# Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.

8 Pages

### Phys 2214 hw2 Solutions

Course: PHYSICS 2214, Fall 2011
School: Cornell
Rating:

Word Count: 1061

#### Document Preview

2214 Phys Homework #2 Solutions September 9, 2011 1. Free oscillations (a) For the initial conditions x(0) = A and v (0) = 0, we have the following results: i. Underdamped: x(t) = Aet/A cos(0 t) ii. Overdamped: x(t) = Aet/A iii. Critically damped: x(t) = A(1 + t/A )et/A These are plotted in the gure. 1.0 0.5 0.0 2 4 6 8 10 Critically Damped 0.5 Underdamped Overdamped 1.0 (b) For the underdamped case,...

Register Now

#### Unformatted Document Excerpt

Coursehero >> New York >> Cornell >> PHYSICS 2214

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
2214 Phys Homework #2 Solutions September 9, 2011 1. Free oscillations (a) For the initial conditions x(0) = A and v (0) = 0, we have the following results: i. Underdamped: x(t) = Aet/A cos(0 t) ii. Overdamped: x(t) = Aet/A iii. Critically damped: x(t) = A(1 + t/A )et/A These are plotted in the gure. 1.0 0.5 0.0 2 4 6 8 10 Critically Damped 0.5 Underdamped Overdamped 1.0 (b) For the underdamped case, since b2 4mk , b b2 4mk = 2m 1 becomes b i 4mk b2 . = 2m The imaginary part of is 4mk b2 2 2 Im() = = 0 1/A . 2m Since decreases when 0 A = 0, write 0 | | = 0 1 1/(0 A )2 . Dividing by 0 , squaring both sides, and solving for 0 A gives 1 0 A = 2| |/0 (| |/0 )2 . i. For | |/0 = 0.001, 0 A = 22.4. ii. For | |/0 = 0.01, 0 A = 7.09. iii. | |/0 = 0.1, 0 A = 2.29. (c) For the underdamped case, the system undergoes many oscillations during the time it takes to relax to its equilibrium position. For the critically damped and the overdamped cases, the system doesnt undergo any oscillations. i. ii. iii. iv. v. vi. vii. Auto suspension: critically damped; slightly underdamped. Door bell: underdamped. Speaker cone: underdamped. Pogo stick: underdamped. Drum head: critical or underdamped. Cymbal: underdamped. Bungee cord + person: underdamped. 2. Driven oscillations A(D ) = 2 0 F0 /m + i2D /A 2 D (a) To nd the magnitude, use A (D )A(D ) F0 1 = 2 m (0 2 )2 + 4 2 / 2 |A(D )| = D 2 D A To nd the phase, rewrite A(D ) as 2 2 D i2D /A 1 F0 0 2 2 2 2 m 0 D + i2D /A 0 D i2D /A 2 2 F0 0 D i2D /A = 2 2 2 2 m (0 D )2 + 4D /A A(D ) = Then tan((D )) = Im(A) 2D /A . =2 2 Re(A) 0 D (b) We can maximize |A(D )| by minimizing the denominator: d 2 2 2 2 (0 D )2 + 4D /A = 0 dD 2 2 2 2(0 D )(2D ) + 8D /A = 0 , so D = 2 2 0 2/A . Since Q = 0 A /2, we get D = 0 1 1/2Q2 . (c) Plug D from part (b) into the complex amplitude to get: |A|max = F0 m 1 = F0 m = F0 2 m 0 1 1/4Q2 2 2 2 2 (0 0 (1 1/2Q2 ))2 + 40 (1 1/2Q2 )/A 1 2 4 (0 /2Q2 )2 + 0 (1 1/2Q2 )/Q2 Q 2 Using m0 = k and A0 = F0 /k gives Amax = A0 Q 11/4Q2 A0 Q for large Q. (d) For large Q, D 0 = k /m. To nd how changes in mass change the frequency, dierentiate with respect to m: d d 1 1 =k (m1/2 ) = k ( )m3/2 = dm dm 2 2 Then | |/ m/2m. But | |/ 4 2/Q = 2/10 = 0.0002. 3 k1 1 = . mm 2m 1/Q, so m/m (e) When D = 0 , F0 /m F0 F0 m A = |A(0 )| = = =. 20 /A b0 bk Dierentiating A with respect to m gives dA F0 1 = dm 2b k m A m = . A 2m and So, for a change in mass of 1 part in 106 , A/A = 5 107 . (f) Suppose Q = 1/ 2. Then max = 0 1 1/2Q2 = 0. This means F0 Q F0 Amax = = A0 . 2 2 m0 m0 1 1/4Q2 So, Amax = A0 , which means that the resonance peak equals the amplitude at D = 0 and the peak goes away. 3. Free and driven oscillations For t = 3A , the amplitude decreases by e3 = 0.05, which is close to zero. (a) Underdamped, since you are told that the oscillations die away. (b) Given f0 = 4Hz , we have 0 = 25.13 rad/s. Since 3A = 10s, then A = 10/3 s. So, Q = 0 A /2 = 41.9, and D = 0 1 1/2Q2 = 25.13 rad/s 0 . (c) A = 10/3 s, f0 = 4 Hz 0 = 8 rad/s. A(d ) = F0 /m 2 2 2 2 (0 D )2 + 4D /A , so Amax = (F0 /m)/(2D /A ). Then we set A = 0.7 Amax and solve for D . 0.7 F0 /m = 2D /A F0 /m 2 2 2 2 (0 D )2 + 4D /A . Squaring both sides and inverting 4D gives 2 2 2 2 2 2.041 /A = (0 D )2 + 4d /A . 4 2 Rearranging terms gives a quadratic equation for D whose solu2 2 2 tions are D = (6.476 10 , 6.161 10 ). Then D = (24.82, 25.45) and D = 0.63 rad/s. (d) At low frequencies, |A|max /A0 = Q = 41.9. At high frequencies, |A|max /|A(D ) . (e) Removing the cashews decreases the mass of the oscillator. Then 0 = k /m increases. The damping constant doesnt change, so Q = 0 A /2 = k m/b decreases. Since 0 /0 = 1/Q, 0 = 0 /Q = b/m, so 0 increases. 4. Power in driven oscillators If the driving force is F0 cos(t) and the position is given by x(t) = A cos(t + ), then v (t) = A sin(t + ) and the instantaneous power delivered to the oscillator by the driving force is PD (t) = F (t) v (t) . Some useful integrals for this problem are: 2 2 sin(x) cos(x)dx = 0 and 0 0 2 sin2 (x)dx = cos2 (x)dx = . 0 (a) The average power delivered by the driving force in one cycle (the period is 2/ ) is: 2/ F0 cos(t)(A) sin(t + )dt 2 0 AF0 2 = cos y sin(y + )dy 2 0 AF0 2 = [cos y sin y cos + cos2 y sin ]dy 2 0 AF0 = sin . 2 avg PD = (b) The instantaneous power dissipated by the damping force is Pdamp (t) = bv 2 . The average power delivered per cycle is: 2/ (A )2 sin2 (t + )dt 2 0 1 = b 2 A2 2 avg Pdamp = b 5 (c) The frequency at which the input power is maximized is obtained by maximizing the answer in part (b): dPdamp d 2 2 d 2 d 2 ( 2 0 )2 + b2 2 /m2 2 2 [( 2 0 )2 + b2 2 /m2 ] 2 [2( 2 0 ) + b2 /m2 ] 2 [( 2 0 )2 + b2 2 /m2 ]2 =0 Setting the numerator equal to 0 gives 2 2 ( 2 0 )2 2 2 ( 2 0 ) = 0 4 2 2 4 + 0 2 2 0 2 4 + 2 2 0 = 0 avg avg Then Pdamp , and thus PD , is a maximum when = 0 . The phase ( ) is given by tan = 2 b = . 2 2 2) A (0 m(0 2 ) Then ( = 0 ) = |i/2 and v (t) = A sin(t /2) = A cos t. So, when = 0 , the driving force and the velocity are in phase. 5. Energy in driven oscillations For a system udergoing forced oscillations at an angular frequency D , x(t) = |A(D )| cos(D t + ) , where |A(D )| = F0 m 1 2 2 2 2 (0 D )2 + 4D /A = 2 A0 0 2 2 2 2 (0 D )2 + 4D /A ; In the last step we used F0 /m = (F0 /m) (k/k ) = (F0 /k ) (k/m) = 2 A0 0 . Also, v (t) = D |A(D )| sin(D t + ) . 6 (a) The kinetic energy is K (t) = mv (t)2 /2, so 1 2 K (t) = mD |A(D )|2 sin2 (D t + ) 2 4 0 1 2 mD A2 2 sin2 (D t + ) = 0 22 2 2 2 (0 D ) + 4D /A (b) The potential energy is U (t) = k (x x0 )2 /2, so 2 1 A0 0 U= A0 k2 2 2 0 D + 2iD /a 2 2 1 0 2 = m0 A2 2 1 0 2 2 0 D + 2iD /A 1 2 2iD /A 2 = m0 A2 2 D 2 0 2 0 D + 2iD /A 2 2 Then 4 1 D + (2D /A )2 22 |U| = m0 A0 2 2 2 (0 D )2 + (2D /A )2 and U (t) = |U| cos2 (D t + ). , (c) 22 2 Kmax 0 D 0 1 =4 , =2 =2 2 2 2 Umax D + 4D /a D + 4/A x + 1/Q2 where x = D /0 and Q = 0 A /2. (d) Kmax = Umax when the ratio in part (c) is equal to 1. Then 2 2 2 0 = D + 4/A , or D = 2 2 0 4/A = 0 1 1/Q2 . Then the total energy, which is equal to the maximum kinetic energy, is: 4 0 1 2 2 m(0 4/A )A2 2 0 2 2 2 2 2 2 [0 (0 4/A )2 + 4(0 4/A )/A 1 4 ( 2 4/ 2 ) = mA2 0 0 2 2 A 0 2 40 /A 1 2 = mA2 0 Q2 (1 1/Q2 ) 0 2 1 2 = m0 A2 (Q2 1) 0 2 E = Kmax = 7 (e) For low driving frequencies, D Q2 1, so Umax Kmax . 22 0 and Kmax /Umax 0 A /2 = (f) For high driving frequencies, D and Kmax /Umax Umax Kmax . 8 1, so
Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

Cornell - PHYSICS - 2214
Phys 2214 Homework #3 SolutionsSeptember 15, 20111. The disturbance travels down the rubber tube with the wave speed,which, for a one-dimensional transverse wave with constant tensionand mass/length, is given byv=.The tension is equal to the weight
Cornell - PHYSICS - 2214
Phys 2214 Homework #4 SolutionsSeptember 30, 20111. For a long, thin rod, the compressions created by a longitudinal wavetraveling along the rod will change the length to a greater extent thanthe diameter. So, we can replace the equation for the bulk
Cornell - PHYSICS - 2214
Phys 2214Assignment #5 Solutions1. Optical standing waves in a laser(a) standing wave frequencies: 1 2 L, so f n n f1 n (b) (i) n nmax nmin [ f 0 f ( f 0 f )] (ii) For n = 1, L c1nc, n = 1,2,3, 2L4L f and n = 6cc 0.075m 7.5cm4 f3. Superpo
Cornell - PHYSICS - 2214
Phys 2214 Homework #6 SolutionsOctober 26, 20111. EM Waves in VacuumGiven Gauss law, E = / 0 , assume = 0, since were in a vacuum.Also, we take the plane wave to be traveling in the +x direction, soE = f (x ct) + g (x ct) + h(x ct) .xyzNote that
Cornell - PHYSICS - 2214
Phys 2214 Homework #7 SolutionsNovember 1, 20111. EM Waves in Conductors.The current density is J = E/, where is the electrical resistivity.Amperes law in dierential form isB =EE+.tIf is small (a good conductor), then the rst term on the right h
Cornell - PHYSICS - 2214
PHYS 2214 Homework #8 Solutions1. Deriving Snells Law.(a) From the gure in the book,d1 d2t = t1 + t2 =+=v1 v2h2 + x21+v1h2 + (l x)22v2.(b) The independent variable here is x, since the others are just parameters. So, we nd the minimum time
Conestoga - EE - 1244
Welcome to 2G03:Problem Solving &amp;Technical CommunicationDr. E. Cranston and Mr. V. Leung2G03 Lecture 1: Introduction (1)Welcome Instructors: Emily Cranston and Vince Leung TAs: Kevin Kan and Sarah Charlong Avenue to Learn Courseware: 2011 updated
Conestoga - EE - 1244
The Writing Process2G03: Problem Solving &amp;Technical CommunicationDr. E. Cranston and Mr. V. Leung2G03 Lecture 2: The Writing Process (1)Presentations: ChE Professors Prof. M. Thompson(Polymers) Prof. C. Filipe(Bioengineering) Prof. T. Adams(Pro
Conestoga - EE - 1244
The Writing Process (Continued)2G03: Problem Solving &amp;Technical CommunicationDr. E. Cranston and Mr. V. Leung2G03 Lecture 3: The Writing Process Continued (1)Housekeeping Emily Cranston: ecranst@mcmaster.ca Vince Leung: leungv@mcmaster.ca Hand in
Conestoga - EE - 1244
The Job Search2G03: Problem Solving &amp;Technical CommunicationDr. E. Cranston and Mr. V. Leung2G03 Lecture 4: The Job Search and Formatting (1)Housekeeping Emily Cranston: ecranst@mcmaster.ca Vince Leung: leungv@mcmaster.ca Hand in assignments in cl
Conestoga - EE - 1244
Boss-Quality Documents,Audience Analysisand Interview SkillsDr. E. Cranston and Mr. V. Leung2G03 Lecture 5: Boss-Quality, Audience Analysis, Interview Skills (1)Housekeeping Emily Cranston: ecranst@mcmaster.ca OFFICE HOURS: Wednesday 1 pm to 3 pm
Conestoga - EE - 1244
Ethics, Paraphrasing and WhatIndustry Expects of YouDr. E. Cranston and Mr. V. Leung2G03 Lecture 6: Ethics, Paraphrasing and What Industry Expects of You (1)Housekeeping Emily Cranston: ecranst@mcmaster.ca OFFICE HOURS: not this week but otherwise
Conestoga - EE - 1244
Awareness inProblem SolvingDr. E. Cranston and Mr. V. Leung2G03 Lecture 7: Awareness in Problem Solving (1)Housekeeping Emily Cranston: ecranst@mcmaster.ca OFFICE HOURS: Wednesdays 1 pm to 3 pm (JHE A412) Vince Leung: leungv@mcmaster.ca Due today
Conestoga - EE - 1244
Problem Solving Strategyand Self-AssessmentDr. E. Cranston and Mr. V. Leung2G03 Lecture 8: Strategy and Self-Assessment (1)Housekeeping Emily Cranston: ecranst@mcmaster.ca OFFICE HOURS: Wednesdays 1 pm to 3 pm (JHE A412) Vince Leung: leungv@mcmast
Conestoga - EE - 1244
Oral Presentation Skills andthe Unique YouMs. S. Charlong, Dr. E. Cranston,and Mr. V. Leung2G03 Lecture 9: Oral Presentations and the Unique You (1)Housekeeping Emily Cranston: ecranst@mcmaster.ca OFFICE HOURS: Wednesdays 1 pm to 3 pm (JHE A412)
Conestoga - EE - 1244
Time and StressManagementDr. E. Cranston and Mr. V. Leung2G03 Lecture 10: Time and Stress Management (1)Housekeeping Emily Cranston: ecranst@mcmaster.ca OFFICE HOURS: Wednesdays 1 pm to 3 pm (JHE A412) Vince Leung: leungv@mcmaster.ca Due today: N
Conestoga - EE - 1244
Analysis and ClassificationDr. E. Cranston and Mr. V. Leung2G03 Lecture 11: Analysis and Classification (1)Housekeeping Emily Cranston: ecranst@mcmaster.ca OFFICE HOURS: Wednesdays 1 pm to 3 pm (JHE A412) Vince Leung: leungv@mcmaster.ca Due today:
Conestoga - EE - 1244
CreativityDr. E. Cranstonand Mr. V. Leung2G03 Lecture 12: Creativity (1)Housekeeping Emily Cranston: ecranst@mcmaster.ca OFFICE HOURS: Dec 7th only: 1 pm to 3 pm (JHE A412) Vince Leung: leungv@mcmaster.ca Due today:Corrected term paper (with abs
Conestoga - EE - 1244
Conestoga - EE - 1244
Conestoga - EE - 1244
Conestoga - EE - 1244
CORRELATIONData arise in pairs(xi, yi),i = 1 , 2, . . . , nResponse y : outcome of experimentExplanatory x: explains outcomeScatterplotPlot observed pairs (xi, yi) in x-y plane1Example: Cancer Mortality in Oregonvs Radioactive ContaminationCoun
Conestoga - EE - 1244
Meaning of r2 - Worked ExampleExamplex = explanatory variable= temperature (degrees Celsius)y = response variable= yield (kg)x181920212223242526272829y76.177.978.178.278.879.779.981.181.281.882.883.51First check that a reg
Conestoga - EE - 1244
PROBABILITYChapter 2.1Deterministic: dx = bx.dtStatistical:observation = true value + error error dierent each time experimentis performed.1Example. Quality control. Sample n,X = proportion defective, p = true defective rate.X = p + errorIn pr
Conestoga - EE - 1244
Conditional ProbabilityProbabilities do not live in a vacuum.They are specied by conditions whichmay change or about which additionalinformation may become available. Consider A, B with P (A), P (B ) given. Suppose have knowledge that B occurred.How
Conestoga - EE - 1244
RANDOM VARIABLESMeasurements and observations are calledrandom variables. Each results from anoutcome of an experiment, so:Denition: Random variable X is realfunction on sample space S.Start with discrete r.v. RX is nite orcountable.1For any real
Conestoga - EE - 1244
Binomial DistributionAssumptions n independent trials trial i has dichotomous outcome (0 or 1, no or yes, failure or success, etc.) denotedby Xi. Called Bernoulli trials. probability of success on anytrial is the same p.1Let Y = # of successes.Y
Conestoga - EE - 1244
Example:- 19 19 square Go board (Japanese boardgame)- throw dart repeatedly and at random- mark square hit111. P (specic square hit) = 19 1= 1193612. How many throws, until highprobability of hitting some squareagain?3. After N throws, what
Conestoga - EE - 1244
Poisson DistributionFlaws occur at random alonglength of oil pipeline. Average per unit length. Y = number ofaws in a randomly selected section of pipeline of length 1. Canwe determine the distribution ofY?Generic situation is incidentshappening o
Conestoga - EE - 1244
MULTIVARIATEDISTRIBUTIONS, DATARarely only one random variable.Usually many cfw_X1, . . . , Xn &amp; function u(X1, . . . , Xn).distribution.Need jointn = 2. Notationcfw_X, Y . Joint p.m.f.f (x, y ) = P (X = x and Y = y ) P (X = x, Y = y )1.Exerci
Conestoga - EE - 1244
Continuous RandomVariablesContinuous random variables are random variables that arise from measurements can take any value in intervalExample.Random numbers in [0, 1]used in random number generators.1.Think of a circle with circumference 1 and a
Conestoga - EE - 1244
NORMAL DistributionMost important distribution in statistics112f (x) = e 2x 21E [X ] =Var X =P (X x) = F (x)21x 1 t= t= e 2 dt 2No formula for this integral. Tables exist C4 p568-9 for standard normalStandard Normal = 0, = 1.Notation:
Conestoga - EE - 1244
Multiple Continuous RandomVariablesJoint probability density functionfX,Y (x, y ) describes pair (X, Y ).fX,Y (x, y ) 0f(x, y )dydxx= y = X,Y=1P (X, Y ) A) =Cumulative joint distribution functionFX,Y (x, y ) =1Marginal DensitiesThe p.d.f. of
Conestoga - EE - 1244
Data and Random VariablesAt the start of this course we discussed variation in data. Suppose that there are two methods of making concrete aggregates, an existingone (denoted by symbol A), and a new one,(denoted by B ), which contains a new bindingage
Conestoga - EE - 1244
Central Limit TheoremNotation. N (, 2) refers to normal with mean and variance 2.Have just seen that when X1, . . . , Xn is a random sample from N (, 2) then sample meansX N (, 2/n). This follows from a more general result that:2If Xi N (i, i ) are
Conestoga - EE - 1244
Introduction to EstimationData arise from measurements whose valuesdepend on characteristics of the process or experiment being considered. Some features ofthe process can be controlled while others cannot and result in variation in the data. If wecou
Conestoga - EE - 1244
Condence IntervalsDenition. Level C condence interval (C.I.) for is interval (L, U ) whereL, U obtained from data, such thatP (L U ) = Cno matter what .Typically, C = 0.90, 0.95, 0.99.C.I. is interval estimator, represents setof plausible estimates
Conestoga - EE - 1244
Condence Intervals for 1 2 Compare responses in two groups Can arise in two ways1.2.12Population12SampleSizen1n2VariableMeanSt. dev.XY1212Parameter of interest=Natural estimator=ThenE[X Y ] =3Assume independent samples. Then
Conestoga - EE - 1244
C.I. for from NormalPopulation with Unknown Previously assumed either:Now is unknown, but sample sizenot large enough for central limit theorem. Inference needs some additionalassumptions.Assume N (, 2) population. Intereston parameter but unknown.
Conestoga - EE - 1244
This following material covers Section 4.4: condence intervals forproportions and variances. Youare not responsible for the derivation (shown in purple) of thesecondence intervals on a test ornal exam. I have shown thedetails for those who are intere
Conestoga - EE - 1244
Testing a Single ParameterSignicance tests are dierent type of statistical inference. Want to choose between competing hypotheses about a parameter. Denition:An hypothesis is an assertionabout a population or about a parameter.We distinguish two kind
Conestoga - EE - 1244
Using OC Curve to DetermineSample SizeExample 4.5-1, p 255. Standard production setup. Average time = 30minutes to complete task. Assume X N (30, 1). Change suggested. New X N (, 1) where hoped &lt; 30. Samplesize n workers to testH0 : = 30H1 : &lt; 30
Conestoga - EE - 1244
Testing a Normal MeanEarlier found sample size to achievespecied Type I error and TypeII error at particular value ofalternative parameter.Here ngiven. Cant pick it. Stuck withwhatever emerges. Test at thelevel of signicance H0 : = 0H 1 : &lt; 01
Conestoga - EE - 1244
Tests Comparing 1 with 2Independent random samples cfw_X1, . . . , Xn1 and22cfw_Y1, . . . , Yn2 from N (1, 1 ), N (2, 2 ) populations respectively. Null hypothesis is:H0 : 1 = 222If 1 , 2 known then test based on normal standardized scoreZ=22I
Conestoga - EE - 1244
Paired t-testEarlier tested 1 = 2 using two independent random samples cfw_X1, . . . , Xn1 and cfw_Y1, . . . , Yn2 from22N (1, 1 ), N (2, 2 ) populations.Basis:X Y N 1 221,n122 +n22-sample Z or 2-sample t, dependingon whether variances w
Conestoga - EE - 1244
Paired t-testEarlier tested 1 = 2 using two independent random samples cfw_X1, . . . , Xn1 and cfw_Y1, . . . , Yn2 from22N (1, 1 ), N (2, 2 ) populations.Basis:X Y 2-sample Z or 2-sample t, dependingwhether variances were known or not.1Varianc
Conestoga - EE - 1244
Testing Equality of Two Proportionsp1 = p2Testing H0 : p1 = p2. = p1 p 2Independent samples Y1, Y2.Y1 binomial(n1, p1), Y2 binomial(n2, p2)YY = P1 P2 = n1 n212Large sample sizes n1, n2.Use normal approximation.Pi N pi, pi(1pi , i = 1, 2ni N
Conestoga - EE - 1244
Conestoga - EE - 1244
Engineering Design &amp; GraphicsEngineering 1C03Dr. Thomas E. Doyle, P.EngDepartment of Electrical and Computer EngineeringETB/106Ofce Hours are Mondays 2:00 - 4:00pmWelcome!IntroductionOverviewPeopleSyllabusSoftwareLecture, Laboratory, Tutorial
Conestoga - EE - 1244
Introduction toTechnical Sketches andEngineering DrawingsDr. Thomas E. Doyle, P.EngDepartment of Electrical and Computer EngineeringETB/106Ofce Hours are Mondays 2:00 - 4:00pmLecture Objectives Geometry of Engineering Graphics Dene projection typ
Conestoga - EE - 1244
Engineering Design &amp;Graphics Week 04Dr. T. E. DoyleOverview Continue example from last lecture Thanksgiving holiday and makeup labs/tutorials Rigid bodies in free space (3-Drelationships)Thanksgiving ReschedulingIf you normally have a lab or tut
Conestoga - EE - 1244
Engineering Design &amp;Graphics Week 06Dr. T. E. DoyleAnnouncements Exams - this week! (worth 10% each) Project groups - must be decided for nextweek Maple/MapleSim - installed for next week Software available on campus in KTH/B123OverviewContinue
Conestoga - EE - 1244
Simple Mechanisms andSystem ModellingEngineering Design &amp; GraphicsEngineering 1C03Dr. T. E. DoyleOverview Simple Mechanisms Introduce System Modelling via MapleSim Introduce ProjectMechanicsMechanics deals with motion, time, and forces.M
Conestoga - EE - 1244
Introduction to ProjectProduct &amp; MapleSimEngineering Design &amp; GraphicsEngineering 1C03Dr. T. E. DoyleOverview Review project specication (noteextended deadline for part-1) Discuss the product, disassemble, reviewmechanism Introduce System
Conestoga - EE - 1244
Introduction to GearDesign: Spur &amp; Rack GearsEngineering Design &amp; Graphics Engineering 1C03 Dr. T. E. Doyle Overview Course Evaluation Project Update Introduce Ideal Spur Gear Design Design Equations Inventor Design Accelerator
Conestoga - EE - 1244
McMaster University Engineering Design &amp; Graphics Engineering 1C03 Dr. T. E. Doyle, P.Eng Dept. of Electrical and Computer Engineering Overview Recall spur gear terminology Examples of gear ratios Gear design formulae
Conestoga - EE - 1244
Ideal Worm Gear Design and Modelling McMaster University Engineering Design &amp; Graphics Engineering 1C03 Dr. T. E. Doyle, P.Eng Dept. of Electrical and Computer Engineering Worm Arrangements Worm Worm Ge
Conestoga - EE - 1244
Course ConclusionDesign &amp; GraphicsEngineering 1Dr. T. E. DoyleWeek Overview Final Exam Announcement DuraBon: 2 hours Material includes topics covered from week 1 through week 12, inclusive There will be no quesBons rega
Conestoga - EE - 1244
Introduction to Professional Engineering Lecture 1COURSE &amp; PROJECT INTRODUCTIONENG 1P03 Lecture 1September 13/15, 2010GREATEST ENGINEERING ACHIEVEMENTS OF THE 20TH CENTURYElectrificationAutomobileAirplaneWater Supply andDistributionElectronicsR
Conestoga - EE - 1244
Engineers WithoutBorders McMasterFirst General MeetingThursday, September 23rd7 8 PM in JHE 328 (Grad Lounge)ORFriday, September 24th1:30 2:30 PM in JHE A114Introduction to Professional Engineering Lecture 2DESIGN PROCESS &amp; BRAINSTORMINGENG 1P03
Conestoga - EE - 1244
McMaster Engineering SocietyWho we are, and what we do!Monday September 27thWednesday September 29th1P03 Lecture PresentationMcMaster Engineering SocietyOur MissionThe McMaster Engineering Society will fosterthe development of well roundedundergr