#$&*
course MTH 279
3/23 730
Query 18 Differential Equations*********************************************
Question: A 10 kg mass stretches a spring 30 mm beyond its unloaded position. The spring is pulled down to a position 70 mm below its unloaded position and released.
Write and solve the differential equation for its motion.
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Your solution:
DE: y''+ (k/m)y = 0
Has roots +/-sqrt(k/m) * i
which gives the general solution : y = Acos[sqrt(k/m) t] + Bsin[sqrt(k/m) t]
Using the given information we can find k = 3266 (approx) using Newton's 2nd law.
Imposing initinal conditions y(0) = .07 and y'(0) = 0, we find A = .07 and B = 0
FINAL EQUATION: y = .07cos(18t) using standard units of length and mass.
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The spring is pulled to .07 m below the unloaded position, which is .04 m below its equilibrium position when the weight is attached.
So the .07 in your solution should be .04 m.
Otherwise all is well.
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary): OK
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Self-critique rating: OK
Question: The graph below represents position vs. clock time for a mass m oscillating on a spring with force constant k. The position function is y = R cos(omega t - delta).
Find delta, omega and R.
Give the initial conditions on the y and y '.
Determine the mass and the force constant.
Describe how you would achieve these initial conditions, given the appropriate mass and spring in front of you.
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Your solution:
R = 3
delta = pi/4
omega = 3
y(0) = 3sqrt(2)/2
y'(0) = 9sqrt(2)/2
???? I didn't know how to recover k and m from the given information.
To achieve the initial conditions, one would want to displace the spring 3 units of measurement and release the spring . To get the delta, I would think of starting the clock at a time t = pi/4 s before the spring reaches its maximum displacement.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary): I was unsure how to recover k and m from the given information. I know that omega is related to these quantities (omega = sqrt(k/m), but that leaves me with one equation and 2 unknowns, and I didn't know what other equation to use. The other equation that came to mind was mg = -kdy, but I didn't find it very useful.
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The period of this function is 2, so omega = 2 pi / 2 = pi.
The amplitude is 3 so the function is of the form y = 3 cos(2 t - delta).
The graph appears to be shifted about 1/4 unit to the right of the graph of y = 3 cos(pi t); replacing t by t - pi/4 will accomplish the shift to the right. Our function is therefore
y = 3 cos( 2 ( t - pi/4) ) ) = 3 cos( 2 t - pi/2).
More accurately, 3 cos(delta) = 2 so delta = arcCos(2/3); which is approximately 0.84, a bit greater than pi/4.
Thus delta = pi/4 (or more accurately arcCos)2/3), omega = pi and R = 3.
I agree that there is not enough info to find k and m: The mass is m, and omega = sqrt(k / m) so k = m omega^2 = 4 s^-1 * m.
The initial condition on y is y(0) = 2.
y ' (0) = -6 sin( -pi/4) = 4.3, approx..
or more accurately
y ' (0) = -6 sin(-arcCos(2/3)) = -6 * sqrt(5) / 3 = -2 sqrt(5)..
*@
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Self-critique rating: OK
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Question: This problem isn't in your text, but is related to the first problem in this set, and to one of the problems that does appear in your text. Repeating the first problem:
'A 10 kg mass stretches a spring 30 mm beyond its unloaded position. The spring is pulled down to a position 70 mm below its unloaded position and released.
Write and solve the differential equation for its motion.'
Suppose that instead of simply releasing it from its 70 mm position, you deliver a sharp blow to get it moving at downward at 40 mm / second.
What initial conditions apply to this situation?
Apply the initial conditions to the general solution of the differential equation, and give the resulting function.
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Your solution:
y(0) = .07
y'(0) = .04
y = .07cos18t + .0022sin18t
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary): OK
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Self-critique rating: OK
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Question: The 32-pound weight at its equilibrium position in a critically-damped spring-and-dashpot system is given a 4 ft / sec downward initial velocity, and attains a maximum displacement of 6 inches from equilibrium. What are the values of the drag force constant gamma and the spring constant k?
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Your solution:
k = 170.7 lb/in
gamma = ????
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
I was confused as to how to turn the initial conditions into something useful that could help me solve for k and gamma. I used Newton's Second Law ro find k, but I'm not even sure if that is right, because I don't know how giving the system an initial velocity would affect the maximum amplitude of the system.
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Self-critique rating: 3
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You need to simply write the differential equation and its initial conditions, then follow the usual methods. Of course interpretation is a challenge.
The equation is
y '' + gamma / m * y ' + k / m * y = 0.
Our conditions are
y(0) = 0
y ' (0) = -4 ft / sec
y(t) = -1/2 ft when | y(t) | is maximized)
The characteristic equation leads to the solution. However the nature of the solution depends on the relative values of the given parameters.
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Question: The motion of a system is governed by the equation m y '' + gamma y ' + k y = 0, with y(0) = 0 and y ' (0) = v_0.
Give the solutions which correspond to the critically damped, overdamped and underdamped cases.
Show that as gamma approaches 2 sqrt(k * m) from above, the solution to the underdamped case approaches the solution to the critically damped case.
Show that as gamma approaches 2 sqrt(k * m) from below, the solution to the overdamped case approaches the solution to the critically damped case.
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Your solution:
critically damped: y = Ae^(lambda_1t) + Bte^(lambda_2t)
overdamped: y = Ae^(lambda_1t) + Be^(lambda_2t)
underdamped: y = e^(alphat)[Acos(beta*t) + Bsin(beta*t)]
In each case, solutions are found using the characterisitic equation to give the roots:
[-gamma +/- sqrt(gamma^2 -4mk) ] / 2m
Focusing in on the important part that tells us if we will have a critically, over-, or underdamped case is the sqrt(gamma^2 -4mk). We know that in a critically damped case, gamma^2 = 4mk, which makes our entire result 0. In an overdamped case, gamma^2 > 4km, and in an underdamped case, gamma^2 < 4km.
So, if we want an under- or overdamped system to mimic a critically damped system, we need gamma to approach 2sqrt(km). Doing this results in an arguement approaching 0 underneath the radical in the expression sqrt(gamma^2 -4mk), which is neccessary to model the behavor of a critically damped system.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
I didn't know exactly how to explain the behavior that was asked for.
@&
One case is oscillatory with decreasing amplitude.
One case approaches zero displacement without the displacement ever changing sign.
In one case the displacement can change sign once before approaching zero without another sign change.
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Self-critique rating: 3"
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#*&!
&#Good responses. See my notes and let me know if you have questions. &#