Reference no: EM132366949
Problem Set -
Background: In an introductory study of differential equations second order ODEs having constant coefficients are explored quite extensively. One reason for this is the wide range of applications that occur where harmonic motion is found. For instance, throughout the applied sciences instances arise where mechanical effects of vibration become very important. This may involve the design of a bridge or some other structure or it may involve the study of matter itself where oscillations of molecules and atoms need to be examined, say, to predict the effects of heat. In such instances often the first model used to examine the phenomena is a second order ODE having constant coefficients.
The Problem: Imagine that you are part of a team that is developing an advanced printed circuit board for use in computer video signal processing. Certain electronic packages containing multitudes of transistors, diodes, and capacitors must be mounted to a board where they electrically interface with other electronic packages via metallic tracks and solder connections. Should the solder fracture, or just develop hairline cracks while in use, the electrical conductivity could be interrupted either intermittently, or catastrophically. This could cause all kind of adverse video signal effects. Imagine surgeons depending on such a printed circuit board as the source of its video image bank during surgery. Our goal is to examine several models for such products. As such the electronic packages may be thought of as isolated masses. The materials attaching them to the board (epoxy and the like) can be modeled as a spring having a spring constant k. The differential equation for such a system is:
m d2y/dt2 + ky = 0, y(t0) = y0, y'(t0) = y'0
As a designer you have some control over m and k.
A more general second order linear equation with constant coefficients and initial conditions is given by
ay'' + by' + c = g(t), y(t0) = y0, y'(t0) = y'0
Physically the initial conditions describe the "initial position" (zero if the mass is at the equilibrium position, positive if it is below the equilibrium position, and negative if it is above the equilibrium position) and the "initial velocity" (0 if the mas is at rest, positive if it is directed downward, and negative if it is directed upward).
To solve such a differential equation, we use dsolve: >>dsolve('a*D2y+b*Dy+cy=g(t)', 'y(t0) = y0', 'Dy(t0) = y0prime')
Problem 1: Consider the IVP below
2y'' + 8y = 0, y(0) = 5, y'(0) = 0.
Solve using MATLAB and obtain a graph of y(t) on the interval [0, 4π].
From an engineering perspective, what units of measure might 2 and 8 represent in the given ODE? What might be typical units for t? What are the amplitude, the period, and the frequency of y(t)?
Problem 2: (a) Use MATLAB to obtain solutions to the initial value problems given below. Describe what is changing in the physical system and what is changing in the system response. Also, plot all solutions on a single graph.
4y'' + 8y = 0, y(0) = 5, y'(0) = 0
6y'' + 8y = 0, y(0) = 5, y'(0) = 0
8y'' + 8y = 0, y(0) = 5, y'(0) = 0
What is changing in the physical system and what is changing in the system response?
(b) How would you expect the graph of y(t) to appear for the following?
12y'' + 8y = 0, y(0) = 5, y'(0) = 0
Graph the solution for the ODE. It should confirm your expectation.
Problem 3: Consider again the system
2y'' + 8y = 0, y(0) = 5, y'(0) = 0
Imagine that the spring coefficient 8 increases to 14 in steps of three, namely, 8, 11, 14. What does this mean physically in terms of the overall mass spring system?
On a single graph plot solutions for each of the steps, i.e., for the three cases where k is 8, then 11, and then 14. How do you expect solution graphs to look for the case where k is 17?
Graph it and see.
Problem 4: To study the effect of varying initial conditions on
2y'' + 8y = 0,
use the same graph to plot all solutions subject to the following initial conditions.
y(0) = 5, y'(0) = 10
y(0) = 5, y'(0) = -10
y(0) = 5, y'(0) = 4
y(0) = -10, y'(0) = 0
y(0) = 5, y'(0) = 10
Describe the physical meaning of the conditions and explain the effects that they have on the given system.
How would you expect the solution for the system
2y'' + 8y = 0, y(0) = 4, y'(0) = 2
to compare with the solution for the system
2y'' + 8y = 0, y(0) = 4, y'(0) = -10?
Give your answer, and then plot the two systems on the same graph. The graph should support your conjecture.
Problem 5: Another factor to consider in investigating vibration of electronic components on circuit cards is the effect of damping. Engineering design often has some control over this as well. Consider the following ODE and the given initial values.
2y'' + 0.5y' + 8y = 0, y(0) = 5, y'(0) = 0
Solve the system. Describe the result.
Next, suppose that the damping coefficients are modified. Besides including 0.5 it also includes 1.0, 1.5, and 2.0. Solve for these four cases and plot the solutions on a single graph. Explain below how you would expect the solution of the system or ODE given below to look. Then check it with a graph.
2y'' + 0.5y' + 8y = 0, y(0) = 5, y'(0) = -10
Problem 6: For shock and vibration testing various profiles need to be evaluated. These may be low frequency, or mid-range frequency vibrations depending on the specifications of the customer or the design team. Amplitude and time duration of these are also variables that engineering teams must address. Even thermal considerations can play a role as material properties are temperature sensitive.
To address these needs the physical system is subjected to a forcing function. The differential equation for this model may take the following form.
y'' + p(t)y' q(t) = g(t) ≠ 0
Notice that such an ODE is non-homogeneous.
As an example suppose that a circuit board is modeled as follows.
2y'' + 0.5y' + 8y = 4 sin t, y(0) = 5, y'(0) = 0
What is the frequency of the function that drives the vibration?
Solve the system with MATLAB. Describe the solution and graph the solution.
Next plot solutions for each of the following on the same graph.
2y'' + 0.5y' + 8y = 4 sin t, y(0) = 5, y'(0) = 0
2y'' + 0.5y' + 8y = 16 sin t, y(0) = 5, y'(0) = 0
2y'' + 0.5y' + 8y = 4 sin 2t, y(0) = 5, y'(0) = 0
Compare and contrast both transient (short term) and steady-state (longer term) responses.