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Initial Value Problems

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Initial Value Problems. MATH 224. General Linear Spring Model. General Model for Spring/Mass system ... Initial value problems (IVP) DE one initial condition ... – PowerPoint PPT presentation

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Title: Initial Value Problems


1
Initial Value Problems
  • MATH 224

2
General Linear Spring Model
  • General Model for Spring/Mass system

3
Particular System
  • k 90
  • m 0.5
  • Assume c 0 (no drag/friction)

4
Review of Solution
  • Solve auxiliary equation
  • Use m values to construct fundamental solution
    set
  • Construct general solution

5
What does it mean?
  • From general solution, we can tell that

6
Particular solutions
  • Two unknown constants in general sol'n
  • Determined by either
  • initial conditions or
  • boundary conditions
  • Determine the predicted behaviour in one
    particular scenario

7
Initial value problems (IVP)
  • DE one initial condition per unknown constant
  • e.g. initial position and velocity
  • Let y(0) -2, and v(0) -0.5

8
Finding particular solution
  • Use general solution, with given initial values
  • Set up one equation for each condition
  • Solve equations for c1, c2

9
Continued
10
Interpreting Particular Solution
  • Solution is a prediction of behaviour
  • We've got y , so we're predicting position of
    mass (y) over time (t) based on DE rule, starting
    at initial conditions
  • Particular solution, based on initial conditions,
    is

11
Graphing analytic solution in MATLAB
  • Graphing just like any other function

t linspace(0, 2, 1000) w sqrt(90/0.5) c1
-0.02 c2 -0.5/w y c1 cos(w t) c2
sin(w t) plot(t, y)
  • Graph shows predicted position (y) over time (t)
  • check satisfies given intial conditions?

12
Cantilevered beam
  • Diving board
  • Cantilevered structure
  • Tip of beam behaves like spring/mass system
  • to a first approximation!

13
Modelling
  • Finding parameters can be done analytically or
    experimentally
  • we'll use k 100, m 0.04, c 0.03
  • DE is

14
General Solution
  • Auxiliary equation is
  • Gives roots of
  • A fundamental solution set is
  • General solution is

15
Comments on General Solution
  • Basic structure
  • Periodicity

16
Initial Conditions
  • Take simple initial conditions
  • y(0) - 0.02 m , v(0) 0 m/s
  • Set up initial condition equations

17
Particular Solution
  • Solve for c1, c2
  • MATLAB, or by hand

18
MATLAB Assistance and Plot
Cantilevered beam example m roots(.04 .03
100) a real(m(1)) b imag(m(1))
initial conditions, using on-paper work c1
-0.02 c2 -c1 a / b t linspace(0, 10,
1000) y c1 exp(at) .cos(bt) ... c2
exp(at) . sin(bt) plot(t, y) xlabel('Time
(seconds)') ylabel('Height of beam tip
(m)') title('Motion of beam after release')
19
Analytic solutions for IVPs
  • Find general solution to DE
  • ignore initial conditions
  • general solution applies to any starting point
    for the system
  • Use initial values and general solution to find
    values of undetermined constants
  • set up system of n equations for c1, cn

20
Just for fun
  • There are other bending modes for cantilevered
    beams, rather than simple sping model
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