Lecture 23 Damped Motion

Size: px
Start display at page:

Download "Lecture 23 Damped Motion"

Transcription

1 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 body is no realisic, because There always eiss a leas a resising force due o surrounding ediu. For eaple a ass can be suspended in a viscous ediu. Hence, he daping forces need o be included in a realisic analysis. Daping Force In he sudy of echanics, he daping forces acing on a body are considered o be d proporional o a power of he insananeous velociy. In he hydro dynaical probles, he daping force is proporional o Daping fo rce ( d / ) d -β. So ha in hese probles Where β is a posiive daping consan and negaive sign indicaes ha he daping force acs in a direcion opposie o he direcion of oion. In he presen discussion, we shall assue ha he daping force is proporional o he d insananeous velociy. Thus for us The Differenial Equaion Suppose Tha Daping fo rce d -β A body of ass is aached o a spring. The spring sreches by an aoun s o aain he equilibriu posiion. The ass is furher displaced by an aoun and hen released. No eernal forces are ipressed on he syse. Therefore, here are hree forces acing on he ass, naely: a) Weigh g of he body b) Resoring force k ( s + ) d c) Daping force -β 7

2 Differenial Equaions (MTH40) Therefore, oal force acing on he ass g k is d ( s + ) β So ha by Newon s second law of oion, we have d g k Since in he equilibriu posiion Therefore g ks 0 d ( s + ) β d d k β Dividing wih, we obain he differenial equaion of free daped oion d β d k For algebraic convenience, we suppose ha Then he equaion becoes: β λ, ω k d d + λ + ω 0 Soluion of he Differenial Equaion Consider he equaion of he free daped oion Pu e, Then he auiliary equaion is: d d + λ + ω 0 d d e, e + λ + ω 0 Solving by use of quadraic forula, we obain λ± λ ω Thus he roos of he auiliary equaion are λ + λ ω, λ λ ω 8

3 Differenial Equaions (MTH40) Depending upon he sign of he quaniy λ ω, we can now disinguish hree possible cases of he roos of he auiliary equaion. Case Real and disinc roos If λ ω > 0 hen β > k and he syse is said o be over-daped. The soluion of he equaion of free daped oion is () c e c e + λ λ ω λ ω or () e [ ce + ce ] This equaion represens sooh and non oscillaory oion. Case Real and equal roos If λ ω 0, hen β k and he syse is said o be criically daped, because any sligh decrease in he daping force would resul in oscillaory oion. The general soluion of he differenial equaion of free daped force is () c e c e + or () e λ ( c c ) Case Cople roos + If λ w < 0, hen β < k and he syse is said o be under-daped. We need o rewrie he roos of he auiliary equaion as: λ + ω λ i, λ ω λ i Thus, he general soluion of he equaion of free daped oion is () λ e c λ + c λ cos ω sin ω This represens an oscillaory oion; bu apliude of vibraion 0 as because of e λ he coefficien. Noe ha Each of he hree soluions conain he daping facor e he ass becoe negligible for larger ies. λ, λ > 0, he displaceens of 9

4 Differenial Equaions (MTH40) Alernaive for of he Soluion When λ ω < is () 0, he soluion of he differenial equaion of free daped oion d d + λ + ω 0 λ e c λ + c λ cos ω sin ω Suppose ha A and φ are wo real nubers such ha So ha c c sin φ, cosφ A A c A c + c, anφ c The nuber φ is known as he phase angle. Then he soluion of he equaion becoes: () Ae λ sin ω λ cosφ + cos λ or () Ae sin( ω λ + φ ) Noe ha ω λ sinφ The coefficien Ae λ is called he daped apliude of vibraions. The ie inerval beween wo successive aia of ( ) is called quasi period, and is given by he nuber π ω λ The following nuber is known as he quasi frequency. ω λ π The graph of he soluion λ () Ae sin ( ω λ +φ ) crosses posiive -ais, i.e he line 0, a ies ha are given by ω λ + φ nπ Where n,,,. For eaple, if we have () 0.5 π e sin 0

5 Differenial Equaions (MTH40) Then π nπ or π π π 0, π, π, or π 4π 7π,,, We noice ha difference beween wo successive roos is π k k quasi period π Since quasi period π. Therefore π k k quasi period λ Since () Ae when sin ω λ + φ, he graph of he soluion λ () Ae sin( ω λ + φ ) ouches he graphs of he eponenial funcions ± Ae λ a he values of for which sin( ω λ + φ ) ± This eans hose values of for which ( n ) π ω λ + φ + ( n + ) or ( π / ) φ where n 0,,,, ω λ Again, if we consider () 0.5 π e sin Then * π π * π π * π 5π,,, Or * 5 π * π 7,, π, Again, we noice ha he difference beween successive values is Eaple * * π k k The values of for which he graph of he soluion λ () Ae sin( ω λ + φ ) ouches he eponenial graph are no he values for which he funcion aains is relaive ereu.

6 Differenial Equaions (MTH40) Inerpre and solve he iniial value proble d d ( 0 ), ( 0) Find eree values of he soluion and check wheher he graph crosses he equilibriu posiion. Inerpreaion Coparing he given differenial equaion d d wih he general equaion of he free daped oion we see ha d d + λ + ω 0 λ 5, ω 4 so ha λ ω > 0 Therefore, he proble represens he over-daped oion of a ass on a spring. Inspecion of he boundary condiions ( 0 ), ( 0) reveals ha he ass sars uni below he equilibriu posiion wih a downward velociy of f/sec. Soluion To solve he differenial equaion d d

7 Differenial Equaions (MTH40) d d We pu e, e, Then he auiliary equaion is ( + 4 )( + ) 0, 4, Therefore, he auiliary equaion has disinc real roos, 4 Thus he soluion of he differenial equaion is: 4 () c e + c e 4 So ha () c e c e 4 Now, we apply he boundary condiions Thus ( ) c.+ c. 0 ( ) c 4 c 0 c c + c 4c Solving hese wo equaions, we have. c 5, c Therefore, soluion of he iniial value proble is 4 () 5 e e Ereu 4 Since () 5 e e Therefore d 5 8 e + e 4 So ha () 0 or e e e + e or ln 8 5

8 Differenial Equaions (MTH40) Since Therefore a 0.57, d 4 5 e e we have () d e e < 0 So ha he soluion has a aiu a and aiu value of is: ( 0.57). 069 Hence he ass aains an eree displaceen of posiion. Check ().069 f below he equilibriu Suppose ha he graph of does cross he ais, ha is, he ass passes hrough he equilibriu posiion. Then a value of eiss for which () 0 5 i.e 4 e e 0 e or ln This value of is physically irrelevan because ie can never be negaive. Hence, he ass never passes hrough he equilibriu posiion. Eaple An 8-lb weigh sreches a spring f. Assuing ha a daping force nuerically equals o wo ies he insananeous velociy acs on he syse. Deerine he equaion of oion if he weigh is released fro he equilibriu posiion wih an upward velociy of f / sec. Soluion Since Therefore, by Hook s law Weigh 8 lbs, Srech s f 8 k 4 k lb / f 4

9 Differenial Equaions (MTH40) Since Therefore β Also d Daping force Weigh ass g 8 4 slugs Thus, he differenial equaion of oion of he free daped oion is given by or d d k β d d 4 4 d d or Since he ass is released fro equilibriu posiion wih an upward velociy f / s. Therefore he iniial condiions are: ( 0) 0, ( 0) Thus we need o solve he iniial value proble d d Solve Subjec o ( 0) 0, ( 0) Pu Thus he auiliary equaion is e, d e , d e or ( + 4) 0 4, 4 So ha roos of he auiliary equaion are real and equal. 4 Hence he syse is criically daped and he soluion of he governing differenial equaion is 4 4 () c e + c e Moreover, he syse is criically daped. 5

10 Differenial Equaions (MTH40) We now apply he boundary condiions. ( ) 0 c.+ c c 0 Thus () c e d 4 c e c e 4 4 c So ha ( 0). 0 4 c Thus soluion of he iniial value proble is Ereu () e 4 Since () e 4 Therefore Thus d 4 4 e ( ) e 4 4 d 0 The corresponding eree displaceen is e 4 4 Thus he weigh reaches a aiu heigh of Eaple + e f 0.76 f above he equilibriu posiion. A 6-lb weigh is aached o a 5 - f long spring. A equilibriu he spring easures 8.f.If he weigh is pushed up and released fro res a a poin - f above he equilibriu posiion. Find he displaceen ( ) if i is furher known ha he surrounding ediu offers a resisance nuerically equal o he insananeous velociy. Soluion Lengh of un - sreched spring 5 f Lengh of spring a equilibriu 8. f Thus Elongaion of spring s By Hook s law, we have. f 6

11 Differenial Equaions (MTH40) Furher Since (.) 5 lb / f 6 k k Weigh ass g Daping force Therefore β d 6 slugs Thus he differenial equaion of he free daped oion is given by or d k d 5 d β d d d or Since he spring is released fro res a a poin f above he equilibriu posiion. The iniial condiions are: ( 0 ), ( 0) 0 Hence we need o solve he iniial value proble d d ( 0 ), ( 0) 0 To solve he differenial equaion, we pu e Then he auiliary equaion is or, d e ± i So ha he auiliary equaion has cople roos, d + i, i The syse is under-daped and he soluion of he differenial equaion is: () e ( c c sin ) cos + e. 7

12 Differenial Equaions (MTH40) Now we apply he boundary condiions ( ) c.+ c.0 0 c Thus () e ( cos + c sin ) e ( sin + c cos ) e ( cos c sin d Therefore ( 0) ) c c Hence, soluion of he iniial value proble is () e cos sin Eaple 4 Wrie he soluion of he iniial value proble d d ( 0 ), ( 0) 0 in he alernaive for () Ae sin ( + φ ) Soluion We know fro previous eaple ha he soluion of he iniial value proble is () e cos sin Suppose ha A and φ are real nubers such ha / sinφ, cosφ A A 4 Then A Also an φ / Therefore an ( ).49radian Since sin φ < 0, cosφ < 0, he phase angle φ us be in rd quadran. Therefore φ π radians Hence 8

13 Differenial Equaions (MTH40) () 0e sin( + 4.9) The values of where he graph of he soluion crosses posiive - ais and he values * γ γ given in he following able. where he graph of he soluion ouches he graphs of ± 0e are Quasi Period γ γ * ( ) * γ γ Since () 0e sin( + 4.9) Therefore λ ω So ha he quasi period is given by π π seconds λ ω Hence, difference beween he successive * γ and γ is π unis. 9

14 Differenial Equaions (MTH40) Eercise Give a possible inerpreaion of he given iniial value probles , ( 0) 0, ( 0) , ( 0), ( 0). A 4-lb weigh is aached o a spring whose consan is lb /f. The ediu offers a resisance o he oion of he weigh nuerically equal o he insananeous velociy. If he weigh is released fro a poin f above he equilibriu posiion wih a downward velociy of 8 f / s, deerine he ie ha he weigh passes hrough he equilibriu posiion. Find he ie for which he weigh aains is eree displaceen fro he equilibriu posiion. Wha is he posiion of he weigh a his insan? 4. A 4-f spring easures 8 f long afer an 8-lb weigh is aached o i. The ediu hrough which he weigh oves offers a resisance nuerically equal o ies he insananeous velociy. Find he equaion of oion if he weigh is released fro he equilibriu posiion wih a downward velociy of 5 f / s. Find he ie for which he weigh aains is eree displaceen fro he equilibriu posiion. Wha is he posiion of he weigh a his insan? 5. A -kg ass is aached o a spring whose consan is 6 N / and he enire syse is hen suberged in o a liquid ha ipars a daping force nuerically equal o 0 ies he insananeous velociy. Deerine he equaions of oion if a. The weigh is released fro res below he equilibriu posiion; and b. The weigh is released below he equilibriu posiion wih and upward velociy of /s. 6. A force of -lb sreches a spring f. A.-lb weigh is aached o he spring and he syse is hen iersed in a ediu ha ipars daping force nuerically equal o 0.4 ies he insananeous velociy. a. Find he equaion of oion if he weigh is released fro res f above he equilibriu posiion. λ Ae sin ω λ + φ b. Epress he equaion of oion in he for () ( ) c. Find he firs ies for which he weigh passes hrough he equilibriu posiion heading upward. 0

15 Differenial Equaions (MTH40) 7. Afer a 0-lb weigh is aached o a 5-f spring, he spring easures 7-f long. The 0-lb weigh is reoved and replaced wih an 8-lb weigh and he enire syse is placed in a ediu offering a resisance nuerically equal o he insananeous velociy. a. Find he equaion of oion if he weigh is released / f below he equilibriu posiion wih a downward velociy of f / s. λ Ae sin ω λ + φ b. Epress he equaion of oion in he for () ( ) c. Find he ie for which he weigh passes hrough he equilibriu posiion heading downward. 8. A 0-lb weigh aached o a spring sreches i f. The weigh is aached o a dashpo-daping device ha offers a resisance nuerically equal o β ( β > 0) ies he insananeous velociy. Deerine he values of he daping consan β so ha he subsequen oion is a. Over-daped b. Criically daped c. Under-daped 9. A ass of 40 g. sreches a spring 0c. A daping device ipars a resisance o oion nuerically equal o 560 (easured in dynes /(c / s)) ies he insananeous velociy. Find he equaion of oion if he ass is released fro he equilibriu posiion wih downward velociy of c / s. 0. The quasi period of an under-daped, vibraing -slugs ass of a spring isπ / seconds. If he spring consan is 5 lb / f, find he daping consan β.

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

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

Some Basic Information about M-S-D Systems

Some Basic Information about M-S-D Systems 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,

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

Practice Problems - Week #4 Higher-Order DEs, Applications Solutions

Practice Problems - Week #4 Higher-Order DEs, Applications Solutions Pracice Probles - Wee #4 Higher-Orer DEs, Applicaions Soluions 1. Solve he iniial value proble where y y = 0, y0 = 0, y 0 = 1, an y 0 =. r r = rr 1 = rr 1r + 1, so he general soluion is C 1 + C e x + C

More information

Viscous Damping Summary Sheet No Damping Case: Damped behaviour depends on the relative size of ω o and b/2m 3 Cases: 1.

Viscous Damping Summary Sheet No Damping Case: Damped behaviour depends on the relative size of ω o and b/2m 3 Cases: 1. Viscous Daping: && + & + ω Viscous Daping Suary Shee No Daping Case: & + ω solve A ( ω + α ) Daped ehaviour depends on he relaive size of ω o and / 3 Cases:. Criical Daping Wee 5 Lecure solve sae BC s

More information

MEI Mechanics 1 General motion. Section 1: Using calculus

MEI Mechanics 1 General motion. Section 1: Using calculus Soluions o Exercise MEI Mechanics General moion Secion : Using calculus. s 4 v a 6 4 4 When =, v 4 a 6 4 6. (i) When = 0, s = -, so he iniial displacemen = - m. s v 4 When = 0, v = so he iniial velociy

More information

AP Calculus BC Chapter 10 Part 1 AP Exam Problems

AP Calculus BC Chapter 10 Part 1 AP Exam Problems AP Calculus BC Chaper Par AP Eam Problems All problems are NO CALCULATOR unless oherwise indicaed Parameric Curves and Derivaives In he y plane, he graph of he parameric equaions = 5 + and y= for, is a

More information

Homework 2 Solutions

Homework 2 Solutions Mah 308 Differenial Equaions Fall 2002 & 2. See he las page. Hoework 2 Soluions 3a). Newon s secon law of oion says ha a = F, an we know a =, so we have = F. One par of he force is graviy, g. However,

More information

Physics 221 Fall 2008 Homework #2 Solutions Ch. 2 Due Tues, Sept 9, 2008

Physics 221 Fall 2008 Homework #2 Solutions Ch. 2 Due Tues, Sept 9, 2008 Physics 221 Fall 28 Homework #2 Soluions Ch. 2 Due Tues, Sep 9, 28 2.1 A paricle moving along he x-axis moves direcly from posiion x =. m a ime =. s o posiion x = 1. m by ime = 1. s, and hen moves direcly

More information

Introduction to Numerical Analysis. In this lesson you will be taken through a pair of techniques that will be used to solve the equations of.

Introduction to Numerical Analysis. In this lesson you will be taken through a pair of techniques that will be used to solve the equations of. Inroducion o Nuerical Analysis oion In his lesson you will be aen hrough a pair of echniques ha will be used o solve he equaions of and v dx d a F d for siuaions in which F is well nown, and he iniial

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

Introduction to Mechanical Vibrations and Structural Dynamics

Introduction to Mechanical Vibrations and Structural Dynamics Inroducion o Mechanical Viraions and Srucural Dynaics The one seeser schedule :. Viraion - classificaion. ree undaped single DO iraion, equaion of oion, soluion, inegraional consans, iniial condiions..

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

ANALYSIS OF LINEAR AND NONLINEAR EQUATION FOR OSCILLATING MOVEMENT

ANALYSIS OF LINEAR AND NONLINEAR EQUATION FOR OSCILLATING MOVEMENT УД 69486 Anna Macurová arol Vasilko Faculy of Manufacuring Tecnologies Tecnical Universiy of ošice Prešov Slovakia ANALYSIS OF LINEAR AND NONLINEAR EQUATION FOR OSCILLATING MOVEMENT Macurová Anna Vasilko

More information

M x t = K x F t x t = A x M 1 F t. M x t = K x cos t G 0. x t = A x cos t F 0

M x t = K x F t x t = A x M 1 F t. M x t = K x cos t G 0. x t = A x cos t F 0 Forced oscillaions (sill undaped): If he forcing is sinusoidal, M = K F = A M F M = K cos G wih F = M G = A cos F Fro he fundaenal heore for linear ransforaions we now ha he general soluion o his inhoogeneous

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

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

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

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

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

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

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

x y θ = 31.8 = 48.0 N. a 3.00 m/s

x y θ = 31.8 = 48.0 N. a 3.00 m/s 4.5.IDENTIY: Vecor addiion. SET UP: Use a coordinae sse where he dog A. The forces are skeched in igure 4.5. EXECUTE: + -ais is in he direcion of, A he force applied b =+ 70 N, = 0 A B B A = cos60.0 =

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

Motion along a Straight Line

Motion along a Straight Line chaper 2 Moion along a Sraigh Line verage speed and average velociy (Secion 2.2) 1. Velociy versus speed Cone in he ebook: fer Eample 2. Insananeous velociy and insananeous acceleraion (Secions 2.3, 2.4)

More information

SHM SHM. T is the period or time it takes to complete 1 cycle. T = = 2π. f is the frequency or the number of cycles completed per unit time.

SHM SHM. T is the period or time it takes to complete 1 cycle. T = = 2π. f is the frequency or the number of cycles completed per unit time. SHM A ω = k d x x = Acos ( ω +) dx v = = ω Asin( ω + ) vax = ± ωa dv a = = ω Acos + k + x Apliude ( ω ) = 0 a ax = ± ω A SHM x = Acos is he period or ie i akes o coplee cycle. ω = π ( ω +) π = = π ω k

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

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

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

2.1: What is physics? Ch02: Motion along a straight line. 2.2: Motion. 2.3: Position, Displacement, Distance

2.1: What is physics? Ch02: Motion along a straight line. 2.2: Motion. 2.3: Position, Displacement, Distance Ch: Moion along a sraigh line Moion Posiion and Displacemen Average Velociy and Average Speed Insananeous Velociy and Speed Acceleraion Consan Acceleraion: A Special Case Anoher Look a Consan Acceleraion

More information

3.6 Derivatives as Rates of Change

3.6 Derivatives as Rates of Change 3.6 Derivaives as Raes of Change Problem 1 John is walking along a sraigh pah. His posiion a he ime >0 is given by s = f(). He sars a =0from his house (f(0) = 0) and he graph of f is given below. (a) Describe

More information

Physics 20 Lesson 5 Graphical Analysis Acceleration

Physics 20 Lesson 5 Graphical Analysis Acceleration Physics 2 Lesson 5 Graphical Analysis Acceleraion I. Insananeous Velociy From our previous work wih consan speed and consan velociy, we know ha he slope of a posiion-ime graph is equal o he velociy of

More information

CHAPTER 12 DIRECT CURRENT CIRCUITS

CHAPTER 12 DIRECT CURRENT CIRCUITS CHAPTER 12 DIRECT CURRENT CIUITS DIRECT CURRENT CIUITS 257 12.1 RESISTORS IN SERIES AND IN PARALLEL When wo resisors are conneced ogeher as shown in Figure 12.1 we said ha hey are conneced in series. As

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

Ground Rules. PC1221 Fundamentals of Physics I. Kinematics. Position. Lectures 3 and 4 Motion in One Dimension. A/Prof Tay Seng Chuan

Ground Rules. PC1221 Fundamentals of Physics I. Kinematics. Position. Lectures 3 and 4 Motion in One Dimension. A/Prof Tay Seng Chuan Ground Rules PC11 Fundamenals of Physics I Lecures 3 and 4 Moion in One Dimension A/Prof Tay Seng Chuan 1 Swich off your handphone and pager Swich off your lapop compuer and keep i No alking while lecure

More information

Transient State Analysis of a damped & forced oscillator

Transient State Analysis of a damped & forced oscillator 1 DSRC Transien Sae Analysis of a daped & forced oscillaor P.K.Shara 1* 1. M-56, Fla No-3, Madhusudan Nagar, Jagannah coplex, Uni-4, Odisha, India. Asrac: This paper deals wih he ehaviour of an oscillaor

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

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

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

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

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

Exam I. Name. Answer: a. W B > W A if the volume of the ice cubes is greater than the volume of the water.

Exam I. Name. Answer: a. W B > W A if the volume of the ice cubes is greater than the volume of the water. Name Exam I 1) A hole is punched in a full milk caron, 10 cm below he op. Wha is he iniial veloci of ouflow? a. 1.4 m/s b. 2.0 m/s c. 2.8 m/s d. 3.9 m/s e. 2.8 m/s Answer: a 2) In a wind unnel he pressure

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

2.1 Harmonic excitation of undamped systems

2.1 Harmonic excitation of undamped systems 2.1 Haronic exciaion of undaped syses Sanaoja 2_1.1 2.1 Haronic exciaion of undaped syses (Vaienaaoan syseein haroninen heräe) The following syse is sudied: y x F() Free-body diagra f x g x() N F() In

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

PHYSICS 220 Lecture 02 Motion, Forces, and Newton s Laws Textbook Sections

PHYSICS 220 Lecture 02 Motion, Forces, and Newton s Laws Textbook Sections PHYSICS 220 Lecure 02 Moion, Forces, and Newon s Laws Texbook Secions 2.2-2.4 Lecure 2 Purdue Universiy, Physics 220 1 Overview Las Lecure Unis Scienific Noaion Significan Figures Moion Displacemen: Δx

More information

Displacement ( x) x x x

Displacement ( x) x x x Kinemaics Kinemaics is he branch of mechanics ha describes he moion of objecs wihou necessarily discussing wha causes he moion. 1-Dimensional Kinemaics (or 1- Dimensional moion) refers o moion in a sraigh

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

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 Vocabulary. Kinematics and One Dimensional Motion. Position. Coordinate System in One Dimension. Kinema means movement 8.

Kinematics Vocabulary. Kinematics and One Dimensional Motion. Position. Coordinate System in One Dimension. Kinema means movement 8. Kinemaics Vocabulary Kinemaics and One Dimensional Moion 8.1 WD1 Kinema means movemen Mahemaical descripion of moion Posiion Time Inerval Displacemen Velociy; absolue value: speed Acceleraion Averages

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

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

Solution: b All the terms must have the dimension of acceleration. We see that, indeed, each term has the units of acceleration

Solution: b All the terms must have the dimension of acceleration. We see that, indeed, each term has the units of acceleration PHYS 54 Tes Pracice Soluions Spring 8 Q: [4] Knowing ha in he ne epression a is acceleraion, v is speed, is posiion and is ime, from a dimensional v poin of view, he equaion a is a) incorrec b) correc

More information

Suggested Practice Problems (set #2) for the Physics Placement Test

Suggested Practice Problems (set #2) for the Physics Placement Test Deparmen of Physics College of Ars and Sciences American Universiy of Sharjah (AUS) Fall 014 Suggesed Pracice Problems (se #) for he Physics Placemen Tes This documen conains 5 suggesed problems ha are

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

dp dt For the time interval t, approximately, we can write,

dp dt For the time interval t, approximately, we can write, PHYSICS OCUS 58 So far we hae deal only wih syses in which he oal ass of he syse, sys, reained consan wih ie. Now, we will consider syses in which ass eners or leaes he syse while we are obsering i. The

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

MA 214 Calculus IV (Spring 2016) Section 2. Homework Assignment 1 Solutions

MA 214 Calculus IV (Spring 2016) Section 2. Homework Assignment 1 Solutions MA 14 Calculus IV (Spring 016) Secion Homework Assignmen 1 Soluions 1 Boyce and DiPrima, p 40, Problem 10 (c) Soluion: In sandard form he given firs-order linear ODE is: An inegraing facor is given by

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

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

Reading from Young & Freedman: For this topic, read sections 25.4 & 25.5, the introduction to chapter 26 and sections 26.1 to 26.2 & 26.4.

Reading from Young & Freedman: For this topic, read sections 25.4 & 25.5, the introduction to chapter 26 and sections 26.1 to 26.2 & 26.4. PHY1 Elecriciy Topic 7 (Lecures 1 & 11) Elecric Circuis n his opic, we will cover: 1) Elecromoive Force (EMF) ) Series and parallel resisor combinaions 3) Kirchhoff s rules for circuis 4) Time dependence

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

8. Basic RL and RC Circuits

8. Basic RL and RC Circuits 8. Basic L and C Circuis This chaper deals wih he soluions of he responses of L and C circuis The analysis of C and L circuis leads o a linear differenial equaion This chaper covers he following opics

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

Q2.1 This is the x t graph of the motion of a particle. Of the four points P, Q, R, and S, the velocity v x is greatest (most positive) at

Q2.1 This is the x t graph of the motion of a particle. Of the four points P, Q, R, and S, the velocity v x is greatest (most positive) at Q2.1 This is he x graph of he moion of a paricle. Of he four poins P, Q, R, and S, he velociy is greaes (mos posiive) a A. poin P. B. poin Q. C. poin R. D. poin S. E. no enough informaion in he graph o

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

قسم: العلوم. This test includes three mandatory exercises. The use of non-programmable calculators is allowed.

قسم: العلوم. This test includes three mandatory exercises. The use of non-programmable calculators is allowed. الهيئة األكاديمي ة المشتركة قسم: العلوم نموذج مسابقة )يراعي تعليق الدروس والتوصيف المعد ل للعام الدراسي 017-016 المادة: الفيزياء الشهادة: الثانوية العام ة الفرع: علوم الحياة نموذج رقم 1 المد ة: ساعتان

More information

SPH3U1 Lesson 03 Kinematics

SPH3U1 Lesson 03 Kinematics SPH3U1 Lesson 03 Kinemaics GRAPHICAL ANALYSIS LEARNING GOALS Sudens will Learn how o read values, find slopes and calculae areas on graphs. Learn wha hese values mean on boh posiion-ime and velociy-ime

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

MTH Feburary 2012 Final term PAPER SOLVED TODAY s Paper

MTH Feburary 2012 Final term PAPER SOLVED TODAY s Paper MTH401 7 Feburary 01 Final erm PAPER SOLVED TODAY s Paper Toal Quesion: 5 Mcqz: 40 Subjecive quesion: 1 4 q of 5 marks 4 q of 3 marks 4 q of marks Guidelines: Prepare his file as I included all pas papers

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 2214 Solution Test 1B Fall 2017

Math 2214 Solution Test 1B Fall 2017 Mah 14 Soluion Tes 1B Fall 017 Problem 1: A ank has a capaci for 500 gallons and conains 0 gallons of waer wih lbs of sal iniiall. A soluion conaining of 8 lbsgal of sal is pumped ino he ank a 10 galsmin.

More information

Chapter 2. Motion in One-Dimension I

Chapter 2. Motion in One-Dimension I Chaper 2. Moion in One-Dimension I Level : AP Physics Insrucor : Kim 1. Average Rae of Change and Insananeous Velociy To find he average velociy(v ) of a paricle, we need o find he paricle s displacemen

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

MOMENTUM CONSERVATION LAW

MOMENTUM CONSERVATION LAW 1 AAST/AEDT AP PHYSICS B: Impulse and Momenum Le us run an experimen: The ball is moving wih a velociy of V o and a force of F is applied on i for he ime inerval of. As he resul he ball s velociy changes

More information

( ) ( ) ( ) ( u) ( u) = are shown in Figure =, it is reasonable to speculate that. = cos u ) and the inside function ( ( t) du

( ) ( ) ( ) ( u) ( u) = are shown in Figure =, it is reasonable to speculate that. = cos u ) and the inside function ( ( t) du Porlan Communiy College MTH 51 Lab Manual The Chain Rule Aciviy 38 The funcions f ( = sin ( an k( sin( 3 38.1. Since f ( cos( k ( = cos( 3. Bu his woul imply ha k ( f ( = are shown in Figure =, i is reasonable

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

x(m) t(sec ) Homework #2. Ph 231 Introductory Physics, Sp-03 Page 1 of 4

x(m) t(sec ) Homework #2. Ph 231 Introductory Physics, Sp-03 Page 1 of 4 Homework #2. Ph 231 Inroducory Physics, Sp-03 Page 1 of 4 2-1A. A person walks 2 miles Eas (E) in 40 minues and hen back 1 mile Wes (W) in 20 minues. Wha are her average speed and average velociy (in ha

More information

Physics 101 Fall 2006: Exam #1- PROBLEM #1

Physics 101 Fall 2006: Exam #1- PROBLEM #1 Physics 101 Fall 2006: Exam #1- PROBLEM #1 1. Problem 1. (+20 ps) (a) (+10 ps) i. +5 ps graph for x of he rain vs. ime. The graph needs o be parabolic and concave upward. ii. +3 ps graph for x of he person

More information

Speed and Velocity. Overview. Velocity & Speed. Speed & Velocity. Instantaneous Velocity. Instantaneous and Average

Speed and Velocity. Overview. Velocity & Speed. Speed & Velocity. Instantaneous Velocity. Instantaneous and Average Overview Kinemaics: Descripion of Moion Posiion and displacemen velociy»insananeous acceleraion»insananeous Speed Velociy Speed and Velociy Speed & Velociy Velociy & Speed A physics eacher walks 4 meers

More information

Let us consider equation (6.16) once again. We have, can be found by the following equation

Let us consider equation (6.16) once again. We have, can be found by the following equation 41 Le us consider equaion (6.16) once again. We hae, dp d Therefore, d p d Here dp is he change in oenu caused by he force in he ie ineral d. Change in oenu caused by he force for a ie ineral 1, can be

More information

Practicing Problem Solving and Graphing

Practicing Problem Solving and Graphing Pracicing Problem Solving and Graphing Tes 1: Jan 30, 7pm, Ming Hsieh G20 The Bes Ways To Pracice for Tes Bes If need more, ry suggesed problems from each new opic: Suden Response Examples A pas opic ha

More information

ODEs II, Lecture 1: Homogeneous Linear Systems - I. Mike Raugh 1. March 8, 2004

ODEs II, Lecture 1: Homogeneous Linear Systems - I. Mike Raugh 1. March 8, 2004 ODEs II, Lecure : Homogeneous Linear Sysems - I Mike Raugh March 8, 4 Inroducion. In he firs lecure we discussed a sysem of linear ODEs for modeling he excreion of lead from he human body, saw how o ransform

More information

Physics 240: Worksheet 16 Name

Physics 240: Worksheet 16 Name Phyic 4: Workhee 16 Nae Non-unifor circular oion Each of hee proble involve non-unifor circular oion wih a conan α. (1) Obain each of he equaion of oion for non-unifor circular oion under a conan acceleraion,

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

EXERCISES FOR SECTION 1.5

EXERCISES FOR SECTION 1.5 1.5 Exisence and Uniqueness of Soluions 43 20. 1 v c 21. 1 v c 1 2 4 6 8 10 1 2 2 4 6 8 10 Graph of approximae soluion obained using Euler s mehod wih = 0.1. Graph of approximae soluion obained using Euler

More information

Sections 2.2 & 2.3 Limit of a Function and Limit Laws

Sections 2.2 & 2.3 Limit of a Function and Limit Laws Mah 80 www.imeodare.com Secions. &. Limi of a Funcion and Limi Laws In secion. we saw how is arise when we wan o find he angen o a curve or he velociy of an objec. Now we urn our aenion o is in general

More information

Predator - Prey Model Trajectories and the nonlinear conservation law

Predator - Prey Model Trajectories and the nonlinear conservation law Predaor - Prey Model Trajecories and he nonlinear conservaion law James K. Peerson Deparmen of Biological Sciences and Deparmen of Mahemaical Sciences Clemson Universiy Ocober 28, 213 Ouline Drawing Trajecories

More information

PHYSICS 149: Lecture 9

PHYSICS 149: Lecture 9 PHYSICS 149: Lecure 9 Chaper 3 3.2 Velociy and Acceleraion 3.3 Newon s Second Law of Moion 3.4 Applying Newon s Second Law 3.5 Relaive Velociy Lecure 9 Purdue Universiy, Physics 149 1 Velociy (m/s) The

More information

Physics 131- Fundamentals of Physics for Biologists I

Physics 131- Fundamentals of Physics for Biologists I 10/3/2012 - Fundamenals of Physics for iologiss I Professor: Wolfgang Loser 10/3/2012 Miderm review -How can we describe moion (Kinemaics) - Wha is responsible for moion (Dynamics) wloser@umd.edu Movie

More information

Dynamics. Option topic: Dynamics

Dynamics. Option topic: Dynamics Dynamics 11 syllabusref Opion opic: Dynamics eferenceence In his cha chaper 11A Differeniaion and displacemen, velociy and acceleraion 11B Inerpreing graphs 11C Algebraic links beween displacemen, velociy

More information

3 at MAC 1140 TEST 3 NOTES. 5.1 and 5.2. Exponential Functions. Form I: P is the y-intercept. (0, P) When a > 1: a = growth factor = 1 + growth rate

3 at MAC 1140 TEST 3 NOTES. 5.1 and 5.2. Exponential Functions. Form I: P is the y-intercept. (0, P) When a > 1: a = growth factor = 1 + growth rate 1 5.1 and 5. Eponenial Funcions Form I: Y Pa, a 1, a > 0 P is he y-inercep. (0, P) When a > 1: a = growh facor = 1 + growh rae The equaion can be wrien as The larger a is, he seeper he graph is. Y P( 1

More information

2002 November 14 Exam III Physics 191

2002 November 14 Exam III Physics 191 November 4 Exam III Physics 9 Physical onsans: Earh s free-fall acceleraion = g = 9.8 m/s ircle he leer of he single bes answer. quesion is worh poin Each 3. Four differen objecs wih masses: m = kg, m

More information

SIGNALS AND SYSTEMS LABORATORY 8: State Variable Feedback Control Systems

SIGNALS AND SYSTEMS LABORATORY 8: State Variable Feedback Control Systems SIGNALS AND SYSTEMS LABORATORY 8: Sae Variable Feedback Conrol Syses INTRODUCTION Sae variable descripions for dynaical syses describe he evoluion of he sae vecor, as a funcion of he sae and he inpu. There

More information

Today: Graphing. Note: I hope this joke will be funnier (or at least make you roll your eyes and say ugh ) after class. v (miles per hour ) Time

Today: Graphing. Note: I hope this joke will be funnier (or at least make you roll your eyes and say ugh ) after class. v (miles per hour ) Time +v Today: Graphing v (miles per hour ) 9 8 7 6 5 4 - - Time Noe: I hope his joke will be funnier (or a leas make you roll your eyes and say ugh ) afer class. Do yourself a favor! Prof Sarah s fail-safe

More information

Parametrics and Vectors (BC Only)

Parametrics and Vectors (BC Only) Paramerics and Vecors (BC Only) The following relaionships should be learned and memorized. The paricle s posiion vecor is r() x(), y(). The velociy vecor is v(),. The speed is he magniude of he velociy

More information

dy dx = xey (a) y(0) = 2 (b) y(1) = 2.5 SOLUTION: See next page

dy dx = xey (a) y(0) = 2 (b) y(1) = 2.5 SOLUTION: See next page Assignmen 1 MATH 2270 SOLUTION Please wrie ou complee soluions for each of he following 6 problems (one more will sill be added). You may, of course, consul wih your classmaes, he exbook or oher resources,

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

INSTANTANEOUS VELOCITY

INSTANTANEOUS VELOCITY INSTANTANEOUS VELOCITY I claim ha ha if acceleraion is consan, hen he elociy is a linear funcion of ime and he posiion a quadraic funcion of ime. We wan o inesigae hose claims, and a he same ime, work

More information