Section 4.9; Section 5.6. June 30, Free Mechanical Vibrations/Couple Mass-Spring System
|
|
- Rosaline Chambers
- 5 years ago
- Views:
Transcription
1 Section 4.9; Section 5.6 Free Mechanical Vibrations/Couple Mass-Spring System June 30, 2009
2 Today s Session
3 Today s Session A Summary of This Session:
4 Today s Session A Summary of This Session: (1) Free Mechanical Vibration (no forcing term).
5 Today s Session A Summary of This Session: (1) Free Mechanical Vibration (no forcing term). (2) Coupled Mass-Spring systems
6 Today s Session A Summary of This Session: (1) Free Mechanical Vibration (no forcing term). (2) Coupled Mass-Spring systems (3) Our first exposure to systems of differential equations
7 Today s Session A Summary of This Session: (1) Free Mechanical Vibration (no forcing term). (2) Coupled Mass-Spring systems (3) Our first exposure to systems of differential equations (4) Eigenvalues and Eigenvectors
8 Today s Session A Summary of This Session: (1) Free Mechanical Vibration (no forcing term). (2) Coupled Mass-Spring systems (3) Our first exposure to systems of differential equations (4) Eigenvalues and Eigenvectors
9 Free Mechanical Vibrations Free mechanical vibration=no forcing function, so f (t) = 0. We are dealing with my + by + ky = 0 where m = mass attached to a spring of stiffness k, subject to friction (or damping) proportional to speed with damping constant b.
10 Free Mechanical Vibrations Free mechanical vibration=no forcing function, so f (t) = 0. We are dealing with my + by + ky = 0 where m = mass attached to a spring of stiffness k, subject to friction (or damping) proportional to speed with damping constant b. Four cases: (1) undamped free case; (2) underdamped case; (3) overdamped case; (4) critical damped case.
11 Free Mechanical Vibrations Free mechanical vibration=no forcing function, so f (t) = 0. We are dealing with my + by + ky = 0 where m = mass attached to a spring of stiffness k, subject to friction (or damping) proportional to speed with damping constant b. Four cases: (1) undamped free case; (2) underdamped case; (3) overdamped case; (4) critical damped case.
12 Case 1: Undamped mass-spring system: b = 0. The equation is given by: my + ky = 0
13 Case 1: Undamped mass-spring system: b = 0. The equation is given by: my + ky = 0 or y + k m y = 0
14 Case 1: Undamped mass-spring system: b = 0. The equation is given by: or my + ky = 0 y + k m y = 0 Let ω 2 = k m. The quantity ω = k m is called the angular frequency (measured in radians per second)
15 Case 1: Undamped mass-spring system: b = 0. The equation is given by: or my + ky = 0 y + k m y = 0 Let ω 2 = k m. The quantity ω = k m is called the angular frequency (measured in radians per second) Period: T = 2π ω (measured in seconds)
16 Case 1: Undamped mass-spring system: b = 0. The equation is given by: or my + ky = 0 y + k m y = 0 Let ω 2 = k m. The quantity ω = k m is called the angular frequency (measured in radians per second) Period: T = 2π ω (measured in seconds) Frequency: f = 1 T = ω 2π (measured in Hertz=1/seconds=# cycles per second)
17 Case 1: Undamped mass-spring system: b = 0. The equation is given by: or my + ky = 0 y + k m y = 0 Let ω 2 = k m. The quantity ω = k m is called the angular frequency (measured in radians per second) Period: T = 2π ω (measured in seconds) Frequency: f = 1 T = ω 2π (measured in Hertz=1/seconds=# cycles per second) The solution is given by: y = C 1 cos ωt +C 2 sinωt = A sin(ωt +φ).
18 Undamped case, cont d Here A is called the amplitude, and φ is called the phase. They are given by A = C1 2 + C2 2 φ = arctan C 1. C 2
19 Undamped case, cont d Here A is called the amplitude, and φ is called the phase. They are given by A = C1 2 + C2 2 φ = arctan C 1. C 2 Example A 1/8 kg manss is attached to a spring with stiffness k = 16N/m. The mass is displaed 0.5 m to the right of the equilibrium point and given an outward initial velocity of 2 m/s. (a) Neglecting damping, find a formula of the displacement as a function of time. Display the values of ω, T, f, A and φ.
20 Undamped case, cont d Here A is called the amplitude, and φ is called the phase. They are given by A = C1 2 + C2 2 φ = arctan C 1. C 2 Example A 1/8 kg manss is attached to a spring with stiffness k = 16N/m. The mass is displaed 0.5 m to the right of the equilibrium point and given an outward initial velocity of 2 m/s. (a) Neglecting damping, find a formula of the displacement as a function of time. Display the values of ω, T, f, A and φ. (b) How long after release does the mass pass first through the equilibrium position?
21 Undamped case, cont d Answers: (a) ω = 8 2 rad/sec; f = 1.8 cycles/sec (or Hz); T = sec.
22 Undamped case, cont d Answers: (a) ω = 8 2 rad/sec; f = 1.8 cycles/sec (or Hz); T = sec. y(t) = 1 2 cos 8 2t sin8 2t
23 Undamped case, cont d Answers: (a) ω = 8 2 rad/sec; f = 1.8 cycles/sec (or Hz); T = sec. A = y(t) = 1 2 cos 8 2t sin8 2t ( 1 2 )2 + ( 1 8 )2 = 17 8 and tanφ = 1/2 1/8 = 4 so φ = rad. y(t) = 17 8 sin(8 2t )
24 Undamped case, cont d Answers: (a) ω = 8 2 rad/sec; f = 1.8 cycles/sec (or Hz); T = sec. A = y(t) = 1 2 cos 8 2t sin8 2t ( 1 2 )2 + ( 1 8 )2 = 17 8 and tanφ = 1/2 1/8 = 4 so φ = rad. y(t) = 17 8 sin(8 2t ) (b) y(t) = 0 means sin(8 2t + φ) = 0 so 8 2t + φ = kπ for k integer. Solving gives: t = kπ φ ω = 0.16 sec. Every 1/2 period (or 0.28 sec) the mass goes through the equilibrium point.
25 Overdamped case: b 2 4k m > 0 The equation is my + by + ky = 0
26 Overdamped case: b 2 4k m > 0 The equation is my + by + ky = 0 The characteristic equation is given by mr 2 + br + k = 0
27 Overdamped case: b 2 4k m > 0 The equation is my + by + ky = 0 The characteristic equation is given by mr 2 + br + k = 0 so r = b b 2m ± 2 4k m 2m The solution is given by: y(t) = C 1 e r1t + C 2 e r 2t
28 Overdamped case: b 2 4k m > 0 The equation is my + by + ky = 0 The characteristic equation is given by mr 2 + br + k = 0 so r = b b 2m ± 2 4k m 2m The solution is given by: y(t) = C 1 e r 1t + C 2 e r 2t Note that both r 1 and r 2 are negative (why?), so as t, y(t) 0.
29 Underdamped case: b 2 4k m < 0 With ω = 4k m b 2 2m, the roots of the characteristic equation are: r = b 2m ± i ω So the solution is given by: y(t) = C 1 e b 2m t cos ωt + C 2 e b 2m t sinωt.
30 Underdamped case: b 2 4k m < 0 With ω = 4k m b 2 2m, the roots of the characteristic equation are: r = b 2m ± i ω So the solution is given by: y(t) = C 1 e b 2m t cos ωt + C 2 e b 2m t sinωt. Over time y(t) dies in an oscillatory fashion. ω is a quasi-angular frequency, T = 2π/ω is called the quasiperiod and f = 1/T is called the quasi-frequency.
31 Underdamped case: b 2 4k m < 0 With ω = 4k m b 2 2m, the roots of the characteristic equation are: r = b 2m ± i ω So the solution is given by: y(t) = C 1 e b 2m t cos ωt + C 2 e b 2m t sinωt. Over time y(t) dies in an oscillatory fashion. ω is a quasi-angular frequency, T = 2π/ω is called the quasiperiod and f = 1/T is called the quasi-frequency. The solution can be put in the form where (as before): A = y(t) = A e b 2m t sin(ωt + φ) C C2 2 φ = arctan C 1 C 2.
32 Underdamped case: b 2 4k m < 0 With ω = 4k m b 2 2m, the roots of the characteristic equation are: r = b 2m ± i ω So the solution is given by: y(t) = C 1 e b 2m t cos ωt + C 2 e b 2m t sinωt. Over time y(t) dies in an oscillatory fashion. ω is a quasi-angular frequency, T = 2π/ω is called the quasiperiod and f = 1/T is called the quasi-frequency. The solution can be put in the form where (as before): A = y(t) = A e b 2m t sin(ωt + φ) C C2 2 φ = arctan C 1 C 2.
33 Critically damped case: b 2 4k m = 0 In this case, the characteristic equation has a double root r = b 2m, So the solution is y(t) = C 1 e b 2m t + C 2 t e b 2m t
34 Critically damped case: b 2 4k m = 0 In this case, the characteristic equation has a double root r = b 2m, So the solution is y(t) = C 1 e b 2m t + C 2 t e b 2m t Example: (related to webassign question and example p. 234)
35 Critically damped case: b 2 4k m = 0 In this case, the characteristic equation has a double root r = b 2m, So the solution is y(t) = C 1 e b 2m t + C 2 t e b 2m t Example: (related to webassign question and example p. 234) Find the value of b for which y + b y + 25y = 0 y(0) = 1, y (0) = 0 is critically damped. Solve for y(t) in this case and sketch it.
36 Critically damped case: b 2 4k m = 0 In this case, the characteristic equation has a double root r = b 2m, So the solution is y(t) = C 1 e b 2m t + C 2 t e b 2m t Example: (related to webassign question and example p. 234) Find the value of b for which y + b y + 25y = 0 y(0) = 1, y (0) = 0 is critically damped. Solve for y(t) in this case and sketch it. Answer: b = 10; y(t) = (1 t)e 5t
37 5.6: Coupled Mass-Spring System: Example 1, p. 308 See the graph and description of the couple mass-spring system on p. 308.
38 5.6: Coupled Mass-Spring System: Example 1, p. 308 See the graph and description of the couple mass-spring system on p I will show the derivation of the following equations in class:
39 5.6: Coupled Mass-Spring System: Example 1, p. 308 See the graph and description of the couple mass-spring system on p I will show the derivation of the following equations in class: We wish to solve: { m1 x = k 1 x + k 2 (y x) m 2 y = k 2 (y x)
40 5.6: Coupled Mass-Spring System: Example 1, p. 308 See the graph and description of the couple mass-spring system on p I will show the derivation of the following equations in class: We wish to solve: { m1 x = k 1 x + k 2 (y x) m 2 y = k 2 (y x) In our example: m 1 = 2kg ; k 1 = 4N/m; m 2 = 1kg ; and k 2 = 2N/m. So: { 2x = 6x + 2y y = 2x 2y
41 5.6: Coupled Mass-Spring System: Example 1, p. 308 See the graph and description of the couple mass-spring system on p I will show the derivation of the following equations in class: We wish to solve: { m1 x = k 1 x + k 2 (y x) m 2 y = k 2 (y x) In our example: m 1 = 2kg ; k 1 = 4N/m; m 2 = 1kg ; and k 2 = 2N/m. So: { 2x = 6x + 2y y = 2x 2y
42 Example 1, p. 308, cont d Let s clean this up a little bit: { x = 3x + y y = 2x 2y
43 Example 1, p. 308, cont d Let s clean this up a little bit: { x = 3x + y y = 2x 2y This is a linear system of differential equations of second order (which is homogeneous; why?) We can put it in the form: { (D 2 + 3)x y = 0 2x + (D 2 + 2)y = 0
44 Example 1, p. 308, cont d Let s clean this up a little bit: { x = 3x + y y = 2x 2y This is a linear system of differential equations of second order (which is homogeneous; why?) We can put it in the form: { (D 2 + 3)x y = 0 2x + (D 2 + 2)y = 0 Therefore (D 2 + 2)(D 2 + 3)x (D 2 + 2)y = 0
45 Example 1, p. 308, cont d Let s clean this up a little bit: { x = 3x + y y = 2x 2y This is a linear system of differential equations of second order (which is homogeneous; why?) We can put it in the form: { (D 2 + 3)x y = 0 2x + (D 2 + 2)y = 0 Therefore (D 2 + 2)(D 2 + 3)x (D 2 + 2)y = 0 We note that: (D 2 + 2)y = 2x.
46 Example 1, p. 308, cont d Let s clean this up a little bit: { x = 3x + y y = 2x 2y This is a linear system of differential equations of second order (which is homogeneous; why?) We can put it in the form: { (D 2 + 3)x y = 0 2x + (D 2 + 2)y = 0 Therefore (D 2 + 2)(D 2 + 3)x (D 2 + 2)y = 0 We note that: (D 2 + 2)y = 2x. So: (D 2 + 2)(D 2 + 3)x 2x = 0.
47 Example 1, p. 308, cont d Let s clean this up a little bit: { x = 3x + y y = 2x 2y This is a linear system of differential equations of second order (which is homogeneous; why?) We can put it in the form: { (D 2 + 3)x y = 0 2x + (D 2 + 2)y = 0 Therefore (D 2 + 2)(D 2 + 3)x (D 2 + 2)y = 0 We note that: (D 2 + 2)y = 2x. So: (D 2 + 2)(D 2 + 3)x 2x = 0. We clean this up a little bit: (D 4 + 5D 2 + 4)x = 0.
48 Example 1, p. 308, cont d Let s clean this up a little bit: { x = 3x + y y = 2x 2y This is a linear system of differential equations of second order (which is homogeneous; why?) We can put it in the form: { (D 2 + 3)x y = 0 2x + (D 2 + 2)y = 0 Therefore (D 2 + 2)(D 2 + 3)x (D 2 + 2)y = 0 We note that: (D 2 + 2)y = 2x. So: (D 2 + 2)(D 2 + 3)x 2x = 0. We clean this up a little bit: (D 4 + 5D 2 + 4)x = 0.
49 Example 1, p. 308, cont d Then factor: (D 2 + 1)(D 2 + 4)x = 0
50 Example 1, p. 308, cont d Then factor: (D 2 + 1)(D 2 + 4)x = 0 Therefore: x(t) = C 1 cos t + C 2 sint + C 3 cos 2t + C 4 sin2t and y = (D 2 + 3)x = 2C 1 cos t + 2C 2 sint C 3 cos 2t C 4 sin2t
51 Example 2 Use the method of the previous example (the annihilator method) to solve the first order (homogeneous) linear system: { x = 3x + 4y y = 4x + 3y
52 Example 2 Use the method of the previous example (the annihilator method) to solve the first order (homogeneous) linear system: { x = 3x + 4y y = 4x + 3y Answer: x(t) = C 1 e 7t + C 2 e t and y(t) = C 1 e t C 2 e t
Section 3.7: Mechanical and Electrical Vibrations
Section 3.7: Mechanical and Electrical Vibrations Second order linear equations with constant coefficients serve as mathematical models for mechanical and electrical oscillations. For example, the motion
More information4.9 Free Mechanical Vibrations
4.9 Free Mechanical Vibrations Spring-Mass Oscillator When the spring is not stretched and the mass m is at rest, the system is at equilibrium. Forces Acting in the System When the mass m is displaced
More informationDifferential Equations: Homework 8
Differential Equations: Homework 8 Alvin Lin January 08 - May 08 Section.6 Exercise Find a general solution to the differential equation using the method of variation of parameters. y + y = tan(t) r +
More informationOscillations. Simple Harmonic Motion of a Mass on a Spring The equation of motion for a mass m is attached to a spring of constant k is
Dr. Alain Brizard College Physics I (PY 10) Oscillations Textbook Reference: Chapter 14 sections 1-8. Simple Harmonic Motion of a Mass on a Spring The equation of motion for a mass m is attached to a spring
More informationF = ma, F R + F S = mx.
Mechanical Vibrations As we mentioned in Section 3.1, linear equations with constant coefficients come up in many applications; in this section, we will specifically study spring and shock absorber systems
More informationMATH2351 Introduction to Ordinary Differential Equations, Fall Hints to Week 07 Worksheet: Mechanical Vibrations
MATH351 Introduction to Ordinary Differential Equations, Fall 11-1 Hints to Week 7 Worksheet: Mechanical Vibrations 1. (Demonstration) ( 3.8, page 3, Q. 5) A mass weighing lb streches a spring by 6 in.
More informationMATH 23 Exam 2 Review Solutions
MATH 23 Exam 2 Review Solutions Problem 1. Use the method of reduction of order to find a second solution of the given differential equation x 2 y (x 0.1875)y = 0, x > 0, y 1 (x) = x 1/4 e 2 x Solution
More informationfor non-homogeneous linear differential equations L y = f y H
Tues March 13: 5.4-5.5 Finish Monday's notes on 5.4, Then begin 5.5: Finding y P for non-homogeneous linear differential equations (so that you can use the general solution y = y P y = y x in this section...
More informationNotes on the Periodically Forced Harmonic Oscillator
Notes on the Periodically orced Harmonic Oscillator Warren Weckesser Math 38 - Differential Equations 1 The Periodically orced Harmonic Oscillator. By periodically forced harmonic oscillator, we mean the
More informationUnforced Mechanical Vibrations
Unforced Mechanical Vibrations Today we begin to consider applications of second order ordinary differential equations. 1. Spring-Mass Systems 2. Unforced Systems: Damped Motion 1 Spring-Mass Systems We
More informationSprings: Part I Modeling the Action The Mass/Spring System
17 Springs: Part I Second-order differential equations arise in a number of applications We saw one involving a falling object at the beginning of this text (the falling frozen duck example in section
More information2. Determine whether the following pair of functions are linearly dependent, or linearly independent:
Topics to be covered on the exam include: Recognizing, and verifying solutions to homogeneous second-order linear differential equations, and their corresponding Initial Value Problems Recognizing and
More informationMath 240: Spring/Mass Systems II
Math 240: Spring/Mass Systems II Ryan Blair University of Pennsylvania Monday, March 26, 2012 Ryan Blair (U Penn) Math 240: Spring/Mass Systems II Monday, March 26, 2012 1 / 12 Outline 1 Today s Goals
More information3.4 Application-Spring Mass Systems (Unforced and frictionless systems)
3.4. APPLICATION-SPRING MASS SYSTEMS (UNFORCED AND FRICTIONLESS SYSTEMS)73 3.4 Application-Spring Mass Systems (Unforced and frictionless systems) Second order differential equations arise naturally when
More information4.2 Homogeneous Linear Equations
4.2 Homogeneous Linear Equations Homogeneous Linear Equations with Constant Coefficients Consider the first-order linear differential equation with constant coefficients a 0 and b. If f(t) = 0 then this
More informationVibrations and Waves MP205, Assignment 4 Solutions
Vibrations and Waves MP205, Assignment Solutions 1. Verify that x = Ae αt cos ωt is a possible solution of the equation and find α and ω in terms of γ and ω 0. [20] dt 2 + γ dx dt + ω2 0x = 0, Given x
More informationWork sheet / Things to know. Chapter 3
MATH 251 Work sheet / Things to know 1. Second order linear differential equation Standard form: Chapter 3 What makes it homogeneous? We will, for the most part, work with equations with constant coefficients
More informationChapter a. Spring constant, k : The change in the force per unit length change of the spring. b. Coefficient of subgrade reaction, k:
Principles of Soil Dynamics 3rd Edition Das SOLUTIONS MANUAL Full clear download (no formatting errors) at: https://testbankreal.com/download/principles-soil-dynamics-3rd-editiondas-solutions-manual/ Chapter
More informationMath 308 Exam II Practice Problems
Math 38 Exam II Practice Problems This review should not be used as your sole source for preparation for the exam. You should also re-work all examples given in lecture and all suggested homework problems..
More informationAnswers and Hints to Review Questions for Test (a) Find the general solution to the linear system of differential equations Y = 2 ± 3i.
Answers and Hints to Review Questions for Test 3 (a) Find the general solution to the linear system of differential equations [ dy 3 Y 3 [ (b) Find the specific solution that satisfies Y (0) = (c) What
More informationMath 240: Spring-mass Systems
Math 240: Spring-mass Systems Ryan Blair University of Pennsylvania Tuesday March 1, 2011 Ryan Blair (U Penn) Math 240: Spring-mass Systems Tuesday March 1, 2011 1 / 15 Outline 1 Review 2 Today s Goals
More informationFinal Exam April 30, 2013
Final Exam Instructions: You have 120 minutes to complete this exam. This is a closed-book, closed-notes exam. You are allowed to use a calculator during the exam. Usage of mobile phones and other electronic
More informationForced Mechanical Vibrations
Forced Mechanical Vibrations Today we use methods for solving nonhomogeneous second order linear differential equations to study the behavior of mechanical systems.. Forcing: Transient and Steady State
More informationSolutions for homework 5
1 Section 4.3 Solutions for homework 5 17. The following equation has repeated, real, characteristic roots. Find the general solution. y 4y + 4y = 0. The characteristic equation is λ 4λ + 4 = 0 which has
More informationMAT187H1F Lec0101 Burbulla
Spring 2017 Second Order Linear Homogeneous Differential Equation DE: A(x) d 2 y dx 2 + B(x)dy dx + C(x)y = 0 This equation is called second order because it includes the second derivative of y; it is
More informationDate: 1 April (1) The only reference material you may use is one 8½x11 crib sheet and a calculator.
PH1140: Oscillations and Waves Name: Solutions Conference: Date: 1 April 2005 EXAM #1: D2005 INSTRUCTIONS: (1) The only reference material you may use is one 8½x11 crib sheet and a calculator. (2) Show
More informationCh 3.7: Mechanical & Electrical Vibrations
Ch 3.7: Mechanical & Electrical Vibrations Two important areas of application for second order linear equations with constant coefficients are in modeling mechanical and electrical oscillations. We will
More informationDate: 31 March (1) The only reference material you may use is one 8½x11 crib sheet and a calculator.
PH1140: Oscillations and Waves Name: SOLUTIONS AT END Conference: Date: 31 March 2005 EXAM #1: D2006 INSTRUCTIONS: (1) The only reference material you may use is one 8½x11 crib sheet and a calculator.
More informationImaginary. Axis. Real. Axis
Name ME6 Final. I certify that I upheld the Stanford Honor code during this exam Monday December 2, 25 3:3-6:3 p.m. ffl Print your name and sign the honor code statement ffl You may use your course notes,
More informationAPPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS
APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS Second-order linear differential equations have a variety of applications in science and engineering. In this section we explore two of them: the vibration
More informationFINAL EXAM SOLUTIONS, MATH 123
FINAL EXAM SOLUTIONS, MATH 23. Find the eigenvalues of the matrix ( 9 4 3 ) So λ = or 6. = λ 9 4 3 λ = ( λ)( 3 λ) + 36 = λ 2 7λ + 6 = (λ 6)(λ ) 2. Compute the matrix inverse: ( ) 3 3 = 3 4 ( 4/3 ) 3. Let
More informationThe Harmonic Oscillator
The Harmonic Oscillator Math 4: Ordinary Differential Equations Chris Meyer May 3, 008 Introduction The harmonic oscillator is a common model used in physics because of the wide range of problems it can
More informationDynamics of structures
Dynamics of structures 1.2 Viscous damping Luc St-Pierre October 30, 2017 1 / 22 Summary so far We analysed the spring-mass system and found that its motion is governed by: mẍ(t) + kx(t) = 0 k y m x x
More informationImaginary. Axis. Real. Axis
Name ME6 Final. I certify that I upheld the Stanford Honor code during this exam Monday December 2, 2005 3:30-6:30 p.m. ffl Print your name and sign the honor code statement ffl You may use your course
More informationApplications of Second-Order Differential Equations
Applications of Second-Order Differential Equations ymy/013 Building Intuition Even though there are an infinite number of differential equations, they all share common characteristics that allow intuition
More informationApplication of Second Order Linear ODEs: Mechanical Vibrations
Application of Second Order Linear ODEs: October 23 27, 2017 Application of Second Order Linear ODEs Consider a vertical spring of original length l > 0 [m or ft] that exhibits a stiffness of κ > 0 [N/m
More informationWeek 9 solutions. k = mg/l = /5 = 3920 g/s 2. 20u + 400u u = 0,
Week 9 solutions ASSIGNMENT 20. (Assignment 19 had no hand-graded component.) 3.7.9. A mass of 20 g stretches a spring 5 cm. Suppose that the mass is also attached to a viscous damper with a damping constant
More information2.3 Oscillation. The harmonic oscillator equation is the differential equation. d 2 y dt 2 r y (r > 0). Its solutions have the form
2. Oscillation So far, we have used differential equations to describe functions that grow or decay over time. The next most common behavior for a function is to oscillate, meaning that it increases and
More informationIntroduction to Vibration. Professor Mike Brennan
Introduction to Vibration Professor Mie Brennan Introduction to Vibration Nature of vibration of mechanical systems Free and forced vibrations Frequency response functions Fundamentals For free vibration
More informationTopic 5 Notes Jeremy Orloff. 5 Homogeneous, linear, constant coefficient differential equations
Topic 5 Notes Jeremy Orloff 5 Homogeneous, linear, constant coefficient differential equations 5.1 Goals 1. Be able to solve homogeneous constant coefficient linear differential equations using the method
More information2.4 Harmonic Oscillator Models
2.4 Harmonic Oscillator Models In this section we give three important examples from physics of harmonic oscillator models. Such models are ubiquitous in physics, but are also used in chemistry, biology,
More informationDifferential Equations and Linear Algebra Exercises. Department of Mathematics, Heriot-Watt University, Edinburgh EH14 4AS
Differential Equations and Linear Algebra Exercises Department of Mathematics, Heriot-Watt University, Edinburgh EH14 4AS CHAPTER 1 Linear second order ODEs Exercises 1.1. (*) 1 The following differential
More informationThursday, August 4, 2011
Chapter 16 Thursday, August 4, 2011 16.1 Springs in Motion: Hooke s Law and the Second-Order ODE We have seen alrealdy that differential equations are powerful tools for understanding mechanics and electro-magnetism.
More informationSecond Order Linear ODEs, Part II
Craig J. Sutton craig.j.sutton@dartmouth.edu Department of Mathematics Dartmouth College Math 23 Differential Equations Winter 2013 Outline Non-homogeneous Linear Equations 1 Non-homogeneous Linear Equations
More informationMATH 246: Chapter 2 Section 8 Motion Justin Wyss-Gallifent
MATH 46: Chapter Section 8 Motion Justin Wyss-Gallifent 1. Introduction Important: Positive is up and negative is down. Imagine a spring hanging with no weight on it. We then attach a mass m which stretches
More informationEven-Numbered Homework Solutions
-6 Even-Numbered Homework Solutions Suppose that the matric B has λ = + 5i as an eigenvalue with eigenvector Y 0 = solution to dy = BY Using Euler s formula, we can write the complex-valued solution Y
More informationOscillations. PHYS 101 Previous Exam Problems CHAPTER. Simple harmonic motion Mass-spring system Energy in SHM Pendulums
PHYS 101 Previous Exam Problems CHAPTER 15 Oscillations Simple harmonic motion Mass-spring system Energy in SHM Pendulums 1. The displacement of a particle oscillating along the x axis is given as a function
More information2.4 Models of Oscillation
2.4 Models of Oscillation In this section we give three examples of oscillating physical systems that can be modeled by the harmonic oscillator equation. Such models are ubiquitous in physics, but are
More informationLecture 6: Differential Equations Describing Vibrations
Lecture 6: Differential Equations Describing Vibrations In Chapter 3 of the Benson textbook, we will look at how various types of musical instruments produce sound, focusing on issues like how the construction
More informationKEELE UNIVERSITY PHYSICS/ASTROPHYSICS MODULE PHY OSCILLATIONS AND WAVES PRACTICE EXAM
KEELE UNIVERSITY PHYSICS/ASTROPHYSICS MODULE PHY-10012 OSCILLATIONS AND WAVES PRACTICE EXAM Candidates should attempt ALL of PARTS A and B, and TWO questions from PART C. PARTS A and B should be answered
More informationVibrations Qualifying Exam Study Material
Vibrations Qualifying Exam Study Material The candidate is expected to have a thorough understanding of engineering vibrations topics. These topics are listed below for clarification. Not all instructors
More information3! + 4! + Binomial series: if α is a nonnegative integer, the series terminates. Otherwise, the series converges if x < 1 but diverges if x > 1.
Page 1 Name: ID: Section: This exam has 16 questions: 14 multiple choice questions worth 5 points each. hand graded questions worth 15 points each. Important: No graphing calculators! Any non-graphing
More informationMath Assignment 5
Math 2280 - Assignment 5 Dylan Zwick Fall 2013 Section 3.4-1, 5, 18, 21 Section 3.5-1, 11, 23, 28, 35, 47, 56 Section 3.6-1, 2, 9, 17, 24 1 Section 3.4 - Mechanical Vibrations 3.4.1 - Determine the period
More informationPhysics 2101 S c e t c i cti n o 3 n 3 March 31st Announcements: Quiz today about Ch. 14 Class Website:
Physics 2101 Section 3 March 31 st Announcements: Quiz today about Ch. 14 Class Website: http://www.phys.lsu.edu/classes/spring2010/phys2101 3/ http://www.phys.lsu.edu/~jzhang/teaching.html Simple Harmonic
More information8. What is the period of a pendulum consisting of a 6-kg object oscillating on a 4-m string?
1. In the produce section of a supermarket, five pears are placed on a spring scale. The placement of the pears stretches the spring and causes the dial to move from zero to a reading of 2.0 kg. If the
More informationWeek #9 : DEs with Non-Constant Coefficients, Laplace Resonance
Week #9 : DEs with Non-Constant Coefficients, Laplace Resonance Goals: Solving DEs with Non-Constant Coefficients Resonance with Laplace Laplace with Periodic Functions 1 Solving Equations with Non-Constant
More informationspring magnet Fig. 7.1 One end of the magnet hangs inside a coil of wire. The coil is connected in series with a resistor R.
1 A magnet is suspended vertically from a fixed point by means of a spring, as shown in Fig. 7.1. spring magnet coil R Fig. 7.1 One end of the magnet hangs inside a coil of wire. The coil is connected
More informationSimple Harmonic Motion
Pendula Simple Harmonic Motion diff. eq. d 2 y dt 2 =!Ky 1. Know frequency (& period) immediately from diff. eq.! = K 2. Initial conditions: they will be of 2 kinds A. at rest initially y(0) = y o v y
More information4.5. Applications of Trigonometry to Waves. Introduction. Prerequisites. Learning Outcomes
Applications of Trigonometry to Waves 4.5 Introduction Waves and vibrations occur in many contexts. The water waves on the sea and the vibrations of a stringed musical instrument are just two everyday
More informationOscillatory Motion SHM
Chapter 15 Oscillatory Motion SHM Dr. Armen Kocharian Periodic Motion Periodic motion is motion of an object that regularly repeats The object returns to a given position after a fixed time interval A
More informationChapter 14 Oscillations. Copyright 2009 Pearson Education, Inc.
Chapter 14 Oscillations Oscillations of a Spring Simple Harmonic Motion Energy in the Simple Harmonic Oscillator Simple Harmonic Motion Related to Uniform Circular Motion The Simple Pendulum The Physical
More informationChapter 14 Oscillations. Copyright 2009 Pearson Education, Inc.
Chapter 14 Oscillations 14-1 Oscillations of a Spring If an object vibrates or oscillates back and forth over the same path, each cycle taking the same amount of time, the motion is called periodic. The
More informationSinusoids. Amplitude and Magnitude. Phase and Period. CMPT 889: Lecture 2 Sinusoids, Complex Exponentials, Spectrum Representation
Sinusoids CMPT 889: Lecture Sinusoids, Complex Exponentials, Spectrum Representation Tamara Smyth, tamaras@cs.sfu.ca School of Computing Science, Simon Fraser University September 6, 005 Sinusoids are
More informationCMPT 889: Lecture 2 Sinusoids, Complex Exponentials, Spectrum Representation
CMPT 889: Lecture 2 Sinusoids, Complex Exponentials, Spectrum Representation Tamara Smyth, tamaras@cs.sfu.ca School of Computing Science, Simon Fraser University September 26, 2005 1 Sinusoids Sinusoids
More informationChapter 14 Oscillations
Chapter 14 Oscillations If an object vibrates or oscillates back and forth over the same path, each cycle taking the same amount of time, the motion is called periodic. The mass and spring system is a
More informationLinear Second Order ODEs
Chapter 3 Linear Second Order ODEs In this chapter we study ODEs of the form (3.1) y + p(t)y + q(t)y = f(t), where p, q, and f are given functions. Since there are two derivatives, we might expect that
More information3! + 4! + Binomial series: if α is a nonnegative integer, the series terminates. Otherwise, the series converges if x < 1 but diverges if x > 1.
Page 1 Name: ID: Section: This exam has 16 questions: 14 multiple choice questions worth 5 points each. hand graded questions worth 15 points each. Important: No graphing calculators! Any non-graphing
More informationChapter 15 Oscillations
Chapter 15 Oscillations Summary Simple harmonic motion Hook s Law Energy F = kx Pendulums: Simple. Physical, Meter stick Simple Picture of an Oscillation x Frictionless surface F = -kx x SHM in vertical
More informationMATH 251 Week 6 Not collected, however you are encouraged to approach all problems to prepare for exam
MATH 51 Week 6 Not collected, however you are encouraged to approach all problems to prepare for exam A collection of previous exams could be found at the coordinator s web: http://www.math.psu.edu/tseng/class/m51samples.html
More informationMath 211. Substitute Lecture. November 20, 2000
1 Math 211 Substitute Lecture November 20, 2000 2 Solutions to y + py + qy =0. Look for exponential solutions y(t) =e λt. Characteristic equation: λ 2 + pλ + q =0. Characteristic polynomial: λ 2 + pλ +
More informationOrdinary Differential Equations
Ordinary Differential Equations for Engineers and Scientists Gregg Waterman Oregon Institute of Technology c 2017 Gregg Waterman This work is licensed under the Creative Commons Attribution 4.0 International
More informationLecture XXVI. Morris Swartz Dept. of Physics and Astronomy Johns Hopkins University November 5, 2003
Lecture XXVI Morris Swartz Dept. of Physics and Astronomy Johns Hopins University morris@jhu.edu November 5, 2003 Lecture XXVI: Oscillations Oscillations are periodic motions. There are many examples of
More informationBSc/MSci MidTerm Test
BSc/MSci MidTerm Test PHY-217 Vibrations and Waves Time Allowed: 40 minutes Date: 18 th Nov, 2011 Time: 9:10-9:50 Instructions: Answer ALL questions in section A. Answer ONLY ONE questions from section
More informationSection 4.3 version November 3, 2011 at 23:18 Exercises 2. The equation to be solved is
Section 4.3 version November 3, 011 at 3:18 Exercises. The equation to be solved is y (t)+y (t)+6y(t) = 0. The characteristic equation is λ +λ+6 = 0. The solutions are λ 1 = and λ = 3. Therefore, y 1 (t)
More informationDouble Spring Harmonic Oscillator Lab
Dylan Humenik and Benjamin Daily Double Spring Harmonic Oscillator Lab Objectives: -Experimentally determine harmonic equations for a double spring system using various methods Part 1 Determining k of
More informationPhysics 207 Lecture 25. Lecture 25. HW11, Due Tuesday, May 6 th For Thursday, read through all of Chapter 18. Angular Momentum Exercise
Lecture 5 Today Review: Exam covers Chapters 14-17 17 plus angular momentum, rolling motion & torque Assignment HW11, Due Tuesday, May 6 th For Thursday, read through all of Chapter 18 Physics 07: Lecture
More informationUndetermined Coefficents, Resonance, Applications
Undetermined Coefficents, Resonance, Applications An Undetermined Coefficients Illustration Phase-amplitude conversion I Phase-amplitude conversion II Cafe door Pet door Cafe Door Model Pet Door Model
More informationMath 216 Second Midterm 20 March, 2017
Math 216 Second Midterm 20 March, 2017 This sample exam is provided to serve as one component of your studying for this exam in this course. Please note that it is not guaranteed to cover the material
More information1. (10 points) Find the general solution to the following second-order differential equation:
Math 307A, Winter 014 Midterm Solutions Page 1 of 8 1. (10 points) Find the general solution to the following second-order differential equation: 4y 1y + 9y = 9t. To find the general solution to this nonhomogeneous
More informationEDEXCEL NATIONAL CERTIFICATE UNIT 28 FURTHER MATHEMATICS FOR TECHNICIANS OUTCOME 3 TUTORIAL 1 - TRIGONOMETRICAL GRAPHS
EDEXCEL NATIONAL CERTIFICATE UNIT 28 FURTHER MATHEMATICS FOR TECHNICIANS OUTCOME 3 TUTORIAL 1 - TRIGONOMETRICAL GRAPHS CONTENTS 3 Be able to understand how to manipulate trigonometric expressions and apply
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall 2007
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering.4 Dynamics and Control II Fall 7 Problem Set #9 Solution Posted: Sunday, Dec., 7. The.4 Tower system. The system parameters are
More informationUndamped Free Vibrations (Simple Harmonic Motion; SHM also called Simple Harmonic Oscillator)
Section 3. 7 Mass-Spring Systems (no damping) Key Terms/ Ideas: Hooke s Law of Springs Undamped Free Vibrations (Simple Harmonic Motion; SHM also called Simple Harmonic Oscillator) Amplitude Natural Frequency
More informationChapter 14 (Oscillations) Key concept: Downloaded from
Chapter 14 (Oscillations) Multiple Choice Questions Single Correct Answer Type Q1. The displacement of a particle is represented by the equation. The motion of the particle is (a) simple harmonic with
More informationApplications of Linear Higher-Order DEs
Week #4 : Applications of Linear Higher-Order DEs Goals: Solving Homogeneous DEs with Constant Coefficients - Second and Higher Order Applications - Damped Spring/Mass system Applications - Pendulum 1
More informationDifferential Equations
Differential Equations A differential equation (DE) is an equation which involves an unknown function f (x) as well as some of its derivatives. To solve a differential equation means to find the unknown
More informationSTRUCTURAL DYNAMICS BASICS:
BASICS: STRUCTURAL DYNAMICS Real-life structures are subjected to loads which vary with time Except self weight of the structure, all other loads vary with time In many cases, this variation of the load
More informationMath 4B Notes. Written by Victoria Kala SH 6432u Office Hours: T 12:45 1:45pm Last updated 7/24/2016
Math 4B Notes Written by Victoria Kala vtkala@math.ucsb.edu SH 6432u Office Hours: T 2:45 :45pm Last updated 7/24/206 Classification of Differential Equations The order of a differential equation is the
More informationODE Math 3331 (Summer 2014) June 16, 2014
Page 1 of 12 Please go to the next page... Sample Midterm 1 ODE Math 3331 (Summer 2014) June 16, 2014 50 points 1. Find the solution of the following initial-value problem 1. Solution (S.O.V) dt = ty2,
More informationSECTION When each term in the differential equation is multiplied by this factor, the result is
CHAPTER 4 SECTION 4 99 APPLICATIONS OF LINEAR ALGEBRA In this chapter we explore various mathematical problems, solutions of which can be simplified with what we have learned about linear algebra 4 Linear,
More informationChap. 15: Simple Harmonic Motion
Chap. 15: Simple Harmonic Motion Announcements: CAPA is due next Tuesday and next Friday. Web page: http://www.colorado.edu/physics/phys1110/phys1110_sp12/ Examples of periodic motion vibrating guitar
More information1. [30] Y&F a) Assuming a small angle displacement θ max < 0.1 rad, the period is very nearly
PH1140 D09 Homework 3 Solution 1. [30] Y&F 13.48. a) Assuming a small angle displacement θ max < 0.1 rad, the period is very nearly T = π L =.84 s. g b) For the displacement θ max = 30 = 0.54 rad we use
More informationOSCILLATIONS ABOUT EQUILIBRIUM
OSCILLATIONS ABOUT EQUILIBRIUM Chapter 13 Units of Chapter 13 Periodic Motion Simple Harmonic Motion Connections between Uniform Circular Motion and Simple Harmonic Motion The Period of a Mass on a Spring
More informationM A : Ordinary Differential Equations
M A 2 0 5 1: Ordinary Differential Equations Essential Class Notes & Graphics D 19 * 2018-2019 Sections D07 D11 & D14 1 1. INTRODUCTION CLASS 1 ODE: Course s Overarching Functions An introduction to the
More informationMATHEMATICS FOR ENGINEERING TRIGONOMETRY TUTORIAL 3 PERIODIC FUNCTIONS
MATHEMATICS FOR ENGINEERING TRIGONOMETRY TUTORIAL 3 PERIODIC FUNCTIONS This is the one of a series of basic tutorials in mathematics aimed at beginners or anyone wanting to refresh themselves on fundamentals.
More informationChapter 15. Oscillatory Motion
Chapter 15 Oscillatory Motion Part 2 Oscillations and Mechanical Waves Periodic motion is the repeating motion of an object in which it continues to return to a given position after a fixed time interval.
More informationSection 5.4 (Systems of Linear Differential Equation); 9.5 Eigenvalues and Eigenvectors, cont d
Section 5.4 (Systems of Linear Differential Equation); 9.5 Eigenvalues and Eigenvectors, cont d July 6, 2009 Today s Session Today s Session A Summary of This Session: Today s Session A Summary of This
More informationFeedback Control Systems
ME Homework #0 Feedback Control Systems Last Updated November 06 Text problem 67 (Revised Chapter 6 Homework Problems- attached) 65 Chapter 6 Homework Problems 65 Transient Response of a Second Order Model
More informationInductance, Inductors, RL Circuits & RC Circuits, LC, and RLC Circuits
Inductance, Inductors, RL Circuits & RC Circuits, LC, and RLC Circuits Self-inductance A time-varying current in a circuit produces an induced emf opposing the emf that initially set up the timevarying
More informationIntroduction to Vibration. Mike Brennan UNESP, Ilha Solteira São Paulo Brazil
Introduction to Vibration Mike Brennan UNESP, Ilha Solteira São Paulo Brazil Vibration Most vibrations are undesirable, but there are many instances where vibrations are useful Ultrasonic (very high
More informationM A : Ordinary Differential Equations
M A 2 0 5 1: Ordinary Differential Equations Essential Class Notes & Graphics C 17 * Sections C11-C18, C20 2016-2017 1 Required Background 1. INTRODUCTION CLASS 1 The definition of the derivative, Derivative
More information