Query_18

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course Mth 279

3/ 25 17

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:

Newton’s law

F = m a , with a = y’’

F = m y’’

F = - k y

- k y = m y’’

y’’ = - k / m y

With a mass of 10kg the object weighs (10 kg) (9.8 m / s^2)

weight = 98 N

k = 98 N / 30 mm = 3300 N / m

y’(0) is the initial velocity which is zero.

Initial conditions : y(0) = -40 mm , y’(0) = 0

The general solution to this type of problem is

y = A cos(omega t) + B sin(omega t)

With omega = sqrt( k / m) or sqrt ( 3300 / 10)= 18 rad / sec

y’ = - omega A sin(omega t) + omega B cos(omega t)

y(0) = A = -40 mm

y’(0) = B = 0

y = -40 cos(omga t)

confidence rating #$&*:232; 3

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Given Solution:

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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.

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Your solution:

These values will be approximate…

The period of the graph is 2.

So that omega = 2 pi / 2 or pi

The amplitude of the graph is 3.

An oscillating graph has the standard form of

y = A sin (omega t + phi)

or

y = A cos (omega t + phi)

This graph uses the cosine form.

The graph appears to be shifted to the right about 1 / 4 of a unit.

Playing with graphing programs online, this can be accounted for by changing the t term in the standard form to a (t - pi / 4)

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You should base this on algebraic analysis rather than a graphing program, though it's even better to do the analysis then play with the program to see that it works out.

Replacing t by t - 1/4 shifts a graph 1/4 unit to the right, since to get a given value of y the value of t would have to be 1/4 of a unit greater.

The rules for stretching, shifting and reflecting graphs are part of basic precalculus and you should review them in such a way that you can not only apply them, but understand why they are as they are. That way you'll always be able to reason them out if you need them.

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Our equation becomes

y = 3 cos ( 2 (t - pi / 4))

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Omega is pi, not 2.

The reason omega is pi is that omega * t should change by 2 pi (which drives us through a complete cycle of the sine function) when t changes by one period. This is why in this case, where the period is 2, we have 2 pi = omega * 2, as you have above, with the result (which you also have above) that omega = pi.

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y = 3 cos (2 t - pi / 2 )

In this case, delta = pi / 2

Initial condition is y(0) = 2

y’ = -6 sin (2t - pi / 2)

y’(0) = - 6 sin ( - pi / 2)

From the given graph, I cannot determine what the slope of the point, y’(0) , is…

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You've missed a few details, but you're on the right track. Be sure you think more about the meanings of the associated transformations.

The correct function would be

y = 3 cos( pi ( t - 1/4) ) ) = 3 cos( pi t - pi/4),

and I believe you'll see exactly why. If not, feel free to ask.

*@

confidence rating #$&*:232; 2

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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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:

The sharp blow changes the initial condition which is now

y(0) = -30 mm

y’(0) = -40 mm / sec

The previous equations are the same:

y’’ = -k / m y

y(t) = A cos (omega t) + B sin ( omega t)

With omega = sqrt( k / m)

y(0) = A cos (omega * 0) = -30 mm

A = -30 mm

y’(t) = - omega A sin (omega t) + omega B cos (omega t)

y’(0) = omega B = -40 mm

B = -2.3 mm

The resulting function is :

y(t) = -30 cos (18 t) - 2.3 sin (18 t)

confidence rating #$&*:232; 3

@&

Good.

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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:

F_damp = - gamma y’

F_spring = - k y

F_net = F_damp + F_spring

F_net = - gamma y’ - k y

Newtons Law: F = m a = m y’’

m y’’ = - gamma y’ - k y

m y’’ + gamma y’ + k y = 0

y’’ + (gamma / m) y’ + (k / m) y = 0

This creates our characteristic equation:

r^2 + (gamma / m) r + k / m = 0

Using the quadratic formula to solve for r:

r = - gamma / m +- sqrt( (gamma^2 / m^2) - 4 k / m) / 2

Given: The spring is critically damped

Critically damped means that gamma^2 / m^2 - 4 k / m = 0

Solving for gamma:

gamma = sqrt( 4 km) = 2 sqrt(k m)

This is positive due to context.

Solving the quadratic formula for r

r = - gamma / 2m = - 2 sqrt( km) / 2m

r = - sqrt (k / m)

From the previous problems, r is equal to omega

The provides the fundamental set

[ e ^ - omega t , t e ^ -omga t ]

The general solution is

y(t) = A e ^ ( - omega t) + B t e ^ ( - omega t)

Initial condition of y(0) = 0

y(t) = A

A = 0

y(t) = B t e ^ ( -omega t)

Initial condition of y’(0) = -4

y’(t) = B e ^ (-omeag t) - B t omega e ^ (-omega t)

y’(0) = B = -4

B = -4

y(t) = -4 t e ^ ( -omega t)

y’(t) = -4 e ^ (-omega t) + 4 t omega e ^ (-omega t)

????? I am confused as where to go from here

@&

Good start. I'm going to give you the rest of the solution:

Our function 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

*@

confidence rating #$&*:232; 2

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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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:

From the previous problem:

r = (- gamma / m + sqrt( (gamma^2 / m^2) - 4 k / m ) ) / 2

When:

gamma = 2 sqrt( k m ) → 0 , critically damped

gamma < 2 sqrt ( k m ) → negative , overdamped

gamma > 2 sqrt ( k m ) → positive , underdamped

????? Not sure if the above assumptions are correct or if this is where I should be heading witht this problem.

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This is a good start.

What now are the solution sets, and the general solutions, for the three different cases?

What happens to the general solution for the underdamped case when gamma approaches 2 sqrt( k m ) from below?

What happens to the general solution for the overdamped case when gamma approaches 2 sqrt( k m ) from above?

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confidence rating #$&*:232; 1

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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#*&!

&#This looks good. See my notes. Let me know if you have any questions. &#