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course mth 279
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Question: A cylinder floating vertically in the water has radius 50 cm, height 100 cm and uniform density 700 kg / m^3. It is raised 10 cm from its equilibrium position and released. Set up and solve a differential which describes its position as a function of clock time.
The same cylinder, originally stationary in its equilibrium position, is struck from above, hard enough to cause its top to come just to but not below the level of the water. Solve the differential equation for its motion with this condition, and use a particular solution to determine its velocity just after being struck.
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Your solution:
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According to my book the equation appears to be:
y"" + (omega)^2 * y = 0
omega = (liquid density * gravity)/(cylinder density * cylinder height)
General solution:
y(t) = a*sin(omega*t) + b*cos(omega * t)
We are told in the problem that y(0) = 10
I don't see that we are given a y'(0) so I am not sure how to finish the problem. Particular solutions aren't mentioned in this section of the book.
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We will assume the upward direction to be positive.
The net force on the system when its displacement from equilibrium is y is equal to the change in the buoyant force. If the cylinder rises above equilibrium the buoyant force is less than that at equilibrium and the net force is negative; at positions below equilibrium the buoyant force is greater than that at equilibrium and the net force is positive.
At position y relative to equilibrium the buoyant force differs from that at equilibrium by and amount equal in magnitude to the difference in the weights of displaced water. The difference in the volumes of displaced water is
volume difference = - cross-sectional area * position relative to equilibrium = - pi r^2 y,
so the difference in the mass of displaced water is - rho r^2 y and the net force is the product of this difference in mass and the acceleration of gravity:
F_net = - rho_water * pi r^2 y * g.
The mass of the entire cylinder is
mass = rho_cylinder * pi r^2 h,
where h is the altitude of the cylinder.
Since F_net = mass * y '' we have
rho_cyl * pi r^2 h * y '' = - rho_water * pi r^2 y * g
which we simplify to obtain
y '' = -rho_water * g / (h * rho_cyl) * y.
This is of the form y '' = - c * y with c = -rho_water * g / (h * rho_cyl) = -1000 kg / m^3 * 9.8 m/s^2 / (1.00 m * 700 kg / m^3) = -14 s^-2, approx..
The mass is that of the cylinder
Solving the equation
y '' = - c y
we obtain
y = B cos(sqrt(c) * t) + C sin(sqrt(c) * t),
which can be expressed in the form
y = A cos(sqrt(c) * t + phi).
A and phi are arbitrary constants, while sqrt(c) = sqrt( 14 s^-2) = 3.7 s^-1., approx..
If the cylinder is raised 10 cm and released then y(0) = .10 meters and y ' (0) = 0 so that
y(0) = A cos( sqrt(c) * 0 + phi) = .10 m and
y ' (0) = -sqrt(c) A sin(sqrt(c) * 0 + phi) = 0.
Thus
A cos(phi) = .10 m and
A sin(phi) = 0.
The latter indicates that phi = 0 or pi radians. The former dictates that phi = 0 radians rather than pi radians, with A = .10 m.
Thus our solution is
y(t) = 10 cm * cos(sqrt(c) * t ) or approximately
y(t) = 10 cm * cos( 3.7 t ).
The cylinder is easily seen to float in equilibrium when 70 cm is below and 30 cm above the water line.
If the cylinder is struck from above at equilibrium and comes just to and not below the surface of the water, then we know that its maximum displacement from equilibrium has magnitude 30 cm. So we can let A = 30 cm.
The velocity of the cylinder at the initial instant is negative and its position is zero, so
y(0) = A cos(sqrt(c) * 0 + phi) = 0
and
y ' (0) < 0.
The first condition requires that cos(phi ) = 0, so that phi = pi/2 or 3 pi / 2.
The second condition requires that sin(phi) < 0, so that phi = 3 pi / 2.
The solution is thus
y(t) = 30 cm * cos(sqrt(c) * t + 3 pi / 2) or approximately
y(t) = 30 cm * cos( 3.7 t + 3 pi / 2).
The velocity function wasn't requects, but it is
v(t) = y ' (t) = -sqrt(c) * 30 cm sin (sqrt(c) t + 3 pi / 2) = -110 cm/s * sin(3.7 t + 3 pi/2).
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Given Solution:
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Question:
For each of the equations and initial conditions below, find the largest t interval in which a solution is known to exist:
• y '' + y ' + 3 t y = tan(t), y(pi) = 1, y ' (pi) = -1
• t y '' + sin(2 t) / (t^2 - 9) y ' + 2 y = 0, y(1) = 0, y ' (1) = 1.
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Your solution:
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y '' + y ' + 3 t y = tan(t)
tan(t) is undefined at -pi/2, pi/2, 3*pi/2, etc
largest solution that includes pi is (pi/2, 3*pi/2)
t y '' + sin(2 t) / (t^2 - 9) y ' + 2 y = 0
y"" + sin(2t)/(t(t^2 - 9) y' + 2y/t = 0
t cannot = 0, -3, 3
Largest interval that includes 1 is (-3, 3)
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Confidence rating:
****
ok #$&*
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Given Solution:
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Self-critique (if necessary):
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ok
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Question: Decide whether the solution of each of the following equations is increasing or decreasing, and whether each is concave up or concave down in the vicinity of the given initial point:
• y '' + y = - 2 t, y(0) = 1, y ' (0) = -1
• y '' + y = - 2 t, y(0) = 1, y ' (0) = -1
• y '' - y = t^2, y(0) = 1, y ' (0) = 1
• y '' - y = - 2 cos(t), y (0) = 1, y ' (0) = 1.
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Your solution:
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y '' + y = - 2 t
Decreasing because first derivative y'(0) is negative
Concave down because y"" is negative when you plug in y(0) = 1. I am not sure if I am doing this right.
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So far, so good.
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y '' - y = t^2
Decreasing because first derivative y'(0) is negative
Concave up because y"" is positive
y '' - y = - 2 cos(t)
Increasing because first derivative y'(0) is positive
Concave down because y"" is negative when you plug in y(0) = 1
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Confidence rating: ****
I used the sign of the given y' to determine increasing and decreasing. I plugged in the y value to see if the y"" was concave of not.
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Given Solution:
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&#Good responses. See my notes and let me know if you have questions. &#