Solving Linear Damping Model Homework

[bobred]

Homework Statement

I am asked to write down the equation of motion of m with respect to O, hence show that x(t) satisfies the differential equation.

m\ddot{x}+r\dot{x}+kx=mg+kl_0+m\ddot{y}

See attached diagram.

Homework Equations

I have the force of the spring H, the Damping R and the weight W as

\bold{H}=k(x-y-l_0)\bold{i}
\bold{R}=r\dot{x}\bold{i}
\bold{W}=-mg\bold{i}

The Attempt at a Solution

Using the above forces and \bold{F}=m\ddot{x}, what I get is
m\ddot{x}-r\dot{x}-kx=-mg-kl_0-ky

I take it I am asked to find the general solution? Where does the m\ddot{y} come from? Do I have the forces correct?

Thanks, James

 

[zzzoak]

I think that y(t) is known function of time.
So coordinate of m is
X=y(t)+d-x
and
mX''=F
m(y''(t)-x'')=F
as y(t) is known function
-mx''=F-my''(t).

 

[bobred]

Thanks, so this would give me

-m\ddot{x}-r\dot{x}-kx=-mg-kl_0-m\ddot{y} and multiplying both sides by -1

m\ddot{x}+r\dot{x}+kx=mg+kl_0+m\ddot{y}

Ok, so how would I find the particular integral seeing that the m\ddot{y} term is present? Would it be the standard trial solution of P\cos(\Omega t) + Q\sin(\Omega t)?

Thanks, James

 

[hikaru1221]

\ddot{y} or further conditions of the motion must be given. Otherwise, it is unsolvable.

 

[bobred]

Can we obtain a general solution? Or because we don't know exactly the roots of the complementary function we cannot get the particular solution?

Also is multiplying the equation of motion by -1 valid?

Thanks

 

[hikaru1221]

Can we obtain a general solution? Or because we don't know exactly the roots of the complementary function we cannot get the particular solution?

Also is multiplying the equation of motion by -1 valid?

Thanks

I'm not sure if there is a general solution.
If \ddot{y}=const then x=Ae^{at}cos(\omega t+\phi)+C.
If \ddot{y}=Ycos(\Omega t) then x=Ae^{at}cos(\omega t+\phi)+Bcos(\Omega t + \Phi)+C.
So the solution varies depending on \ddot{y}. Anyway the problem only asks you to deduce the equation, not to solve it. And multipyling the equation by any number is always valid.

 

[bobred]

Hi

I was pre-empting things a little. The course text has a similar example, I have adapted it slightly.

I am given y=a\cos(\Omega t) so m\ddot{y}=-ma\Omega^{2}\cos(\Omega t)? So let P=ma\Omega^{2} changing the origin to the equilibrium position we have

m\ddot{x}+r\dot{x}+kx=-P\cos(\Omega t)

The text goes on to find the coefficients of the trial solution B\cos(\Omega t)+C\sin(\Omega t) in terms of r, k, m etc starting out with m\ddot{x}+r\dot{x}+kx=P\cos(\Omega t) Is what I have done so far ok?

Thanks, James

 

[hikaru1221]

The solution given by the text only describes the motion in steady state. You may look at my previous post for the general solution :)

 

[bobred]

Thanks

Just one last thing, is what I have done with

m\ddot{x}+r\dot{x}+kx=y(t)

to

m\ddot{x}+r\dot{x}+kx=\ddot{y}(t)

replacing y(t) with \ddot{y}(t) valid?

 

[hikaru1221]

No :(
For example, if y=at^2 then \ddot{y}=2a. They're obviously different, right? :)

 

[bobred]

Ok, sorry if I'm being slow, thanks for your patience.

This is the equation of motion from equilibrium.

m\ddot{x}+r\dot{x}+kx=m\ddot{y}

I am told that y=P\cos(\Omega t)

but as the rhs has the \ddot{y} so should it now be

m\ddot{x}+r\dot{x}+kx=-mP\Omega^{2}\cos(\Omega t) which is the second derivative of y=P\cos(\Omega t).

Thanks again

 

[hikaru1221]

I'm not sure what you were asking about when substituting y with y" in post #9, but you're right at post #11 :)

 

[bobred]

Hi

I was trying to ask the same question as post 11, obviously not very wel :-)

Thank you very much for all of you help.

James