#$&*
course Mth 279
7:58 pm 7/13I started seeing a tutor and I feel much better about the material already.
Query 12 Differential Equations
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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:
y'' + omega^2* y = 0
Lambda^2 + omega^2* lambda = 0
Lambda = sqrt(omega^2) * I
Lambda = omega*I
Where lambda_1 = -omega *i, lambda_1 = omega*i
y_1 = e^(0*t)cos(omega*t)
y_2 = e^(0*t)sin(omega*t)
y = Asin(omega*t) + B cos(omega*t)
I don't really know how the above turns into the below equation
y(t) = y_0'/omega * sin(omega*t) + y_0 cos(omega*t)
That's what the book has
The block starts at equilibrium
y(0) = 0
0 = y_0'/omega sin(0) + y_0 cos(0)
y_0 = 0
y(t) = y_0'/omega * sin(omega*t)
To find the velocity, I need to solve for y_0'
When the cylinder is at max displacement, it has gone through 1/4 of it's oscillation. 1/4 of 2pi = pi/2.
But to use the time, I need to solve for how far under water it goes.
Y = rho/rho_l * L
Y = 700/1000*1 = 70 cm
If 70 cm is already in the water, then that means the change in y = 30
30 = y_0'/ omega * sin(pi/2)
y_0' = 30 *omega
Omega = sqrt(rho_l *g/(rho*L) = sqrt(1000*9.8/(700*1)) = sqrt(14)
y_0' = 30* sqrt(14)
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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:
1. Because p(t), q(t) and g(t) must be continuous on the interval, tan(t) needs to be evaluated for where it does not exist. Which is any multiple of pi/2
The points that need to be considered are:
Multiples of Pi/2 and pi
The largest interval is (pi/2, 2)
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The tangent is undefined at pi / 2, but it is defined (with value 0) at pi.
Between 0 and 2 pi, the tangent is undefined at pi/2 and 3 pi / 2. It is also undefined at any angle which differs from either of these by an integer multiple of 2 pi.
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2. the equation cannot exist where t = 3, -3 but must include 1
The largest interval is (-3, 3)
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Which of these intervals contains pi? It is within that interval that we are guaranteed of a solution.
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Given Solution:
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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- sin(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:
***I changed your first equation to match what was in the book. Originally you had the same question twice
1. y'' + 1 = 2-0
y''(0) = 3
Because the first derivative is negative, that means the equation is decreasing. Because the second derivative is positive, the equation is concave up
2. y'' + 1 = 0
y'' = -1
The equation is decreasing because the derivative is negative and it is concave down because the second derivative is negative
3. y'' - 1 = 0
y'' = 1
Increasing because derivative is positive, concave up because second derivative is positive
4. y'' - 1 = -2 (1)
y'' = -1
Increasing because derivative is positive, concave down because the second derivative is negative
confidence rating #$&*:
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Given Solution:
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&#Good responses. See my notes and let me know if you have questions. &#