4 Pulling on a spool of threat

Problem 8. (8.47) A cylindrical spool of thread of mass m and moment of inertia about its centerI is free to roll on its side on a table, without slipping. It has inner radius r, i.e. the thread formsa cylinder of radius r, and outer radius R (the plastic part that contacts the table). If you pullon the string with tension T and at an angle from the horizontal, what is the acceleration of thespool? Which direction does it move?

Hint: without slipping is defined in the first paragraph of the solution

Solution We begin by clarifying what it means for the spool to roll without slipping. Thismeans first, that if the spool moves say some distance sideways, all of that movement must bedue to (correlated with) the spool spinning that much, not being dragged without spinning. Theimportant consequences are that we can relate the horizontal motion (distance, speed, and acceler-ation) of the spool with the angular motion (, the angular velocity, , the angular acceleration).Second, while friction does act on the spool, it does no work (this will not be relevant to thisproblem, but it is often important)

Moving on, we set up our problem. Lets say that if the spool accelerates to the right, theacceleration x is positive. Likewise, if it accelerates to the right, lets say its angular acceleration is positive. Finally, let the force of friction be F and point to the right.

It seems like we cant say yet which direction F points. If the spool rolls to the right, frictionpoints left, and vice versa. So we shouldnt be able to say which direction F points until we know thedirection the spool moves, and we wont be able to do those calculations until we know somethingabout F . We can get around this by saying that, if F is negative, it points to the left. Likewise ifx is negative, the spool moves to the right, and must be negative as well. The downside is thatwell have to keep careful track of our minus signs, because... well now that weve defined positiveand negative directions, we cant just write down scalar quantities and fiddle with their signs later.This will be true in general of more complicated problems.

Remembering that = I and that should have the same sign as , we know that a torquecausing the spool to roll right is positive, and one causing it to roll left is negative. The torque dueto the pulling on the thread is Tr then. The torque due to the rightward pointing friction vectoris FR. (Check: if F is negative, the torque is positive, as it should be). Then we have

= Tr FR = I (24)

Newtons law gives, remembering a = x =d2x

dt2,

mx = F + T cos (25)

Where we ignore the forces in the y direction, and the only two forces acting in the x directionare shown. Note that both forces are positive. Finally, we use the relation between and x thatarises because the spool does not slip. The way radians are defined, a spin by an angle causes thecircumference to travel R. (Note: if youre ever unsure about this relation, just check the unitson both sides to make sure). This gives, taking double time derivatives of both sides:

R = x (26)



There are only three unknowns at this point: , x, and F . We solve for F in terms of x in (25),then substitute (24) into (26) followed by substituting in F . The resulting equation is

R

(Tr +R(mx T cos )

I

)= x (27)

Multiplying through by I then expanding gives

RT (R cos r) = (I +mR2)x (28)

We now turn to the question of the direction in which the spool moves. Note that I + mR2 ispositive, as is RT . Then the sign of x is just the sign of R cos r, which is positive for below acertain critical angle, negative for above a critical angle, and 0 for = cos1

( rR

).

Note that the sign does not depend on T ! This means that at the critical angle, no matterhow hard you pull the spool wont move, at least until you overpower the force of friction, which isbounded. If is the coefficient of static friction, we must have T cos < F < (mg T sin ). Thefact that, depending on the angle you can make the spool go either should not be surprising: athigh angles, you are barely pulling sideways, while the string tries to unwind and push the spoolleft. At low angles, your pulling overpowers this force.

Hopefully this is an interesting result and shows that you really benefit from defining a positivedirection for quantities and sticking to it.

Heres a fun puzzle

Problem 9. A Puzzle

Problem 10. (4.21) A mass hangs from a ceiling. A piece of paper is held up between the massand the ceiling obscuring three strings and two springs. All you see is two other strings protrudingfrom the top and bottom of the paper, connecting to the ceiling and mass respectively. How shouldthe other strings and springs be attached to each other so that, if the system starts at equilibriumand one of the springs is cut, the mass will rise up. Note that objects can only be attatched attheir endpoints

Solution Ah that would ruin the fun wouldnt it? You can probably think of some silly waysto do it, like hanging a spring so that when you cut the string, it falls off and there is less weightpulling down, so the system rises up. Think of something cooler instead.

We conclude with more technical material. If you really want to learn the binomial theorem,heres some good practice:

5 Binomial Theorem and Pendula

This section assumes an understanding of Taylor series, though it relies only on knowing the Taylorseries for sin.

Here is the extended binomial theorem: Define, for r a real number and k an integer(r

k

)=r(r 1) (r k + 1)

k!(29)



This makes sense if we regardr! = r(r 1)(r 2) (30)

as an infinite product, which is not well defined when r is not a positive integer. However, if wereturn to the normal definition of a binomial coefficient,(

r

k

)=

r!

k!(r k)! =1

k!

r(r 1) (r k + 1)(r k)(r k 1) (r k)(r k 1) =

r(r 1) (r k + 1)k!

(31)The following are the most common uses of binomial theorem in physics:

Problem 11. Calculate

(1/2

k

)and

(1k

)for k = 0, 1, 2, 3

Solution Noting 0! is defined to be 1,(1/2

0

)= 1,

(1/2

1

)=

1

2,

(1/2

2

)=

(1/2)(1/2)2!

= 18,

(1/2

3

)=

(1/2)(1/2)(3/2)3!

=3

48(32)

(10

)= 1,

(11

)= 1,

(12

)=

(1)(2)2!

= 1,

(13

)=

(1)(2)(3)3!

= 1 (33)

The binomial theorem is as follows:

(1 + x)r =n=0

(r

n

)xn (34)

Note that in mathematics we usually state this for (x + y)r. In physics, this is almost alwaysonly useful when it is in the form (1 +x)r. In physics the two most important examples are r = 1and r = 1/2.

Problem 12. Expand the first few terms (1 + x)1 and (1 + x)1/2 using the binomial theorem

Solution(1 + x)1 = 1 x+ x2 x3 + (35)

Note that if we replace x by x, we get:

(1 x)1 = 11 x = 1 + x+ x

2 + x3 + (36)

Which you should recognize either as the sum of a geometric sequence, or more generally, as aTaylor series. In fact, the extended binomial theorem just amounts to computing Taylor series.

(1 + x)1/2 =

1 + x = 1 +x

2 x

2

8+

3x3

48 (37)

So why is this useful in physics? Well in physics, we often deal with situations where quantitiesare small. In the past this has largely meant sin . Why is this? Study the Taylor expansionfor sin

sinx = x x3

3!+x5

5! (38)



If is small, the x3/6 term is really negligible compared to x. If x is on the order of 101 (.1or so), then x3 is .001 which isnt going to make much of a difference to our answer in most cases.Even if x is one half, x = .5, then x3/6 = 1/48 .02 which shouldnt have too much of an effect(often 4% is smaller than other errors involved in measurements etc).

Likewise, in the two expansions above, if x is small, the x2 terms are negligible (note that theyhave more effect than the x3 in the sin expansion though, which is why the latter expansion is useda ton, while expansion for (1 + x)1 is only used in its full form, and the other expansion is notoften used at introductory levels).

From a mathematical perspective, we can make this rigorous by writing sinx = x+O(x3) whereO(x3) is big O notation, and means that the remainder of the term varies at least as x3. Notethat if this term later gets multiplied by an x, xO(x3) = O(x2). Also note the addition properties:O(x3) +O(x3) = O(x3) and O(x2) +O(x3) = O(x2). Composition gives O(O(x)) = O(x). Finallymultiplication by a scalar has no effect: cO(x4) = O(x4). Then we can operate with the big O termalgebraicly.

The way this is used in physics: Suppose that at the end of our solution we end up with ananswer like 3 +x+ 5x2 +O(x3). Then we can say that we have second order approximation for theanswer. In other words, the real answer will only vary from our answer by something that dependson x3 (or higher degree). If x is small, this variation is very tiny (and since x2 is very small, ouranswer is extremely precise). Note that we must be careful because if at some point we get a termlike 1 + x + 3x2 + xO(x3) then we have to swallow the x2 term into the new O(x2) = xO(x3) sothat we can only write 1 +x+O(x2). If we we are too eager to make approximations, then the bigO notation will swallow up too many terms, and our answer wont be very useful.

So whenever a physicst says first order approximation, what they mean is that when theywrite 3+2x, they really mean 3+2x+O(x2). Often in physics it is more useful to make argumentsthat something varies second order in x verbally and not actually write it formally, but this notationallows to see what physicist really mean when they say to first order in x.

Problem 13. (4.23a) For small oscillations, the period of a pendulum is approximately T 2pil/g independent of the amplitude 0. Use dt = dx/v to show that the exact expression for T

is

T =

8l

g

00

dcos cos 0

(39)

Solution It is strongly recommended you attempt to solve this problem without reading the solutionConsidering that the problem tells you the desired answer, it shouldnt be too bad.

We focus on the time it takes the pendulum to fall from its initial angle 0 to the bottom of itsarc, and multiply by 4. If the pendulum is at an angle with the vertical, then its height belowthe pivot of the arc is l cos and it has fallen l cos l cos 0. Using conservation of energy, we canwrite

mgh = mgl(cos cos 0) = 12mv2 v =

2gl(cos cos 0) (40)

We can see why dt = dx/v: the change in time is equal to the change in distance divided by thevelocity (start by checking units). Since we wish to calculate (4 times) the change in time over a



given distance, we wish to calculate

4

00

dt = 4

00

dx2gl(cos cos 0)

(41)

Since dx = ld this is just the integral desired.

Problem 14. (4.23b) Find an approximation to T up to second order in 0, by writing cos =1 2 sin2(/2) (a trig identity) then make the change of variables sinx = sin(/2)/ sin(0/2).Expand the integrand in terms of 0 and integrate to show

T 2pil

g(1 +

202

+O(4)) (42)

Solution Again, it is strongly recommended you attempt to solve this problem without readingthe solution Considering that the problem not only tells you the desired answer, but also tells youhow to do the solution, it cant be too bad. Just remember your calculus, be careful, and dontforget the binomial theorem or the idea of making approximations (apply these last two at the end)

Using the given trig identity, we get the integrand to be

d

2

sin2(0/2) sin2(/2)=

d

2

sin2(0/2) (sinx sin(0/2))2(43)

Where in the last equality we perform the suggested change of variables (though we havent changedthe variable of integration yet) Since we have changed variables, we differentiate the change ofvariables to get

cosx dx =1

2cos(/2)/ sin(0/2)d (44)

(remember 0 is constant). Our variables of integration change to

sinx = sin(0/2)/ sin(0/2) = 0 x = 0 and sinx = sin(0/2)/ sin(0/2) = 1 x = pi/2 (45)

Then, pulling the 1/

2 out of the integral, changing limits, and substituting for d gives

T = 2

l

g

pi/20

2cosx sin(0/2)

cos(/2)dx

(sin(0/2))2(1 sin2 x)

=

4

l

g

pi/20

dx

cos(/2)= 4

l

g

pi/20

dx1 (sinx sin(0/2))2

(46)

Where we have rewritten the change of variables as sin(/2) = sinx sin(0/2) in some locations.Which brings us to the binomial theorem. The expansion (use r = 1/2, binomial coefficients



1,1/2, 3/8,5/16, . . .) is:

T = 4

l

g

pi/20

dx(1 sin2(0/2) sin2 x

)1/2=

4

l

g

pi/20

dx

(1 +

1

2sin2 x sin2(0/2) +

3

8sin4 x sin4(0/2) +

)(47)

Note that all the signs are positive, because the second term is 12x where x is actually that entire

term from the top line, which is negative already.

The first thing we want to note is that

pi/20

sin2 x dx =pi

4. This integral comes up so much in

physics that we strongly suggest you know the trig identity mentioned in the problem statement,which allows you to evaluate this and similar integrals. So we have:

T = 4

l

g

(pi

2+

1

2

pi

4sin2(0/2) +

pi/20

(3

8sin4 x sin4(/2) +

5

16sin6 x sin6(/2) +

)dx

)(48)

Now if you recall from the Taylor expansion of sin from (38), we can make the following arguments.We justify each step involving big O notation very carefully. You should check each equality andremember why the simplifications we make hold.

sin(0/2) =02

+O

((02

)3)=02

+O(30)

(49)

Verbally, all we have said is that we have computed a second order approximation for sin. Next, wesqaure sin. To calculate the big O notation, we just square the right hand side of (49) and computewith big O notation:

sin2 0/2 =

(02

+O(30))2

=204

+0O

(30)

2+O

(30)2

=204

+O(40)

+O(60)

=204

+O(40)

(50)

Again, we could make this argument verbally by saying that a 30 term will only appear as fourthorder once sin is squared. But its good to see why this is true formally. At this point, hopefullyyou will see why this allows us to compute the second term in (48). Well perform that calculationafter seeing how to deal with the third and final term.

Looking at the Taylor expansion for sin, we see that we can also write

sin(0/2) = O(0/2) = O(0) (51)

Now lets look at the third term in (48):

pi/20

(3

8sin4 x sin4(/2) +

5

16sin6 x sin6(/2) +

)dx =

sin4(0/2)

pi/20

(3

8sin4 x

)dx+ sin6(0/2)

pi/20

(5

16sin6 x

)dx+ =

O(40)

+O(60)

+ = O (40) (52)MIT ESP Splash 11/19/2010 13 of 18


We can get rid of the integrals because the integrals are just constants, they dont vary as 0/2.Since we have big O notation multiplied by a constant, we can just absorb the constant into thebig O notation. Hopefully you can see that the big O terms came from the equality in (51).We emphasize equality because, even though the equalities are approximations, they are precisesstatements: sin(/2) is in fact in some sense equal to O(0) insofar as we can perform thatsubstitution. They are not equal in that O(0) is not well defined: we lose some information in oursubstitution. But thats fine. In order to find an approximation, we wish to throw out information.Big O notation lets us do this.

Also, note that the are in fact O(80) in (52) (though we can only see this in hindsightafter performing the calculations). To be completely rigorous, we have to also say that the omittedterms are higher order in 0. To do this, we can make use of this fact:(1 sin2(0/2) sin2 x

)1/2= 1 +

1

2sin2 x sin2 0/2 +O

(sin4(0/2)

)= 1 +

1

2sin2 x sin2 0/2 +O

(40)

(53)Hopefully you can see for yourself where the last equality comes from. The last argument we wouldneed to make is that integration by a different variable, with constant limits, will not affect theorder of the result in 0. This would have saved us some work in dealing with the third term.

So, finally, we can write (48) as

T = 4

l

g

(pi

2+pi

8

(204

+O(40))

+O(40))

= 2pi

l

g

(1 +

2016

)+O(40) (54)

We will spare the tedious details of dealing with the two big O terms. It should be pretty clear atthis point that they will migrate outside everything else unscathed and then combine.

So we have in fact, gotten a bonus term: this is third order approximation in 0.

In fact, the fourth order term is11

307240. If you want to show this, beware that two terms

contribute to this fraction.So what does this say? It says that the period of a pendulum is 2pi

lg as normal, but that

in addition there is a correction based on 20/16. If 0 is, say, 90, the correction to the period is,

pi2/64, which is about 15%.Note that, in fact the fourth order term is very tiny, even though 40 is larger than

20, because

0 > 1. Now this seems like a problem. There are two ways to realize why this isnt important.First, the coefficients get vanishingly small. Second, we can change units: define a new unit thec. elegans. We will define 1 c. elegans to be equal to 2pi radians. Then 0 is .25 c. elegans. Sinceit is perfectly valid to perform our calculations in any unit system we like (though we should becareful about derivatives of sin), then 40 is vanishingly small. Of course, we have to change ourcoefficients slightly to reflect this, which ultimately forces us to return to an argument about ourcoefficients being small, but hopefully this solves the problem of, say, the 600000 term being really,really large.

And well stick the remainder of the math topics in this last section



6 A Brief Mathematical Appendix

6.1 Taylor Series

If you want to really learn about Taylor series, read a textbook on single variable calculus, or doanalysis. In the meantime, well cover what you need to know. The Taylor series for a function fthat is infinitely differentiable at a point a is

p(x) =n=0

f (n)(a)

n!(x a)n (55)

Where f (n) is the nth derivative of f . Taking nth derivatives and plugging in x = a shows that phas all of the same derivatives as f at a. This is left as an exercise. Intuitively, it should be clearthat if we track the value of f , how fast f is changing, how fast the rate of change of f changes, andso on, ad infinitum, then weve essentially recorded everything we need to know about the functionto calculate its value.

Well, this would be ideal. Unfortunately, some functions die at certain places. The mostimportant examples are 1x , and lnx. First, we cant take a MacLaurin series (which is the termfor a Taylor series evaluated at 0), because it isnt even defined at 0. So we can take its Taylorseries at 1. Equivalently, we can take the MacLaurin series for 11+x , and the MacLaurin series forln(x 1). Note that, unfortunately,

The following are the important Taylor series:

1

1 + x1 x+ x2 x3 +

ln(1 + x) 1 + x+x2

2+x3

3

ex 1 + x+x2

2!+x3

3!+

sin(x) x x3

3!+x5

5!+

cos(x) 1 x2

2!+x4

4!

Note that there is no particular reason why we cant put imaginary arguments. So lets calculateeix:

eix = 1+ix+(ix)2

2! (ix)

3

3!+ =

(1 x

2

2!+x4

4!

)+i

(x x

3

3!+x5

5!

)= cos(x)i sin(x) (56)

Also note that eipi = 1, which many people find very interesting for some reason.Well use these formulas in our brief treatment of ordinary linear homogeneous differential

equations:

6.2 Slightly more treatment of Linear Differential Equations

The first thing to note is thatd

dtekt = kekt (57)



for k a constant. When solving differential equations, this fact will be very useful. To see why(intuitively), we realize that ordinary linear homogenous differential equations are something of theform x = kx or a more complicated sort of expression (well define it fully in a moment). Then insome sense, the derivative of the exponential function gives a very minimal change which allows usmaximum control over the result. Keep this in mind during the following discussion:

An ordinary, linear, homogeneous, differential equation is of the form:

dnx

dtn+ an1

dn1xdtn1

+ + a2d2x

dt2+ a1

dx

dt+ a0x = 0 (58)

Where x is a function of t, x(t), and the ai are real scalar constants. (We can put a constant on thefirst term, but equivalently we can leave it off). You wish to find all possible solutions, that is, allfunctions that x could be. The equation is ordinary because it does not involve partial derivatives(in essence, you are only solving for one function of one variable). It is linear because none of thederivatives of x(t) are raised to a power or multiplied by each other. Nor are any of them multipliedby any functions of t. It is homogeneous because there is no constant term: that is, the RHS ofthe equation is 0, not, say, 6. And it is a differential equation because it is an equation involvingderivatives of an unknown function.

The study of non-homogeneous differential equations depends very highly on the study of ho-mogeneous equations. This is because the derivative operator is linear. If you study linear algebra,youll see why this is important.

The characteristic polynomial of a differential equation is

xn + an1xn1 + + a2x2 + a1x+ a0 (59)

If the roots of a characteristic polynomial of a differential equation (from here on out, we willassume that it is homogeneous, linear, and ordinary) are r1, r2, r3, , rn, and they are all distinct,the possible solutions for x are all functions of the form:

A1er1t +A2e

r2t + +Anernt (60)

If there is a double root, say r1 = r2, then we replace the first two terms with A1er1t + A2te

r1t.If there is a triple root, the next term is A3t

2er1t and so forth. The Ai are arbitrary constants:choosing any combination of constants yields a valid solution.

The reason this is true is as follows: consider just one term, say Aert. Its nth derivative isrn(Aert

). Then looking at our original differential equation, and computing all the derivatives, we

have on the LHS,Aert

(rn + an1rn1 + + a 2r2 + a1r + a0

)(61)

which is definitely 0, since r is a root of the characteristic equation. Moreover, suppose we havetwo terms: Aer1t and Ber2t. Since the derivative is linear (it will be hard to stress how importantlinearity is), ddt (x1 + x2) =

ddtx1 +

ddtx12 (or rather, it is linear because of this property. Thats

what linearity means/is defined to be). Then the LHS of our differential equation will become

Aer1t(rn1 + an1r

n11 + + a 2r21 + a1r1 + a0

)+Ber2t

(rn2 + an1r

n12 + + a 2r22 + a1r2 + a0

)(62)

We can see that this is still a solution. Then by induction, we can add any combination of singleterms which satisfy the homogeneous differential equation, and still get 0.



For equations with one main condition(Those linear), we give you permissionTo take you solutions,With firm resolutions,And add them in superposition.

Moreover, if we have a non-homogeneous differential equation, then, suppose that xp (read: xparticular) solves the non-homogeneous differential equation, and suppose that xh (read: x homo-geneous) solves the equivalent homogeneous equation (where we replace the non-zero constant c by0). Then since adding xh to xp still yields a solution to the non-homogeneous equation, becauseyouve just added 0 to the LHS. Anyways, we return to the application to physics.

The degree of a differential equation is the degree of its characteristic polynomial, n. The mostimportant case well study is second degree. Consider, noting we have reverted to the dot notation:

x+ a1x+ a2x = 0 (63)

Suppose the distinct roots are r1 and r2. Then the solutions are of the form Aer1t + Ber2t. If

r1 = r2, then the solution is Aer1t + Bter2t. You can check this by hand. Then, in the case where

r1 and r2 are real, were pretty much done.On the other hand, we might be a little concerned with the possibility that the roots are not

real. You might also be concerned when you realize that the equation which gives harmonic motionis

x = 2x x+ 2x = 0 (64)which is linear, and its solution is, for arbitrary A (amplitude) and (phase shift)

A sin(t+ ) (65)

Which looks nothing like what we have claimed is the most general solution to ordinary linearhomogeneous differential equations.

If, at this point, you havent read the section about Taylor polynomials, you should do so.We rectify this apparent contradiction: ff the roots are not real, then let r1 = a + bi and we

must have r2 = a bi (fundamental theorem of algebra). The solution is

x(t) = Ae(a+bi)t +Be(abi)t = Aeate(bt)i +Beate(bt)i =eat (A(cos(bt) + i sin(bt)) +B(cos(bt) + i sin(bt))) =eat (A(cos(bt) + i sin(bt)) +B(cos(bt) i sin(bt))) =

eat ((A+B) cos(bt) + i(AB) sin(bt)) (66)Now pick some arbitrary (probably real) constants A and B. Try to solve

A = A+B and B = i(AB) iB = (AB) (67)Clearly, if we allow A and B to be complext, we can find A and B for any A and B. There is noreason why A and B cant be complex: we never specified that x must be a real valued function.And if we did, we can temporarily relax that restriction and show that the function ends up beingreal anyways. If A and B are real, the x is going to end up being real anyways. So we can make Aand B be whatever we want. Conclusion: 1) dont be afraid of complex numbers. They work like



other numbers too, just sometimes better. 2) We can replace (A + B) and i(A B) by arbitraryconstants, A and B. In fact, for convenience of notation, we can drop the primes too, and justwrite

x(t) = (A sin(bt) +B cos(bt)) eat (68)

We cheated slightly and switched A and B and switched the order of terms. But anyways, this isa common and more general form for harmonic motion. Note that a controls the decay or growthof the amplitude. If a = 0, the differential equation was of the form x = 2x with characteristicequation x2 + 2 = 0, and we get simply harmonic motion because the amplitude doesnt change.If a is negative, then eat decays, which is called damped motion. Frankly, this involves a lot ofugly mathematics, so its not worth going into here. Youll probably learn about it eventually. Inthe meantime, well show why our form in (68) is equivalent to the originally proposed solution forharmonic motion in (65). Apply the sin addition formula:

A sin(t+ ) = A sin(t) cos() +A cos(t) sin() (69)

If A and B are the arbitrary constants in (68) then if we can solve

A = A cos() and B = A sin() (70)

for A and , we can write anything of the form (68) in the form (65) simply by finding a suitable Aand . All we have to do is note B/A = tan() which can be solved (uniquely for pi/2 < < pi/2)for for all ratios, and then we can easily calculate the necessary A. So the two forms are equivalent.Phew.

So it actually turns out, as mentioned much earlier, that harmonic motion is just exponentialwith imaginary exponent... go figure.

Finally, because we mentioned it briefly, we will state, but not prove AM-GM.

6.3 AM-GM

AM-GM stands for Arithmetic Mean-Geometric Mean. What is states is that for positive quantitiesa1, a2, , an, the arithmetic mean is greater than or equal to the geometric mean:

a1 + a2 + ann

na1a2 an (71)

Moreover it states that there is equality (AM = GM) if and only if all the terms are equal. Itsstatement is easiest to use and see for n = 2, and it can be proved very quickly by hand as well:

a1 + a22

a1a2 and LHS = RHS iff a1 = a2 (72)

Overall, we apologize for not including more material (we assume that if you made it to thispoint, youd be interested in more). There are plenty more good problems we would have liked toinclude. But deadlines have a bad habit of sneaking up on you.


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