When a mass-less spring with one end attached to a ceiling and the other end attached to a point mass m, the displacement z of the mass off the equilibrium position is modeled by $mz''+\mu z'+kz=f$ where mu is the damping coefficient, k is the spring constant, and f is the external force. Assume that m=2kg mu=1Ns/m k=2N/m and f(t)=sin 2t N Find the steady state solution. ie: the function z*(t) s.t \(lim_{t\rightarrow \infty}|z(t)-z*(t)|=0

Question

When a mass-less spring with one end attached to a ceiling and the other end attached to a point mass m, the displacement z of the mass off the equilibrium position is modeled by $mz''+\mu z'+kz=f$ where  mu is the damping coefficient, k is the spring constant, and f is the external force.  Assume that m=2kg mu=1Ns/m k=2N/m and f(t)=sin 2t N  Find the steady state solution.  ie: the function z*(t) s.t \(lim_{t\rightarrow \infty}|z(t)-z*(t)|=0

fibonaccichick666 · Answer

@Michele_Laino I need a hint/general example.  Do you have one?

michele_laino · Answer

we have to know the formula for the function f

michele_laino · Answer

I can give you an example, please wait...

fibonaccichick666 · Answer

ok, thanks, that is all that is given

michele_laino · Answer

here is my example:
$$\Large  mx'' + \beta x' + kx = {F_0}\sin \left( {n\omega t} \right),\quad n \in \mathbb{N}$$

fibonaccichick666 · Answer

f=sin(2t) to answer your previous.  Also, I mean, I need something so that I can figure out how to do it.

fibonaccichick666 · Answer

like should I use the laplace transform?

irishboy123 · Answer

complex exponentials is the easiest way to go.

or laplace will work....

fibonaccichick666 · Answer

I just don't even know where to start.  So I need a bit of a hint or something I can follow that is similar

fibonaccichick666 · Answer

so I got a little hint from a classmate, and I am here $$2s^2 L(z)-2as-2b+2sL(z)-2a+2L(z)=\frac{2}{s^2+4}$$

fibonaccichick666 · Answer

where y(0)=a  and y'(0)=b

irishboy123 · Answer

the equation you have to solve is
$$2z'' + z' + 2z = \sin 2t$$ yes?

imagine instead that you are solving 
$$2Z'' + Z' + 2Z = e^{i 2t}$$


the imaginary (sin) part of the solution is the bit you are looking for.

this is just like solving a 2nd order with a simple exponential RHS, except you have a bit of algebra to do to separate the real and complex parts of the solution.

BUT go with the Laplace if you can do that now and this need urgent attention.  

i would be more than happy to walk through both approaches with you later tomorrow.

fibonaccichick666 · Answer

so, could I use method of undertemined coefficients, or is laplace the best way?  I'm having issues doing the inverse laplace

irishboy123 · Answer

mu is 1 so your Laplace is already going AWOL.  also, not having, or guessing IV's, can be a PITA with Laplace transforms.

go with the complex solution.  it's clever and intuitive.

 i have to go now.  good luck.  maybe back again tomorrow.....;)

fibonaccichick666 · Answer

alright, thanks

michele_laino · Answer

we can use the Fourier transform too

fibonaccichick666 · Answer

I'm using method of undetermined coefficients

anonymous · Answer

Have you tried $A\sin 2t+B\cos2t$ as a trial solution?

fibonaccichick666 · Answer

ok, so tried that, now got that

fibonaccichick666 · Answer

issue, what is a steady state solution

fibonaccichick666 · Answer

I found $$z*=(-3/20)sin2t-(1/20)cos2t $$

fibonaccichick666 · Answer

is that right @SithsAndGiggles , @Michele_Laino , or @IrishBoy123 ?

michele_laino · Answer

what are the initial condition?

fibonaccichick666 · Answer

we don't have any, that's all that we are given

fibonaccichick666 · Answer

I used method of undertermined coefficients then solved and plugged it into the limit

anonymous · Answer

Yes I think it's correct.

fibonaccichick666 · Answer

thank you siths!

anonymous · Answer

You're welcome!

fibonaccichick666 · Answer

can you look at that other one I tagged you in?