#$&*
course Mth 277
4-10 11
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:
w =mg is wt of mass, when added to spring we have Hooke’s Law
W + F_R = mg - kY = 0, where Y = (mg)/k
Now when spring is pulled down and released it’s motion is represented by
m*(d^2y/dt^2) = W + F_R + F_D = mg - k(Y + y) - `lambda(dy/dt)
=my’’ + lambda(y’) + ky = 0
So we have
m = W/g = 10kg/9.8m/s^2 = approx. 1.02 (kg*s^2)/m
10kg = k(30 mm), k = (1/3) kg/mm
Init value problem governing the motion is
1.02 (kg*s^2)/m*y’’ + (1/3) kg/mm*y = 0, y(0) = 70mm, y’(0) = 0
y’’ + .98*(1/3)y = 0 y’’ + (49/150)y = 0, in this form it is easy to see solution will be nonreal when plugged into charistic equationwe get
r = 0 +- `sqrt(49/150)i
So general solution is
y(t) = c1 cos(`sqrt(49/150)t) + c2 sin(`sqrt(49/150)t)
Imposing init condition we get
y(t) = 70*cos(`sqrt(49/150)t)
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Given Solution:
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Self-critique (if necessary):
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@&
A 10 kg mass weighs 98 Newtons. Thus a 98 N change in force will change the length of the spring by 30 mm. Assuming the force to be a linear function of the spring we find that the force constant is
k = 98 N / (30 mm) = 3.3 N / mm, approx., or about 3300 N / m.
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.
y should be position relative to equilibrium.
The equilibrium position is 30 mm below the unloaded position. So the 70 mm position is at y = 40 mm or y = -40 mm, depending on which direction you take as positive.
Thus, if y is the position relative to the unloaded position, y(0) = -40 mm and since the spring is released from rest, y ' (0) = 0.
The net force on the system is
F = -k y so that
m y '' = - k y and
y '' = -k/m * y.
The general solution to this equation is
y = A cos(omega t) + B sin(omega t)
with omega = sqrt(k / m) = sqrt(3300 N/m / (10 kg) ) = 18 rad / sec, approx..
We get
y ' = -omega A sin(omega t) + omega B cos(omega t)
The initial conditions give us
y(0) = A = -40 mm
and
y ' (0) = omega B = 0
so that
A = -40 mm, B = 0 and our solution is
y = -40 mm * cos(omega * t).
*@
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, `omega= 2, `delta = 1/4
Init conditions
y= 3cos(-1/4)
I confused myself with this problem, which is embarrassing because this is something we have worked thru several times.
You would find init cond by finding the values at t = 0 of y’ and y right??
@& Right idea, but omega isn't the period and delta doesn't work quite the way you assum.
The graph is shifted 1/4 unit to the right, so t will be replaced by t - 1/4. This results in delta = pi / 4, not 1/4:
Check the following:
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).
Thus delta = pi/2, omega = 2 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 ' (0) = -6 sin( -pi/2) = 6.
*@
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Given Solution:
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Self-critique (if necessary):
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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. 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 cm / 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:
Init value problem governing the motion is
1.02 (kg*s^2)/m*y’’ + (1/3) kg/mm*y = 0, y(0) = 70mm, y’(0) = -400mm/s
y’’ + .98*(1/3)y = 0 y’’ + (49/150)y = 0, in this form it is easy to see solution will be nonreal when plugged into charistic equation we get
r = 0 +- `sqrt(49/150)i
So general solution is
y(t) = c1 cos(`sqrt(49/150)t) + c2 sin(`sqrt(49/150)t)
Imposing init condition we get
y(t) = 70*cos(`sqrt(49/150)t)
y’(t) = -c1*`sqrt(49/150)sin(`sqrt(49/150)t) + c2`sqrt(49/150)cos(`sqrt(49/150)t)
y’(0) = c2`sqrt(49/150) = -40
c2 = approx. -70
So our solution would be
y(t) = 70 cos(`sqrt(49/150)t) - 70 sin(`sqrt(49/150)t)
@&
A 10 kg mass weighs 98 Newtons. Thus a 98 N change in force will change the length of the spring by 30 mm. Assuming the force to be a linear function of the spring we find that the force constant is
k = 98 N / (30 mm) = 3.3 N / mm, approx., or about 3300 N / m.
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.
y should be position relative to equilibrium.
The equilibrium position is 30 mm below the unloaded position. So the 70 mm position is at y = 40 mm or y = -40 mm, depending on which direction you take as positive.
Thus, if y is the position relative to the unloaded position, y(0) = -40 mm and since the spring is released from rest, y ' (0) = 0.
The net force on the system is
F = -k y so that
m y '' = - k y and
y '' = -k/m * y.
The general solution to this equation is
y = A cos(omega t) + B sin(omega t)
with omega = sqrt(k / m) = sqrt(3300 N/m / (10 kg) ) = 18 rad / sec, approx..
We get
y ' = -omega A sin(omega t) + omega B cos(omega t)
The initial conditions give us
y(0) = A = -40 mm
and
y ' (0) = omega B = 0
so that
A = -40 mm, B = 0 and our solution is
y = -40 mm * cos(omega * t).
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).
*@
????I believe I did this correctly????
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
@& Right idea and good procedure on everything, but there are errors in detail so check the notes on each problem.*@