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Main points of this exam paper are: Piecewise Continuous Function, Laplace Transform, Ned, Critically Damped, Spring Mass System, Oscillate, Rest Position, Non Homogeneous Equation, Extra Multiplicative Factor, Associated Homogeneous
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Midterm Exam II
April 1, 2004
Name: Student Number: Instructor: Section:
This exam has 9 questions for a total of 100 points.
In order to obtain full credit for partial credit problems, all work must be shown. Credit will not be given for an answer not supported by work.
The last page of this examination contains a table of Laplace transforms for your use.
Do not write in this box.
Total:
(a) (2 points) True or False: The Laplace transform of a piecewise continuous function f (t) is defined by:
L{f (t)} =
0
e−stf (t)dt
(b) (2 points) True or False: A critically damped spring-mass system cannot oscillate, but the mass may cross its rest position as t → +∞.
(c) (2 points) True or False: When choosing the form for a particular solution to a second or- der linear non-homogeneous equation, an extra multiplicative factor of t is always needed when the characteristic equation of the associated homogeneous problem has complex roots.
(d) (2 points) True or False: The Laplace transform of a product of two functions is obtained by taking the product of the Laplace transforms of each function, i.e. L{f (t)g(t)} = L{f (t)}L{g(t)}.
(a) (2 points) Give a value of ω at which resonance will occur.
(b) (2 points) If ω = 4.5, what is the beat frequency (i.e., the frequency of observing successive maxima or successive minima in the solution u(t))?
(b) (5 points) For m = 1 kg and an initial displacement of 10 cm (i.e., 101 m), set up and solve an initial value problem for the equation of motion if the mass is simply released at t = 0.
(c) (2 points) What is the quasi-frequency of the motion you determined in (b)?
(d) (2 points) If the same 1 kg mass were attached to a weaker spring with Hooke’s constant 1 3 N/m, would the system be overdamped, underdamped, or critically damped?
y′′^ − 2 y′^ − 15 y = e^2 t^ + 2δ(t − 1) ; y(0) = 0, y′(0) = 0.
f (t) =
2 t 0 ≤ t < 2; t^2 t ≥ 2.
(a) (8 points) (s + 4) s^2 − 4 s + 13
(b) (8 points) e−^5 s s^2 (s − 2)
