Math 2250 Lab 9
Due Date: 3/26/2015
Name:
1. Archimedes’ principle states that a body partially or totally submerged in a liquid
is buoyed up by a force equal to the weight of the liquid displaced. The fact that the
body is resting or floating on water implies that the downward force due to the weight
W = mg of the body must be the same as the upward buoyant force of the water, namely
the weight of the water displaced. Call y = 0 the equilibrium position of the lower end
of a body when it is floating in a liquid. If the body is now depressed a distance y
from its equilibrium position, an additional upward force will act on the body due to
the additional liquid displaced. If the body’s cross-sectional area is A, then the volume
of additional liquid displaced is (Ay)ρ, where ρ is the weight per unit volume of liquid.
Since this additional upward force is the net force acting on the depressed body, we have,
by Newton’s second law of motion, with the positive direction downward
m
d2 y
= −ρAy.
dt2
A floating body therefore executes simple harmonic motion when depressed from its
equilibrium position.
With this in mind, suppose a body of mass m has a cross-sectional area A sq ft. It
is depressed y0 ft from its equilibrium position in a liquid whose weight per ft3 is ρ and
released.
(a) Show details for the following result: The model is
d2 y
m 2 + ρAy = 0,
dt
(1)
y(0) = y0 , y 0 (0) = 0,
p
with unique solution y(t) = y0 cos ( ρA/mt).
(b) Assume the body is a right circular cylinder with vertical axis (i.e. along the y-axis)
and radius r, show that the period T is given by
r
2 mπ
T =
.
r
ρ
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2. Consider the following series RLC circuit:
R
V
L
C
Suppose a periodic voltage source v(t) = V0 cos(ωt) is applied to the circuit. Using
Kirchoff’s laws, the following differential equation gives the charge Q(t) of the capacitor.
L · Q00 (t) + R · Q0 (t) +
1
· Q(t) = V0 cos(ωt)
C
Take L = 1 V · s · A−1 , R = 0 Ω, C −1 = 0.25 V · C−1 , and V0 = 4 V. Note: the left sides
of these equations involve variables, while the right sides involve units.
Assume the following initial conditions for Q(t):
Q(0) = 0
Q0 (0) = 0
(a) Angular frequency is the coefficient ω in a trigonometry term cos(ωt), cos(ωt − α),
sin(ωt), sin(ωt − α). This is different than the (usual) frequency used in physics
and egineering courses. What is the natural angular frequency ω0 of this system?
Hint: ω0 is the angular frequency of the solutions to the unforced equation
L · Q00 (t) + R · Q0 (t) +
1
· Q(t) = 0
C
(b) Assume that ω 6= ω0 . Use the method of undetermined coefficients to solve for
the particular solution QP (t). Then use Q(t) = QP (t) + QH (t) to solve the initial
value problem for Q(t), where QH (t) is a solution to the homogeneous (unforced)
equation. You may wish to use a computer to check that this solution is correct.
(c) Write down the specific case of the solution Q(t) (found in part b) for ω = 0.6.
Compute the period of this solution, which is a superposition of two cosine functions.
Use Matlab, Maple, or other technology to graph one period of the solution.
(d) Now let ω = ω0 . Use the method of undetermined coefficients to solve for a new
particular solution QP (t). Then use Q(t) = QP (t) + QH (t) to solve the initial value
problem for Q(t). Graph this solution over the interval 0 ≤ t ≤ 50 seconds. What
phenomenon does this solution exhibit?
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