dynamic response spring mass system 3

[zekeman]

I have a design problem that I can do numerically but I'm looking for a more elegant method.

Here goes:

A spring, S is interposed between 2 masses (M-S-m aligned vertically) and compressed by the gravity of the much larger mass, M which sits on top. The smaller mass, m which is much smaller than the spring mass is at the bottom and is initially restrained. Now the restraint is suddenly removed. The system is now moving vertically under gravity.
What is the response?

I am after the dynamic equations that include the spring mass.


Thanks for any ideas.

 

[msquared48]

So, if I understand the problem correctly, after the restraint of the smaller mass is removed, there is no support for the system? The whole system is falling in space due to gravity as the three masses respond internally to the lack of restraint to the system?

What is the real world application here?

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask

 

[msquared48]

Also, do you want to integrate the frictional damping effect due to wind resistance as the system falls?

Mike McCann
MMC Engineering
Motto: KISS
Motivation: Don't ask

 

[Twoballcane]

Response to what? It seems there is no Q to relate. Are you asking for the mode of a spring with two different masses on each end?

Tobalcane
"If you avoid failure, you also avoid success."

 

[hydromech]

Zekeman...

Show us a picture that will make us believe this is a real situation and not an undergraduate student question.

 

[SNORGY]

I estimate that the first response is that the assembly begins to fall.

Regards,

SNORGY.

 

[IRstuff]

Wouldn't it be a supersposition of a moving reference frame and an oscillating mass-spring-mass system?

TTFN

FAQ731-376

 

[PNachtwey]

I have a design problem that I can do numerically but I'm looking for a more elegant method.

What is a more elegant method? Do you want a symbolic solution? An equation as a function of time? Will the exact solution be significantly better so it is worth the effort?

What are the equations you used for your numerical solution?
( This will let me know if you have worked out a numerical solution ).

There may not be an exact solution if the formula is non-linear or the solution may be very complex.

Have you tried wxMaxima?

A similar problem is stacked servo controlled hydraulic cylinders. That is a real problem.


Peter Nachtwey
Delta Computer Systems

http://www.deltamotion.com

 

[zekeman]

"Zekeman...

Show us a picture that will make us believe this is a real situation and not an undergraduate student question."

It's been about 45 years since I was in academia. This is a REAL application.

 

[zekeman]

"Also, do you want to integrate the frictional damping effect due to wind resistance as the system falls? "

Ignore friction and windage.

In fact, the larger mass is >> than the spring mass and the end mass.
So the problem really reduces to the usual spring-mass dynamic problem except that the mass<< spring mass.And the large mass, M is out of the equation.












, of the end mass of a compressed spring with the end mass<< spring mass.

 

[zekeman]

"Wouldn't it be a supersposition of a moving reference frame and an oscillating mass-spring-mass system? "

Yes, but the greater and only concern is that

spring mass>> the end mass so the distributed mass of the spring comes into play.

Also, I have simplified the problem ignoring the free fall and am looking at a spring single mass system where

m<< spring mass.

 

[zekeman]

"What is a more elegant method? Do you want a symbolic solution? An equation as a function of time? Will the exact solution be significantly better so it is worth the effort?"


I was thinking the wave equation with the end mass which has the distributed mass embedded.

My socalled numerical method is not really numerical but would be modeling the spring into a number of discrete masses separated by ideal springs which can be solved but the acuracy of this model is in question.

 

[zekeman]

I am repeating the simplified version of my original problem.

Ignore friction and windage.

In fact, the larger mass is >> than the spring mass and the end mass.
So the problem really reduces to the usual spring-mass dynamic problem except that the mass<< spring mass.And the large mass, M is out of the equation.

 

[zekeman]

"A similar problem is stacked servo controlled hydraulic cylinders."

I DOUBT IT.

 

[israelkk]

When the restraint is suddenly removed there is no longer an assurance that the contract between the small mass and the spring exists all the time it may separate. Since "m which is much smaller than the spring mass" to my opinion you may actually need to divide the springs into small segments and calculate the internal dynamics of the spring coils movement to find where the spring is no longer acting a force on the small mass.

 

[ione]

Let’s call:

k = spring constant
Ms =spring mass
m = mass << Ms
l = spring length

We can write the energy E of the system :

E = 1/2* m* (dl/dt)^2 + 1/2* k * l^2 + 1/2*(Ms/3)*l^2

Note that the term ½*(Ms/3)*l^2 represents the spring kinetic energy.

Rearranging:

E = 1/2* (m + Ms/3) * (dl/dt)^2 + 1/2*k * l^2

And this corresponds to a harmonic oscillator with frequency

f = 1/(2*PI) *SQRT[k/(m + Ms/3)]

Now being Ms>>m it will also be Ms/3 >> m and so

f = 1/(2*PI) *SQRT[k/(Ms/3)]

 

[IRstuff]

but why Ms/3? Sounds sort of plausible, but is there a mathematical derivation?

TTFN

FAQ731-376

 

[israelkk]

If you take the spring as a distributed mass and integrate over its length the equivalent spring mass/inertia for natural frequency calculations is 1/3 of the spring mass.

 

[SNORGY]

First, apologies for my earlier "tongue in cheek" post.

The derivation of M/3 can be found here:

http://en.wikipedia.org/wiki/Effective_mass_(spring-mass_system)

A real world application might be a motocross bike with a broken shock absorber, or conversely, sizing a shock absorber for a motocross bike based on undamped dynamic response.


Regards,

SNORGY.

 

[ione]

IRstuff,

Yes there is and israelkk is right. It has been reported by Lawrence Ruby in its “Equivalent mass of a coil spring” The Physics Teacher – March 2000 -- Volume 38, Issue 3, pp. 140-141 Issue Date: March 2000.

Mr. Ruby’s work assumes the spring expands in a uniform manner. I’ll try to report the way the kinetic energy of the spring has been calculated (this is trick).
Each element of the spring having a mass dMs moves with a velocity Vx proportional to the distance x from the spring edge:

Vx = Vm*(x/l)

Being:
l = spring length
Vm = velocity of the suspended mass m

Moreover:
dMs = (Ms/l)*dx

So the kinetic energy Ks of the spring, integrating between [0,l] is (INT stands for integral):

Ks = 1/2 * INT[(Vx)^2*dm]= 1/2 * INT[(Vm)^2* (x^2)/(l^2)*Ms/l*dx ]= 1/2 * Ms*(Vm)^2/(2*l^3)*iNT[ x^2 dx]= 1/2 * Ms*(Vm)^2/(2*l^3)*[l^3/3] = 1/2 * Ms/3*(Vm)^2