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#1 2006-06-19 13:03:26

indiansnipr
Guest

Differential Equations

Hi i need some help with a differential equations problem. The question is:-
(w/g)y(double prime) + ky(single prime)+cy=F(t)
w= the weight of the object attached to the spring
g=32
k= damping factor
c=spring constant
F(t)= the external force on the system.
A 64 pound weight is attached to the end of the spring. After reaching the equilibrium position the spring is stretched one foot beyond the equilibrium position. The weight is then released and as it is released it is struck a downward blow giving it an initial velocity of 2 ft/sec. Take the moment the weight is released and struck as time zero. At time zero a periodic external force given by F(t) = (1/2)cos(4t) pounds begins acting on the system. t is time in seconds. Consider the damping factor to be negligible, i.e., take k to be zero. The spring constant is 32. Find the function giving y, the position of the bottom of the weight as a function of time given in seconds.
if some one could help me with this problem i would really appreciate it.
Thanks

#2 2006-06-19 21:20:39

numen
Member
Registered: 2006-05-03
Posts: 115

Re: Differential Equations

Are you sure k is supposed to be zero? Then the y' would be removed completely from the equation, how about 1? Then it wouldn't change anything in the equation.

I'm having some problems with understanding the problem, english is not my native language, but I guess the equation becomes (with k being zero):

Could we simply take F(t) = (1/2)cos(4t) above? If so, I could solve for y. I just want to make sure things gets right.

Bang postponed. Not big enough. Reboot.

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#3 2006-06-19 23:53:50

indiansnipr
Guest

Re: Differential Equations

Yeah i think that is what happens and that was what i got to but i was stuck from that point on i tried solving the on homogenous equation but i guess i was making a mistake because it was becoming really nasty. Thanks

#4 2006-06-19 23:55:50

indiansinpr
Guest

Re: Differential Equations

In the problem it says that k is zero so i guess it is.

#5 2006-06-20 00:13:01

numen
Member
Registered: 2006-05-03
Posts: 115

Re: Differential Equations

Ok. But the homogenous is the simplest:

, which gives
and

So the homogenous solution is

Ok, I see it. I doubt you're supposed to get imaginary numbers here. Even if we neglect k and set it as 1 instead of 0, we'd get imaginary numbers anyway. have you encountered imaginary numbers such as i before?
I could try solve it anyway if you want, but if you haven't encountered numbers such as i before, I don't think this is correct. It depends on who gave you this problem though

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#6 2006-06-20 00:15:58

ptyindian
Member
Registered: 2006-06-20
Posts: 2

Re: Differential Equations

yes we have encounterd imaginary numbers and we used sine and cosine with the imaginary numbers.

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#7 2006-06-20 00:32:25

numen
Member
Registered: 2006-05-03
Posts: 115

Re: Differential Equations

Ok, well then you could find the particular solution by using the fact that

So you'd get the equation

Where the particular solution can be solved with the approach

And you could always rewrite the homogenous solution with exponentials and sin/cos instead of expoentials with i if that makes it easier.

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#8 2006-06-20 01:17:39

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Differential Equations

Using Euler's identity, we can make:

Where D_1 and D_2 are real numbers.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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