# Spring-Mass system (with damping force AND impressed force). Find Position fcn.

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• Jul 23rd 2010, 07:58 AM
EmilyL
Spring-Mass system (with damping force AND impressed force). Find Position fcn.
Here is the problem:

A spring is such that a 2 pound weight stretches it 6 inches. There is a damping force present with magnitude the same as the magnitude of the instantaneous velocity. An impressed force f(t) = 2 sin 8t, acts on the spring. At t= 0, the weight is released from a point 3 inches below equilibrium. Find its position function.

I'm not sure if this is correct, but as far as setting up the initial differential equation, I think it is like this:

(1/16)x'' + x' + (1/3)x = 2sin8t

(Speechless) Is this differential equation correct? If so, can someone show step by step how to solve this using differential operators (the annihilator method) or LaPlace transforms, and how to use that to get the position function?

Thank You! (Happy)
• Jul 23rd 2010, 08:56 AM
Ackbeet
I don't agree with the DE. Try to set up the homogeneous case (the one without the $2 \sin(8t)$ impressed force) first:

$F = m a = m \ddot{x}$. You've got two forces present: the spring force and the damping force. Spring forces look like the following (at least in the regime of Hooke's Law): $F = - k x$. You're told that a force of 2 pounds stretches the spring 6 inches. That tells you that $k = 1/3$, as I think you've already computed. $k$ is always positive in these cases. The mass is 2 pounds. The damping force acts in a direction opposite to the spring force, correct? It's going to have to have the opposite sign.

So let's say that $2 \ddot{x} = - k x + \dot{x}$. How does that look to you? Can you finish writing the DE, including the impressed force?
• Jul 23rd 2010, 09:14 AM
EmilyL
Hmm, yes I found the k=1/3 the same way you did, and put that as the coefficient for x in my equation. For the mass, I thought I would have to divide the pounds by acceleration since F=ma, so I did (2 pounds)/(32 ft/sec/sec) to get a mass of 1/16 slugs. I'm not completely sure how to include the impressed force, but I thought you're just supposed to take all the homogenous parts of the equation on one side (so that it would equal zero if there was no impressed force), and then put the impressed force on the right side instead of the 0.
• Jul 23rd 2010, 09:43 AM
Ackbeet
You don't divide by the acceleration. The acceleration is the second derivative of the displacement, so you just leave it where it is. Your method of putting in the impressed force is correct. So the DE looks like this:

$2\ddot{x}-\dot{x}+kx=2\sin(8t).$

If you're up for it, the Laplace Transform (LT) method is probably the easiest method of solution, because you can take care of the DE itself, the initial conditions, and the inhomogeneous part all in one fell swoop. So, take the LT of the entire equation. What do you get?
• Jul 23rd 2010, 10:59 AM
EmilyL
Are you sure? In some of the examples, she converted the pounds to slugs by dividing it by 32 before putting it into the equation. Since it is supposed to be ma (or mx''), and pounds is a unit of force (not a mass), don't you need to convert it?
• Jul 23rd 2010, 11:09 AM
Ackbeet
Ok. You're right and you're not right. Dang English system! The metric system works so much better.

You're correct in that you do need to divide by the acceleration. I was wrong there. However, the correct constant to divide by is not 32, because the units are not the same as for the rest of the problem. You've been given pounds and inches. Therefore, your final answer must be in those units. So you need to convert:

$\displaystyle{32\,\frac{\text{ft}}{\text{sec}^{2}} \times\frac{12\,\text{in}}{1\,\text{ft}}=384\,\fra c{\text{in}}{\text{sec}^{2}}.}$

$\frac{1}{192}\,\ddot{x}-\dot{x}+kx=2\sin(8t).$

Are we on the same page now?
• Jul 23rd 2010, 11:17 AM
EmilyL
Ah alright, yes, thanks. I'm not very good with the LaPlace Transforms yet, do you mind helping me through with the Annihilator method? With that, I believe it would be like this:

D^3 = (D^2 + 64)
homogeneous part: D= 0,0,0

Nonhomogenous part: D= +/- 8i

Now I'm a little rusty as far as getting Y(c) and Y(p)
• Jul 23rd 2010, 11:21 AM
Ackbeet
Hmm. I'd probably do the equivalent guess-and-check method. First, the homogeneous solution:

Assume $x_{c}(t)=e^{ r t}.$ Plug that into the DE. What do you get?
• Jul 23rd 2010, 07:20 PM
EmilyL
I don't think we covered the guess-and-check method. The way we did it was by finding the roots of D after applying the correct annihilators. Then we'd somehow put in the equation x = c(1)e^at + c(2)e^at +...... or we use sin and cos terms if imaginary numbers were involved

If I do it like that, then I think the equation would be:

C(1) + C(2)x + C(3)x^2 = C(4)cos(8^[1/2]) + C(5)sin(8^[1/2])

Does this look OK? From what I understand, the left side would be Y(c) while the right side is Y(p). How can I solve for the coefficients, and then get that to the position function?
• Jul 24th 2010, 03:26 AM
Ackbeet
I'm not sure I followed all of that. The annihilator method is essentially the same as the guess-and-check method. We can go with the annihilator approach. To recap, the DE is

$\frac{1}{192}\,\ddot{x}-\dot{x}+kx=2\sin(8t).$

First step: solve the homogeneous equation, which is

$\frac{1}{192}\,\ddot{x}-\dot{x}+kx=0.$

What do you get?
• Jul 24th 2010, 05:19 AM
EmilyL
I multiplied 192 through the equation to get x'' - 192x + 64 = 0
Then I substituted x'' with m^2, x' with m, and x with 1, and then solved the equation with the quadratic equation.
I got: m= 96 +/- 16[sqrt(143)]

The next step I'm not so sure about, but from what I understand, the solution would be:

x = C(1)e^(96+16sqrt[143]) + C(2)e^(96-16sqrt[143])

the (1) and (2) are supposed to be subscripts, by the way
• Jul 24th 2010, 07:12 AM
Ackbeet
I think the correct roots are $96\pm 8\sqrt{143}.$ Perhaps you forgot to divide by 2 in both places?

So, you also forgot the t's, I think:

$x_{c}(t)=C_{1}e^{(96+8\sqrt{143})t}+C_{2}e^{(96-8\sqrt{143})t}.$

You obviously had the right idea here. Just some small algebra mistakes and notational problems.

So you have the complimentary solution. What's the next step in the annihilator approach?
• Jul 24th 2010, 07:33 AM
EmilyL
Ah yeah, thats right. The next step would be to find the annihilator of the nonhomogeneous side and find the roots. The annihilator would be D^2 + 64, making the roots +/-8i

so y(p) would be Acos(8t) + Bsin(8t)

Now how do I use this to get the position function? Are the coefficients solvable, or do I need to leave them as-is?
• Jul 24th 2010, 08:22 AM
Ackbeet
So far so good. I think what you do now is you take your original non-homogeneous DE, written as $L_{c}x=2\sin(8t),$ where $L_{c}$ is the complimentary operator that annihilates the LHS, and you operate on the whole equation with the operator that annihilates the RHS. That'll give you a higher-order homogeneous ODE that you need to solve. What do you get at that point?
• Jul 25th 2010, 05:44 AM
EmilyL
I'm not sure I know what you mean, would it be like this: (D^2 + 64)(D^3) = D^5 + 64D^3 ?

At some point, you're supposed to take the derivative and second derivative of one of the y values (i'm 70% sure its y(p)) and plug that into the original equation, which helps find some coefficients, but I'm not sure at which point that is done.
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