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course Mth 279
5/14
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:
m = 10kg
Y = 0.03m
y(0) = 0.07m
g = 9.81 m/s^2
my'' + (gamma)y' + ky = 0
Hook's law:
W = mg
W = km
(10kg)(9.81m/s^2) = k (0.03m)
k = 3270
10y'' + 0y' + 3270y = 0
y(0) = 0.07
y ' (0) = 0
y'' + 32.7y = 0
General solution:
y(t) = Acos(5.72t) + Bsin(5.72t)
where sqrt(32.7) = 5.72 approx, from the quadratic equation: (-0+-sqrt(0^2 - 4(32.7)))/2
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3270/10 = 327, so angular frequency is about 18 rad/s, not 5.7 rad/s.
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Applying the initial conditions:
y(0) = Acos(5.72*0) + Bsin(5.72*0) = 0.07
A(1) + B(0) = 0.07
A = 0.07
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The new equilibrium position of the system will be 30 mm below the unloaded position. When pulled down 70 mm the position of the system will therefore be -40 mm.
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y'(0) = -5.72Asin(5.72*0) + 5.72Bcos(5.72*0) = 0
-5.72A(0) + 5.72B(1) = 0
5.72B = 0
B = 0
y(t) = 0.07cos(5.72t)
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This would be
40 mm * cos( 18 t)
per my notes above.
Your procedure was good and your errors here were minor, though you want to make note of them.
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Given Solution:
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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:
y = R cos(omega t - delta)
From the graph:
R = 3
if T = 2, and T = 2pi/omega
omega = 2pi/2 = pi
y(0) = 3cos(pi * 0 - delta) = 2
3cos(-delta) = 2
3cos(delta) =2 due to cosine being an even funtion.
delta = cos^(-1)(2/3) = 0.27pi
y(0) = 2
y'(0) = 0
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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(pi 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 - 1/4 will accomplish the shift to the right. Our function is therefore
y = 3 cos( pi ( t - 1/4) ) ) = 3 cos( pi t - pi/4).
Thus delta = pi/4, omega = pi and R = 3.
**** not enough info: The mass is m, and omega = sqrt(k / m) so k = m omega^2 = 4 m. ****
The initial condition on y is y(0) = 2.
y ' (t) = (3 cos( pi t - pi/4)) ' = -2 pi sin( pi t - pi/4) so
y ' (0) = -3 pi sin( -pi/4) = 3 pi sqrt(2) / 2 = 6.5, approximately.
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Given Solution:
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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:
m = 10kg
Y = 0.03m
y(0) = 0.07m
g = 9.81m/s^2
W = mg
F = ma
k = 3270 by hook's law in previous problem.
40mm/s (1m/1000mm) = 0.04m/s
Assuming this is V_0
which is y'(0) = 0.04m/s
10y'' + 3270y = 0
y'' + 32.7y = 0
y(t) = Acos(5.72t) + Bsin(5.72t)
y(0) = Acos(0) + Bsin(0) = 0.07
A(1) + B(0) = 0.07
A = 0.07
y'(0) = -5.72Asin(0) + 5.72Bcos(0) = 0.04
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This and your assumption of y(0) = .07 are consistent with choosing downward as the positive direction. This is perfectly valid, but note that the solution I've given below assumes upward to be positive.
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-5.72A(0) + 5.72B(1) = 0.04
5.72B = 0.04
B = 0.007
y(t) = 0.07cos(5.72t) + 0.007sin(5.72t)
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Good solution, except for the relatively minor errors you made previously.
Here's a solution consistent with all the given conditions:
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Recall that the equilibrium position for the oscillator is 40 mm below its unloaded position.
The 70 mm position is 30 mm below equilibrium, so that initially y = -30 mm, giving us the initial condition
y(0) = -30 mm.
The initial velocity is 40 mm / sec downward, so that
y ' (0) = -40 mm/sec.
As before the equation of motion is
y '' = -k/m * y.
and its general solution is
y(t) = A cos(omega t) + B sin(omega t)
with omega = sqrt(k / m).
y(0) = -30 mm yields A = -30 mm.
y ' (t) = -omega A sin(omega t) + omega B cos(omega t)
so that
y ' (0) = omega * B.
Recall that omega = 18 rad / sec, approx., so
y ' (0) = 18 rad / sec * B = -40 mm / sec
so that
B = -40 mm/s / (18 rad /s) = -2.3 mm.
The equation of motion for the mass is therefore
y(t) = -30 cos( 18 t) - 2.3 sin(18 t).
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Given Solution:
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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:
W = 32lb
W = mg
g = 32
m = 1
y'(0) = 4
max d = 6 in. = 1/2 ft.
Y + y(0)
by hook's law:
32lb = kY
Y = 6in. - y(0)
k = 32/(6-y(0))
my'' = W + F_R + F_D = mg - k(Y + y) - (gamma)y'
since mg -kY = 0
my'' + (gamma)y' + ky = 0
Critically damped.
gamma^2 = 4km
m = 1
gamma^2 = 4k
general solution:
y(t) = Ae^(r_1*t) + Bte^(r_2*t)
same root
r = -gamma/2
y(t) = Ae^(-gamma/2 * t) + Bte^(-gamma/2 * t)
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Not bad, but you can determine both gamma and k from the given information:
If y(t) is the position of the object, the damping force is of form F_damp = - gamma y ' and the spring force is F_spring = - k y so that
F_net = -gamma y ' - k y
and
m y '' + gamma y ' + k y = 0
or
y '' + gamma / m * y ' + k / m * y = 0.
(The unit of mass in the British system is the slug; the weight of 1 slug is 1 slug * 32 ft / s^2 = 32 slug * ft / s^2 = 32 pounds, so our weight has mass 1 slug.
6 inches is 1/2 foot.
y is in units of ft , y ' in units of ft / s^2 and y '' in units of ft / s^2
k * y is in units of lb, so k is in units of lbs / ft, k / m is in units of (lbs / ft) / slugs, which is reduces to s^2
gamma * y ' is in units of pounds, so gamma is in units of lbs / (ft / s) = kg s^-1, so gamma / m is in units of s^-1.
Our conditions are
y(0) = 0
y ' (0) = -4 ft / sec
y(t) = -1/2 ft when | y(t) | is maximized).
The characteristic equation is
r^2 + gamma / m * r + k / m = 0
with solutions
r = (-gamma / m +- sqrt( gamma^2 / m^2 - 4 k / m) ) / 2.
The system is critically damped so the discriminant is 0:
gamma^2 / m^2 - 4 k / m = 0
with the result that
gamma = +-2 sqrt(k * m).
gamma is positive, so we discard the negative solution.
It follows that
r = -gamma / (2 m) = -2 sqrt(k * m) / (2 m) = -sqrt ( k / m )
and our fundamental set is
{e^(-sqrt( k / m) * t, t e^(-sqrt(k / m) * t) }.
Note that we generally use omega is these solutions, where omega = sqrt(k / m). However omega has a strong connotation with angular frequency, and our solutions in the critically damped (and overdamped) cases are not oscillatory so, to avoid possible confusion, we'll leave the notation as sqrt(k / m).
Our general solution is
y(t) = A e^(-sqrt(k / m) * t) + B t e^(-sqrt(k / m) * t).
Initially y = 0 s0
y(0) = 0
giving us
A e^(-sqrt(k / m) * 0) + B * 0 * e^(-sqrt(k / m) * 0) = 0
with the result that
A = 0
giving us
y(t) = B t e^(-sqrt(k / m) * t).
We also note for future reference that
y ' (t) = B e^(-sqrt(k / m) * t) - B t sqrt(k/m) * e^(-sqrt(k / m) * t).
The initial velocity is 4 ft/sec downward so
y ' (0) = - 4 ft / sec
giving us
B e^(-sqrt(k / m) * 0) = -4
so that
B = -4.
Our function is thus
y = -4 t e^(-sqrt(k / m) * t)
and
y ' = -4 e^(-sqrt(k / m) * t) + 4 t sqrt(k/m) * e^(-sqrt(k / m) * t).
(note on units: it should be clear that B is in units of ft / sec; k/m is in units of s^-2 so sqrt(k/m) is in units of s^-1, which is appropriate since t is in units of s)
We are given the weight of the object, and as noted earlier its mass is easily found; in this case the mass is 1 slug.
Our equation is y = -4 t e^(-sqrt(k / m) * t); the only remaining unknown quantity is k.
To evaluate k we use the last bit of given information:
y = -1/2 ft is the position at which distance from equilibrium is maximized.
This occurs when the mass comes to rest, so we have
y(t) = 1/2 when y ' (t) = 0.
y ' (t) = -4 e^(-sqrt(k / m) * t) + 4 sqrt(k / m) t e^(-sqrt(k / m) * t)
y ' (t) = 0 when
y ' = -4 e^(-sqrt(k / m) * t) + 4 t sqrt(k/m) * e^(-sqrt(k / m) * t)
so we solve
-4 e^(-sqrt(k / m) * t) + 4 t sqrt(k/m) * e^(-sqrt(k / m) * t) = 0
for t, obtaining
(4 - 4 t sqrt(k/m) ) e^(-sqrt(k / m) * t) = 0
so that
t = 1 / sqrt(k / m).
So the maximum displacement from equilibrium occurs when t = 1 / (sqrt(k/m)), and at that instant y = -1/2 foot. Writing this as an equation
y(1 / sqrt(k/m)) = -1/2 so
-4 (1 / sqrt(k/m)) e^(-sqrt(k/m) * (1 / (sqrt(k/m) ) ) = -1/2
-4 e^-1 = -1/2 sqrt(k/m)
sqrt(k/m) = 8 / e
(With units the equation -4 e^-1 = -1/2 sqrt(k/m) reads -4 ft / s * e^-1 = -1/2 ft * sqrt(k/m), so that sqrt(k/m) is in units of (ft / s) / ft = s^-1. This is consistent with k in units of lb / ft and mass in slugs (see the earlier note summarizing units)).
k = 8 m / e
(We note that k is therefore in units of mass / time, appropriate units for a force constant).
As seen before gamma = sqrt( 2 k m) so
gamma = sqrt( 2 * (8 m / e) * m ) * m) = 4 m sqrt(1/e).
(the units of k * m are lbs / ft * slugs = kg^2 s^-2, so the units of sqrt(k m) and therefore gamma are kg s^-1, consistent with the units as noted early in the solution).
We calculate our numerical values of k and gamma:
m = 1 (mass is 1 slug) so
k = 8 / e = 2.9 kg / s^2 = 2.9 lb / ft
gamma = 4 m sqrt(1/e) kg s^-1 = 1.45 kg s^-1, or 1.45 lb / (ft / s).
For every foot of stretch the spring exerts 2.9 pounds, and for every ft / sec of velocity the drag force is .30 lb.
Checking against common sense:
The initial drag force is 1.45 lb / (ft / s) * 4 ft /s = 6 lb, approx.. This would result in an initial acceleration of about 6 ft / s^2 on our 1-slug mass.
The maximum spring force is 2.9 lb * 1/2 ft = 1.45 lb.
The time required to reach max displacement is 1 / sqrt(k / m) = .6 second
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Given Solution:
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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:
my'' + gamma * y' + ky = 0
y(0) = 0
y'(0) = V_0
Critically => y(t) = Ae^(r_1*t) + Bte^(r_1*t)
gamma^2 = 4km
r_1 = -gamma/(2m)
Over => y(t) = Ae^(r_1*t) + Be^(r_2*t)
gamma^2 > 4km
Under => y(t) = e^(alpha*t)(Acos(beta*t)+Bsin(beta*t))
gamma^2 < 4km
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Given Solution:
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Good work. You're on track, but still making some relatively minor errors.
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