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ME2016A, Spring 2005, Dr. Ferri Name: _____ SOLUTION ________ Test 2, March 16
This is a closed-book, closed-notes test. Please show all your work to ensure that you receive full credit. There are 7 problems for a total of 40 points.
Honor Pledge: On my honor, I pledge that I have neither given nor received any inappropriate aid in the preparation of this test.
_______________________________
Signature
Problem 1 (5) ___ 5 _____
Problem 2 ( 5) ___ 5 ______
Problem 3 (6) ___ 6 _____
Problem 4 (2) ___ 2 _____
Problem 5 (2) ___ 2 _____
Problem 6 (10) __ 10 ______
Problem 7 (10) ___ 10 _____
Total (40) ______ 40 ______
Problem 1. (5 points) Consider the function f(x) shown in the figure below with zeroes at x = 2 and x = 5. (a) Sketch 2 iterations of the Newton-Raphson algorithm starting from x0 = 0.
(b) To which root will the Newton Raphson method converge if x0 = 4?
f
1 2 x
NR, 2nd iteration
NR, 2nd iteration
Problem 2. (5 points) Consider the matrix A below and its inverse.
A and
A^1
(a) Calculate the infinity-norm of the matrix A: A (^) ∞= ____ 12 _______
(b) Calculate the infinity-norm of A-1: = ∞
A −^1 ___ 13/8 _____
(c) Based on your answer to parts (a) and (b), calculate cond(A,inf): ___ 39/2=19.5 ___
(d) If the elements of the B vector are perturbed by 1% ( ∆ B / B =0.01) and if the condition
number of A is 1000, what is the maximum value of ∆ x / x , for the system A*x = B?
B
B
cond A x
x = 10
Problem 6. (10 points) We seek the solution to the system of equations A*x = B, where B = [ 4 1 5 ]T^. The A matrix is defined in the Matlab workspace. The Matlab lu command returns the following result:
» [L,U,P]=lu(A) L = 1.0000 0 0 0 1.0000 0 0.5000 -0.5000 1. U = 2 1 0 0 1 0 0 0 2 P = 0 0 1 0 1 0 1 0 0
Solve for the solution x using forward and back-substitutions. Show your work.
LU = PA, therefore LUx = PB = [ 5 1 4 ]T^. Let d = Ux, then Ld = PB**
3
2
1
d
d
d L d. Solution, d1 = 5, d2 = 1, d3 = 4-d1/2+d2/2 = 2
3
2
1
x
x
x U x. Solution, x3 = 1, x2 = 1, x1 = (5-x2)/2 = 2
So, the solution is
x
Problem 7. (10 points) The slider-crank mechanism shown below is frequently used to convert rotary motion to reciprocating motion; eg., the piston-crankshaft systems in your car’s engine. Link 2 (the crank ) is driven at a steady angular velocity of ω2, causing the slider to translate left and right. The position analysis is much simpler than the 4-bar mechanism:
= −^2
3
(^12)
θ 3 sin r sin θ
r (7.1)
The velocity analysis is also simpler, involving the simultaneous solution of 2 equations:
v (^) B + ω 3 r 3 sinθ 3 =−ω 2 r 2 sin θ 2 (7.2)
ω 3 r 3 cosθ 3 = ω 2 r 2 cos θ 2 (7.3)
Write a Matlab function that takes as input r2, r3, and w2, and generates a plot of the angular velocity of link3 (w3) and the velocity of the slider (vB) vs the crank angle theta2. Your function must do the following:
(a) Increment theta2 from 0 to 360O^ in steps of 1O. (b) Find theta3 using (7.1). Note that the Matlab command for sin-1(x) is “asin(x)” (c) Find vB and w3 from equations (7.2) and (7.3), and store them in vectors VB and W (d) Plots vB and w3 vs theta2.
θ 2
r (^2) r (^3)
P
A
θ 3 B
slider
Solution: There are many possibilities. Note that the system of equations that yield vB and w are in a “triangular form.” Thus, no “inversion” is necessary.
function [VB,W3,T2] = slider(r2,r3,w2)
T2 = 0:pi/180:2pi; L2 = length(T2); for k = 1:L2; theta2 = T2(k); theta3 = asin(r2sin(theta2)/r3); w3 = w2r2cos(theta2)/(r3cos(theta3)); vb = -w2r2sin(theta2) - w3r3sin(theta3); vB(k) = vb; W3(k) = w3; end*
plot(T2,vB,T2,W3) xlabel('theta2 (rad)') legend('vB','w3')
