SCIE 001 MATHEMATICS ASSIGNMENT 4 (Due Oct. 15, 2012)
There are two parts to this assignment. The first part is on WeBWorK and is due at 11:59 pm on Sun. Oct.
14. The second part consists of the questions on this page and is due by 10 am on Mon. Oct. 15. For these
questions, you are expected to provide full solutions with complete arguments and justifications. You will be
graded primarily on the correctness, clarity and elegance of your solutions. Your answers must be typeset or
very neatly written. They must be stapled, with your name and student number at the top of each page.
1. Let m > 0 and k > 0 be given constants (m is the mass of a particle attached to a spring that has a
Hooke’s Law constant k), and consider the problem of finding the displacement x(t) that satisfies the
differential equation
k
d2 x
= − x, 0 < t < ∞,
dt2
m
and satisfies at t = 0 the two initial conditions
x(0) = x0 ,
x0 (0) = v0 .
The solution can be expressed as
x(t) = A cos(ωt + φ0 ),
0 ≤ t < ∞.
Suppose m = 2 kg and k = 50 N · m−1 . Find A, ω and φ0 (including their units) if the initial position
and velocity of the mass is:
(a)
(b)
(c)
(d)
x0
x0
x0
x0
= 3 m, and v0 = 1 m · s−1 .
= 3 m, and v0 = −1 m · s−1 .
= −3 m, and v0 = 1 m · s−1 .
= −3 m, and v0 = −1 m · s−1 .
2. Find all strictly positive values of the constant λ such that the boundary value problem
d2 u
= −λ u,
dx2
0 < x < π,
u(0) = 0,
u0 (π) = 0
has nontrivial solutions, and also give the corresponding solutions u(x).
3. For the satellite dish problem done in class, recall that the incident ray was the horizontal straight line
y = b and the reflected ray was the straight line y = b + mr (x − a), where (a, b) is a point on the curve C
in the xy-plane that represents the cross-section of the reflecting surface. The x-intercept of the reflected
ray was denoted x∗ (x∗ = a − (b/mr )).
(a) For a semicircular
curve C we have x2 + y 2 = R2 with −R < x < 0 and we found x∗ = R2 /(2a)(=
√
2
∗
−R /(2 R2 − b2 )), where R > 0 is a constant. Explain
√ why this formula for x is physically reasonable only for −R/2 < a < 0 (or equivalently 0 < b < 3R/2). What happens if −R < a < −R/2 (or
√
3R/2 < b < R)?
(b) For a parabolic curve C we have x − ky 2 = 0, where k > 0 is a constant. Show that in this case
the x-intercept of the reflected ray x∗ is independent of a (or b), so all reflected rays converge on the
same point x∗ on the x-axis: this is a good design.