General Tips on How to Do Well in Physics Exams. 1. Establish a good habit in keeping track of your steps. For example, when you use the equation

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1 General Tps on How to Do Well n Physcs Exams 1. Establsh a good habt n keepng track o your steps. For example when you use the equaton = d d to solve or d o you should rst rewrte t as = d d (1) and then begn to put n numbers such as: d = 5m 6m = 30m Then you'll know that what you have just got s not yet 1/ d. o you must stll do one more step to nd that o d (2) but s only d = 30m (3) as the nal answer. Now you don't have a good habt you would wrte only 1 1 = wthout sayng what t s really equal to. You would then orget that you have one more step to do and smply gve your answer as d = 1 m. 30 Comparng the correct answer (3) wth the wrong answer (5) I hope you can now see where you are lkely to make mstakes. It all bols down to the bad habt o omttng the step gven as Eq. (1) and also beng lazy n not payng attenton to unts n Eq. (4) and not explctly notng that t s actually 1/ d you have calculated there not yet d. That s you should have wrtten Eqs. (1) and (2) nstead o Eq. (4). Wrtng Eq. (4) nstead o Eqs. (1) and (2) means that you have a bad habt and you are lazy. You can not blame others you get wrong answers ths way. (4) (5)

2 (2) I you are to draw a gure to help you gure out the answer avod puttng nto the gure normaton that s not gven by the problem. For example an angle s nvolved avod makng t a rght angle or 45 or 30 or 60 unless the problem says so. These are specal angles wth specal propertes whch are not shared by a general angle. I the problem dd not say that you have such a specal angle then you should not draw such a specal angle. That s n your drawng you should avod drawng any such specal angles. Another example s: I a rectangle s nvolved n the problem you should avod drawng t as a square (unless t s a square). It should clearly be a rectangle. In act you should avod drawng t n such a way that one sde appears to be twce or three tmes etc. as long as the other sde. Ther rato should not be a smple nteger. It can mslead you nto thnkng that you have ths extra normaton n the problem that you really don't have. Now these are just examples. I can not pont out all thngs you are lkely to do. The general gudelne here s that you should avod dong thngs to ool yoursel nto thnkng that you have thngs you really don't have. (3) Do not mprovse. Physcs laws do not allow you to mody them. For example n optcs there s the equaton or double-slt ntererence: = (*) d sn θ mλ. I you are gven one m and one d but two derent λ s then usng the rst λ you can nd the rst θ and usng the second λ you can nd the second θ but you can not put n the derence n λ and hope to drectly compute the derence n θ. That would be a msuse o the ormula. That s what I mean by mprovse! A ormula does not allow you to use t n a way that t really doesn t allow. On the other hand n the example ormula gven when d s much larger than λ so you know that both θ s are very small then you can approxmate sn θ by θ n radans. We then have dθ1= mλ1 and dθ 2 = mλ 2 so takng the derence o the two equatons does gve you d( θ θ ) = m( λ λ ) or d θ = m λ o n ths case you can ndeed put n the derence o λ and obtan drectly the derence n θ. But ths s an excepton. It can be done

3 n ths case only because both λ and θ enter the equaton lnearly. Ths s not true wth the orgnal equaton (*)! I the change n λ and θ are very small then you can derentate both sdes o the orgnal e- quaton and obtan d cosθ dθ = m d λ. From ths equaton you can ndeed put n the very small quantty dλ and obtan drectly the very small quantty dθ and vce versa. But notce that t has cos θ n t whereas the orgnal equaton has sn θ. Another example s the equaton or Doppler sht o a sound wave requency when the source s movng wth the speed v and the lstener s movng wth the speed v :. - = v v v -v where v s the speed o sound s the actual requency o the sound emtted by the source and s the apparent requency o the sound heard by the lstener. When solvng such problems you can not work wth a relatve speed as you are movng wth the source so only the lstener s movng wth the speed v - v or movng wth the lstener (who s not you) so only the source s movng wth the speed v -v. I you do that you must also change v because v s dened wth respect to statonary carrer o the wave and you are movng wth respect to the statonary carrer o the wave then the sound speed s no longer v. I you don t take care o ths pont and stll work wth a relatve speed then you are mprovsng snce you have changed the content o the equaton. I the problem s about the Doppler sht o a lght wave then the ormula changes to: R = c -v c + v where c s the speed o lght and v v -v s the relatve speed. That s only the relatve speed matters n ths equaton. Ths pont

4 actually has very sophstcated reasons behnd t and s the great dscovery o Ensten. Namely lght s movng at the same speed o lght no matter how ast you are movng. It also has to do wth the act that lght waves have no carrer so you can t move wth respect to that non-exstent carrer to change ts speed. (4) Use dmensonal analyss to help you nd some o your mstakes. For example n wave theory we have λ = c. That s the wavelength o the wave tmes the requency o the wave s equal to the speed o the wave. The dmenson or unt o λ s meter or m. The dmenson or unt o s Hz or 1/s. Hence the product on the let hand sde o the equaton has the dmenson or unt o m/s whch s precsely the unt o the wave speed c. Now suppose you care- λ = c lessly put t as. You should be able to nd ths mstake easly by lookng at the dmenson or unt on the two sdes o the equaton. The dmenson or unt o λ s m. But the dmenson or unt o c s (m/s) (1/s) = m/s 2 whch s not what you have on the let hand sde o the equaton Thereore t must be wrong. As a matter o act ths dmensonal analyss mmedately suggests to you that the correct ormula should be λ = c and not λ = c. Every ormula n physcs allows you to analyzng t ths way and the dmenson or unt must be the same on ts two sdes. (5) Many ormulas allow you to have some eelng on why certan quanttes are n the numerator and why certan other quanttes are n the denomnator. Take the ormula or the speed o transverse wave on a strng. F v= µ where F s the tenson n the strng and µ s the lnear mass densty o the strng. In ths ormula t s not dcult to see why F s n the numerator and µ s n the denomnator and not the other way a- round. It s because ncreasng F can clearly make v larger because neghborng molecules n the strng can nluence each other more easly when F s larger and ncreasng µ wll clearly make v smaller because now neghborng molecules n the strng wll have a harder tme to nluence each other when µ s larger because larger µ

5 means larger nerta. o you can now see that you should never wrte t as v = µ / F. (As to why there s a square root sgn dmensonal analyss can help you see t.) (6) I possble you should use a gure to help you get thngs straght. For example take the ormula I = I 2 0 cos ( φ / 2). You should remember the phasor dagram assocated wth t: φ / 2 φ Ths phasor dagram not only let you see why t s φ / 2 and not just φ as the argument o the cosne uncton t also let you see that ths ormula s assocated wth the ntererence o two narrow lght beams wth a phase derence o φ between them so t can t be used or the dracton o a sngle wde slt nor the dracton o N narrow slts wth N > 2. You wll then not use t n the wrong place. On the other hand you wll also be able to see that you can also use ths ormula or the ntererence o two sound or rado waves generated at two pont sources A and B when the recever s at dstance d 1 orm source A and d 2 rom source B. Then all you need s to also see that n ths case φ = 2π(d 1 - d 1 ) / λ. On the other hand two lght beams are ntererng n phase to begn wth and a thn glass plate wth ndex o reracton n and thckness t s nserted nto one beam perpendcular to the beam leavng the other beam undsturbed then the above ormula can also be used except that now t t φ = 2π = 2 π( n 1) t λ / n λ λ because t s the change o phase o one beam (that s a new phase 2 π[ t/( λ / n)] mnus an old phase 2 π[ t / λ ]) wth the phase o the other beam unchanged. o t s also the derence between the phases o the two beams ater the glass plate has been nserted they are n phase beore the glass plate s nserted. Note that n all these cases when the ormula can be used there are always two narrow beams o some wave ntererng whle there s a phase derence between them.

6 Many ormulas allow you to understand them usng some sort o gures. o you should learn such ormulas ths way and not try to just memorze them. (7) I the problem gves you a quantty and you dd not use t to get your answer then you should ask yoursel the ollowng queston: hould the answer be ndependent o ths quantty? Assumng that t s true then you should be able to change the value o ths quantty to say zero or nnty ( allowed). But the correct answer or these cases mght be very easy to see. I t dsagrees wth your answer then you know that your answer must be wrong and the correct answer should depend on ths quantty. For example take the problem o nsertng a glass plate nto one beam and you are asked the ntensty change when t s ntererng wth another beam. I your answer does not depend on the thckness t o the glass plate then you should see that t has to be wrong snce or t equal to zero we have no glass plate and the phase sht must be zero! o how can you nd a nonzero answer that s ndependent o t? (8) Even when a quantty does enter your soluton you can stll ask whether t has entered correctly. et us agan take the glass-platenserton problem as an example. I you gve the phase sht as t t φ = 2π = 2πn λ / n λ then you can ask whether n has entered correctly. Well you can look at the case when n =1. The glass plate has been replaced by a layer o ar. That s nothng s now nserted nto the beam path. The phase sht should clearly be zero n that case. But s your answer zero n that case? It s not φ s gven by the above equaton. o you know t has to be wrong! It mmedately remnds you that you orgot to subtract the phase due to a layer o ar o the same thckness. That s you dd not compute the change. O course you know nothng about what s a phase all these remarks are useless to you. You must learn the general concepts rst.

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Structure and Drive Paul A. Jensen Copyright July 20, 2003 Structure and Drve Paul A. Jensen Copyrght July 20, 2003 A system s made up of several operatons wth flow passng between them. The structure of the system descrbes the flow paths from nputs to outputs.

Please review the following statement: I certify that I have not given unauthorized aid nor have I received aid in the completion of this exam. NME (Last, Frst): Please revew the followng statement: I certfy that I have not gven unauthorzed ad nor have I receved ad n the completon of ths exam. Sgnature: INSTRUCTIONS Begn each problem n the space

THE CHINESE REMAINDER THEOREM. We should thank the Chinese for their wonderful remainder theorem. Glenn Stevens THE CHINESE REMAINDER THEOREM KEITH CONRAD We should thank the Chnese for ther wonderful remander theorem. Glenn Stevens 1. Introducton The Chnese remander theorem says we can unquely solve any par of

Physics 4C. Chapter 19: Conceptual Questions: 6, 8, 10 Problems: 3, 13, 24, 31, 35, 48, 53, 63, 65, 78, 87 Physcs 4C Solutons to Chater 9 HW Chater 9: Concetual Questons: 6, 8, 0 Problems:,, 4,,, 48,, 6, 6, 78, 87 Queston 9-6 (a) 0 (b) 0 (c) negate (d) oste Queston 9-8 (a) 0 (b) 0 (c) negate (d) oste Queston

Prof. Dr. I. Nasser T /16/2017 Pro. Dr. I. Nasser T-171 10/16/017 Chapter Part 1 Moton n one dmenson Sectons -,, 3, 4, 5 - Moton n 1 dmenson We le n a 3-dmensonal world, so why bother analyzng 1-dmensonal stuatons? Bascally, because

PHYS 1101 Practice problem set 12, Chapter 32: 21, 22, 24, 57, 61, 83 Chapter 33: 7, 12, 32, 38, 44, 49, 76 PHYS 1101 Practce problem set 1, Chapter 3: 1,, 4, 57, 61, 83 Chapter 33: 7, 1, 3, 38, 44, 49, 76 3.1. Vsualze: Please reer to Fgure Ex3.1. Solve: Because B s n the same drecton as the ntegraton path s

One Dimensional Axial Deformations One Dmensonal al Deformatons In ths secton, a specfc smple geometr s consdered, that of a long and thn straght component loaded n such a wa that t deforms n the aal drecton onl. The -as s taken as the

Errors for Linear Systems Errors for Lnear Systems When we solve a lnear system Ax b we often do not know A and b exactly, but have only approxmatons Â and ˆb avalable. Then the best thng we can do s to solve Âˆx ˆb exactly whch

MAE140 - Linear Circuits - Winter 16 Final, March 16, 2016 ME140 - Lnear rcuts - Wnter 16 Fnal, March 16, 2016 Instructons () The exam s open book. You may use your class notes and textbook. You may use a hand calculator wth no communcaton capabltes. () You have

Inner Product. Euclidean Space. Orthonormal Basis. Orthogonal Inner Product Defnton 1 () A Eucldean space s a fnte-dmensonal vector space over the reals R, wth an nner product,. Defnton 2 (Inner Product) An nner product, on a real vector space X s a symmetrc, blnear,

Chapter 3. r r. Position, Velocity, and Acceleration Revisited Chapter 3 Poston, Velocty, and Acceleraton Revsted The poston vector of a partcle s a vector drawn from the orgn to the locaton of the partcle. In two dmensons: r = x ˆ+ yj ˆ (1) The dsplacement vector

P A = (P P + P )A = P (I P T (P P ))A = P (A P T (P P )A) Hence if we let E = P T (P P A), We have that Backward Error Analyss for House holder Reectors We want to show that multplcaton by householder reectors s backward stable. In partcular we wsh to show fl(p A) = P (A) = P (A + E where P = I 2vv T s the

MAE140 - Linear Circuits - Winter 16 Midterm, February 5 Instructons ME140 - Lnear Crcuts - Wnter 16 Mdterm, February 5 () Ths exam s open book. You may use whatever wrtten materals you choose, ncludng your class notes and textbook. You may use a hand calculator

Physics 2A Chapter 9 HW Solutions Phscs A Chapter 9 HW Solutons Chapter 9 Conceptual Queston:, 4, 8, 13 Problems: 3, 8, 1, 15, 3, 40, 51, 6 Q9.. Reason: We can nd the change n momentum o the objects b computng the mpulse on them and usng

Physics 53. Rotational Motion 3. Sir, I have found you an argument, but I am not obliged to find you an understanding. Physcs 53 Rotatonal Moton 3 Sr, I have found you an argument, but I am not oblged to fnd you an understandng. Samuel Johnson Angular momentum Wth respect to rotatonal moton of a body, moment of nerta plays

= z 20 z n. (k 20) + 4 z k = 4 Problem Set #7 solutons 7.2.. (a Fnd the coeffcent of z k n (z + z 5 + z 6 + z 7 + 5, k 20. We use the known seres expanson ( n+l ( z l l z n below: (z + z 5 + z 6 + z 7 + 5 (z 5 ( + z + z 2 + z + 5 5

From Biot-Savart Law to Divergence of B (1) From Bot-Savart Law to Dvergence of B (1) Let s prove that Bot-Savart gves us B (r ) = 0 for an arbtrary current densty. Frst take the dvergence of both sdes of Bot-Savart. The dervatve s wth respect to

CHAPTER 10 ROTATIONAL MOTION CHAPTER 0 ROTATONAL MOTON 0. ANGULAR VELOCTY Consder argd body rotates about a fxed axs through pont O n x-y plane as shown. Any partcle at pont P n ths rgd body rotates n a crcle of radus r about O. The

COS 511: Theoretical Machine Learning. Lecturer: Rob Schapire Lecture #16 Scribe: Yannan Wang April 3, 2014 COS 511: Theoretcal Machne Learnng Lecturer: Rob Schapre Lecture #16 Scrbe: Yannan Wang Aprl 3, 014 1 Introducton The goal of our onlne learnng scenaro from last class s C comparng wth best expert and

Answers Problem Set 2 Chem 314A Williamsen Spring 2000 Answers Problem Set Chem 314A Wllamsen Sprng 000 1) Gve me the followng crtcal values from the statstcal tables. a) z-statstc,-sded test, 99.7% confdence lmt ±3 b) t-statstc (Case I), 1-sded test, 95%

8.6 The Complex Number System 8.6 The Complex Number System Earler n the chapter, we mentoned that we cannot have a negatve under a square root, snce the square of any postve or negatve number s always postve. In ths secton we want

Lecture 2 Solution of Nonlinear Equations ( Root Finding Problems ) Lecture Soluton o Nonlnear Equatons Root Fndng Problems Dentons Classcaton o Methods Analytcal Solutons Graphcal Methods Numercal Methods Bracketng Methods Open Methods Convergence Notatons Root Fndng

MEASUREMENT OF MOMENT OF INERTIA 1. measurement MESUREMENT OF MOMENT OF INERTI The am of ths measurement s to determne the moment of nerta of the rotor of an electrc motor. 1. General relatons Rotatng moton and moment of nerta Let us

You will analyze the motion of the block at different moments using the law of conservation of energy. Physcs 00A Homework 7 Chapter 8 Where s the Energy? In ths problem, we wll consder the ollowng stuaton as depcted n the dagram: A block o mass m sldes at a speed v along a horzontal smooth table. It next

Conservation of Energy Lecture 3 Chapter 8 Physcs I 0.3.03 Conservaton o Energy Course webste: http://aculty.uml.edu/andry_danylov/teachng/physcsi Lecture Capture: http://echo360.uml.edu/danylov03/physcsall.html 95.4, Fall 03,

and problem sheet 2 -8 and 5-5 problem sheet Solutons to the followng seven exercses and optonal bonus problem are to be submtted through gradescope by :0PM on Wednesday th September 08. There are also some practce problems,

Physics 240: Worksheet 30 Name: (1) One mole of an deal monatomc gas doubles ts temperature and doubles ts volume. What s the change n entropy of the gas? () 1 kg of ce at 0 0 C melts to become water at 0 0 C. What s the change n entropy

Limited Dependent Variables Lmted Dependent Varables. What f the left-hand sde varable s not a contnuous thng spread from mnus nfnty to plus nfnty? That s, gven a model = f (, β, ε, where a. s bounded below at zero, such as wages

1 (1 + ( )) = 1 8 ( ) = (c) Carrying out the Taylor expansion, in this case, the series truncates at second order: 68A Solutons to Exercses March 05 (a) Usng a Taylor expanson, and notng that n 0 for all n >, ( + ) ( + ( ) + ) We can t nvert / because there s no Taylor expanson around 0 Lets try to calculate the nverse

find (x): given element x, return the canonical element of the set containing x; COS 43 Sprng, 009 Dsjont Set Unon Problem: Mantan a collecton of dsjont sets. Two operatons: fnd the set contanng a gven element; unte two sets nto one (destructvely). Approach: Canoncal element method:

I certify that I have not given unauthorized aid nor have I received aid in the completion of this exam. ME 270 Fall 2012 Fnal Exam Please revew the followng statement: I certfy that I have not gven unauthorzed ad nor have I receved ad n the completon of ths exam. Sgnature: INSTRUCTIONS Begn each problem

Important Instructions to the Examiners: Summer 0 Examnaton Subject & Code: asc Maths (70) Model Answer Page No: / Important Instructons to the Examners: ) The Answers should be examned by key words and not as word-to-word as gven n the model

Complex Numbers Alpha, Round 1 Test #123 Complex Numbers Alpha, Round Test #3. Wrte your 6-dgt ID# n the I.D. NUMBER grd, left-justfed, and bubble. Check that each column has only one number darkened.. In the EXAM NO. grd, wrte the 3-dgt Test

One-sided finite-difference approximations suitable for use with Richardson extrapolation Journal of Computatonal Physcs 219 (2006) 13 20 Short note One-sded fnte-dfference approxmatons sutable for use wth Rchardson extrapolaton Kumar Rahul, S.N. Bhattacharyya * Department of Mechancal Engneerng,

1. Estimation, Approximation and Errors Percentages Polynomials and Formulas Identities and Factorization 52 ontents ommonly Used Formulas. Estmaton, pproxmaton and Errors. Percentages. Polynomals and Formulas 8. Identtes and Factorzaton. Equatons and Inequaltes 66 6. Rate and Rato 8 7. Laws of Integral Indces Georga Tech PHYS 624 Mathematcal Methods of Physcs I Instructor: Predrag Cvtanovć Fall semester 202 Homework Set #7 due October 30 202 == show all your work for maxmum credt == put labels ttle legends Name: PHYS 110 Dr. McGoern Sprng 018 Exam 1 Multple Choce: Crcle the answer that best ealuates the statement or completes the statement. #1 - I the acceleraton o an object s negate, the object must be