I solved it! I'll edit this post with a scan of the sheet of paper i wrote on.

- Jul 25th 2010, 06:16 PMEmilyL
I solved it! I'll edit this post with a scan of the sheet of paper i wrote on.

- Jul 26th 2010, 01:37 AMAckbeet
Great! I'm very willing to double-check your work, if you like. I can wait for your post.

- Jul 26th 2010, 10:31 AMEmilyL
I haven't gotten time to get to my scanner yet, but basically I did it in terms of feet (so that x is 1/2 feet instead of 6 inches) to get k=4 and m=1/16.

I got y(c) = c(1)e^/8 + c2e^/8t.

I found y(p) with the annihilator method, getting y(p)=Asin8t + Bsincos8t

I then found y'(p) and y''(p), and plugged it into the original equation to get coefficients of A=0, B=-1/4

This made my y(p)= (-1/4)cos8t

Then I plugged in the initial conditions for the entire y(t) to get the coefficients c1 and c2, which were c(1)=1/2 and c(2)=4

So my final y(t) was:

y(t) = 1/2(e^-8t) + 4te^-8t - (1/4)cos8t

=] - Jul 26th 2010, 12:13 PMAckbeet
Is this the same problem? I confess myself bewildered by the apparent random changes in method you've used. Here are some specific comments:

1. If the problem gives you pounds and inches, your answer had better be in pounds and inches. Get everything in terms of the given units. This is as true in the industrial world of engineering as it is in academia.

2. I'm not sure about your application of the annihilator approach. Up to Post # 14, I would have said things were good. Your post # 15 was incorrect. You needed to operate on the original DE with the operator that annihilated the RHS. That is, with the following DE:

$\displaystyle x'' - 192x' + 64x = 384 \sin(8t)$ (since you multiplied through by 192, that has to happen to the non-homogeneous side as well!)

This we can write as

$\displaystyle (D^{2}-192 D+64)x(t)=384\sin(8t).$

After left-operating with the annihilator of the RHS, we get

$\displaystyle (D^{2}+64)(D^{2}-192 D+64)x(t)=0.$

What do you get when you solve this homogeneous DE? - Jul 26th 2010, 12:27 PMEmilyL
Darn, thought I had it for sure...

hmm, D=8i, -8i, 96 +8sqrt(143), 96 -8sqrt(183) ?

I'm not sure if this is appropriate or not, but I need some more immediate help with the problems in this thread, if you wouldn't mind taking a look at them: http://www.mathhelpforum.com/math-he...but-stuck.html - Jul 26th 2010, 03:45 PMmantra002
Ackbeet, I believe there is a critical flaw in the setup of this problem. You corrected Emily's DE by flipping the sign of the damper in the problem, this is not correct. In fact, by making this change the system is now unstable and you will not get a meaningful solution. Note that a positive velocity will now create a positive dampening force! As velocity rises the force grows forever and ever.

In fact, this is not what should happen, the dampening force is not opposite the spring, but opposite the velocity, so it is negative when on the RHS. The correct DE to start is:

(W/G)*x'' + B*x'+K*x = 2*sin(8/t) - W (this last W only exists if gravity is counted in the problem)

Where W is the weight in lbs (2 here), G the acceleration due to gravity in In/s^2, B is the dampening coefficient (1 here), K is the spring coefficient (1/3 here). I assumed the forcing function to be positive, but it could very well be negative depending on how the problem is set up (is it 'pushing' or 'pulling').

Also, I apologize for the ugly notation, I don't know how to use TeX (Crying) - Jul 26th 2010, 04:13 PMAckbeet
I think you're right. It's what was going through my mind, but didn't make its way into my typing. I don't have time right now to fix everything, but I'll get to it soon.

Thanks for catching that!