Assignment 20

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course Mth 279

7/22

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 = 10 kg; y = .03 ; displacment = 0.04

F_net = mg - ky = 0

0 = (10*9.8) - k(0.03)

k = 3266

general form

my’’ + ky = 0

10y’’ + 3266y = 0

Now solving

y = Ae^rt

y’ rAe^rt

y’’= r^2Ae^rt

subbing in

10r^2 + 3200 = 0

r^2 = -326.6

r = +-sqrt(-326.6)

r = +-sqrt(326.6)i

general soln to imaginary soln: let w=omega = sqrt(k/m) = sqrt(3266/10) = 18.1

y = Acos(wt) + Bsin(wt)

y’ = -wAsin(wt) + wBcos(wt)

finding values of A and B

y(0) = Acos(18.1*0) + Bsin(18.1*0) = 0.04

0.04 = Acos(0)

A = 0. 04

y’(0) = -18.1Asin(0) + 18.1Bcos(0) = 0

y’(0) = 0 because the speed/velocity is 0 at time t=0 correct??? I just want to make sure my reasoning is correct.

0 = 18.1B

B = 0

y(t) = 0.04cos(18.1t)

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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: Let w = omega and d=delta

y = Rcos(wt-d)

Based on the given graph R is amp so R = 3

Period of graph = T =2

w = 2pi/T = 2pi/2 = pi

y(t) = 3cos(pi*t -d)

graph shifts about 1/4 units to right so t-1/4

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

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

so d = pi/4

initial conditions let t = 0

y(0) = 3cos(pi(0) - pi/4)

y(0) = 3sqrt(2)/2

y’(t) = 3[-pisin(pi*t - pi/4)]

y’(0) = -3pisin(pi(0) - pi/4)

y’(0) = -3pi(-sqrt(2)/2)

y’(0) = 3pi(sqrt(2)/2)

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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:

Soln to first problem:

y(t) = 0.04cos(18.1t)

y(0) = 0.04 m; v=0.04/s; y’(0) = 0.04m/s

my’’ + ky = 0

gen soln:

y(t) = Acos(wt)+Bsin(wt)

w= sqrt(k/m) = 18.1

y(0) = Acos(0) + Bsin(0) = 0.04

A = 0.04

y’(0) = 18.1Asin(0) + B(18.1)cos(0) = 0.04

18.1B = 0.04

B = 0.0022 = 2/905

y(t) = 0.04cos(18.1t) + 2/905*sin(18.1t)

@&

We have assumed the upward is positive. So the downward initial velocity will be -.04 m/s, not .04 m/s.

This will affect your final solution. But your method is sound.

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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:

weight = 32 lb =mg

m = 3.27 kg

@&

Your units are off. You don't get kg by dividing weight in pounds by 9.8 m/s^2. You're mixing units between the old British system and the metric system.

Otherwise your approach on this problem is good.

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v_1 = 4ft/s

y = 6in = .5ft

F_damp = - gamma*y’

F_spring = -ky

my’’ = -gamma*y’ - ky

3.27y’’ + gamma*y’ + ky = 0

so

3.27r^2 + gamma*r + k = 0

r^2 + gamma/3.27 * r + k/3.27 = 0

r = -gamma/3.27 +- sqrt((gamma^2/10.7) - 4k/3.27)/2

being critically damped means discriminant = 0

solve for gamma

gamma^2/10.7 - 4k/3.27 = 0

gamma^2 = 4k/3.7 * 10.7

gamma^2 = 13.08k

gamma = sqrt(13.08k) ???

Not sure if this is correct or how to solve for k???

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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 damped

y(t) = Ae^rt + Bte^rt

gamma^2 = 4km

r = -gamma/2m

overdamped

y(t) = Ae^rt + Be^r_2t

gamma^2 = 4km

underdamped

y(t) = e^(at) (Acos(bt) + Bsin(bt))

gamma^2 < 4km

No idea what is going on. I saw this information from the book but I don’t understand how to answer that questions???

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The key idea is that if there's too much damping the system will approach equilibrium without going past the equilibrium point. With just the right damping it will pass the equilibrium point exactly one time before returning. If the system is underdamped then it will oscillate with exponentially decreasing amplitude.

As in the preceding the characteristic equation has solutions

r = (-gamma / m +- sqrt( gamma^2 / m^2 - 4 k / m) ) / 2 = -gamma / (2 m) +- gamma / (2 m) * sqrt( 1 - 4 k m / (gamma^2) ) = -gamma / (2 m) * (1 +-sqrt( 1 - 4 k m / gamma^2) ).

The discriminant is:

zero when gamma = 2 sqrt(k * m),

negative when gamma < 2 sqrt( k * m )and

positive when gamma > 2 sqrt( k * m ) ,

corresponding respectively to the critically damped, overdamped and underdamped cases.

What solutions follow from these three cases?

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Given Solution:

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&#Good responses. See my notes and let me know if you have questions. &#