Some Basic Information about M-S-D Systems

Size: px
Start display at page:

Download "Some Basic Information about M-S-D Systems"

Transcription

1 Some Basic Informaion abou M-S-D Sysems 1 Inroducion We wan o give some summary of he facs concerning unforced (homogeneous) and forced (non-homogeneous) models for linear oscillaors governed by second-order, consan coefficien sysems. Recall ha he homogeneous differenial equaion ha we are dealing wih may be wrien in he form which has characerisic equaion m d d ) x + γ d d x + ω ox =, or where γ = c/m and ω o m λ + c λ + k =, λ + γλ + ω o =, = k/m. Using hese values of he consans, he roos are λ ± = γ ± 4 γ 4 ωo = γ ± γ ωo ± ωo = ±iω o, γ = = ( ) ωo γ 1 ± 1, γ. γ The Harmonic Oscillaor The firs alernaive leads o he Harmonic Oscillaor which has general soluion We make he folowing remarks: x() = C 1 cos (ω o ) + C sin (ω o ). (a) ω o is called he naural frequency. This is he number of waves passing a fixed observer in uni ime. 1

2 (b) The quaniy π ω o is called he wave lengh. I is usually called λ and is he disance a which he profile repeas iself. (c) The ime aken for one complee wave o pass any paricular poin is called he period, T, of he wave. Since he cosine and sine funcions are periodic of period π, he period is given by T = λ/π = 1/ω o. (d) x() = A cos (ω o δ) wih A = C 1 + C and cos (δ) = C 1 /A, sin (δ) = C /A or an (δ) = C /C 1 is called he phase-ampliude form of he soluion; A is he ampliude and δ is he phase shif. Example.1 : Le m = 1 and k = 36. Then ω o = 6 and he soluion of he iniial value problem ẍ + 36 x = wih iniial condiions x() = 1, ẋ() = 1 can be deermined from he general form of he soluion x() = C 1 cos (6 ) + C sin (6 ) ẋ() = 6C 1 sin (6 ) + 6C cos (6 ) Then x() = 1/ = C 1 and ẋ() = 1 C or C = 1/6. Hence he required soluion is x() = 1 cos (6 ) sin (6 ). (1 ) The AMPLITUDE of he vibraion is A = + ( 1 5 6) =.57. The PHASE 18 SHIFT is compued from he definiion of δ aking care ha he correc branch of he inverse angen is chosen. Here C 1 = 1/ and C = 1/6 so sin (δ) > and cos (δ) > which implies ha δ lies beween and π/. ( 1 ) an (δ) = ( 1 6) = 6 = 3, and we may conclude ha, approximaely, δ = 1.49 radians or, again approximaely, degrees. The quesion of deermining he spring consan is a maer of experimenal measuremen. Consider he following example:

3 Example. : A mass weighing 6 lbs. sreches a spring 6 in. Deermine he funcion x() ha describes he displacemen of he mass if he mass is released from res 18 in. below he equilibrium posiion. From his daa, we need o deermine he spring consan or he raio of he spring consan o he mass. Cauion: You mus be sure ha he unis are consisen! Hooke s Law says ha he spring force is linear and of he form F = k x. Since we have ha he force applied o he spring is 6 lbs., i follows ha he equaion 6 = k.5 so ha k = 1lbs./f.. Nex, he mass mus be deermined from he equaion F = m a. Since F = 6 and g = 3f./sec., we have ha he mass m = 15/8 slugs. Now, ω o = k/m = (1 8)/15 = 64 sec. Wih hese values of he parameers he equaion of moion becomes ẍ + 64 x =, and since he iniial posiion x() = 1.5f. (18 in. is 1.5 f.) and he mass is released from res, ẋ() =, we can solve he iniial value problem o see ha x() = cos (6 ). Now ry he following exercise: Exercise.3 : A mass of 1 gm. sreches a spring 5 cm. If he mass is se in moion from is equilibrium posiion wih a downward velociy if 1 cm./sec., and if here is no damping, deermine he posiion x of he mass a any ime. When does he mass firs reurn o he equilibrium posiion? 3 The Damped Oscillaor If damping is presen, hen γ and here are he usual hree cases which depend on he sign of he descriminan which, in urn, depends on he relaive sizes of he parameers. This relaive dependence is bes seen in he form of he descriminan ha we derived above. Spefically, remembering ha ω o = g/m and γ = c/m, he hree cases are: 3

4 Case Type 1 ( ωo γ ) > overdamping (c > 4mk) 1 ( ωo γ ) = criical damping (c = 4mk) 1 ( ωo γ ) = underdamping (c < 4km) Example 3.1 : Criical Damping An 8 lb. weigh is aached o a spring of lengh 4 fee. A equilibrium, he spring has lengh 6 f. Deermine he displacemen from equilibrium as a funcion of ime, x() if he drag coefficien is and (a) he mass is released from is equilibrium posiion wih a downward iniial velociy of 1 f./sec.; (b) he mass is released 6 in. above he equilibrium wih and iniial velociy of 5 f./sec. in he upward direcion. As before, one can compue boh he mass and he spring consan which are respecively, 1/4 slug and 4. So he differenial equaion is ẍ + 8 ẋ + 16 x =, and is o be solved wih wo ses of iniial condiions: (a) x() =, ẋ() = 1 and (b) x() =.5, ẋ() = 5. For his equaion, we are in he criically damped case since he characerisic equaion is λ +8λ+16 = or (λ+4) =. Hence he general soluion of he homogeneous equaion is x() = C 1 e 4 + C e 4. The firs se of iniial condiions lead o a soluion x a () = e 4. The graph of he soluion is given in he nex figure. For he second se of iniial condiions, he soluion is x b () =.5 e e 4. Again we can plo he soluion Noe ha in his las case here is an overshoo of he equilibrium ha occurs for his paricular choice of iniial condiion. 4

5 x() Figure 1: The Soluion x a x() Figure : The Soluion x b 5

6 Exercise 3. : Assume ha he sysem described by he equaion mẍ + cẋ + kx = is eiher criically damped or overdamped. Show ha he mass can pass hrough he equilibrium posiion a mos once, regardless of he iniial condiions. Example 3.3 : The Overdamped Case A 3 lb. weigh sreches a spring 8 fee. If he resisive force due o damping is 5 ẋ, deermine he displacemen as a funcion of ime if he mass is released from 1 f. below he equilibrium posiion wih (a) an upward velociy of 1 f./sec.; (b) an upward velociy of 6 f./sec. The corresponding differenial equaion is ẍ + 5 ẋ + 4 x =, which has characerisic roos λ 1 = 1 and λ = 4. The general soluion is hen This is he case of overdamping. x() = C 1 e + C e 4. Now, for he firs se of iniial condiions x() = 1, ẋ() = 1 (noe ha he sign indicaes ha he iniial velociy is upward) he corresponding soluion is x a () = e. 1.8 x() Figure 3: The Soluion x a For he second se of iniial condiions, x() = 1, ẋ() = 6, he corresponding soluion is x b () = 5 3 e 4 3 e. 6

7 1.8 x() Figure 4: The Soluion x b We now urn o he case of underdamping. Consider he following example: Example 3.4 : The Underdamped Case A 16 lb. weigh sreches a spring fee. Deermine he displacemen as a funcion of ime if he resisive force due o damping is.5 ẋ and he mass is released from he equilibrium posiion wih a downward velociy of 1 f./sec. In his case he spring consan k = 8 lb./f. while he mass is.5 slugs. The iniial condiions are x() =, ẋ() = 1 and he governing equaion is whose characerisic equaion has roos ẍ + ẋ + 16 x =, λ ± = 1 ± i 3 7, so ha, using Euler s formula, he general soluion of he equaion is x() = e [ C 1 cos C sin 3 ] 7. Using he iniial condiions, we find ha he soluion is x() = 7 1 e sin.

8 This is clearly an oscillaory soluion, bu he ampliude of he oscillaions decreases exponenially..3 x() Figure 5: The Underdamped Soluion Noice ha he soluion x() as. The oscillaion is NOT periodic. Moreover, from our physical inuiion, we would expec ha in his case, he moion would evenually sop in he equilibrium sae. This is NOT he case here; his is only an approximae model for he acual physical siuaion, much of he approximaion being made in our assumpion of he damping forces acing on he sysem. One can see he exponenial decay more clearly in he following figure where he solid lines are he graphs of he funcions ± 7 1 e. Exercise 3.5 : Here is a problem where you can use MAPLE o illusrae he dependence of he soluion of he damped oscillaor on he damping facor. Given he iniial value problem ẍ + c ẋ + 6 x = x() =, ẋ() = 1 find he soluion of he equaion for each of he values c = 6,c = 4 6, and c = 6. Idenify, for each value of he parameer c which of he cases you are in and plo he hree soluions on a single display. 8

9 .3. x() Forced Oscillaions Figure 6: The Exponenial Decay of Oscillaions We now urn o he problem of deermining he behavior of a forced damped oscillaor. The general equaion we will wrie as m ẍ + c ẋ + k x = F(), where F() is he forcing funcion. This problem can be solved for reasonable forcing funcions by he Mehod of Undeermined Coefficiens. We will give some simple examples, bu even hese will reveal some basic behavior. As usual, we begin wih he undamped case, or he Forced Harmonic Oscillaor. m ẍ + k x = F(). In his case, le us ry he simple case in which F() = e. In order o find a paricular soluion of he nonhomogeneous equaion, we ake a rial funcion in he form x() = A e. NOTE: we know ha he general soluion of he harmonic oscillaor is a linear combinaion of cosine and sine funcions so his rial funcion is NOT a soluion of he homogeneous equaion. Differeniaing he rial funcion wice yields ẍ() = 4 A e and subsiuing his resul ino he forced oscillaor equaion yields or m ( 4 A e ) + k A e = A e, 9

10 4 m A + k A = 1, and so, rearranging (4 m + k) A = 1. Hence A = 1/(4m + k). From his calculaion we can conclude ha he general soluion has he form: ( ) 1 x() = C 1 cos (ω o ) + C sin (ω o ) + e. m (4 + ωo) Noice ha, whaever he iniial condiions are, he las erm on he righ-hand side decays rapidly wih ime. Wha we expec o see in any paricular soluion is ha he iniial porion of he soluion will differ from he soluion of an unforced harmonic oscillaor, bu ha as ime increases, he soluion seles down o he pure oscillaions. If we ake a paricular se of iniial condiions, say x() = 1, ẋ() = 1, i.e., we sar from 1 uni below he equilibrium wih velociy 1 downward, and if we choose paricular values for he parameers, say m = 1 and k = 1/, hen he soluion is: x() = 1 ( ) 3 cos ( ) sin + 4 ) 3 e(, From wha is said above, if we ake he homogeneous equaion wih iniial condiions x() = 1/3, ẋ() = 5/3, he soluion of his iniial value problem is jus x() = 1 ( ) 3 cos ( ) sin and we expec he oscillaion of he forced equaion o approach he his oscillaory soluion for large. This phenomenon is illusraed in Figures 7-9 where he doed curve in he las figure represens he ransien soluion. We will see he occurance of he ransien/seady sae soluions again. Now we wan o look a he resul of imposing a sinusoidal forcing erm. Again, we look a he harmonic oscillaor wih ( m ) = 1 and( k = 1/. The general soluion of he homogeneous problem is x() = C 1 cos + C sin ), so ha we see he naural frequency is ω o = 1/. We sar wih he forcing frequency he same, i.e. ( F() = cos ). Our experience wih he Mehod of Undeermined ( Coefficiens ( leads us o expec ha we should find a soluion of he form [A cos )+B sin )] and ha he soluion o he full problem would grow in ampliude as ges large. Le us illusrae by doing he compuaions here. 1

11 Figure 7: Soluion for Forced Oscillaor Figure 8: The Transien Soluion Figure 9: The Seady Sae vs. Full Soluion 11

12 The Mehod of Undeermined Coefficiens ells us ha we should ake a rial soluion of he form x() = [A cos (/sqr) + B sin (/sqr)]. Then differeniaing wo imes yields: ẋ() = A cos(/ ) + B sin(/ ) + [ A sin(/ ) + B cos(/ ] ) ẍ() = [ A sin(/ ) + B cos(/ ] [ ) + A sin(/ ) + B cos(/ ] ) + [ A cos(/ ) + B ] sin(/ ) = [ A sin(/ ) + B cos(/ ] [ ) + A cos(/ ) B ] sin(/ ). Subsiuing back ino he differenial equaion yields: ẍ() + 1 x() = [ A cos(/ ) B sin(/ ] ) + [ A sin(/ ) + B cos(/ ] ) + [ A cos(/ ) + B sin(/ ] ) = [ A sin(/ ) + B cos(/ ] ) = cos (/ ). Equaing coefficiens of like erms, we find ha A = and / B = 1 or B = /, from which we conclude ha a paricular soluion of he non-homogeneous equaion is x nh () = ( ) ( sin ). Now we can wrie down he general soluion of he non-homogeneous problem ( ) ( ) x() = C 1 cos + C sin + ( ) ( sin ). The consans are, as usual, deermined by applying he iniial condiions o his full soluion and NOT o jus he soluion of he homogeneous equaion. In he case ha we ake iniial condiions x() = 1, ẋ() = 1, he soluion is 1

13 ( ) x() = cos + ( ) sin + ( ) ( sin ). Noe he las erm corresponding o x nh has a facor of and we expec he phenomenon of resonance. Here is a graph of he soluion Figure 1: The Phenomenon of Resonance If we now increase he frequency of he forcing funcion, we see anoher phenomenon. Le us consider he equaion ẍ + 1 x = cos ( 3 4 ). This ime, since he forcing funcion does no have he same frequency as he naural frequency of he sysem, he es funcion should be aken in he form A cos (3/4) + B sin (3/4). The usual calculaions lead o a soluion of he non-homogeneous equaion x nh () = 16 cos (3/4) so ha, aking iniial condiions x() =, ẋ() =, i.e. saring from he equilibrium posiion, he soluion of he iniial value problem is ( ) x() = 16 cos 16 cos ( ) 3. 4 The graph of his soluion in Figure 11 illusraes he phenomenon of beas. Exercise 4.1 Use he Mehod of Undeermined Coefficiens o solve he differenial equaion and hen use he general form of he soluion o deermine he unknown coefficiens. In his way, verify ha he given soluion is correc. 13

14 Figure 11: The Phenomenon of Beas If we increase he frequency of he forcing funcion o, say, 1.5, hen he soluion of he same iniial value problem as in he previous case leads o a soluion: x() = 4 ( ) 7 cos 4 ( ) 3 7 cos. Exercise 4. From he general form of he soluion x() = c 1 cos (/ )+sin (/ ) (4/7) cos (3/) find he values of c 1, c for which he soluion saisfies he iniial condiions x() =, ẋ() =. The nex figure shows he soluion. 5 Forced Damped Oscillaions We urn now o he case ha he forced mass-spring sysem is aached o a dashpo. Tha means ha we are dealing wih a sysem wih fricion and he governing equaion is given by m ẍ + γ ẋ + k x = F o cos (ω ). We know ha, wih damping, all soluions of he unforced sysem decay exponenially. We have discussed behavior of he responses of over-damped, criically damped, and underdamped sysems. Thus, in all cases, he soluion of he forced equaion can be wrien 14

15 Figure 1: The Response for Larger Frequency as x() = x h () + x p () where x h is he general soluion of he homogeneous equaion and represens a ransien response while he paricular soluion, x p, of he nonhomogeneous problem persiss wih ime and is called he seady sae response. In order o find he soluion of he nonhomogeneous problem, we again use he Mehod of Undeermined Coefficiens, looking for a rial soluion in he form which has derivaives x p () = A cos (ω ) + B sin (ω ) ẋ() = A ω sin (ω ) + B ω cos (ω ), and ẍ() = A ω cos (ω ) B ω sin (ω ). Subsiuing hese relaions ino he differenial equaion and combining like erms in he usual way, we find ha [(k m ω ) A + γ ω B] cos (ω ) + [(k m ω ) B γ ω A] sin (ω ) = F o cos (ω ). Equaing he coefficiens on boh sides of his equaion we ge he sysem (k m ω ) A + γ ω B = F o γ ω A + (k m ω ) B =. 15

16 Solving his sysem for he undeermined coefficiens A and B, we find ha A = F o (k m ω ) k m ω ) + γ ω and B = F o γ ω) k m ω ) + γ ω. I follows ha he seady sae soluion has he form x p () = F o (k m ω ) + γ ω {(k m ω ) cos (ω ) + γ ω sin (ω )}. Noice ha his soluion involves a facor muliplying a bounded oscillaion and he magniude of his facor is a muliple of he ampliude of he original forcing funcion, namely F o. This muliplicaive facor is 1 ((k m ω ) + γ ω and is referred o as he frequency gain or simply he gain of he sysem since i represens he chance in he magniude of he sinusoidal inpu. Noice ha he frequency gain has he unis of lengh/force. The magniude of he seady sae oscillaion depends, of course, on he forcing frequency and is given by A(ω) = F o M(ω) where M(ω) = 1 ((k m ω ) + γ ω = (1/m) ( k ω) ( m + γ ). ω m 16

Section 3.8, Mechanical and Electrical Vibrations

Section 3.8, Mechanical and Electrical Vibrations Secion 3.8, Mechanical and Elecrical Vibraions Mechanical Unis in he U.S. Cusomary and Meric Sysems Disance Mass Time Force g (Earh) Uni U.S. Cusomary MKS Sysem CGS Sysem fee f slugs seconds sec pounds

More information

Wall. x(t) f(t) x(t = 0) = x 0, t=0. which describes the motion of the mass in absence of any external forcing.

Wall. x(t) f(t) x(t = 0) = x 0, t=0. which describes the motion of the mass in absence of any external forcing. MECHANICS APPLICATIONS OF SECOND-ORDER ODES 7 Mechanics applicaions of second-order ODEs Second-order linear ODEs wih consan coefficiens arise in many physical applicaions. One physical sysems whose behaviour

More information

dt = C exp (3 ln t 4 ). t 4 W = C exp ( ln(4 t) 3) = C(4 t) 3.

dt = C exp (3 ln t 4 ). t 4 W = C exp ( ln(4 t) 3) = C(4 t) 3. Mah Rahman Exam Review Soluions () Consider he IVP: ( 4)y 3y + 4y = ; y(3) = 0, y (3) =. (a) Please deermine he longes inerval for which he IVP is guaraneed o have a unique soluion. Soluion: The disconinuiies

More information

Electrical Circuits. 1. Circuit Laws. Tools Used in Lab 13 Series Circuits Damped Vibrations: Energy Van der Pol Circuit

Electrical Circuits. 1. Circuit Laws. Tools Used in Lab 13 Series Circuits Damped Vibrations: Energy Van der Pol Circuit V() R L C 513 Elecrical Circuis Tools Used in Lab 13 Series Circuis Damped Vibraions: Energy Van der Pol Circui A series circui wih an inducor, resisor, and capacior can be represened by Lq + Rq + 1, a

More information

Differential Equations

Differential Equations Mah 21 (Fall 29) Differenial Equaions Soluion #3 1. Find he paricular soluion of he following differenial equaion by variaion of parameer (a) y + y = csc (b) 2 y + y y = ln, > Soluion: (a) The corresponding

More information

Two Coupled Oscillators / Normal Modes

Two Coupled Oscillators / Normal Modes Lecure 3 Phys 3750 Two Coupled Oscillaors / Normal Modes Overview and Moivaion: Today we ake a small, bu significan, sep owards wave moion. We will no ye observe waves, bu his sep is imporan in is own

More information

Section 7.4 Modeling Changing Amplitude and Midline

Section 7.4 Modeling Changing Amplitude and Midline 488 Chaper 7 Secion 7.4 Modeling Changing Ampliude and Midline While sinusoidal funcions can model a variey of behaviors, i is ofen necessary o combine sinusoidal funcions wih linear and exponenial curves

More information

Section 3.5 Nonhomogeneous Equations; Method of Undetermined Coefficients

Section 3.5 Nonhomogeneous Equations; Method of Undetermined Coefficients Secion 3.5 Nonhomogeneous Equaions; Mehod of Undeermined Coefficiens Key Terms/Ideas: Linear Differenial operaor Nonlinear operaor Second order homogeneous DE Second order nonhomogeneous DE Soluion o homogeneous

More information

Module 3: The Damped Oscillator-II Lecture 3: The Damped Oscillator-II

Module 3: The Damped Oscillator-II Lecture 3: The Damped Oscillator-II Module 3: The Damped Oscillaor-II Lecure 3: The Damped Oscillaor-II 3. Over-damped Oscillaions. This refers o he siuaion where β > ω (3.) The wo roos are and α = β + α 2 = β β 2 ω 2 = (3.2) β 2 ω 2 = 2

More information

d 1 = c 1 b 2 - b 1 c 2 d 2 = c 1 b 3 - b 1 c 3

d 1 = c 1 b 2 - b 1 c 2 d 2 = c 1 b 3 - b 1 c 3 and d = c b - b c c d = c b - b c c This process is coninued unil he nh row has been compleed. The complee array of coefficiens is riangular. Noe ha in developing he array an enire row may be divided or

More information

RC, RL and RLC circuits

RC, RL and RLC circuits Name Dae Time o Complee h m Parner Course/ Secion / Grade RC, RL and RLC circuis Inroducion In his experimen we will invesigae he behavior of circuis conaining combinaions of resisors, capaciors, and inducors.

More information

Lab 10: RC, RL, and RLC Circuits

Lab 10: RC, RL, and RLC Circuits Lab 10: RC, RL, and RLC Circuis In his experimen, we will invesigae he behavior of circuis conaining combinaions of resisors, capaciors, and inducors. We will sudy he way volages and currens change in

More information

Math Week 14 April 16-20: sections first order systems of linear differential equations; 7.4 mass-spring systems.

Math Week 14 April 16-20: sections first order systems of linear differential equations; 7.4 mass-spring systems. Mah 2250-004 Week 4 April 6-20 secions 7.-7.3 firs order sysems of linear differenial equaions; 7.4 mass-spring sysems. Mon Apr 6 7.-7.2 Sysems of differenial equaions (7.), and he vecor Calculus we need

More information

Chapter #1 EEE8013 EEE3001. Linear Controller Design and State Space Analysis

Chapter #1 EEE8013 EEE3001. Linear Controller Design and State Space Analysis Chaper EEE83 EEE3 Chaper # EEE83 EEE3 Linear Conroller Design and Sae Space Analysis Ordinary Differenial Equaions.... Inroducion.... Firs Order ODEs... 3. Second Order ODEs... 7 3. General Maerial...

More information

Math 333 Problem Set #2 Solution 14 February 2003

Math 333 Problem Set #2 Solution 14 February 2003 Mah 333 Problem Se #2 Soluion 14 February 2003 A1. Solve he iniial value problem dy dx = x2 + e 3x ; 2y 4 y(0) = 1. Soluion: This is separable; we wrie 2y 4 dy = x 2 + e x dx and inegrae o ge The iniial

More information

Lecture 23 Damped Motion

Lecture 23 Damped Motion Differenial Equaions (MTH40) Lecure Daped Moion In he previous lecure, we discussed he free haronic oion ha assues no rearding forces acing on he oving ass. However No rearding forces acing on he oving

More information

Oscillations. Periodic Motion. Sinusoidal Motion. PHY oscillations - J. Hedberg

Oscillations. Periodic Motion. Sinusoidal Motion. PHY oscillations - J. Hedberg Oscillaions PHY 207 - oscillaions - J. Hedberg - 2017 1. Periodic Moion 2. Sinusoidal Moion 3. How do we ge his kind of moion? 4. Posiion - Velociy - cceleraion 5. spring wih vecors 6. he reference circle

More information

KINEMATICS IN ONE DIMENSION

KINEMATICS IN ONE DIMENSION KINEMATICS IN ONE DIMENSION PREVIEW Kinemaics is he sudy of how hings move how far (disance and displacemen), how fas (speed and velociy), and how fas ha how fas changes (acceleraion). We say ha an objec

More information

Chapter 15 Oscillatory Motion I

Chapter 15 Oscillatory Motion I Chaper 15 Oscillaory Moion I Level : AP Physics Insrucor : Kim Inroducion A very special kind of moion occurs when he force acing on a body is proporional o he displacemen of he body from some equilibrium

More information

LINEAR MODELS: INITIAL-VALUE PROBLEMS

LINEAR MODELS: INITIAL-VALUE PROBLEMS 5 LINEAR MODELS: INITIAL-VALUE PROBLEMS 9 5 LINEAR MODELS: INITIAL-VALUE PROBLEMS REVIEW MATERIAL Secions 4, 4, and 44 Problems 9 6 in Eercises 4 Problems 7 6 in Eercises 44 INTRODUCTION In his secion

More information

Math Week 15: Section 7.4, mass-spring systems. These are notes for Monday. There will also be course review notes for Tuesday, posted later.

Math Week 15: Section 7.4, mass-spring systems. These are notes for Monday. There will also be course review notes for Tuesday, posted later. Mah 50-004 Week 5: Secion 7.4, mass-spring sysems. These are noes for Monday. There will also be course review noes for Tuesday, posed laer. Mon Apr 3 7.4 mass-spring sysems. Announcemens: Warm up exercise:

More information

Math 2214 Sol Test 2B Spring 2015

Math 2214 Sol Test 2B Spring 2015 Mah 14 Sol Tes B Sring 015 roblem 1: An objec weighing ounds sreches a verical sring 8 fee beond i naural lengh before coming o res a equilibrium The objec is ushed u 6 fee from i s equilibrium osiion

More information

23.2. Representing Periodic Functions by Fourier Series. Introduction. Prerequisites. Learning Outcomes

23.2. Representing Periodic Functions by Fourier Series. Introduction. Prerequisites. Learning Outcomes Represening Periodic Funcions by Fourier Series 3. Inroducion In his Secion we show how a periodic funcion can be expressed as a series of sines and cosines. We begin by obaining some sandard inegrals

More information

Lecture 2-1 Kinematics in One Dimension Displacement, Velocity and Acceleration Everything in the world is moving. Nothing stays still.

Lecture 2-1 Kinematics in One Dimension Displacement, Velocity and Acceleration Everything in the world is moving. Nothing stays still. Lecure - Kinemaics in One Dimension Displacemen, Velociy and Acceleraion Everyhing in he world is moving. Nohing says sill. Moion occurs a all scales of he universe, saring from he moion of elecrons in

More information

!!"#"$%&#'()!"#&'(*%)+,&',-)./0)1-*23)

!!#$%&#'()!#&'(*%)+,&',-)./0)1-*23) "#"$%&#'()"#&'(*%)+,&',-)./)1-*) #$%&'()*+,&',-.%,/)*+,-&1*#$)()5*6$+$%*,7&*-'-&1*(,-&*6&,7.$%$+*&%'(*8$&',-,%'-&1*(,-&*6&,79*(&,%: ;..,*&1$&$.$%&'()*1$$.,'&',-9*(&,%)?%*,('&5

More information

Final Spring 2007

Final Spring 2007 .615 Final Spring 7 Overview The purpose of he final exam is o calculae he MHD β limi in a high-bea oroidal okamak agains he dangerous n = 1 exernal ballooning-kink mode. Effecively, his corresponds o

More information

Traveling Waves. Chapter Introduction

Traveling Waves. Chapter Introduction Chaper 4 Traveling Waves 4.1 Inroducion To dae, we have considered oscillaions, i.e., periodic, ofen harmonic, variaions of a physical characerisic of a sysem. The sysem a one ime is indisinguishable from

More information

2.9 Modeling: Electric Circuits

2.9 Modeling: Electric Circuits SE. 2.9 Modeling: Elecric ircuis 93 2.9 Modeling: Elecric ircuis Designing good models is a ask he compuer canno do. Hence seing up models has become an imporan ask in modern applied mahemaics. The bes

More information

Announcements: Warm-up Exercise:

Announcements: Warm-up Exercise: Fri Apr 13 7.1 Sysems of differenial equaions - o model muli-componen sysems via comparmenal analysis hp//en.wikipedia.org/wiki/muli-comparmen_model Announcemens Warm-up Exercise Here's a relaively simple

More information

k 1 k 2 x (1) x 2 = k 1 x 1 = k 2 k 1 +k 2 x (2) x k series x (3) k 2 x 2 = k 1 k 2 = k 1+k 2 = 1 k k 2 k series

k 1 k 2 x (1) x 2 = k 1 x 1 = k 2 k 1 +k 2 x (2) x k series x (3) k 2 x 2 = k 1 k 2 = k 1+k 2 = 1 k k 2 k series Final Review A Puzzle... Consider wo massless springs wih spring consans k 1 and k and he same equilibrium lengh. 1. If hese springs ac on a mass m in parallel, hey would be equivalen o a single spring

More information

t is a basis for the solution space to this system, then the matrix having these solutions as columns, t x 1 t, x 2 t,... x n t x 2 t...

t is a basis for the solution space to this system, then the matrix having these solutions as columns, t x 1 t, x 2 t,... x n t x 2 t... Mah 228- Fri Mar 24 5.6 Marix exponenials and linear sysems: The analogy beween firs order sysems of linear differenial equaions (Chaper 5) and scalar linear differenial equaions (Chaper ) is much sronger

More information

Math Linear Differential Equations

Math Linear Differential Equations Mah 65 - Linear Differenial Equaions A. J. Meir Copyrigh (C) A. J. Meir. All righs reserved. This workshee is for educaional use only. No par of his publicaion may be reproduced or ransmied for profi in

More information

In this chapter the model of free motion under gravity is extended to objects projected at an angle. When you have completed it, you should

In this chapter the model of free motion under gravity is extended to objects projected at an angle. When you have completed it, you should Cambridge Universiy Press 978--36-60033-7 Cambridge Inernaional AS and A Level Mahemaics: Mechanics Coursebook Excerp More Informaion Chaper The moion of projeciles In his chaper he model of free moion

More information

Week 1 Lecture 2 Problems 2, 5. What if something oscillates with no obvious spring? What is ω? (problem set problem)

Week 1 Lecture 2 Problems 2, 5. What if something oscillates with no obvious spring? What is ω? (problem set problem) Week 1 Lecure Problems, 5 Wha if somehing oscillaes wih no obvious spring? Wha is ω? (problem se problem) Sar wih Try and ge o SHM form E. Full beer can in lake, oscillaing F = m & = ge rearrange: F =

More information

Physics 235 Chapter 2. Chapter 2 Newtonian Mechanics Single Particle

Physics 235 Chapter 2. Chapter 2 Newtonian Mechanics Single Particle Chaper 2 Newonian Mechanics Single Paricle In his Chaper we will review wha Newon s laws of mechanics ell us abou he moion of a single paricle. Newon s laws are only valid in suiable reference frames,

More information

Chapter #1 EEE8013 EEE3001. Linear Controller Design and State Space Analysis

Chapter #1 EEE8013 EEE3001. Linear Controller Design and State Space Analysis Chaper EEE83 EEE3 Chaper # EEE83 EEE3 Linear Conroller Design and Sae Space Analysis Ordinary Differenial Equaions.... Inroducion.... Firs Order ODEs... 3. Second Order ODEs... 7 3. General Maerial...

More information

Voltage/current relationship Stored Energy. RL / RC circuits Steady State / Transient response Natural / Step response

Voltage/current relationship Stored Energy. RL / RC circuits Steady State / Transient response Natural / Step response Review Capaciors/Inducors Volage/curren relaionship Sored Energy s Order Circuis RL / RC circuis Seady Sae / Transien response Naural / Sep response EE4 Summer 5: Lecure 5 Insrucor: Ocavian Florescu Lecure

More information

MEI STRUCTURED MATHEMATICS 4758

MEI STRUCTURED MATHEMATICS 4758 OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced Subsidiary General Cerificae of Educaion Advanced General Cerificae of Educaion MEI STRUCTURED MATHEMATICS 4758 Differenial Equaions Thursday 5 JUNE 006 Afernoon

More information

Sub Module 2.6. Measurement of transient temperature

Sub Module 2.6. Measurement of transient temperature Mechanical Measuremens Prof. S.P.Venkaeshan Sub Module 2.6 Measuremen of ransien emperaure Many processes of engineering relevance involve variaions wih respec o ime. The sysem properies like emperaure,

More information

10. State Space Methods

10. State Space Methods . Sae Space Mehods. Inroducion Sae space modelling was briefly inroduced in chaper. Here more coverage is provided of sae space mehods before some of heir uses in conrol sysem design are covered in he

More information

INDEX. Transient analysis 1 Initial Conditions 1

INDEX. Transient analysis 1 Initial Conditions 1 INDEX Secion Page Transien analysis 1 Iniial Condiions 1 Please inform me of your opinion of he relaive emphasis of he review maerial by simply making commens on his page and sending i o me a: Frank Mera

More information

15. Vector Valued Functions

15. Vector Valued Functions 1. Vecor Valued Funcions Up o his poin, we have presened vecors wih consan componens, for example, 1, and,,4. However, we can allow he componens of a vecor o be funcions of a common variable. For example,

More information

IB Physics Kinematics Worksheet

IB Physics Kinematics Worksheet IB Physics Kinemaics Workshee Wrie full soluions and noes for muliple choice answers. Do no use a calculaor for muliple choice answers. 1. Which of he following is a correc definiion of average acceleraion?

More information

Laplace transfom: t-translation rule , Haynes Miller and Jeremy Orloff

Laplace transfom: t-translation rule , Haynes Miller and Jeremy Orloff Laplace ransfom: -ranslaion rule 8.03, Haynes Miller and Jeremy Orloff Inroducory example Consider he sysem ẋ + 3x = f(, where f is he inpu and x he response. We know is uni impulse response is 0 for

More information

System of Linear Differential Equations

System of Linear Differential Equations Sysem of Linear Differenial Equaions In "Ordinary Differenial Equaions" we've learned how o solve a differenial equaion for a variable, such as: y'k5$e K2$x =0 solve DE yx = K 5 2 ek2 x C_C1 2$y''C7$y

More information

Chapter 2. First Order Scalar Equations

Chapter 2. First Order Scalar Equations Chaper. Firs Order Scalar Equaions We sar our sudy of differenial equaions in he same way he pioneers in his field did. We show paricular echniques o solve paricular ypes of firs order differenial equaions.

More information

Second-Order Differential Equations

Second-Order Differential Equations WWW Problems and Soluions 3.1 Chaper 3 Second-Order Differenial Equaions Secion 3.1 Springs: Linear and Nonlinear Models www m Problem 3. (NonlinearSprings). A bod of mass m is aached o a wall b means

More information

4.5 Constant Acceleration

4.5 Constant Acceleration 4.5 Consan Acceleraion v() v() = v 0 + a a() a a() = a v 0 Area = a (a) (b) Figure 4.8 Consan acceleraion: (a) velociy, (b) acceleraion When he x -componen of he velociy is a linear funcion (Figure 4.8(a)),

More information

Echocardiography Project and Finite Fourier Series

Echocardiography Project and Finite Fourier Series Echocardiography Projec and Finie Fourier Series 1 U M An echocardiagram is a plo of how a porion of he hear moves as he funcion of ime over he one or more hearbea cycles If he hearbea repeas iself every

More information

Position, Velocity, and Acceleration

Position, Velocity, and Acceleration rev 06/2017 Posiion, Velociy, and Acceleraion Equipmen Qy Equipmen Par Number 1 Dynamic Track ME-9493 1 Car ME-9454 1 Fan Accessory ME-9491 1 Moion Sensor II CI-6742A 1 Track Barrier Purpose The purpose

More information

1 1 + x 2 dx. tan 1 (2) = ] ] x 3. Solution: Recall that the given integral is improper because. x 3. 1 x 3. dx = lim dx.

1 1 + x 2 dx. tan 1 (2) = ] ] x 3. Solution: Recall that the given integral is improper because. x 3. 1 x 3. dx = lim dx. . Use Simpson s rule wih n 4 o esimae an () +. Soluion: Since we are using 4 seps, 4 Thus we have [ ( ) f() + 4f + f() + 4f 3 [ + 4 4 6 5 + + 4 4 3 + ] 5 [ + 6 6 5 + + 6 3 + ]. 5. Our funcion is f() +.

More information

Math 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities:

Math 2142 Exam 1 Review Problems. x 2 + f (0) 3! for the 3rd Taylor polynomial at x = 0. To calculate the various quantities: Mah 4 Eam Review Problems Problem. Calculae he 3rd Taylor polynomial for arcsin a =. Soluion. Le f() = arcsin. For his problem, we use he formula f() + f () + f ()! + f () 3! for he 3rd Taylor polynomial

More information

Structural Dynamics and Earthquake Engineering

Structural Dynamics and Earthquake Engineering Srucural Dynamics and Earhquae Engineering Course 1 Inroducion. Single degree of freedom sysems: Equaions of moion, problem saemen, soluion mehods. Course noes are available for download a hp://www.c.up.ro/users/aurelsraan/

More information

WEEK-3 Recitation PHYS 131. of the projectile s velocity remains constant throughout the motion, since the acceleration a x

WEEK-3 Recitation PHYS 131. of the projectile s velocity remains constant throughout the motion, since the acceleration a x WEEK-3 Reciaion PHYS 131 Ch. 3: FOC 1, 3, 4, 6, 14. Problems 9, 37, 41 & 71 and Ch. 4: FOC 1, 3, 5, 8. Problems 3, 5 & 16. Feb 8, 018 Ch. 3: FOC 1, 3, 4, 6, 14. 1. (a) The horizonal componen of he projecile

More information

72 Calculus and Structures

72 Calculus and Structures 72 Calculus and Srucures CHAPTER 5 DISTANCE AND ACCUMULATED CHANGE Calculus and Srucures 73 Copyrigh Chaper 5 DISTANCE AND ACCUMULATED CHANGE 5. DISTANCE a. Consan velociy Le s ake anoher look a Mary s

More information

A. Using Newton s second law in one dimension, F net. , write down the differential equation that governs the motion of the block.

A. Using Newton s second law in one dimension, F net. , write down the differential equation that governs the motion of the block. Simple SIMPLE harmonic HARMONIC moion MOTION I. Differenial equaion of moion A block is conneced o a spring, one end of which is aached o a wall. (Neglec he mass of he spring, and assume he surface is

More information

Physics 1402: Lecture 22 Today s Agenda

Physics 1402: Lecture 22 Today s Agenda Physics 142: ecure 22 Today s Agenda Announcemens: R - RV - R circuis Homework 6: due nex Wednesday Inducion / A curren Inducion Self-Inducance, R ircuis X X X X X X X X X long solenoid Energy and energy

More information

R.#W.#Erickson# Department#of#Electrical,#Computer,#and#Energy#Engineering# University#of#Colorado,#Boulder#

R.#W.#Erickson# Department#of#Electrical,#Computer,#and#Energy#Engineering# University#of#Colorado,#Boulder# .#W.#Erickson# Deparmen#of#Elecrical,#Compuer,#and#Energy#Engineering# Universiy#of#Colorado,#Boulder# Chaper 2 Principles of Seady-Sae Converer Analysis 2.1. Inroducion 2.2. Inducor vol-second balance,

More information

Simulation-Solving Dynamic Models ABE 5646 Week 2, Spring 2010

Simulation-Solving Dynamic Models ABE 5646 Week 2, Spring 2010 Simulaion-Solving Dynamic Models ABE 5646 Week 2, Spring 2010 Week Descripion Reading Maerial 2 Compuer Simulaion of Dynamic Models Finie Difference, coninuous saes, discree ime Simple Mehods Euler Trapezoid

More information

Physics 180A Fall 2008 Test points. Provide the best answer to the following questions and problems. Watch your sig figs.

Physics 180A Fall 2008 Test points. Provide the best answer to the following questions and problems. Watch your sig figs. Physics 180A Fall 2008 Tes 1-120 poins Name Provide he bes answer o he following quesions and problems. Wach your sig figs. 1) The number of meaningful digis in a number is called he number of. When numbers

More information

Chapter 7: Solving Trig Equations

Chapter 7: Solving Trig Equations Haberman MTH Secion I: The Trigonomeric Funcions Chaper 7: Solving Trig Equaions Le s sar by solving a couple of equaions ha involve he sine funcion EXAMPLE a: Solve he equaion sin( ) The inverse funcions

More information

MA Study Guide #1

MA Study Guide #1 MA 66 Su Guide #1 (1) Special Tpes of Firs Order Equaions I. Firs Order Linear Equaion (FOL): + p() = g() Soluion : = 1 µ() [ ] µ()g() + C, where µ() = e p() II. Separable Equaion (SEP): dx = h(x) g()

More information

6.003 Homework #9 Solutions

6.003 Homework #9 Solutions 6.00 Homework #9 Soluions Problems. Fourier varieies a. Deermine he Fourier series coefficiens of he following signal, which is periodic in 0. x () 0 0 a 0 5 a k sin πk 5 sin πk 5 πk for k 0 a k 0 πk j

More information

EE100 Lab 3 Experiment Guide: RC Circuits

EE100 Lab 3 Experiment Guide: RC Circuits I. Inroducion EE100 Lab 3 Experimen Guide: A. apaciors A capacior is a passive elecronic componen ha sores energy in he form of an elecrosaic field. The uni of capaciance is he farad (coulomb/vol). Pracical

More information

LAPLACE TRANSFORM AND TRANSFER FUNCTION

LAPLACE TRANSFORM AND TRANSFER FUNCTION CHBE320 LECTURE V LAPLACE TRANSFORM AND TRANSFER FUNCTION Professor Dae Ryook Yang Spring 2018 Dep. of Chemical and Biological Engineering 5-1 Road Map of he Lecure V Laplace Transform and Transfer funcions

More information

where the coordinate X (t) describes the system motion. X has its origin at the system static equilibrium position (SEP).

where the coordinate X (t) describes the system motion. X has its origin at the system static equilibrium position (SEP). Appendix A: Conservaion of Mechanical Energy = Conservaion of Linear Momenum Consider he moion of a nd order mechanical sysem comprised of he fundamenal mechanical elemens: ineria or mass (M), siffness

More information

Answers to 1 Homework

Answers to 1 Homework Answers o Homework. x + and y x 5 y To eliminae he parameer, solve for x. Subsiue ino y s equaion o ge y x.. x and y, x y x To eliminae he parameer, solve for. Subsiue ino y s equaion o ge x y, x. (Noe:

More information

Phys 221 Fall Chapter 2. Motion in One Dimension. 2014, 2005 A. Dzyubenko Brooks/Cole

Phys 221 Fall Chapter 2. Motion in One Dimension. 2014, 2005 A. Dzyubenko Brooks/Cole Phys 221 Fall 2014 Chaper 2 Moion in One Dimension 2014, 2005 A. Dzyubenko 2004 Brooks/Cole 1 Kinemaics Kinemaics, a par of classical mechanics: Describes moion in erms of space and ime Ignores he agen

More information

ME 391 Mechanical Engineering Analysis

ME 391 Mechanical Engineering Analysis Fall 04 ME 39 Mechanical Engineering Analsis Eam # Soluions Direcions: Open noes (including course web posings). No books, compuers, or phones. An calculaor is fair game. Problem Deermine he posiion of

More information

Math 2214 Solution Test 1A Spring 2016

Math 2214 Solution Test 1A Spring 2016 Mah 14 Soluion Tes 1A Spring 016 sec Problem 1: Wha is he larges -inerval for which ( 4) = has a guaraneed + unique soluion for iniial value (-1) = 3 according o he Exisence Uniqueness Theorem? Soluion

More information

ENGI 9420 Engineering Analysis Assignment 2 Solutions

ENGI 9420 Engineering Analysis Assignment 2 Solutions ENGI 940 Engineering Analysis Assignmen Soluions 0 Fall [Second order ODEs, Laplace ransforms; Secions.0-.09]. Use Laplace ransforms o solve he iniial value problem [0] dy y, y( 0) 4 d + [This was Quesion

More information

Boyce/DiPrima/Meade 11 th ed, Ch 3.1: 2 nd Order Linear Homogeneous Equations-Constant Coefficients

Boyce/DiPrima/Meade 11 th ed, Ch 3.1: 2 nd Order Linear Homogeneous Equations-Constant Coefficients Boce/DiPrima/Meade h ed, Ch 3.: nd Order Linear Homogeneous Equaions-Consan Coefficiens Elemenar Differenial Equaions and Boundar Value Problems, h ediion, b William E. Boce, Richard C. DiPrima, and Doug

More information

Continuous Time. Time-Domain System Analysis. Impulse Response. Impulse Response. Impulse Response. Impulse Response. ( t) + b 0.

Continuous Time. Time-Domain System Analysis. Impulse Response. Impulse Response. Impulse Response. Impulse Response. ( t) + b 0. Time-Domain Sysem Analysis Coninuous Time. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 1. J. Robers - All Righs Reserved. Edied by Dr. Rober Akl 2 Le a sysem be described by a 2 y ( ) + a 1

More information

15. Bicycle Wheel. Graph of height y (cm) above the axle against time t (s) over a 6-second interval. 15 bike wheel

15. Bicycle Wheel. Graph of height y (cm) above the axle against time t (s) over a 6-second interval. 15 bike wheel 15. Biccle Wheel The graph We moun a biccle wheel so ha i is free o roae in a verical plane. In fac, wha works easil is o pu an exension on one of he axles, and ge a suden o sand on one side and hold he

More information

Module 2 F c i k c s la l w a s o s f dif di fusi s o i n

Module 2 F c i k c s la l w a s o s f dif di fusi s o i n Module Fick s laws of diffusion Fick s laws of diffusion and hin film soluion Adolf Fick (1855) proposed: d J α d d d J (mole/m s) flu (m /s) diffusion coefficien and (mole/m 3 ) concenraion of ions, aoms

More information

Challenge Problems. DIS 203 and 210. March 6, (e 2) k. k(k + 2). k=1. f(x) = k(k + 2) = 1 x k

Challenge Problems. DIS 203 and 210. March 6, (e 2) k. k(k + 2). k=1. f(x) = k(k + 2) = 1 x k Challenge Problems DIS 03 and 0 March 6, 05 Choose one of he following problems, and work on i in your group. Your goal is o convince me ha your answer is correc. Even if your answer isn compleely correc,

More information

Mon Apr 9 EP 7.6 Convolutions and Laplace transforms. Announcements: Warm-up Exercise:

Mon Apr 9 EP 7.6 Convolutions and Laplace transforms. Announcements: Warm-up Exercise: Mah 225-4 Week 3 April 9-3 EP 7.6 - convoluions; 6.-6.2 - eigenvalues, eigenvecors and diagonalizabiliy; 7. - sysems of differenial equaions. Mon Apr 9 EP 7.6 Convoluions and Laplace ransforms. Announcemens:

More information

LAB 6: SIMPLE HARMONIC MOTION

LAB 6: SIMPLE HARMONIC MOTION 1 Name Dae Day/Time of Lab Parner(s) Lab TA Objecives LAB 6: SIMPLE HARMONIC MOTION To undersand oscillaion in relaion o equilibrium of conservaive forces To manipulae he independen variables of oscillaion:

More information

u(x) = e x 2 y + 2 ) Integrate and solve for x (1 + x)y + y = cos x Answer: Divide both sides by 1 + x and solve for y. y = x y + cos x

u(x) = e x 2 y + 2 ) Integrate and solve for x (1 + x)y + y = cos x Answer: Divide both sides by 1 + x and solve for y. y = x y + cos x . 1 Mah 211 Homework #3 February 2, 2001 2.4.3. y + (2/x)y = (cos x)/x 2 Answer: Compare y + (2/x) y = (cos x)/x 2 wih y = a(x)x + f(x)and noe ha a(x) = 2/x. Consequenly, an inegraing facor is found wih

More information

HOMEWORK # 2: MATH 211, SPRING Note: This is the last solution set where I will describe the MATLAB I used to make my pictures.

HOMEWORK # 2: MATH 211, SPRING Note: This is the last solution set where I will describe the MATLAB I used to make my pictures. HOMEWORK # 2: MATH 2, SPRING 25 TJ HITCHMAN Noe: This is he las soluion se where I will describe he MATLAB I used o make my picures.. Exercises from he ex.. Chaper 2.. Problem 6. We are o show ha y() =

More information

NEWTON S SECOND LAW OF MOTION

NEWTON S SECOND LAW OF MOTION Course and Secion Dae Names NEWTON S SECOND LAW OF MOTION The acceleraion of an objec is defined as he rae of change of elociy. If he elociy changes by an amoun in a ime, hen he aerage acceleraion during

More information

AP CALCULUS AB 2003 SCORING GUIDELINES (Form B)

AP CALCULUS AB 2003 SCORING GUIDELINES (Form B) SCORING GUIDELINES (Form B) Quesion A blood vessel is 6 millimeers (mm) long Disance wih circular cross secions of varying diameer. x (mm) 6 8 4 6 Diameer The able above gives he measuremens of he B(x)

More information

Lecture 13 RC/RL Circuits, Time Dependent Op Amp Circuits

Lecture 13 RC/RL Circuits, Time Dependent Op Amp Circuits Lecure 13 RC/RL Circuis, Time Dependen Op Amp Circuis RL Circuis The seps involved in solving simple circuis conaining dc sources, resisances, and one energy-sorage elemen (inducance or capaciance) are:

More information

MATH 4330/5330, Fourier Analysis Section 6, Proof of Fourier s Theorem for Pointwise Convergence

MATH 4330/5330, Fourier Analysis Section 6, Proof of Fourier s Theorem for Pointwise Convergence MATH 433/533, Fourier Analysis Secion 6, Proof of Fourier s Theorem for Poinwise Convergence Firs, some commens abou inegraing periodic funcions. If g is a periodic funcion, g(x + ) g(x) for all real x,

More information

Morning Time: 1 hour 30 minutes Additional materials (enclosed):

Morning Time: 1 hour 30 minutes Additional materials (enclosed): ADVANCED GCE 78/0 MATHEMATICS (MEI) Differenial Equaions THURSDAY JANUARY 008 Morning Time: hour 30 minues Addiional maerials (enclosed): None Addiional maerials (required): Answer Bookle (8 pages) Graph

More information

Solutions to Assignment 1

Solutions to Assignment 1 MA 2326 Differenial Equaions Insrucor: Peronela Radu Friday, February 8, 203 Soluions o Assignmen. Find he general soluions of he following ODEs: (a) 2 x = an x Soluion: I is a separable equaion as we

More information

Second Order Linear Differential Equations

Second Order Linear Differential Equations Second Order Linear Differenial Equaions Second order linear equaions wih consan coefficiens; Fundamenal soluions; Wronskian; Exisence and Uniqueness of soluions; he characerisic equaion; soluions of homogeneous

More information

Thus the force is proportional but opposite to the displacement away from equilibrium.

Thus the force is proportional but opposite to the displacement away from equilibrium. Chaper 3 : Siple Haronic Moion Hooe s law saes ha he force (F) eered by an ideal spring is proporional o is elongaion l F= l where is he spring consan. Consider a ass hanging on a he spring. In equilibriu

More information

Robotics I. April 11, The kinematics of a 3R spatial robot is specified by the Denavit-Hartenberg parameters in Tab. 1.

Robotics I. April 11, The kinematics of a 3R spatial robot is specified by the Denavit-Hartenberg parameters in Tab. 1. Roboics I April 11, 017 Exercise 1 he kinemaics of a 3R spaial robo is specified by he Denavi-Harenberg parameers in ab 1 i α i d i a i θ i 1 π/ L 1 0 1 0 0 L 3 0 0 L 3 3 able 1: able of DH parameers of

More information

Notes 04 largely plagiarized by %khc

Notes 04 largely plagiarized by %khc Noes 04 largely plagiarized by %khc Convoluion Recap Some ricks: x() () =x() x() (, 0 )=x(, 0 ) R ț x() u() = x( )d x() () =ẋ() This hen ells us ha an inegraor has impulse response h() =u(), and ha a differeniaor

More information

6.003 Homework #9 Solutions

6.003 Homework #9 Solutions 6.003 Homework #9 Soluions Problems. Fourier varieies a. Deermine he Fourier series coefficiens of he following signal, which is periodic in 0. x () 0 3 0 a 0 5 a k a k 0 πk j3 e 0 e j πk 0 jπk πk e 0

More information

Summary of shear rate kinematics (part 1)

Summary of shear rate kinematics (part 1) InroToMaFuncions.pdf 4 CM465 To proceed o beer-designed consiuive equaions, we need o know more abou maerial behavior, i.e. we need more maerial funcions o predic, and we need measuremens of hese maerial

More information

Decimal moved after first digit = 4.6 x Decimal moves five places left SCIENTIFIC > POSITIONAL. a) g) 5.31 x b) 0.

Decimal moved after first digit = 4.6 x Decimal moves five places left SCIENTIFIC > POSITIONAL. a) g) 5.31 x b) 0. PHYSICS 20 UNIT 1 SCIENCE MATH WORKSHEET NAME: A. Sandard Noaion Very large and very small numbers are easily wrien using scienific (or sandard) noaion, raher han decimal (or posiional) noaion. Sandard

More information

1. VELOCITY AND ACCELERATION

1. VELOCITY AND ACCELERATION 1. VELOCITY AND ACCELERATION 1.1 Kinemaics Equaions s = u + 1 a and s = v 1 a s = 1 (u + v) v = u + as 1. Displacemen-Time Graph Gradien = speed 1.3 Velociy-Time Graph Gradien = acceleraion Area under

More information

AP Chemistry--Chapter 12: Chemical Kinetics

AP Chemistry--Chapter 12: Chemical Kinetics AP Chemisry--Chaper 12: Chemical Kineics I. Reacion Raes A. The area of chemisry ha deals wih reacion raes, or how fas a reacion occurs, is called chemical kineics. B. The rae of reacion depends on he

More information

5.1 - Logarithms and Their Properties

5.1 - Logarithms and Their Properties Chaper 5 Logarihmic Funcions 5.1 - Logarihms and Their Properies Suppose ha a populaion grows according o he formula P 10, where P is he colony size a ime, in hours. When will he populaion be 2500? We

More information

SOLUTIONS TO ECE 3084

SOLUTIONS TO ECE 3084 SOLUTIONS TO ECE 384 PROBLEM 2.. For each sysem below, specify wheher or no i is: (i) memoryless; (ii) causal; (iii) inverible; (iv) linear; (v) ime invarian; Explain your reasoning. If he propery is no

More information

Equations of motion for constant acceleration

Equations of motion for constant acceleration Lecure 3 Chaper 2 Physics I 01.29.2014 Equaions of moion for consan acceleraion Course websie: hp://faculy.uml.edu/andriy_danylo/teaching/physicsi Lecure Capure: hp://echo360.uml.edu/danylo2013/physics1spring.hml

More information

( ) a system of differential equations with continuous parametrization ( T = R + These look like, respectively:

( ) a system of differential equations with continuous parametrization ( T = R + These look like, respectively: XIII. DIFFERENCE AND DIFFERENTIAL EQUATIONS Ofen funcions, or a sysem of funcion, are paramerized in erms of some variable, usually denoed as and inerpreed as ime. The variable is wrien as a funcion of

More information

Unit 1 Test Review Physics Basics, Movement, and Vectors Chapters 1-3

Unit 1 Test Review Physics Basics, Movement, and Vectors Chapters 1-3 A.P. Physics B Uni 1 Tes Reiew Physics Basics, Moemen, and Vecors Chapers 1-3 * In sudying for your es, make sure o sudy his reiew shee along wih your quizzes and homework assignmens. Muliple Choice Reiew:

More information