#$&*
course Mth 279
May 7 around 1:45pm.
Query 18 Differential Equations*********************************************
Question: A 10 kg mass stretches a spring 30 mm beyond its unloaded position. The spring is pulled down to a position 70 mm below its unloaded position and released.
Write and solve the differential equation for its motion.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Mass = 10kg
Displacement = 30mm
Pulled displacement = 70mm
Solve for motion.
ax’’(t) + bx’(t) + cx = d
Could you use this equation? If x were the position of an object and t the time, then the first derivative is the velocity, the second the acceleration, and this would describe the motion of the object?!
@& Set the expression m a = m y '' for the net force equal to the expression - k y for the net force, evaluate k from the given information, then solve the equation.*@
@&
A 10 kg mass weighs 98 Newtons. Thus a 98 N change in force will change the length of the spring by 30 mm. Assuming the force to be a linear function of the spring we find that the force constant is
k = 98 N / (30 mm) = 3.3 N / mm, approx., or about 3300 N / m.
The new equilibrium position of the system will be 30 mm below the unloaded position. When pulled down 70 mm the position of the system will therefore be -40 mm.
Thus, if y is the position relative to the unloaded position, y(0) = -40 mm and since the spring is released from rest, y ' (0) = 0.
The net force on the system is
F = -k y so that
m y '' = - k y and
y '' = -k/m * y.
The general solution to this equation is
y = A cos(omega t) + B sin(omega t)
with omega = sqrt(k / m) = sqrt(3300 N/m / (10 kg) ) = 18 rad / sec, approx..
We get
y ' = -omega A sin(omega t) + omega B cos(omega t)
The initial conditions give us
y(0) = A = -40 mm
and
y ' (0) = omega B = 0
so that
A = -40 mm, B = 0 and our solution is
y = -40 mm * cos(omega * t).
*@
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Honestly, I don’t know how to even start this problem, or well, I’m having a brain cramp. Do I solve it just like a normal differential equation and then look at the graph to find initial conditions?!
@& You need to be able to use the circular model, and/or shifting and stretching transformations, to identify omega, delta and A for any sine or cosine graph of the form y = A sin(omega t + phi) or y = A cos(omega t + phi).
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.
The initial condition on y is y(0) = 2.
y ' (0) = -6 sin( -pi/2) = 6.
*@
@& Check my notes. I'll welcome additional questions if you have them.*@
#$&*
course Mth 279
May 7 around 1:45pm.
Query 18 Differential Equations*********************************************
Question: A 10 kg mass stretches a spring 30 mm beyond its unloaded position. The spring is pulled down to a position 70 mm below its unloaded position and released.
Write and solve the differential equation for its motion.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Mass = 10kg
Displacement = 30mm
Pulled displacement = 70mm
Solve for motion.
ax’’(t) + bx’(t) + cx = d
Could you use this equation? If x were the position of an object and t the time, then the first derivative is the velocity, the second the acceleration, and this would describe the motion of the object?!
@& Set the expression m a = m y '' for the net force equal to the expression - k y for the net force, evaluate k from the given information, then solve the equation.*@
@&
A 10 kg mass weighs 98 Newtons. Thus a 98 N change in force will change the length of the spring by 30 mm. Assuming the force to be a linear function of the spring we find that the force constant is
k = 98 N / (30 mm) = 3.3 N / mm, approx., or about 3300 N / m.
The new equilibrium position of the system will be 30 mm below the unloaded position. When pulled down 70 mm the position of the system will therefore be -40 mm.
Thus, if y is the position relative to the unloaded position, y(0) = -40 mm and since the spring is released from rest, y ' (0) = 0.
The net force on the system is
F = -k y so that
m y '' = - k y and
y '' = -k/m * y.
The general solution to this equation is
y = A cos(omega t) + B sin(omega t)
with omega = sqrt(k / m) = sqrt(3300 N/m / (10 kg) ) = 18 rad / sec, approx..
We get
y ' = -omega A sin(omega t) + omega B cos(omega t)
The initial conditions give us
y(0) = A = -40 mm
and
y ' (0) = omega B = 0
so that
A = -40 mm, B = 0 and our solution is
y = -40 mm * cos(omega * t).
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Honestly, I don’t know how to even start this problem, or well, I’m having a brain cramp. Do I solve it just like a normal differential equation and then look at the graph to find initial conditions?!
@& You need to be able to use the circular model, and/or shifting and stretching transformations, to identify omega, delta and A for any sine or cosine graph of the form y = A sin(omega t + phi) or y = A cos(omega t + phi).
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.
The initial condition on y is y(0) = 2.
y ' (0) = -6 sin( -pi/2) = 6.
*@
@& Check my notes. I'll welcome additional questions if you have them.*@