Question

Consider the initial value problem my^('')+cy^(')+ky=F(t),y(0)=0,y^(')(0)=0 modeling the motion of a spring-mass-dashpot system initially at rest and subjected to an applied force F(t), where the unit of force is the Newton ( N ). Assume that m=2 kilograms, c=8 kilograms per second, k=80 Newtons per meter, and the applied force in Newtons is F(t)={(30 if 0<=t<=(\pi )/(2)),(0 if t>(\pi )/(2)):} a. Solve the initial value problem, using that the displacement y(t) and velocity y^(')(t) remain continuous when the applied force is discontinuous. For 0<=t<=(\pi )/(2),y(t)= help (formulas) For t>(\pi )/(2),y(t)= 1elp (formulas) b. Determine the long-term behavior of the system. Is \lim_(t->\infty )y(t)=0 ? If it is, enter zero. If not, enter a function that approximates y(t) for very large positive values of t. For very large positive values of t,y(t)~~ help (formulas)

          Consider the initial value problem
my^('')+cy^(')+ky=F(t),y(0)=0,y^(')(0)=0
modeling the motion of a spring-mass-dashpot system initially at rest and subjected to an applied force F(t), where the unit of force is the Newton ( N ).
Assume that m=2 kilograms, c=8 kilograms per second, k=80 Newtons per meter, and the applied force in Newtons is
F(t)={(30 if 0<=t<=(\pi )/(2)),(0 if t>(\pi )/(2)):}
a. Solve the initial value problem, using that the displacement y(t) and velocity y^(')(t) remain continuous when the applied force is discontinuous.
For 0<=t<=(\pi )/(2),y(t)=
help (formulas)
For t>(\pi )/(2),y(t)=
1elp (formulas)
b. Determine the long-term behavior of the system. Is \lim_(t->\infty )y(t)=0 ? If it is, enter zero. If not, enter a function that approximates y(t) for very large
positive values of t.
For very large positive values of t,y(t)~~
help (formulas)

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consider the initial value problem mycykyfty00y00 modeling the motion of a spring mass dashpot system initially at rest and subjected to an applied force ft where the unit of force is the ne 99438

Added by Todd R.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Minh Le

04/15/2022

Step 1

Assume that $$m = 2$$ kilograms, $$c = 8$$ kilograms per second, $$k = 80$$ Newtons per meter, and the applied force in Newtons is $$F(t) = \begin{cases} 30 & \text{if } 0 \le t \le \pi/2, \\ 0 & \text{if } t > \pi/2. \end{cases}$$ a. Solve the initial value Show more…

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Consider the initial value problem my^('')+cy^(')+ky=F(t),y(0)=0,y^(')(0)=0 modeling the motion of a spring-mass-dashpot system initially at rest and subjected to an applied force F(t), where the unit of force is the Newton ( N ). Assume that m=2 kilograms, c=8 kilograms per second, k=80 Newtons per meter, and the applied force in Newtons is F(t)={(30 if 0<=t<=(\pi )/(2)),(0 if t>(\pi )/(2)):} a. Solve the initial value problem, using that the displacement y(t) and velocity y^(')(t) remain continuous when the applied force is discontinuous. For 0<=t<=(\pi )/(2),y(t)= help (formulas) For t>(\pi )/(2),y(t)= 1elp (formulas) b. Determine the long-term behavior of the system. Is \lim_(t->\infty )y(t)=0 ? If it is, enter zero. If not, enter a function that approximates y(t) for very large positive values of t. For very large positive values of t,y(t)~~ help (formulas)

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Transcript

00:01 Alright, so let's do this question.

00:03 So you have the equation for the mastering system.

00:07 Right, so i already summarize every necessary information right here.

00:12 Okay, so what is our equation? before going that to the deeper solutions, then i'm going to explain what formula we're going to use.

00:27 Right we're gonna use the following two formula one in two right here in order to solve for this equation right so let let let get started by using the lapage information all right so we have the equation is two because m equal to two right y double prime plus a y prime plus ad y equal to 30 time u to the t minus pi over two what is u to the minus pi over two is due to the means that you to the t minus c which is we call the step function is equal to one whenever t less than or equal to c and then equal to zero when t bigger than c right so as you can see the capital f equal to 30 when t less than zero and bigger than 0 and less than pi over 2 and then ft equal to 0 when t bigger than pi over 2 so that why i have t minus pi over 2 right here.

01:29 All right, so we're going to use the lablation both sides.

01:33 I'm going to take the level information both sides.

01:37 So on the left hand side, i will have 2x squared plus as plus 80, because i have the initial data are 0, right, y0 is 0, why 0 is 0 and y.

01:53 On the right on the right hand side, i'm going to use the first formula and my c equal to pi over 2.

01:59 So it turns out that i have 30 times exponential of negative pi over 2, s divided by s.

02:09 Okay, the next step is i will solve for li...

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