# Thread: Modeling a Mass-Spring System

1. ## Modeling a Mass-Spring System

A particle of mass
m is suspended from a ceiling by a spring of natural length l and

stiffness
k. Assume that there is no damping

If the mass is displaced from equilibrium show that the equation governing its
subsequent motion is given by
mz'' + kz = 0.

My working:If we let e be the extension then N2L is ke-mg=0 so ke=mg.

An engineer measures the mass of the particle to be
m = 2, and furthermore k = 4.
He also finds that internal friction in the spring provides damping that is proportional

to the velocity of the particle with a constant of proportionality
μ = 4.

Show that the equation for the subsequent motion after the displacement of the
mass is now given by
2z''
+ 4z' + 4z = 0

My working: what is the constant of proportionality?

2. For the first problem, I'm not sure your application of N2L is correct. The first step is to assign a coordinate system. I believe z is intended to be the displacement from equilibrium. Do you want z to be positive up or down?

3. up please lol. My reasoning was that the restoring force and the displacement act in opposite directions.

4. Originally Posted by boromir
up please lol. My reasoning was that the restoring force and the displacement act in opposite directions.
They do. So what are all the forces acting on this spring?

5. ke and mg

6. Originally Posted by boromir
ke and mg
What's e? In particular, how does e relate to z?

7. yeah e=z

8. So what does Newton's Second Law say? Where's the acceleration term in your equations?

9. acceleration =0 since it's in equilibrium. You may have missed that. It's the next part i'm stuck

10. Originally Posted by boromir
acceleration =0 since it's in equilibrium. You may have missed that. It's the next part i'm stuck
Incorrect. It's displaced from equilibrium. It is not actually in equilibrium. Another clue: "its subsequent motion". This mass is moving!

11. ok so the acceleration is z'' so I get kz-mg=mz''

12. Much closer. However, there are still two problems.

1. z is measured from the equilibrium point of the mass on the spring, whereas the equation you wrote is really more correct for a different variable y, measured from the equilibrium point of the spring without a mass on it. This point is getting at the heart of the problem here: what's the difference between the system with a mass and without a mass? Another way of phrasing it is this: what is the effect of gravity on such a vertical mass-spring system?

2. The kz term should have the opposite sign (in addition to the change mentioned above), since the spring is a restoring force.

13. I don't know, I suppose you get an mg term. Come on, spill the goods.

14. Originally Posted by boromir
I don't know, I suppose you get an mg term. Come on, spill the goods.
Well, this is the main point of the problem, as I see it, so I'm not going to just tell you the answer. That's not the way MHF works. I'll give you a hint, though:

Compare these two scenarios: the spring without any mass in equilibrium, with the mass on the spring in equilibrium. Just imagine hanging the mass on the spring, and waiting until it stops bobbing. As a thought experiment, what's the result going to be?

15. more extension?

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