# Linear Damping Model

• Jun 13th 2008, 03:53 AM
moolimanj
Linear Damping Model
Hi all.

I have got the following question regarding a seismograph and what I would like to do is to express the forces acting on the particle in terms of the given variables and parameters and the unit vector i. Can some one please help?

Is there any way I can express the displacement vector P of the mass relative to the origin 0 in terms of x,y, d and i. If there is, what would the equation of motion of the particle look like and how can I show that x(t) satisfies the differential equation

md^2x/dt^2 +rdx/dt+kx=mg+kl+md^2y/dt^2
• Jun 13th 2008, 04:11 AM
topsquark
Quote:

Originally Posted by moolimanj
Hi all.

I have got the following question regarding a seismograph and what I would like to do is to express the forces acting on the particle in terms of the given variables and parameters and the unit vector i. Can some one please help?

Is there any way I can express the displacement vector P of the mass relative to the origin 0 in terms of x,y, d and i. If there is, what would the equation of motion of the particle look like and how can I show that x(t) satisfies the differential equation

md^2x/dt^2 +rdx/dt+kx=mg+kl+md^2y/dt^2

I disagree with your equation of motion slightly.

We are applying Newton's second law here, but there is one extra feature: the acceleration of the box comes in as relative motion. So whatever expression we get for the net force, we need to add a $m d^2y / dt^2$ to it. I will simply put that into the equation "by hand."

I have a coordinate system with the positive direction upward. So in the Free Body diagram of the mass there is a weight (w) acting downward, a spring force (f) acting upward, and a resistive force (R) from the piston acting downward.

$f = kl = k(x - l_0)$

$w = mg$

$R = r \frac{dx}{dt}$

So Newton's second in the vertical direction gives:
$m \frac{d^2x}{dt^2} = -mg + k(x - l_0) - r \frac{dx}{dt} + m \frac{d^2y}{dt^2}$
(Your sign on mg was wrong. Typo I presume.)

-Dan
• Jun 17th 2008, 06:35 AM
moolimanj
can someone help with the force diagram? What would it look like?
• Jun 18th 2008, 04:52 AM
topsquark
Quote:

Originally Posted by moolimanj
can someone help with the force diagram? What would it look like?

Quote:

Originally Posted by topsquark
So in the Free Body diagram of the mass there is a weight (w) acting downward, a spring force (f) acting upward, and a resistive force (R) from the piston acting downward.

Be careful not to add a force representing the upward acceleration. This is not due to a contact force on the mass, so it has it's origin in the spring and damping forces. (Or you could argue it's an artifact of the accelerating coordinate system.)

-Dan
• Jun 22nd 2010, 04:59 AM
bobred
I have the same question but the setup is slightly different (see pic)

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

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

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

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

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

Thanks, James